ECE-205 Dynamical Systems

Transcription

1 ECE-5 Dynamical Sysems Course Noes Spring Bob Throne Copyrigh Rober D. Throne

2 Copyrigh Rober D. Throne

3 . Elecrical Sysems The ypes of dynamical sysems we will be sudying can be modeled in erms of algebraic equaions, differenial equaions, or inegral equaions. We will begin by looking a familiar mahemaical models of ideal resisors, ideal capaciors, and ideal inducors. Then we will begin puing hese models ogeher o develop models for RL and RC circuis. Finally, we will review soluion echniques for he firs order differenial equaion we derive o model he sysems.. Ideal Resisors The governing equaion for a resisor wih resisance R is given by Ohm s law, v() = Ri() where v () is he volage across he resisor and i () is he curren hrough he resisor. Here R is measured in Ohms, v () is measured in vols, and i () is measured in amps. The enire expression mus be in vols, so we ge he uni expression. Ideal Capaciors [vols] = [Ohms][amps] The governing equaion for a capacior wih capaciance C is given by dv() i () = C d Here C is measured in farads, and again v () is measured in vols and i () is measured in amps. This expression also helps us wih he unis. The enire expression mus be in erms of curren, so looking a he differenial relaionship we can deermine he uni expression [amps] = [farads][vols]/[seconds] We can inegrae his equaion from an iniial ime up o he curren ime as follows: dv() i () = C d i() d = dv() C Nex, since we wan o inegrae up o a final ime, we need o use a dummy variable in he inegral ha is no. This is an imporan habi o ge ino do no use as he dummy variable of inegraion if we expec a funcion of ime as he oupu! Here we have chosen o use he dummy variable λ. Also we incorporae he fac ha a ime he volage is v ( ), while a ime he volage is v () Copyrigh Rober D. Throne 3

4 C v ( ) i( λ) dλ = dv( λ) v ( o ) Carrying ou he inegraion we ge which we can rearrange as C i( λ) dλ = v () v ( = ) + C v () v ( i( λ) dλ ) This expression ells us here are wo componens o he volage across a capacior, he iniial volage v ( ) and he par due o any curren flowing hrough he capacior afer ha ime, i( λ) dλ C Finally, hese expressions help us deermine some imporan characerisics of our ideal capacior: If he volage across he capacior is consan, hen he curren hrough he capacior mus be zero since he curren is proporional o he rae of change of he volage. Hence, a capacior is an open circui o dc. I is no possible o change he volage across a capacior in zero ime.the volage across a capacior mus be a coninuous funcion of ime, oherwise an infinie amoun of curren would be required..3 Ideal Inducors The governing equaion for an inducor wih inducance L is given by di() v () = L d Here L is measured in henrys, and again v () is measured in vols and i () is measured in amps. This expression also helps us wih he unis. The enire expression mus be in erms of volage, so looking a he differenial relaionship we can deermine he uni expression [vols] = [henrys][amps]/[seconds] We can inegrae his equaion from an iniial ime up o he curren ime as follows: di() v () = L d v() d = di() L Copyrigh Rober D. Throne 4

5 Nex, since we wan o inegrae up o a final ime, so we again have chosen o use he dummy variable λ. Also we incorporae he fac ha a ime he curren is, i ( ) while a ime he curren is i (). Carrying ou he inegraion we ge which we can rearrange as L L i ( ) v( λ) dλ = di( λ) i ( o ) v( λ) dλ = i () i ( = ) + L i () i ( v( λ) dλ This expression ells us here are wo componens o he curren hrough an inducor, he iniial curren i ( ) and he par due o any volage across he inducor afer ha ime, v( λ) dλ L. Finally, hese expressions help us deermine some imporan characerisics of our ideal inducor: If he curren hough an inducor is consan, hen he volage across he inducor mus be zero since he volage is proporional o he rae of change of he curren. Hence, an inducor is a shor circui o dc. I is no possible o change he curren hrough an inducor in zero ime.the curren hrough an inducor mus be a coninuous funcion of ime, oherwise an infinie amoun of volage would be required. ) Copyrigh Rober D. Throne 5

6 Copyrigh Rober D. Throne 6

7 . Firs Order Circuis A firs order circui is a circui wih one effecive energy sorage elemen, eiher an inducor or a capacior. (In some circuis i may be possible o combine muliple capaciors or inducors ino one equivalen capacior or inducor.) We begin his secion wih he derivaion of he governing differenial equaion for various firs order circuis. We will hen pu he firs order equaion ino a sandard form ha allows us o easily deermine physical characerisics of he circui. Nex we show an alernaive mehod for checking some pars of he governing differenial equaions. We hen solve he differenial equaions for he case of piecewise consan inpus, and finish he secion wih an alernaive mehod of solving he differenial equaions using inegraing facors.. Governing Differenial Equaions for Firs Order Circuis In his secion we derive he governing differenial equaions ha model various RL and RC circuis. We hen pu he governing firs order differenial equaions ino a sandard form, which allows us o read off descripive informaion abou he sysem very easily. The sandard form we will use is dy() τ + y() = Kx() d Here we assume he sysem inpu is x () and he sysem oupu is y (). τ is he sysem ime consan, which indicaes how long i will ake he sysem o reach seady sae for a sep (consan) inpu. K is he saic gain of he sysem. For a consan inpu of ampliude A ( x() = Au(), where u () is he uni dy() sep funcion), in seady sae we have = and y() = Kx() = KA. Hence he saic gain les us d easily compue he seady sae value of he oupu. For circuis wih capaciors he differenial equaion will in general be in erms of a volage (he oupu y () will be a volage), while for circuis wih inducors he differenial equaion will in general be in erms of curren (he oupu y () will be a curren). Example... Consider he RC circui shown in Figure.. The volage source is vs (). We sar o derive he governing differenial equaion by deermining he single curren in he loop vs() c() dvc() ir() = v = ic() R = C d or dvc() vs() vc() C = d R where vc () is he volage across he capacior and he curren in he loop is equal o he curren hrough he resisor ir () and he curren hrough he capacior ic (). We can pu his ino a more sandard form by rearranging he erms dvc () RC + vc() = vs() d If we define he ime consanτ = RC, hen we have Copyrigh Rober D. Throne 7

8 Here he saic gain K =. dvc () τ + vc() = vs() d R C Figure.. Circui for Example... Example... Consider he RC circui shown in Figure.. Again he volage source is vs (). We again sar o derive he governing differenial equaion by deermining he curren hrough resisor R a, vs() vc i() () = Ra This curren mus be equal o he sum of he currens hrough he capacior and R b, i vc ( ) dv ( ) ( ) C c = + R d b Equaing hese we ge he governing differenial equaion: Rearranging erms we ge or i ( v () v () vc () dv ( ) = = + C R R d ) s c c a dvc () C + + vc() = vs() d Ra Rb Ra dvc () Ra + Rb C + vc () = vs () d R R R a b RaRC b dvc () Rb + vc () = vs () R + R d R + R a b b a a b Copyrigh Rober D. Throne 8

9 RaRC b Wih ime consanτ = R + R a b and saic gain R = we ge b K R a + R b dvc () τ + vc () = Kvs () d R a + - R b + - C Figure.. Circui used in Example... Example..3. Consider he operaional-amplifier circui shown in Figure.3. The inpu volage is again vs () and he oupu volage (he volage across he load resisor R L ) is he same as he volage across he capacior (since he + erminal of he op amp is assumed o be grounded). We will assume an ideal op amp, which implies he condiions + i () = i () = + v () = v () Le s look a he currens flowing ino he negaive (feedback) erminal of he op-amp using he ideal opamp model. Since for our example he non-invering erminal is ied o ground we have v () =. Wih + hese assumpions our governing differenial equaion becomes vs () vc() dvc() = + + C Ra Rb d Rearranging his gives dvc() vc() vs () C + = d R b R a or dvc () Rb RC b + vc () = vs() d Ra Rb Seing he ime consan τ = RC b and saic gain K = we finally have R a Copyrigh Rober D. Throne 9

10 dvc () τ + vc () = Kvs () d R b C - + R a R L Figure.3. Circui for Example..3. Example..4. Consider he RL circui shown in Figure.4. The single curren in he loop is given by vs() vl() i () = R where d ( ) vl() = L i d Combining and rearranging we ge di() L + Ri() = vs () d or L di() + i () = vs () R d R L Wih ime consan τ = and saic gain K = he governing differenial equaion is R R di() τ + i( ) = Kvs ( ) d Copyrigh Rober D. Throne

11 R L Figure.4. Circui for Example..4. Example..5. Consider he RC circui shown in Figure.5. The single curren source mus be divided beween he curren flowing hrough resisor R b and he curren flowing hrough he capacior C, vc() dvc() is () = + C Rb d Rearranging we ge dvc () RbC + vc() = Ri b s() d Wih ime consan τ = R C and saic gain b K = Rb he governing differenial equaion is dvc () τ + vc() = Kis() d R a R b + - C Figure.5. Circui used in Example..5. Copyrigh Rober D. Throne

12 . Thevenin Resisance, Time Consans, and Saic Gain Alhough we are focusing our aenion on deriving he governing equaions for firs order circuis, i is useful and very convenien o be able o check our equaions as much as possible. Firs of all, for firs order RC circuis he ime consans will be of he formτ = RC h eq where R h is he Thevenin resisance seen from he pors of he equivalen capacior, C eq. For firs order RL circuis he Leq ime consans will be of he form τ = where R h is he Thevenin resisance seen from he pors of Rh he equivalen inducor, L eq. Recall ha when deermining he Thevenin resisance all independen volage sources are reaed as shor circuis, and all independen curren sources are reaed as open circuis. Secondly, if we are looking a consan inpus, hen we use he fac ha a capacior is an open circui o dc and an inducor is a shor circui o dc. In addiion, for consan inpus in seady sae all of he ime derivaives are zero (in seady sae nohing changes in ime). Example... Consider he circui shown in Figure. (Example..). The Thevenin resisance seen from he capacior is equal o R, so he ime consan is τ = RC. For a dc inpu, he capacior looks like an open circui, so in seady sae he volage across he capacior is equal o v s, he inpu volage, so he saic gain is K =. These resuls mach our previous resuls. Example... Consider he circui shown in Figure. (Example..). The Thevenin resisance seen RaRb RaRbC from he capacior is Rh = Ra Rb =, so he ime consan isτ = RC h =. For a dc inpu, Ra + Rb Ra + Rb he capacior looks like an open circui, so in seady sae he volage across he capacior is given by he Rb Rb volage divider relaionship vc = vs, so he saic gain is K =. These resuls mach our Ra + Rb R a + R b previous resuls. Example..3. Consider he circui shown in Figure.3 (Example..3). The Thevenin resisance seen by he capacior is a lile more difficul o deermine, and o do i correcly is beyond he scope of his course. For a dc inpu, he capacior looks like an open circui, so summing he currens ino he negaive vc vs Rb erminal of he op amp we have + =, or in seady sae vc = vs Hence he saic gain is Rb Ra Ra Rb K =. R a Example..4. Consider he circui shown in Figure.4 (Example..4). The Thevenin resisance seen by he inducor is Rh = R. For a dc inpu, he inducor looks like a shor circui. Hence he seady sae curren flowing in he circui for a dc inpu is i = v s, so he saic gain is K =. R R Copyrigh Rober D. Throne

13 Example..5. Consider he circui shown in Figure.5 (Example..5). The Thevenin resisance seen by he capacior is Rh = Rb so he ime consan is τ = RC b. For a dc inpu he capacior looks like an open circui, so in seady sae vc = Ri b, so he saic gain is K = Rb..3 Solving Firs Order Differenial Equaions In his secion we will go over wo mehods for solving firs order differenial equaions. We will iniially solve he equaions by breaking he soluion ino he naural response (he response wih no inpu) and hen he forced response (he response when he inpu is urned on). We will apply his mehod o problems where he inpu is a consan value, or is swiched beween consan values. This mehod will also work wih any inpu, and we will examine he resuls for a sinusoidal inpu laer. In he las secion we will go over a differen mehod of soluion using inegraing facors, which will work for any ype of inpu, and is an imporan mehod in helping us characerize how a sysem will respond o any ype of inpu..3. Soluion using Naural and Forced Responses Consider a sysem described by he firs order differenial equaion dy() τ + y() = Kx() d In his equaion, τ is he ime consan and K is he saic gain. We will solve his equaion in wo pars. We will firs deermine he naural response, yn(). The naural response is he when he inpu is zero. Then we will deermine he forced response, yf (). The forced response is he response due o he inpu only. The oal response is hen he sum of he naural and forced responses, y () = y() + y (). Naural Response: To deermine he naural response we assume here is no inpu in he sysem, so we have he equaion dyn() τ + yn() = d r Le s assume a soluion of he form yn() = ce, where c and r are parameers o be deermined. Subsiuing his assumpion ino he differenial equaion we ge r r r τrce + ce = ce [ τr + ] = If c = hen we are done, and he naural response will be yn() =. This soluion cerainly saisfies he r differenial equaion. However, if c, and since e can never be zero, we mus haveτ r + =, or r =. In his case he naural response will be τ / y () = ce τ. n Forced Response: To deermine he forced response we mus know he sysem inpu, x (). We will iniially assume an inpu ha is zero before = and hen has consan ampliude A for, n f Copyrigh Rober D. Throne 3

14 Then for we have he equaion < x () = A dy f () τ + y f () = KA d Since his is a linear ordinary differenial equaion we only need o find one soluion. One obvious dy f () soluion o his equaion is he soluion in seady sae, when =. In seady sae we have d y () = KA f Noe ha for a consan inpu, he seady sae oupu is he produc of he saic gain and he ampliude of he inpu. Toal Soluion: The oal soluion o he problem is he sum of hese wo soluions / y() = yn() + yf () = ce τ + KA Now assume he iniial ime is and a his ime he oupu is denoed y ( ). Subsiuing his ino our equaion we have y( ) ce / τ = + KA, or c = ( y( ) KA) e / τ, and our oal soluion is () ( ( ) ) ( )/ τ y = y KA e + KA For simpliciy, le s wrie our seady sae value explicily, so y( ) = KA and we have he soluion ( ( ) ( ( )/ τ = ) ( )) e + ( ) y y y y Significance of he Time Consan In much of wha we do, we will be concerned wih he ime consans of a sysem in one way or anoher. Le s look a he response of our firs order sysem assuming he sysem is iniially a res ( y () = ) and he final value is one ( y( ) = ). Le s look a he response of our sysem as he ime akes on he values of ineger number of ime consans: Time ( ) / τ / y () = e τ τ.63 τ.865 3τ τ τ Copyrigh Rober D. Throne 4

15 Figure.6 show his resul graphically, The way his informaion is usually inerpreed is ha a sysem is wihin 5% of is final value in 3 ime consans, wihin % of is final value in 4 ime consans, and wihin % of is final value in 5 ime consans. Hence he use of ime consans gives us a quick way o describe one aspec of he behavior of a sysem. As we will see, as he sysems become more complex, he use of ime consans indicaes which par of he soluion is he mos imporan and how he sysem responds o periodic inpus (sines and cosines) y() Number of Time Consans / Figure.6. Graph of y () = e τ for = τ up o = 7τ. y () is wihin 5% of is final value in 3 ime consan, wihin % of is final value in 4 ime consans, and wihin % of is final value in 5 ime consans. Example.3.. Consider he circui in Figure. (Example..). Le s firs assume Ra = Rb = kω and C = µ F. Then Rh= kω, τ = ms, and K =.5. Nex we will assume he iniial volage on he capacior is zero ( vc( ) = vc() = ) and he inpu is as follows: Copyrigh Rober D. Throne 5

16 < 8 vs () = 8 < 6 > 6 Here he inpu is in vols and he ime is measured in milliseconds. We now wan o deermine he oupu. We will do his by looking a he iniial and final values for each ime inerval, where he ime inervals are deermined by he imes during which he inpu volage is consan. The differenial equaion is dvc () τ + vc() = Kvs() d Clearly y () = v() and x () = v(). The soluion in each inerval will be of he form c s [ ] y = y y e + y ( )/ τ ( ) ( ) ( ) ( ) A his poin we jus need o be able o deermine wha y ( ) and y( ) mean for each inerval. Firs inerval ( < 8 ms) : We have he iniial value in his inerval y() = v c () = vols. To deermine he final value, we use he saic gain and he ampliude of he inpu for his inerval. vc ( ) = y( ) = K = = Hence for his inerval, we have he soluion v() = y () = e + = e c / τ /. Before we go on o he nex inerval we need o figure ou he value of y () a he end of his inerval, his value will be he iniial poin during he nex inerval. A he end of he inerval we will have y e.8/. 8 (.8) = = e = Second inerval (8 < 6 ms) : The iniial value for his inerval will be he end poin of he previous inerval, so y ( ) =. To deermine he final value we again use he saic gain y( ) = K ( ) = ( ) = We now have almos everyhing we need, however, our soluion assumed a ime of zero was measured a he beginning of he inerval. Hence o use our previous soluion we need o subrac he ime a he beginning of he inerval from our acual ime in our form of he soluion, so our ime will be measured from he beginning of he inerval. Our soluion for his inerval is hen y( ) = [ ( ) ( ) (.8)/ τ (.8)/. ] e + = e A he end of his inerval we will have Copyrigh Rober D. Throne 6

17 y. 6) = = =. (.6.8)/. 8 ( e e Third inerval ( > 6 ms) : The iniial value for his inerval will be he end poin of he previous inerval, so y ( ) =. To deermine he final value we have y( ) = K = K = Again we mus scale our soluion so ime is measured from he beginning of he inerval, so we have (.6)/ τ (.6)/. y ( ) = [.5] e +. 5 =.5 e +. 5 Toal soluion: To ge he oal soluion, we lis he soluions during each ime inerval: < /. e < 8ms vc () = y() = (.8)/. e 8 < 6 ms (.6)/..5e +.5 6ms To ge he curren hrough he capacior, we use he relaionship () dv () c ic = C for each ime inerval d above. Doing his we ge < /. dv (). 8 c e < ms ic () = C = (.8)/. d.e 8 < 6 ms (.6)/..5e 6ms Here ic () is measured in amps. Figure.7 shows he inpu volage, he volage across he capacior, and he curren hrough he capacior as a funcion of ime. Noe ha he volage across he capacior is coninuous, as i mus be. However for his inpu, which is disconinuous, he curren hrough he capacior is disconinuous. Le s also look a he answer o see if we can check our resuls and if he answer makes sense. When he source volage is iniially urned on, he volage across he capacior is zero and all of he volage generaed by he source is equal o he volage across resisor R a. If here were any volage drop across R b a he iniial ime, here would also be a volage drop across he capacior since hey are in parallel. In seady sae, he capacior looks like an open circui, so here is no curren flowing hrough he capacior and he maximum possible volage a his ime is half he volage of he source, which agrees wih our resuls. In his example he inpu was held consan for an equivalen of eigh ime consans, so he volage across he capacior had essenially reached seady sae. Finally is useful o poin ou ha if he volage across he capacior is described by he relaionship v / ( ) [ () ( )] τ c = vc vc e + vc( ) Then he curren hrough he capacior is given by Copyrigh Rober D. Throne 7

18 dvc () C ic( ) = C = [ vc() vc( )] e d τ Wha his means is ha if he volage across a capacior is growing exponenially, hen he curren hrough he capacior is decreasing exponenially. Similarly, if he volage across a capacior is decreasing exponenially, he curren hrough he capacior will be growing exponenially. This is also behavior our resuls show. Similar resuls also hold for inducors. Example.3.. Consider he circui in Figure.4 (Example..4). Le s firs assume R= R h = Ω L. and L = mh. Thenτ = = =. = µ s and K=.. Nex we will assume he iniial R curren hrough he inducor is i() = maand he inpu is as follows: < <. vs () = 3. < Here he inpu is in vols and he ime is measured in milliseconds. We now wan o deermine he oupu. We will do his by looking a he iniial and final values for each ime inerval, where he ime inervals are deermined by he imes during which he inpu volage is consan. The differenial equaion for his sysem is again di() τ + i() = Kvs () d Clearly y () = i () and x () = v(). The soluion in each inerval will be of he form s [ ] y = y y e + y ( )/ τ ( ) ( ) ( ) ( ) A his poin we jus need o be able o deermine wha y () and y( ) mean for each inerval. / τ Copyrigh Rober D. Throne 8

19 V s () (vols) V c () (vols) i c () (ma) Time (ms) Figure.7. Resuls for Example.3.. Firs inerval ( <. ms ) : We have he iniial value y() = i() =.amps in his inerval. To deermine he final value, we use he saic gain and he ampliude of he inpu for his inerval i( ) = y( ) = K = =. Hence for his inerval, we have he soluion / τ / τ /. [ ] y ( ) = y() y( ) e + y( ) = [..] e +. =.e +. Before we go on o he nex inerval we need o figure ou he value of y () a he end of his inerval, his value will be he iniial poin during he nex inerval. A he end of he inerval we will have y e e./. (.) =. +. =. +. =.63 Copyrigh Rober D. Throne 9

20 Third inerval (. <. 5 ms ) : The iniial value for his inerval will be he end poin of he previous inerval, so y ( ) =.63. To deermine he final value we again use he saic gain y( ) = K ( 3) =. ( 3) =.3 We again need o subrac he ime a he beginning of he inerval from our acual ime in our form of he soluion, so our ime will be measured from he beginning of he inerval. Our soluion for his inerval is hen y ( ) = [.63 (.3)] e + (.3) =.463e.3 (.)/ τ (.)/. A he end of his inerval we will have y = = = (.5.)/..5 (.5).463e.3.463e Fourh inerval (.5 ms ) : The iniial value for his inerval will be he end poin of he previous inerval, so y ( ) =.966. To deermine he final value we have y( ) = K 4 =.4 Again we mus scale our soluion so ime is measured from he beginning of he inerval, so we have (.5)/ τ (.5)/. y ( ) = [.966.4] e +. 4 =.5966e +.4 Toal soluion: To ge he oal soluion, we lis he soluions during each ime inerval: < /..e +. <. ms il () = y() = (.)/..463e.3. <.5 ms (.5)/..5966e ms To ge he volage across he inducor we use he relaionship () di () L vl = L and compue he volage d for each ime inerval. Doing his we ge < /. di (). L e < ms vl() = L = (.)/. d 4.63e. <.5 ms (.5)/ e.5 ms Figure.8 shows he inpu volage, he curren hrough he inducor, and he volage across he inducor as a funcion of ime. Noe ha he curren hrough he inducor is coninuous, as i mus be, while in his case he volage across he inducor is no coninuous. Again le s look a our soluion o see if i makes Copyrigh Rober D. Throne

21 sense. Firs of all, he volage/curren relaionships for he inducor are consisen wih wha we expec. The iniial curren in he inducor is ma, as we require, and he iniial volage from he source is vols. Applying Kirchhoff s laws around he loop, we expec he iniial volage drop across he inducor o be given by vs () i() R= (.)() = vols, which is wha we have. In seady sae he inducor looks like a shor circui, so here should be no volage drop across he inducor once he sysem reaches seady sae, which again maches our resuls. Noe ha he sysem only reaches seady sae near.7 or.8 ms. In addiion, in seady sae he volage drop across he resisor mus mach he volage supplied by he source, or vs ( ) i( ) R= 4 (.4)() = vols, which again maches our resuls. Le s look a he resuls a one oher convenien poin in ime, say =. ms. Using he equaions we derived above (and he known inpu) we have vs (.) = 3 vols il (.) =.96 ma vl(.) =.7 vols Applying Kirchhoff s laws around he loop we have vs(.) is(.) R vl(.) = 3 (.96)() (.7). We can obviously check as many poins in ime as we wan in his way. This ype of checking does no guaranee our answer is correc, bu i does help find obvious errors..3. Soluion Using Inegraing Facors An alernaive mehod of soluion of firs order differenial equaions is by he use of inegraing facors. This mehod of soluion is imporan o undersand because as we sar o analyze differen ypes of sysems, we need o be able o undersand how we would solve for he oupu when we don acually know wha he inpu is. This helps us characerize sysems independen of he acual (specific) inpu. The use of inegraing facors for solving firs order differenial equaions is based on he fac ha when we differeniae an exponenial, we ge he same exponenial back muliplied by some oher erm. For () example, if x () = e φ, hen d d φ() φ() dφ() dφ() x () = e = e = x () d d d d In wha follows, he mehod looks fairly lenghy, bu wih pracice mos of he seps can be done in your head. Le s apply his idea o our equaion dy() τ + y() = Kx() d Copyrigh Rober D. Throne

22 4 V s () (vols) i L () (ma) v L () (vols) Time (ms) Figure.8. Resuls for Example.3.. This mehod will work beer if we rearrange our equaion a bi o he form dy() K + y () = x () d τ τ () Nex, we look a differeniaing he produc ye () φ, where φ () will be deermined by he differenial equaion we are rying o solve. This leads o he equaion d φ() dy() φ() dφ() φ() φ() dy() dφ( ) ye ( ) e y() e e y ( ) d = + = d d + d d Nex, we equae he erm in brackes o he lef hand side of our original differenial equaion, Copyrigh Rober D. Throne

23 dy() dφ() dy() + y () = + y () d d d τ Clearly his means ha dφ( ) = d τ Solving his simple equaion we ge φ() = τ Now we pu his back ino our equaion above o ge d / dy() / / / dy() ye () τ e τ y() e τ e τ y () d = + = + d τ d τ The erm on he far righ is he same as he lef hand side of our differenial equaion muliplied by so his mus equal he righ hand side of our differenial equaion muliplied by he same hing, d / τ / τ dy() / τ K ye () e y () e x () d = + = d τ τ Nex we eliminae he middle erm o ge he exac differenial we wan / e τ, d / τ / τ K ye () e x () d = τ Finally we inegrae from and iniial ime wih iniial value y ( ) o final ime wih value y (), The lef hand side can be inegraed as λτ / λτ / K ( ) x( ) d y λ e dλ e λ dλ dλ = τ d / / / y( / e τ dλ = = K λ) e λτ dλ ye () τ y ( ) e λτ x( λ) dλ τ or This is he general soluion, for any inpu x (). ( )/ τ ( λ)/ τk () = ( ) + ( λ) τ y y e e x dλ Copyrigh Rober D. Throne 3

24 Example.3.. Le s now look a he same inpu as before, x () = A for wih iniial condiion = and y ( ) = y(). The soluion o he differenial equaion becomes / τ ( λ)/ τk y( ) = y() e + e Adλ τ / / / K τ τ λτ y( ) = y() e + e e Adλ τ / τ / τ λτ / λ= = e + λ= y ( ) y() e KA e / τ / τ / τ y( ) = y() e + e KA e Wih he subsiuion y( ) = KA, we ge ( ) = () + / / y y e τ KA e τ or he same soluion as before. () = ( ) + ( ) / / y y e τ y e τ / [ ] τ y ( ) = y() y( ) e + y( ) Example.3.. Le s use inegraion facors o deermine he soluion o he differenial equaion dy() = ay() + bx() d The firs hing we need o do is pu all of he y erms on he lef hand side, Then we need or Then we have dy() ay() = bx() d dφ() = a d φ () = a d ye a a () e = b x () d Inegraing boh sides we ge Copyrigh Rober D. Throne 4

25 or d a λ ( ) a a ( ) ( ) a λ b ( ) dλ = = e yλ dλ e y e y e xλ dλ a ( ) a a () = ( ) + λ b ( λ) y e y e e x dλ Example.3.3. Le s use inegraion facors o deermine he soluion o he differenial equaion dy() y() = x() d Then we need dφ() = d or φ () = Then we have d ye () = e x () d Inegraing boh sides we ge e y( λ) dλ e y () e y ( ) e x( λ) dλ λ λ d = = dλ or λ ( ) y () = e y ( ) + e e x( λ) dλ Copyrigh Rober D. Throne 5

26 Chaper Problems.) For each of he circuis below: i) Deermine he governing differenial equaion using Kirchhoff s Laws and wrie i in sandard form. For par F he oupu is he volage across resisor RC and his canno be wrien in sandard form. ii) Deermine he ime consan and saic gain from he differenial equaion you derive in (i) iii) For all circuis excep F, deermine he Thevenin resisance from he pors of he capacor or inducor and verify he ime consans. iv) For all circuis excep F, deermine he saic gain by deermining he DC volage across he capacior or he curren hrough and inducor. Copyrigh Rober D. Throne 6

27 Answers: L dil () dvc () + il () = iin (), C( Ra + Rb ) + vc () = Rbiin () Rb d d L dil () R a RaR b dvc() R b + il () = iin (), C + vc () = vin () Ra + Rb d Ra + Rb Ra + Rb d Ra + Rb RR a b + RR c b + RR a c dvc() R c L dvin () R a C + vc () = vin (), + vin () = vc () Ra + Rc d Ra + Rc Rb d Rb.) For a simple series RC circui he response of he sysem when he inpu is a uni sep is / RC / τ y() = e = e The -9% rise ime, r, as shown below. The rise ime is simply he amoun of ime necessary for he oupu o rise from % o 9% of is final value. Show ha for his sysem he rise ime is given by = τ ln(9) r Copyrigh Rober D. Throne 7

28 .3) For each of he circuis below: i) Deermine he governing differenial equaion using Kirchhoff s Laws and wrie i in sandard form. ii) Deermine he ime consan and saic gain from he differenial equaion you derive in (i) iii) For all circuis excep F, deermine he Thevenin resisance from he pors of he capacor or inducor and verify he ime consans. iv) For all circuis excep F, deermine he saic gain by deermining he DC volage across he capacior or he curren hrough and inducor. Copyrigh Rober D. Throne 8

29 Answers: ( Ra + Rb) L dil() L dil() + il () = Vin (), + il () = Vin () R R d R R + R d R + R a b a a b a b dvc() ( Ra + Rb) L dil () ( Ra + Rb ) C + vc () = Ri a in(), + il() = vin () d R R d R a b a ( Ra + Rc) L dil() Rc + il( ) = vin ( ), CR RR + RR + RR d RR + RR + RR a c b c a b a c b c a b dvc() Rb + v () = v () d R c C in a.4) Consider a firs order syem described by he differenial equaion τ y () + y () = Kx (). a) If he iniial value is y () = and he final value is y( ) =, wha is y(4τ )? b) If he iniial value is y () = and he final value is y( ) = 8, wha is y(4τ )? c) If he iniial value is y () = and he final value is y( ) = 4, wha is y(4τ )? Answers: 9.8, 7.8, ) For he following circui, find an expression for v() for. Scrambled answers: 5 ms, 3 V.6) Consider he circui shown in he figure below: For boh pars of his problem he circui is iniially a res (no charge on he capacior) and he swich is conneced o he lef par of he circui for < 4 ms, and is conneced o he righ par of he circui for > 4ms. For boh pars of his problem, you are o skech he volage across he capacior from o ms. You need o primarily deermine he appropriae ime consans and seady sae values, and use he following able as a guide. Copyrigh Rober D. Throne 9

30 Time ( ) / τ y () y ss τ.63 y ss τ.865 y ss 3τ 3.95 y ss 4τ 4.98 y ss 5τ y ss a) For Ra = kω, Rb = 5 kω, Rc = Ω, Rd = 6 Ω, C =. µ F, Vin = 6V skech he volage across he capacior. b) For Ra = 4 kω, Rb = 4 kω, Rc = kω, Rd = kω, C = µ F, Vin = 6V skech he volage across he capacior..7) Consider he circui shown in he figure below: For boh pars of his problem he circui is iniially a res (no charge on he capacior) and he swich is conneced o he lef par of he circui for < 4 ms, and is conneced o he righ par of he circui for > 4ms. For boh pars of his problem, you are o skech he volage across he capacior from o ms. You need o primarily deermine he appropriae ime consans and seady sae values, and use he following able as a guide. Time ( ) / τ y () y ss τ.63 y ss τ.865 y ss 3τ 3.95 y ss 4τ 4.98 y ss 5τ y ss a) For Ra = kω, Rb = kω, Rc = kω, Rd = kω, C = µ F, Vin = 6V skech he volage across he capacior. b) For Ra = kω, Rb = 4 kω, Rc = kω, Rd = kω, C = µ F, Vin = 6V skech he volage across he capacior. c) Copyrigh Rober D. Throne 3

31 .8) Consider he circui shown in he figure below: a) Deermine an expression for he ime consan of he circui for he ime when he capacior is charging ( <. seconds) and discharging ( >. seconds) in erms of he parameers C, R a, Rb, Rc and R d. (Do no use numbers). b) Deermine an expression for he saic gain of he circui for <. seconds in erms of he parameers C, R a, Rb, Rc and R d. (Do no use numbers). c) For Ra = kω, Rb = kω, Rc = kω, Rd = kω, C = µ F, Vin = 6V accuraely skech he volage across he capacior from o ms. You need o specifically lablel he volages a =. seconds and =. seconds. You need o primarily deermine he appropriae ime consans and seady sae values, and use he following able as a guide. Time ( ) / τ y () y ss τ.63 y ss τ.865 y ss 3τ 3.95 y ss 4τ 4.98 y ss 5τ y ss Answers:.6 vols and.3 vols.9) An RC circui has paramers equaion R = R R h a b, τ = R C h, K= Rh and is described by he firs ordeer For his circui R a = Ω, R b = Ω, and C = mf dvc () τ + vc () = Kiin () d.the inpu curren is Copyrigh Rober D. Throne 3

32 ma <. sec iin () = ma. <.4 sec 3 ma.4 sec a) Deermine an analyical expression for he volage across he capacior in each of hese regions b) Ener your analyical expression as an anonymous funcion in Malab, and simulae he sysem using Simulink. Show ha all hree of your answers are idenical. Turn in your Malab (driver file) code and your nealy labeled plos. Be sure o se he iniial value of he inegraor o zero..) An RL circui has paramers R h = R a + R b + R c, τ = L/ Rh, K= ( R a + R b )/ R h and is described by he firs ordeer equaion dil () τ + il () = Kiin () d For his circui, Ra = R b = Rc = Ω and L = 8 mh.the inpu curren is.a <.4 ms in() =.3 A.4 <. ms. 5 A. ms a) Deermine an analyical expression for he curren hrough he inducor in each of hese regions b) Ener your analyical expression as an anonymous funcion in Malab, and simulae he sysem using Simulink. Show ha all hree of your answers are idenical. Turn in your Malab (driver file) code and your nealy labeled plos..) An RC circui has paramers Rh= Ra Rb, τ = R C, h K= Rh and is described by he firs ordeer equaion dvc () τ + vc () = Kiin () d For his circui R = kω, R = 5 kω, C =. mf and he iniial volage is v () =.5V The inpu curren is a b ma <. sec iin () = 4 ma. <.3 sec 3 ma.3 sec a) Deermine an analyical expression for he volage across he capacior in each of hese regions c Copyrigh Rober D. Throne 3

33 b) Ener your analyical expression as an anonymous funcion in Malab, and simulae he sysem using Simulink. Run he simulaion for second (from o second). Show ha all hree of your answers are idenical. Turn in your Malab (driver file) code and your nealy labeled plos. Be sure o se he iniial value of he inegraor o he correc value..) An RL circui has paramers Rh= Ra Rb, τ = L/ Rh, = and is described by he firs ordeer K Ra equaion For his circui, R a = Ω, R b = 4Ω, L = 3 mh inpu curren is v dil () τ + il () = Kvin () d in and he iniial curren hrough he inducor is zero. The V <.6 ms () = V.6 ms < ms 6V ms a) Deermine an analyical expression for he curren hrough he inducor in each of hese regions b) Ener your analyical expression as an anonymous funcion in Malab, and simulae he sysem using Simulink. Run he simulaion for 5 ms. Show ha all hree of your answers are idenical. Turn in your Malab (driver file) code and your nealy labeled plos. Be sure o se he iniial value of he inegraor o he correc value..3) For each of he following firs order differenial equaions, use an inegraing facor o wrie y () as a funcion of is iniial value y ( ) and an inegral of he inpu (plus some oher funcions) ay () + by () = cx () y () y () = x ( + ) y () + y () = x () y () + cos() y () = ex () Answers: Copyrigh Rober D. Throne 33

34 b b ( ) c ( λ ) a a ( λ ) y ( e y ( e a y () = e ) + x( λ) dλ y () = e ) + x( λ+ ) dλ y xλ dλ y e e xλ dλ λ sin( ) + sin( ) sin( ) λ+ sin( λ) () = y ( ) + ( ) () y ( ) e ( ) = +.4) For each of he following firs order differenial equaions, use an inegraing facor o wrie y () as a funcion of is iniial value y ( ) and an inegral of he inpu (plus some oher funcions) y () = ay() + bx() y () = ay() + bx() y () + y () = x () y () + y () = x ( ) Answers: a a ( ) ( λ ) a ( ) a ( λ ) y( be y( be y () = e ) + x( λ) dλ y () = e ) + x( λ) dλ λ λ () = y ( ) + ( ) () y ( ) e ( ) = + y xλ dλ y e xλ dλ Copyrigh Rober D. Throne 34

35 3. Second Order Circuis A second order circui is a circui wih wo effecive energy sorage elemens, eiher wo capaciors, wo inducors, or one of each. (In some circuis i may be possible o combine muliple capaciors or inducors ino one equivalen capacior or inducor.) We begin his secion wih he derivaion of he governing differenial equaion for various second order circuis. A his poin we will focus on circuis ha we can pu ino a sandard form. Once we have covered Laplace ransforms we will analyze differen ypes of second order circuis. This sandard second order form will again allow us o easily deermine physical characerisics of he circui and predic he ime response. We hen solve he differenial equaions for he case of a consan inpu. U3. Governing Differenial Equaions for Second Order Circuis: Sandard FormU In his secion we derive he governing differenial equaions ha model various RL, RC, and RLC circuis. We hen pu he governing second order differenial equaions ino a sandard form, which allows us o read off descripive informaion abou he sysem very easily. The sandard form we will use is d y () dy( ) + ζωn + ωny ( ) = Kωnx ( ) d d or d y () ζ dy () + + y () = Kx () ωn d ωn d Here we assume he sysem inpu is x () and he sysem oupu is y (). ω n is he sysem naural frequency, which indicaes he frequency a which he sysem will oscillae if here is no dampling. The naural frequencyω n has unis of radians/second. ζ is he damping raio, which indicaes how much damping here is in he sysem. A damping raio of zero indicaes here is no damping a all. The damping raioζ is dimensionless. K is he saic gain of he sysem. For a consan inpu of ampliude A dy() d y () ( x() = Au(), where u () is he uni sep funcion), in seady sae we have =, =, and d d y() = Kx() = KA. Hence he saic gain les us easily compue he seady sae value of he oupu. To deermine he unis of he saic gain we use or [unis of y] = [unis of K][unis of x] [unis of K] = [unis of y]/[unis of x] Noe ha no all second order circuis can be modeled by a differenial equaion of his form. While we can always wrie he lef hand side of he differenial equaion in his form, for some circuis he righ hand side of he differenial equaion may conain derivaives of he inpus. In addiion, his form may no always be he bes way o wrie he differenial equaion. Copyrigh Rober D. Throne 35

36 UExample 3... UConsider he RLC circui shown in Figure 3.. The inpu, x (), is he applied volage and he oupu, y (), is he volage across he capacior. If we denoe he curren flowing in he circui as i (), hen applying Kirchhoff s volage law around he single loop gives us he equaion di() x ( ) + L + y () + ir () = d We can also relae he volage across he capacior wih he curren flowing hrough he capacior dy() i () = C d Subsiuing his equaion ino our firs expression we ge or d y () dy () x() + LC + y() + RC = d d LC d y () dy() + RC + y() = x () d d Comparing his expression wih our sandard form we ge naural frequency: LC ω =, or ω n = LC n damping raio: ζ RC ω =, or R ζ = n C L saic gain: K = Figure 3.. Circui for Example 3... Example 3... UConsider he RLC circui shown in Figure 3.. The inpu, x (), is he applied curren and he oupu, y (), is he curren hrough he inducor. If we denoe he node volage a he op of he circui as v * (), hen applying Kirchhoff s curren law give us Copyrigh Rober D. Throne 36

37 * * v () dv () x ( ) = + y() + C R d We can also relae he volage across he inducor wih he curren flowing hrough he inducor * dy() v () = L d Subsiuing his equaion ino our firs expression we ge or Ldy x ( ) = + y() + LC R d d LC () d y () d y d R d () Ldy () + + y() = x() Comparing his expression wih our sandard form we ge naural frequency: LC ω =, or ω n = LC n damping raio: ζ L ω = n R, or ζ = R L C saic gain: K = Figure 3.. Circui used in Example 3... Example UConsider he RLC circui shown in Figure 3.3. The inpu, x (), is he applied curren and he oupu, y (), is he curren hrough he inducor. If we denoe he node volage a he op of he circui as v * (), hen applyingkirchhoff s curren law give us * () x ( ) = y() + C dv d Copyrigh Rober D. Throne 37

38 We can hen deermine he node volage v * () as * dy() v () = Ry() + L d Subsiuing his equaion ino our firs expression we ge or d dy dy d y x () = y () + C [ Ry () + L ] = y () + RC + LC d d d d () () () d y () dy () LC + RC + y() = x() d d Comparing his expression wih our sandard form we ge naural frequency: LC ω =, or ω n = LC n damping raio: ζ RC ω =, or R ζ = n C L saic gain: K = Figure 3.3. Circui used in Example Example UConsider he RLC circui shown in Figure 3.4. The inpu, x (), is he applied volage and he oupu, y (), is he volage across resisor R b. Node volages va () and vb () are as shown in he figure. Applying Kirchhoff s curren law gives a node va () gives va() x() va() vb() va() dva() Ca = R R R d Copyrigh Rober D. Throne 38

39 which we can simplify as dva 3 () () () () va x vb = RCa d Summing he currens ino he negaive erminal of he op amp gives us va() dvb() + Cb = R d or dvb () () va = RCb d Subsiuing his expression ino our simplified equaion above we ge or Finally, we have or () () 3 dvb d dvb RC b x() vb() RC a RC b d = d d d vb () dvb () R a b + b + b = C C 3 RC v () x() d d Rb y( ) = vb ( ) R + R Ra + Rb vb () = y () R b a b resuling in he differenial equaion d y dy a b b () () R C C + 3 RC + y() = x() d d R + R a R b b Comparing his expression wih our sandard form we ge naural frequency: RCC a b ω =, or ω n = R CC n a b damping raio: ζ 3RCb ω =, or 3 ζ = n C C b a saic gain: K Rb = R + R a b Copyrigh Rober D. Throne 39

40 Figure 3.4. Circui used in Example Solving Second Order Differenial Equaions in Sandard Form In his secion we will solve second order differenial equaions he sandard form d y () dy( ) + ζω ( ) () n + ωny = Kωnx d d for a consan (sep) inpu. We will solve his equaion in wo pars. We will firs deermine he naural response, yn(). The naural response is he response due only o iniial condiions when no inpus are presen. Then we will deermine he forced response, yf (). The forced response is he response due o he inpu only, assuming all iniial condiions are zero. The oal response is hen he sum of he naural and forced responses, y () = yn() + yf () Naural Response. To deermine he naural response we assume here is no inpu in he sysem, so we have he equaion d yn() dyn( ) + ζω ( ) n + ωnyn = d d r Le s assume a soluion of he form yn() = ce, where c and r are parameers o be deermined. Subsiuing his assumpion ino he differenial equaion we ge or r ce + ζω rce + ω ce = r r r n n ce r [ r ζωnr ωn] + + = Copyrigh Rober D. Throne 4

41 If c = hen we are done, and he naural response will be y () =. This soluion cerainly saisfies he r differenial equaion. However, if c, and since e can never be zero, we mus have r + ζωnr + ωn = Using he quadraic formula, he roos of his equaion are n ζωn ± (ζωn) 4ωn r = = ζωn ± ζ ωn ωn = ζωn ± ωn ζ We now have four cases o consider depending on he value of he damping raio ζ. These four cases are: over damped ( ζ > ), criically damped ( ζ = ), undamped ( ζ = ), and under damped ( < ζ < ). We will consider each of hese in urn. Overdamped ( ζ > ). In his case we have wo real and disinc roos, The naural response is hen r = n + n r ζω ω ζ ζω ω ζ = n n y () = ce + ce n r r where c and c are consans o be deermined by he iniial condiions. Noe ha boh r and r are always negaive, since ζ > ζ. Criically Damped ( ζ = ). In his case we iniially appear o have only one soluion, r = r = r = ωn From differenial equaions, we know ha in his siuaion we should look for an addiional soluion of he form r yn() = ce Le s check o see if his works. Wih ζ =, he differenial equaion becomes d yn() dyn( ) + ω ( ) n + ωnyn = d d We have hen dy () n d ( ce r ) ce r cre r d = d = + d y () n d ( ce ) d ce cre cre cr e cr e cre cr e d = d = d + = + + = + r r r r r r r r Copyrigh Rober D. Throne 4

42 Subsiuing hese ino he differenial equaion we ge [ ] [ ] [ [ cre r + cr e r + ω r r r ] r ] r nce + cre + ωn cre = ce r + ωnr + ωn + ce [ r + ωn] = Since we know r = ωn his equaion is clearly saisfied. Hence our naural soluion in his case will be of he form r r y () = c e + c e n where c and c are consans o be deermined by he iniial condiions. Undamped ( ζ = ). In his case here is no damping, and he sysem oscillaes a frequency ω n. The naural response is of he form y ( ) = c sin( ω + φ ) where c and φ are consans o be deermined by he iniial condiions. n n Under Damped ( < ζ < ). In his case will have wo complex conjugae roos, which we can wrie as r = ζω ± jω ζ = ζω ± jω n n n d ω ω ζ on, i will be usually easier o remember he roos of his equaion as d = n is he damped frequency. This is he frequency his sysem will oscillae wih. As we go r = ζω ± jω = σ ± jω n d d I is also useful a his poin o examine he roos of his equaion in he complex plane and see wha we can deermine. Figure 3.5 illusraes he relaionships we will be discussing. Firs of all, since our equaion for he roos is real we mus have complex conjugae roos o he equaion, which he figure shows. If we look a he magniude of he roos, we ge * r = r r = ( ζωn) + ( ωn ζ ) = ζ ωn + ωn ωnζ = ωn So he roos will all lie on a circle wih magniude r = ωn. Secondly, if we look a he angle made wih he negaive real axis, we can see ha ζωn cos( θ) = = ζ ω n Copyrigh Rober D. Throne 4

43 + + Figure 3.5. Relaionship beween he locaion of he complex roos(+) and he naural frequency ( ω, he magniude of he roos), he damped frequency ( ωd = ωn ζ, he imaginary par of he roo and he frequency of oscillaion), and he damping raio (ζ, cos( θ) = ζ ). When ζ = (undamped ) he angleθ = 9 o and he roos are purely imaginary. In his case ωd = ωn and he sysem jus oscillaes a he naural frequency. When ζ = (criically damped) he angleθ = o and he roos are purely real and are repeaed. If ζ boh roos are real, no repeaed, and are on he real axis. In his caseω d = and here is no oscillaion. Beween hese exremes we have an under damped sysem ( < ζ < ). When ζ = (undamped ) he angleθ = 9 o and he poles are purely imaginary. In his case ωd = ωn and he sysem jus oscillaes a he naural frequency. When ζ = (criically damped) he angleθ = o and he poles are purely real. In his caseω d = and here is no oscillaion. Beween hese wo exremes we have an under damped sysem ( < ζ < ). Our soluion a his poin for he naural response o he under damped sysem is hen n ( ζωn+ jωd ) ( ζωn jωd ) ζωn jωd jωd yn() = ce + ce = e ce + ce Copyrigh Rober D. Throne 43

44 Since we wan a real valued soluion, le s make an assumpion abou he relaionship beween he wo unknown consans. Assume c j c e φ = j Then we have c c y n() = e j c = e j e jφ e ζωn j( ωd+ φ) j( ωd+ φ) Finally we expand his ou using Euler s ideniy o ge c ζωn y ( ) n = e {cos( ωd+ φ) + jsin( ωd+ φ)} { cos( ωd+ φ) jsin( ωd + φ)} j or [ ] ζωn y () = ce sin( ω + φ) where c andφ are consans o be deermined by he iniial condiions. n d 3.3. Forced Response. To deermine he forced response we mus know he sysem inpu, x (). For now we will assume an inpu ha is zero before = and hen has consan ampliude A for, < x () = A Then for we have he equaion d yn() dyn() + ω ( ) n + ωnyn = KA d d Since his is a linear ordinary differenial equaion we only need o find one soluion. One obvious d yf () dyf () soluion o his equaion is he soluion in seady sae, when = =. In seady sae we d d have y f () = KA Noe ha for a consan inpu, he seady sae oupu is he produc of he saic gain and he ampliude of he inpu. Copyrigh Rober D. Throne 44

45 3.3.3 Toal Soluion. The oal soluion o our differenial equaion is he sum of he naural and forced responses, which is summarized below: ζ = undamped imaginary roos y( ) = KA + csin( ω + φ) n ζωn n d ωn ωn < ζ < under damped complex conjugae roos y( ) = KA + ce si ( ω + φ) ζ = criially damped real, repeaed roos y() = KA + c e + c e ζ > ( ζωn+ ωn ζ ) ( ζωn ω n ζ ) ) over damped real, disinc roos y( = KA + c e + c e 3.4 Response of Under Damped Sysems a Res For he under damped case, we have he soluion ζωn y() = KA + ce sin( ω + φ) We will deermine he soluion assuming he sysem is iniially a res ( y () = and y () = ). Le s look firs a he derivaive erm, dy() ζωn ζωn = ce ( ζωn)sin( ωd + φ) + ce ( ωd)cos( ωd + φ) d A he iniial ime ( = ) we will have or Hence y () = ζω sin( φ) + ω cos( φ) = n sin( φ) ωd ωn ζ ζ an( φ) = = = = cos( φ) ζω ζω ζ n d n d ζ φ = an ζ From his we can deermine ha he hypoenuse of he riangle is r =, so ha snφ i ( ) = ζ and cos( φ) = ζ. Nex we use he iniial condiion ha he iniial posiion is zero, y() = = KA + csin( φ) or KA KA c = = sin( φ) ζ Finally, our soluion for y () is Copyrigh Rober D. Throne 45

46 ζωn y( ) = KA e sin( ω ) d + φ ζ ζ φ = an ζ Figure 3.6 shows he response (oupu) of a sysem iniially a res. For his sysem KA = and ω n = rad/sec. There are four responses for under damped sysems ( < ζ <. ) and one response for a criically damped sysem ( ζ =.). The only hing changing in he responses shown in Figure 3.6 is he damping raio, so i should be clear ha he damping raio can affec he response of he sysem quie a lo. For example, he peak ampliude and ime a which he sysem reaches he peak ampliude is differen for he differen responses. Similarly, he frequency a which he sysem oscillaes, he damped frequency, is differen for he differen responses. Finally, he ime i akes he sysem o reach seady sae is differen for he differen responses. For sysems which fi ino our sandard second order form, we can predic he response of he sysem and characerize he response in erms of our parameers ( ωn, ζ, K ). The mos common characerizaions are depiced in Figure 3.7, which shows he response of a second order sysem ( ωn = rad / sec, ζ =.5, K =.) iniially a res o a uni sep inpu (an inpu of consan ampliude. saring a ime zero). Typical characerizaions of second order sysems shown in he figure are () he ime o peak ( T p ), he ime i akes he oupu o reach is peak value; () he percen overshoo ( PO ), which indicaes he amoun he larges peak of he oupu overshoos he final (seady sae) value of he oupu; and (3) he seling ime ( T s ), which indicaes he ime i akes for he ransiens in he oupu o seling ou. Afer he seling ime he sysem oupu remains wihin ± % of he final (seady sae) value. (This is a % definiion of seling ime, % definiions are also used, hough he % definiion is no as commonly used.) In he following secions we indicae how we can deermine hese quaniies in erms of he parameers ha characerize our sysem Time o Peak. From our soluion o he response of our under damped second order sysem o a sep inpu, we can deermine he ime a which y () reaches is peak value by aking he derivaive of y () and seing i equal o zero. This will give us he maximum value of y () and he ime his occurs a will be called he ime o peak, T p. dy() KA = [ ζω sin( ) cos( )] n ωd+ φ + ωd ωd+ φ = d ζ ζω sin( ω + φ) = ω cos( ω + φ) n d d d ωd ωn ζ ζ an( ωd + φ) = = = ζω ζω ζ n n Copyrigh Rober D. Throne 46

47 ζ ωd + φ = an = φ ζ Hence ω a he ime o peak, d = Tp, mus equal one period of he angen, which is π, so π Tp = ω Remember haω is equal o he imaginary par of he complex roos of d d r + ζω r + ω = n n ζ =. ζ =. ζ =.5 ζ =.7 ζ =.. y() Time (sec) Figure 3.6. The response (oupu) of a sysem iniially a res. For his sysem KA = and ω n = rad/sec. There are four responses for under damped sysems ( < ζ <. ) and one response for a criically damped sysem ( ζ =.). The only hing changing in he responses is he damping raio. Copyrigh Rober D. Throne 47

48 3.4. Percen Overshoo. Evaluaing y () a he peak ime T p we ge he maximum value of y (), ζωntp yt ( p) = KA e sin( ω ) dtp + φ ζ y( T p π ζωn d ) KA e ω π = sin( ω ) d + φ ζ ωd ζπ ζ yt ( p ) = KA + e sin( φ) ζ We ge he las equaion by using he fac ha sin( φ+ π) = sin( φ). Finally, since we have previously deermined ha sin( φ) = ζ, The percen overshoo is defined as yt ( p ) = KA + e ζπ ζ yt ( p ) y( ) Percen Overshoo = PO = % y( ) Noe ha his is a sandard definiion for percen overshoo, independen of he sysem order or ype of sysem we are analyzing. Noe also ha he reference level is he value of he funcion in seady sae. For our under damped second order sysem we have y( ) = KA, so we have PO = KA ζπ ζ + e KA KA % or PO = e ζπ ζ % Seling Time. The seling ime of a sysem is defined as he ime i akes for he oupu of a sysem wih a sep inpu o say wihin a given percenage of is final value. We will use he % seling ime crieria, which is generally four ime consans, T = 4τ. For any exponenial decay, he general form is wrien as e / τ, where τ is he ime consan. Funcions of he form s e / τ cos( ω + φ) or d Copyrigh Rober D. Throne 48

49 e / τ sin( ω + φ) such as we have in our soluion will oscillae, bu will sill decay a he same rae as he d exponenial alone. For our sysem we have τ = or σ = ζωn, where σ is he absolue value of he ζω n real pars of he soluions o r + ζω r + ω =. Hence for our sysem we esimae he seling ime as n n 4 4 Ts = 4τ = = ζω σ n Figure 3.7. The response (oupu) of a sysem iniially a res. For his sysem KA =, ω n = rad/sec, and ζ =.5. Typical characerizaions of second order sysems () he ime o peak ( T p ), he ime i akes he oupu o reach is peak value; () he percen overshoo ( PO ), which indicaes he amoun he larges peak of he oupu overshoos he final (seady sae) value of he oupu ; and (3) he seling ime ( T s ), which indicaes he ime i akes for he ransiens in he oupu o seling ou. Afer he seling ime he sysem oupu remains wihin ± % of he final (seady sae) value. Copyrigh Rober D. Throne 49

50 3.5 Second Order Sysem Examples Example Consider he circui used in Example 3.. wih parameer values, L = mh, C = µ F, R = 4 Ω. Assume he inpu is.5 amp sep, x ( ) =.5 u ( ). We can hen deermine he parameers: ω = n 3,6. rad / sec 3 6 LC = ( )( ) = 3 L ζ = = = R C ()(4) K = π π π Tp = = = =.sec ω d ω ζ 3, n T s 4 4 = = =.3sec ζω (.395)(36.) n ζπ.395π ζ.395 PO = e % = e % 6% The ime domain expression for he oupu is hen given by ζ.395 φ an = = an = an [.357] =.647 rad ζ.395 ζωn 49.7 ( ) y = KA e sin( ω ).5[.9 sin( )] d + φ = e + ζ The response for his sysem is shown in Figure 3.8. The ime o peak, seling ime, and percen overshoo are displayed on he figure. Copyrigh Rober D. Throne 5

51 Figure 3.8. Response of he circui used in Example 3.5. wih parameer values, R = 4 Ω. Assume he inpu is.5 amp sep. L = mh, C = µ F Example Consider he circui used in Example 3..4 wih parameer values, C = µ F, Cb = µ F, R= kω, R = 3 kω, R = kω. Assume he inpu is a vol sep, x () = u (). We can a hen deermine he parameers: ω = n 36. rad / sec R CC = ( )( ) = a b 6 3 Cb 3 ζ = = =.47 6 C a 3 Rb K = = = Ra + Rb 3 + π π π Tp = = = =.sec ω d ω ζ n b a Copyrigh Rober D. Throne 5

52 T s 4 4 = = =.7 sec ζω (.47)(36.) n ζπ.47π ζ.47 PO = e % = e % 8.5% The ime domain expression for he oupu is hen given by ζ.47 φ an an = = = an [.657] =.85 rad ζ.47 ζωn 48.6 ( ) y = KA e sin( ω ).4[.39 sin(79..85)] d + φ = e + ζ The response for his sysem is shown in Figure 3.9. Noe ha alhough he inpu is posiive, he oupu has a negaive seady sae value. This is because he saic gain is negaive. However, all of he parameers we are ineresed in (ime o peak, seling ime, or percen overshoo) are sill measured in he same way. Example Consider he response of an unknown second order sysem, wih sep inpu 4 vols, x () = 4 u (). The measured oupu is also in vols. The response of his unknown sysem is shown in Figure 3.. We wan o ry and deermine he sysem parameers from he sysem oupu. For his sysem, he seady sae value is yss = y( ) = b =.5. Since we know he inpu was 4, we have y.5 KA = yss = K(4) =.5, or K = ss = =.375. The % seling ime T s occurs when 4 4 y ( ) y ss <. for all Ts. Based on his graph, his occurs somewhere near.6 seconds, so Ts.6 sec. The ime o peak T p can be measured off he graph o be approximaely. seconds, so Tp.sec.To deermine he percen overshoo, we firs need he seady sae value of he oupu, which we have deermined, and hen how much beyond his he sysem has ravelled. For his sysem we have y( ) = yss = b=.5 and yt ( p ) y( ) = a a.5 Hence he percen overshoo is given by PO = % = % 7%. I is imporan o b.5 remember ha he percen overshoo is measured from he seady sae value of he oupu, no from he value of he inpu. We can now use hese parameers, or various combinaions of hem, o deermine he damping raio and naural frequency. Copyrigh Rober D. Throne 5

53 Figure 3.9. Response of he circui used in Example 3.5. wih parameer values C = µ F, Cb = µ F, R= kω, R = 3 kω, R = kω. The inpu is a vol sep. Noe ha alhough he inpu o a b he circui is a posiive volage, he oupu is negaive because he saic gain is negaive. a Copyrigh Rober D. Throne 53

54 Figure 3.. Oupu of an unknown second order sysem, analyzed in Example Example Consider he circui used in Example 3..3 wih parameer values, C = mf, L = mh, R = Ω. Assume he sysem is iniially a res and he inpu is a amp sep, x () = u (). We wan o deermine and characerize he oupu of he sysem. We sar by deermining he sysem parameers: ω = n 36. rad / sec 3 3 LC = ( )( ) = 3 R C ζ = = =.58 3 L Since ζ >. we have an over damped sysem. We know from our previous soluion we will have a soluion of he form y() = KA + c e + ce r r Copyrigh Rober D. Throne 54

55 where and r ζω ω ζ = n + n = (.58)(36.) + (36.).58.7 r ζω ω ζ (.58)(36.) (36.) = n n = In order o deermine he complee response we will use he general form of he soluion in wha follows. Because he sysem sars a zero we have he condiion y() KA c c = = + + or c c + = KA Because he sysem sars a res we need o look a he slope A he iniial ime hen we also have y () = rce + rc e r r r y () = rc + rc = or c = c r Combining hese condiions we have r r r r c+ c = c + c = c = c = KA r r r or c r = KA and c r r r = r r For our sysem his becomes KA. Our soluion is hen r r y() = KA e + e r r r r r r y ( ) =.45e +.45e The response of his sysem is displayed in Figure 3.. As his figure shows, here is no overshoo, so deermining he ime o peak or percen overshoo is meaningless. We can, however, deermine he 4 seling ime. However, we canno use our previous formula Ts =, since his was derived for an ζωn under damped sysem. Wha we need is o use he more general form ha he seling ime is equal o / four ime consans, Ts = 4τ. Recall ha he general form of a decaying exponenial is e τ, whereτ is he sysem ime consan. For his sysem we are going o have wo ime consans since we have wo exponenial erms,.7 / τ e = e, τ = =.887 sec / τ e = e, τ = =.3 sec 887. Copyrigh Rober D. Throne 55

56 The sysem response and a plo of hese exponenials is shown in Figure 3.. As his figure shows, he response of he exponenial wih he smaller ime consan is much more rapid han he response of he exponenial wih he larger ime consan. The response of he sysem is more nearly like he response of he exponenial wih he larger ime consan. Hence, o deermine he seling ime, we use he larges ime consan Ts = 4 τ (4)(.887) =.355sec This is a general resul ha we will use laer, he response of he sysem is dominaed by he response of he par wih he larges ime consan. Figure 3.. Response of sysem analyzed in Example This is an over damped sysem. The seling ime of he sysem is esimaed o be T =.355 sec. s Copyrigh Rober D. Throne 56

57 Figure 3.. Response of sysem analyzed in Example and plos of he exponenials ha make up he response. For his sysem, ( ).45 y = e +.45e. This sysem has componens wih ime consans τ = =.887 sec and τ = =.3sec. The response of he sysem is clearly dominaed by he componen of he response wih he larges ime consan. Copyrigh Rober D. Throne 57

58 Chaper 3 Problems 3.) Show ha he governing differenial equaion for he following circui is given by Rb R bccy a b () + Rb( Ca + Cb) y () + y () = x () Ra Ca + Cb Ra and ha for his circuiω n =, ζ =, and K = R CC CC R b a b a b b Copyrigh Rober D. Throne 58

59 3.) Show ha he governing differenial equaion for he following circui is given by LCR a L + RaRbC y () + y () + y () = x () Ra + Rb Ra + Rb Ra + Rb Ra + Rb and ha for his circuiωn =, ζ = LCR a L+ RRC a LCR R + R b a a b, and K = R + R a b. 3.3) For his problem, consider six second order sysems described by he following differenial equaions: y () + 9 y () + y () = Kx () y () + y () + 5 y () = 5 Kx () y () + 4 y () + 3 y () = 3 Kx () y () + 6 y () + 8 y () = 8 Kx () y () + 6 y () + 9 y () = 9 Kx () y () + 6 y () + 3 y () = 3 Kx () a) Assume he sysems are iniially a res and inpu is a sep, x() = Au(), deermine expressions for he sysem oupu by finding he forced and unforced responses and hen solving for he unknown coefficiens jus as we did in class. b) For he sysems wih real roos, show ha your soluion mees he wo iniial condiions ( y () = y ( ) = ). For he sysems wih complex roos, deermine ζ and ωn from he governing differenial equaion, and show ha your soluion agrees wih he form ( ) ζω n e y KA sin( ) = ω d + θ θ = cos ( ζ) ωd = ωn ζ ζ Do no assume his is he form of he soluion, bu use i o check your answer. Copyrigh Rober D. Throne 59

60 Answers: y() = KA 5 KAe + 4KAe, y() = KA KAe 5KAe y( ) = KA[.e sin( )], y( ) = KA + KAe KAe o o () A KAe KAe, y( KA e ) y = K 3 ) = [.83 sin( ] 3.4) The response of a second order sysem is y ( ) =.5e sin( rad) a) Take he derivaive of his funcion o deermine he ime a which he maximum occurs (he ime o peak) b) Deermine he maximum value of his funcion (he value a he ime o peak) c) Deermine he percen overshoo using your answer o (b) d) For his response deermine ζ, ω d, and ω n e) Compue he percen overshoo using he formula and verify your answer o (c). PO = e ζπ ζ % 3.5) The response of a second order sysem is ( ) [ y = 38e sin( rad)] a) Take he derivaive of his funcion o deermine he ime a which he maximum occurs (he ime o peak) b) Deermine he maximum value of his funcion (he value a he ime o peak) c) Deermine he percen overshoo using your answer o (b) d) For his response deermine ζ, ω d, and ω n e) Compue he percen overshoo using he formula and verify your answer o (c). PO = e ζπ ζ % Copyrigh Rober D. Throne 6

61 3.6) One of he mehods ha can be used o idenify ζ and ωn for mechanical sysems he log-decremen mehod, which we will derive in his problem. If our sysem is a res and we provide he mass wih an iniial displacemen away from equilibrium, he response due o his displacemen can be wrien ( ) ζωn x cos( ) = Ae ωd + θ where x () = displacemen of he mass as a funcion of ime ζ = damping raio ω n = naural frequency ω d = damped frequency = ωn ζ π Afer he mass is released, he mass will oscillae back and forh wih period given by Td =, so if we ωd measure he period of he oscillaion ( T d ) we can esimae ω d.le's assume is he ime of one peak of he cosine. Since he cosine is periodic, subsequen peaks will occur a imes given by n = + ntd, where n is an ineger. a) Show ha x( ) ζωntn d = e x ( n) b) If we define he log decremen as x( ) δ = ln x ( n) show ha we can compue he damping raio as δ ζ = 4n π + δ c) Given he iniial condiion response shown in he Figures on he nex page, esimae he damping raio and naural frequency using he log-decremen mehod. (You should ge answers ha include he numbers 5,.,. and 5, approximaely.) Copyrigh Rober D. Throne 6

62 Figure. Iniial condiion response for second order sysem A. Figure. Iniial condiion response for second order sysem B. Copyrigh Rober D. Throne 6

63 4. Sysem Properies In his chaper we will sar looking a various properies ha can be used o characerize a sysem. We will iniially illusrae hese conceps as much as possible wih examples from circuis and mechanical sysems. However, since hese are general conceps we will begin o explore absrac sysems described only by algebraic, inegral, or differenial equaions. Our goal is o be able o be able o deermine wheher or no a mahemaical model of a sysem possesses hese properies, and o develop he necessary vocabulary. U4. Linear (L) Sysems Le s assume we have a sysem wih an inpu x() producing and oupu y(). We can wrie his graphically as x ( ) y ( ). A sysem possesses he scaling or homogeneiy propery if αx () αy () for any consan α and any inpu x (). In paricular, if a sysem possesses he homogeneiy propery, if he inpu is zero he oupu will be zero ( α = ), if he inpu is doubled he oupu is doubled ( α = ), and if we change he sign of he inpu we also change he sign of he oupu ( α = ). These are very simple and common ess ha can quickly be used o deermine if a sysem does no possess he homogeneiy propery. Nex le s assume we name possible wo inpus as x () and x (), and we name he corresponding oupus y () and y (). Hence we know x() y() and x() y(). A sysem possesses he addiiviy propery if α x () + α x () α y () + α y () for all consans α and α, and all inpus x () and x (). Definiion: A linear sysem is any sysem ha possesses boh he homogeneiy and he addiiviy properies. Example 4... Consider he simple resisive circui shown in Figure 4., wih he sysem inpu defined in ou as he inpu volage v () and he sysem oupu defined as he curren flowing in he circui, i (). For his simple sysem we have he mahemaical model in ou v () i () = R Clearly his model saisfies he homogeneiy requiremen, since if we scale he inpu by a consanα we also scale he oupu by α, in ou α v () α i () = R Copyrigh Rober D. Throne 63

64 in Le s nex assume we have wo inpus v () in and v (), as shown in Figure 4.. If we use superposiion, we replace each volage source wih a shor circui and deermine he resuling oupu curren for each inpu volage source acing alone. This gives us in in ou v () ou v () i () = and i () = R R Adding hese we clearly have in in in ou ou ou v () + v () v () i () = i () + i () = = R R If should be clear for his example ha if we scale boh of he inpus, we would also scale boh of he oupus, so in in ou ou α v () + αv() α i () + αi () = R Since he sysem has boh he propery of homogeneiy and addiiviy, he sysem is linear. Figure 4.. Circui used in Example 4... Figure 4.. Second circui used in Example 4... Copyrigh Rober D. Throne 64

65 Example 4... Consider he simple resisive circui shown in Figure 4.3, wih sysem inpu equal o he in ou inpu volage v () and he sysem oupu equal o he volage v () measured across he resisor R b. We assume he volage v. The curren flowing hrough he circui is in v ( ) v i( ) = Ra + Rb and he oupu volage is in ou v () v v ( ) = i ( ) Rb = Rb Ra R + b Now le s check o see if he homogeneiy condiion is saisfied. If he sysem inpu (he inpu volage) in is zero, v () =, hen we expec he sysem oupu (he oupu volage) o be zero. However, i is clear ha under hese circumsances he sysem oupu will be ou v v () = Rb Ra + Rb Hence he homogeneiy condiion is no saisfied, and hus he sysem is no linear. I is imporan o noe in his example ha we need o look carefully a he sysem inpu and he sysem oupu. Figure 4.3. Circui used in Example Example Consider he circui shown in Figure 4.4. The sysem inpu is he inpu volage v () and he sysem oupu is he volage across he capacior, v ( ). The curren flowing in he circui is given by ou in We hen have ou d ( ) i( ) = C v d Copyrigh Rober D. Throne 65

66 or ou in dv () ou v () C R= v () d ou dv () ou in + v () = v () d RC RC We can solve his using inegraing facors, as before, ( ) d v ou / / () e RC = v in () e RC d Nex we inegrae from some iniial ime up o he curren ime,, v ou / RC in λ/ RC ( e λ ) d v ( λ) d λ = v ( λ) e dλ dλ Finally, we have he inpu-oupu relaionship ou / RC ou / RC in λ/ RC () e v ( ) e = v ( λ) e o dλ v ou ou ( )/ RC in ( λ) () = ( ) + v ( λ) v e e dλ Figure 4.4. Circui used in Example Firs le s check for homogeneiy. If he inpu is zero, hen we expec he oupu o be zero. However, in his case, if he inpu is zero, he oupu will be ou ou ( )/ RC v () = v ( ) e Hence, in order for he sysem o possess he homogeneiy propery, he iniial condiions mus be zero. This is a general requiremen for all sysems. Le s hen assume he iniial condiions are zero, hen we have Copyrigh Rober D. Throne 66

67 If we scale he inpu, we scale he oupu, Finally, if and v ou in ( λ) () = v ( λ) e dλ ou in ( λ) αv () αv ( λ) = e dλ ou in = v v () ( λ) e ( λ) dλ hen v ou in = v () ( λ) e ( λ) dλ ou ou ou in v + v = + v ( λ) α () α () α v () α ( λ) e dλ Hence he sysem also mees he addiiviy condiion and is hus a linear sysem. Example The following models of sysems, wih sysem inpu x (), and sysem oupu y (), do no saisfy he homogeneiy condiion, and hence are no linear models: 4. Tesing for Linear Sysems y () = x () + y() = x ( ) y ( ) = e x () y ( ) = sin( x ( )) y ( ) = x () We will presen wo differen, hough equivalen, mehods for esing for lineariy of a mahemaical model of a sysem. A mahemaical model of a sysem mus pass one of hese (or an equivalen) es for he model o be linear. I is much easier o show a mahemaical model of a sysem is no linear han i is linear. The firs mehod we will demonsrae assumes we have an algebraic or inegral relaionship beween he sysem inpu and he sysem oupu. This es is sraighforward, bu we do no wan o have o solve a differenial equaion o use i. Thus, we will presen a second mehod o use wih differenial equaions. In order o simplify noaion, we will assume we have a sysem operaor, which we will denoe as. The oupu of a sysem is he resul of he sysem operaor operaing on he inpu. Hence if he sysem inpu is x () and he sysem oupu is () y ( ) = x( ) y, hen we would have { } Copyrigh Rober D. Throne 67

68 Consider he wo signal flow diagrams shown in Figure 4.5. In he op figure, we examine he oupu of he sysem, z (),when he inpu o he sysem is he inpu, αx() + αx(), z () = { α x () + α x () }. In he boom figure, we examine he oupu of he sysem o inpu () x, y () { x () } () =, hen form he linear combinaion of hese, x, y () { x () } { } { } z () = α y () + α y () = α x () + α x (). =, and inpu If he oupu is he same for boh pahs, i.e., if z() = z(), hen he sysem is linear. If his is no rue, hen he sysem is no linear. Le s illusrae he mehod wih a few examples. In he following examples, we assume he sysem inpu is x () and he sysem oupu is y (). Example 4... Consider he mahemaical model Figure 4.5. Signal flow graph of lineariy es. sysem? For his sysem, he linear operaor is { } y( ) = s in ( x ) ( ). Does his represen a linear x () = sin () x (). Along he op pah we have { α α } { α α } z () = x () + x () = sin () x () + x () Along he boom pah we have z () = α x () + α x () = α sin () x () + α sin () x () = sin () α x ( ) + α x () { } { } { } Since z () = z () he mahemaical model is linear. Copyrigh Rober D. Throne 68

69 Example 4... Consider he mahemaical model y () = x () + b. Does his represen a linear sysem? Wihou even really rying, we know his equaion does no saisfy he homogeneiy condiion, so i does no represen a linear sysem. However, le s see wha happens wih our new es. For his sysem, he x () = x () + b. Along he op pah we have linear operaor is { } { α α } { α α } z() = x() + x() = x() + x() + b Along he boom pah we have z() = α { x() } + α { x() } = α{ x() + b} + α{ x() + b} = { αx() + αx() } + b{ α+ α} Now if we compare hese, we do no have z() = z() for all possibleα andα, hence he model is no linear. ( λ) Example Consider he mahemaical model y ( ) = e x( λ) dλ. Does his represen a linear sysem? Along he op pah we have Along he boom pah we have { } { } z () = α x () + α x () = α x ( λ) + α x ( λ) dλ { } α { } z () = α x () + x () ( λ) ( λ) ( λ) = e + e = + { } z ( ) α x ( λ) dλ α x ( λ) dλ e α x ( λ) α x ( λ) dλ Since z () = z (), he mahemaical model is linear. The signal flow graph mehod we have presened works well for deermining if a mahemaical model of a sysem is linear, provided we have he oupu of he sysem wrien as an algebraic or inegral funcion of he inpu. However, i is very common o model sysems in erms of differenial equaions, and we would like o be able o deermine if a sysem modeled by a differenial equaion is linear wihou having o solve he differenial equaion, as fun as ha migh be. The second mehod we will presen for deermining if a mahemaical descripion of a sysem is linear is easier o demonsrae hen explain, bu he general idea is as follows:. Wrie wo differenial equaions, one wih sysem inpu x () and sysem oupu y (), he second wih sysem inpu x () and sysem oupu y ().. Muliply he x() y() equaion by α and he x() y() equaion by α 3. Add he equaions ogeher and regroup, we wan o wrie he resuling differenial equaion in erms of αx() + αx() and αy() + αy(). 4. Make he subsiuions X () = αx( ) + αx( ) and Y ( ) = αy( ) + αy( ) in he differenial equaion. Copyrigh Rober D. Throne 69

70 5. If he resuling differenial equaion is he same as he original differenial equaion, wih x () replaced by X () and y () replaced by Y, () hen we have shown ha αx() + αx() αy() + αy() and we can conclude ha he sysem is linear. If his is no rue, hen he sysem is no linear. Noe ha in order o saisfy he homogeneiy condiions, we mus assume he iniial condiions for he sysem are all zero. Example Consider he mahemaical model y () + sin () y () = e x (). Does his represen a linear sysem? We have y () + sin () y() = e x() and y () + sin () y() = e x() Muliplying byα and α we have αy() αsin () y() αe + = x() and αy () + α sin () y() = αe x() Adding and regrouping we have α y () + α y () + sin () α y () + α y () = e α x () + α x () [ ] [ ] [ ] Subsiuing we have Y () + sin () Y () = e X () Thus, his sysem represens a linear sysem. Example Consider he mahemaical model y () + yx () () = x (). Does his represen a linear sysem? We have y () + y() x() = x() and y () + y() x() = x() Muliplying byα and α we have αy () + αy() x() = αx() and αy () + αy() x() = αx() Adding and regrouping we have [ αy () + αy ( ) ] + αy() x() + αy() x() = [ αx( ) + αx( ) ] Subsiuing we have Y () + αy() x() + αy() x() = X () A his poins, i is clear ha we canno wrie his resuling differenial equaion jus in erms of X () and Y, () so he sysem is no linear. Example Consider he mahemaical model y ( ) + y ( ) = x ( ) +. Does his represen a linear sysem? This one is easy if we use he homogeneiy condiion. If he inpu is zero he oupu should also be zero. However, in his model, if he inpu is zero, we sill have he oupu being nonzero. Hence his model is nonlinear. We should poin ou ha his echnique can be used for sysems ha are no differenial equaions, bu i may someimes be more difficul han he flow-graph echniques. Copyrigh Rober D. Throne 7

71 4. Time-Invarian (TI) Sysems A ime-invarian (TI) sysem is one in which, if he inpu x () is delayed by an amoun T hen he oupu y () is delayed by he same amoun, wihou changing shape. Symbolically, if for a sysem we have x ( ) y ( ), hen if he sysem is also ime-invarian we will have x( T) y ( T ). Figure 4.6 presens a signal flow graph es for ime-invariance, assuming we can wrie he oupu as an algebraic funcion or inegral of he inpu. Along he op pah we delay he inpu and hen deermine he oupu. Along he boom pah, we pu he usual inpu ino he sysem and hen delay he usual oupu. If he resuls of hese wo pahs are idenical, hen he sysem is ime-invarian. There are a few subleies involved in his es, so read hrough he following examples carefully. Figure 4.6. Signal flow graph of ime invariance es. Example 4... Consider a sysem wih he mahemaical model y () = α() x (). Does his model represen a linear sysem? Along he op pah, delaying he inpu we have { } z() = x ( T) = α () x ( T) Along he second pah (delaying he oupu) we have { } z () = x () = α( T ) x ( T ) = T Clearly z() z(), so he model is no ime-invarian. Example 4... Consider he RC circui shown in Figure 4.7, wih an iniial charge on he capacior. The sysem inpu is he applied volage x () and he sysem oupu is he volage across he capacior y (). We need o wrie he oupu as a funcion of he inpu. The curren in he loop is given by x () y () dy () = C R d or Copyrigh Rober D. Throne 7

72 dy() + y () = x () d RC RC d ye / RC / RC = e x d Inegraing boh sides and rearranging we ge () () λ/ RC / RC / RC λ/ RC d = y e d y( λ) e λ y( ) e ( ) e x( λ) dλ dλ = () ( )/ ( )/ ) RC λ RC ( y = y e + e x ( λ ) dλ Along he op pah of he signal flow graph, delaying he inpu, we have { x ( T) } ) e ( )/ RC ( λ)/ RC z() = = y ( + e x( λ T) dλ Noe ha we only delay he inpu, we do no change any oher funcions. Along he second pah of he signal flow graph we only delay he oupu, which means we ake he oupu and replace all insances of wih T. This leads o { } = T ( ( T )/ RC ( T λ)/ RC z ( ) = x () y ) e + e x ( λ ) dλ = T Now we wan o see if z() = z(). Since we do no know wha he inpu is, we will change variables in he inegrals so boh of hem are simple funcions of he dummy inegral variable. z () is already in he correc form, so we need o change variables in z ( ). Le s le σ = λ T, or λ = σ + T. Then can rewrie z ( ) as T ( )/ ( )/ ) RC T σ RC = T z () ( e + x ( σ ) y e dσ Now le s compare z ( ) and z (), and deermine if hey are equal or if we can make hem equal. Firs of all, he erms associaed wih he iniial condiions canno be made equal, so if his sysem is going o be ime invarian he iniial condiions mus be zero. Boh of hem have he same inegrand, and boh have he same upper limi on he inegral. The only difference in he inegral erm is he lower limi. In order for he wo inegrals o be equal here are wo choices. The mos obvious choice is o assume ha he inpu is zero for all imes before he iniial ime. Then he lower limi on he inegral in z ( ) is effecively sill effecively, Copyrigh Rober D. Throne 7

73 T ( T σ )/ RC z () = e x( σ) dσ T T ( T σ)/ RC ( T σ)/ RC = e x( σ) dσ + e x( σ) dσ = T T e x( σ ) = i nhis range ( T σ )/ RC x( σ) dσ The oher opion is for =. Wih his choice of iniial value, he firs erms are equal and we have T ( ) = ( T σ )/ RC ( ) = ( σ) z z e x dσ Noe ha in order for a sysem described by a differenial equaion o be ime invarian, we mus generally also be aware of he iniial condiions. Figure 4.7. Circui used in Example 4... Example Consider he model of a sysem y () = x. Is his model ime-invarian? This example is somewha ricky, since we have o inerpre he ess correcly. Along he op pah, we are supposed o subrac T from he argumen of x (), so we have () { ( ) z x T} x = = T This is no he resul you expec, bu i is he correc way o inerpre he op pah of our signal flow graph. Along he boom pah we delay he oupu byt so we have () { () T z = x } = x = T Clearly z() z(), so he model is no ime-invarian. Copyrigh Rober D. Throne 73

74 Example Consider he model of a sysem y ( ) = x( ). Is his model ime-invarian? We have { } z ( ) = x( T) = x( T) and z () = { x ()} = x( + T) = T Clearly z() z(), so he model is no ime-invarian. Tesing differenial equaions for ime-invariance in general is somewha more difficul, so we will jus sae a resul: If he differenial equaion is jus a funcion of he inpu x () and he oupu y (), and hese are boh jus simple funcions of (e.g. here are no erms like x( ), y ( / ), x( ) ), hen he differenial equaion is ime-invarian if here are no oher funcions of ime oher han he inpu and oupu funcions. Example The following models of sysems are no ime invarian (hough hey are linear!): The following models of sysems are ime-invarian: y () = ex () y+ sin() y () = cos() x () y () + y () = x () y () + x () y () = x () y () + y () = x () y () + y () = x () 4.3 Causal Sysems If a sysem is causal, hen he sysem oupu y ( ) a some arbirary ime can only depend on he sysem inpu x () up unil (and including) ime. Anoher way of describing a causal sysem is ha i is nonanicipaive, he oupu does no anicipae he inpu. While mos of he sysems we hink of are causal, his is because hey work in real-ime. Tha is, we are collecing or soring daa o be processed as he daa comes in, no a a laer ime. However, many discree-ime sysems, such as an ipod or MP3 player, have daa or music sored in advance. When he music is played i is possible for he sysem o look a fuure values of he inpu (he discree-ime signal represening he music) and make adjusmens based on he fuure. This is possible because hese sysems do no need o work in real-ime. In wha follows, we again assume he sysem inpu is x (), he sysem oupu is y (), and x ( ) y ( ). Example Consider he mahemaical model of a sysem y ( ) = x ( + ). Does his model represen a causal sysem? Ofen he easies way o analyze problems like his is o pu in various values of. Copyrigh Rober D. Throne 74

75 Thus for = we have y() = x(), and clearly he oupu a ime zero depends on he inpu a ime one, and he sysem is no causal. Example Consider he mahemaical model of a sysem y( ) = e x( ). Does his model represen a causal sysem? This is a causal sysem, since he oupu y () a any ime depends only on he sysem inpu a he same ime. Remember ha we are only concerned wih inpu-oupu relaionships. The exponenial erm does no affec he causaliy of he sysem. Example Consider he mahemaical model of a sysem y () = x( ). Does his model represen a causal sysem? This sysem is no causal, since for = we have y( ) = x(), and he oupu depends on a fuure value of he inpu. Example Consider he mahemaical model of a sysem y ( ) = x. Does his model represen a causal sysem? This sysem is a bi more difficul o analyze han he previous sysems. If his sysem is causal, hen we mus have or 3. This will no be rue for all ime, so he sysem is no causal. Example Consider he mahemaical model of a sysem y () + y () = x ( + ). Does his model represen a causal sysem? In order o answer his, we will solve for he oupu as a funcion of he inpu (and review inegraion facors as an added bonus!). We have d ye () = ye () + e y () e [ y () y ()] e [ x ( ) ] d = + = + or d ye = e x [ + ] ( ) ( ) d Inegraing boh sides from some iniial ime up o an arbirary final ime we have Rearranging his we have λ = dλ ( + ) d [ y ( ) e ] d ye ( ) y ( ) e e λ λ λ = x ( λ+ ) d λ () ( ) + y y e e + λ = + x( λ+ ) dλ Now we can deermine ha he oupu a any ime, y (), depends on he inpu from ime + up unil ime +. Thus he sysem is no causal. Copyrigh Rober D. Throne 75

76 4.4 Memoryless Sysems A sysem is memoryless or insananeous if he oupu a any ime does no depend on pas or fuure values of he inpu. If he oupu does direcly depend on he inpu, hen y () mus be an algebraic funcion of x (). Example Consider he mahemaical model of a sysem y ( ) = x ( + ). Does his model represen a memoryless sysem? No, he sysem model is no memoryless, since he oupu a any ime depends on a fuure inpu. Example Consider he mahemaical model of a sysem y ( ) = x ( ). Does his model represen a memoryless sysem? No, he sysem model is no memoryless, since he oupu a any ime depends on a pas inpu. Example Consider he mahemaical model of a sysem y () = x () +. Does his model represen a memoryless sysem? Yes, he sysem model is memoryless, since he oupu a any ime depends on he inpu only a ha ime. Example Consider he mahemaical model of a sysem y () + y () = x ( + ). Does his model represen a memoryless sysem? No, he sysem model is no memoryless, since he oupu a any ime depends on pas and fuure values of he inpu. 4.5 Inverible Sysems An inverible sysem a sysem in which each oupu is associaed wih a unique inpu. Tha is, here is a one-o-one relaionship beween he sysem inpu and he sysem oupu. Example The mahemaical models of sysems y () = cos( x ()) and y () = x ) are no inverible, since here is more han one inpu ha produces he same oupu. ( 4.6 Bounded Inpu Bounded Oupu (BIBO) Sable Sysems A mahemaical model of a sysem ha produces a bounded oupu for every bounded inpu is a bounded-inpu bounded-oupu sable sysem. Noe ha we do no need o know wha he oupu is for every inpu, only ha i is bounded. In addiion, since we are looking a he inpu oupu relaionship we assume all of he iniial condiions are zero. Mahemaically, if x ( ) M for some finie consan M means y ( ) < Nfor some finie consan N, hen he sysem is BIBO sable. Example Is he mahemaical model of a sysem y () = e BIBO sable? If we assume M x ( ) M hen we have y () e = Nand he model is BIBO sable. x () Copyrigh Rober D. Throne 76

77 Example Is he mahemaical model of a sysem y ( ) = cos BIBO sable? The answer is x( ) yes, since we know ha he cosine is always bounded beween - and, so he oupu is always bounded, even if we do no know wha i is. ( λ) Example Is he mahemaical model of a sysem y () = e x( λ) dλbibo sable? Since we assume he inpu is bounded, x ( ) M, we can wrie ( λ) ( λ) λ y() = e x( λ) dλ M e dλ = Me e dλ = Me ( e ) = M ( e ) M Hence y ( ) M and he oupu is bounded, so his is a BIBO sable model. 4.7 Linear Time-Invarian (LTI) Sysems In his course we will focus our aenion on sysems ha are boh linear and ime-invarian, commonly referred o as LTI sysems. If we have an LTI sysem and know he response of he sysem o specific inpu, hen we can deermine he response of he sysem o any linear and ime shifed combinaion of hose inpus. For example, assume we know he inpu/oupu relaionships xi() yi() for various inpus xi () and he corresponding oupus yi (). Then, since he sysem is LTI, we know αixi() αiyi(), α x () α y () x y, and α x ( ) α y ( ), i( i) i( i) i i i i i i implicaions of his wih a few examples.. Le s illusrae he i i i i i i i i Example Assume ha we know ha if he inpu is x () = u (), a uni sep (Heaviside) funcion, hen he oupu of an LTI sysem will be y ( ) = e u ( ). Now assume we wan o use his informaion o deermine he response of he sysem o a pulse of widh (duraion) T and ampliude A. The firs hing we need o do is o wrie ou new inpu xnew() in erms of our known inpu. We can wrie he pulse inpu as xnew () = Au() Au( T ) = Ax() Ax( T ) Since he sysem is LTI we know Ax() Ay() Ax( T ) Ay( T ) and he new oupu will be ( T) y () = Ay() Ay( T ) = Ae u() Ae u( T ) new Copyrigh Rober D. Throne 77

78 4.8 Linearizing Nonlinear Sysem Models While we will be focusing on linear sysems, many sysems we commonly use are acually nonlinear. However, if we operae hem only over a limied range of inpus, he assumpion of lineariy is reasonably accurae. This is very common wih elecronic circuis, such as BJT and MOSFET ransisors, where we use small signal models and assume hese devices can be modeled as linear for small enough inpu signals. In general, if we have an inpu-oupu relaionship of he form, y () = f( x ()), hen we can use a Taylor series approximaion of f ha we can use for small x (). I may seem odd ha we are allowing x o be a funcion of ime, bu he idea is he same. When he inpu is zero we have y() = f(), and his provides our nominal operaing poin. We can hen approximae he oupu for small inpus as df ( x) df ( x) y () f() + x x ( ) y() x x ( ) dx = = + dx = If we look a he deviaions from he operaing poin, we have df ( x) y () y() = y () x= x() = mx() dx df ( x) where we have defined m as he slope near x =, m = x =. Wih his approximaion we have he dx linear relaionship beween y () and x (), y() = mx(). Example Consider he nonlinear sysem model y () = + x (). Deermine a linear model for small signals. We can wrie his as y() = f( ) = y ( ) = f( x) = ( + x) so df ( x ) x= = ( + x) x= = dx 4 y () x () 4 Looking a he deviaion abou he operaing poin we ge he small signal linear model y () = y () x() 4 Copyrigh Rober D. Throne 78

79 Example Consider he nonlinear sysem model small signals. We can wrie his as () y () = + 3e x. Deermine a linear model for y ( ) = f() = 4 y () = f( x) = + 3e x df ( x) 3 x x= = e x= = 3 dx so y () 4+ 3 x () We hen have he small signal linear model y () = y () 4= 3 x () Copyrigh Rober D. Throne 79

80 Chaper 4 Problems ) For each of he following mehemaical descripion of a sysem, deermine if he sysem is linear, ime invarian, causal, and memoryless and fill in he following able. Unless he sysem does no mee he homogeneiy condiion, you mus use a formal echnique o show he sysem is linear. If i does no mee he homogeneiy condiion, you mus give an example. For he memoryless and causal you can jus say i s obvious (assuming i is). You can assume y ( ) = and x () = for <. Sysem Linea (Y/N) Time Invarian (Y/N) Memoryless (Y/N) y ( ) = x() y () = + x () y () + e y ()= c o() s x( ) + y ( ) = e λ x( λ) dλ y () = e λ x( λ+ ) dλ y ( ) = e λ λx( λ) dλ y () + yx () () = x () Causal (Y/N) Answers: 5 are L, 4 are TI, is memoryless, 6 are causal ) For each of he following mahemaical descripions of a sysem, deermine if he sysem is BIBO sable. a) y () = + x () b) () y () = e x c) y ( ) = cos( x( ) ) ( λ ) ( ) ( λ) d) y = e x dλ e) y Answers: 3 are BIBO sable ( λ ) () e x( λ) = dλ f) y() = x() ( ) 3) Assume we have an LTI sysem wih he inpu-oupu relaionship x () = u () y () = e u ( ). This means ha if he inpu is a uni sep saring a zero, he oupu is a decaying exponenial saring a =. Deermnine an expression for oupu he following inpus: a) x ( ) = u ( ) b) x ( ) 6 u ( ) x ( ) = u ( ) u ( ) The answer o c is = + c) [ ] y e ( ) ( 3) () = u ( ) e u ( 3) Copyrigh Rober D. Throne 8

81 4) Malab Problem One propery of linear sysems ha we have no discussed is ha if a sysem is linear, hen if he inpu is a sinusoid, hen he oupu mus be a sinusoid. The magniude and phase can be differen beween he inpu and oupu signals, bu hey mus boh be sinusoids of he same frequency or he sysem is no linear. a) Creae a new Malab m-file. Use linspace o creae a ime vecor from o seconds using poins. Creae he inpu sinusoid x = cos(); You can jus ype his command ino your m-file, you should no use an ananymous funcion here. The variables and x will be an array of he form = [... ] = [... ] x= [cos( ) cos( )... cos( )] The sysems we wan o look a are represened by he following relaionships sysem A: y = an( x) sysem B: y = exp( x) sysem C : y = / ( + x) sysem D: y = x^ Use Malab o plo he sysem inpu and sysem oupu on four graphs (one a single page). Use he subplo command. Before your firs plo, ype orien all, his will use more of he page. Your graph should look like ha on he nex page. You will also need he commands xlabel, ylabel, grid, and legend (among ohers). You may also need o use he commands./ and.* for elemen by elemen division and muliplicaion. All of hese sysems are clearly non-linear. Turn in your code and your plo. b) Someimes, alhough a sysem is nonlinear, we can approximae he sysem as linear if we only consider small signals. If he inpu signal is small enough, we know from calculus ha we can approximae he oupu using he Taylor series (abou x =, since we assume a small inpu), dy( x) yx ( ) y() + x dx x= Even hough x is a funcion of ime, we can sill use his approximaion for funcions. We can rewrie his as dy( x) y = y( x) y( ) mx = x dx x= where dy( x) m = dx x= If his firs order approximaion is valid, hen we have he linear relaionship y mx. Now change your code so he inpu o he sysem is x = cos()/. For each of your sysems, deermine y () and plo ime () versus y-y() using he same forma as in par a. Do no plo he inpu signal, we only wan o see he oupu signal. The firs hree sysem should now approximaely be linear, while he fourh is sill nonlinear. Noe ha if you have done his correcly, he oupu signal should be cenered a zero. Turn in your code and your plo. Copyrigh Rober D. Throne 8

82 Sysem A - Inpu Oupu Sysem B 3 Inpu Oupu Sysem C Inpu Oupu Sysem D Inpu Oupu Time (sec) Copyrigh Rober D. Throne 8

83 5) Malab Problem Read he Appendix a he end of he problem secion and hen do he following: a) Creae an m-files o se up and solve marix equaions for he node values for he following circuis. Use boh mehods from he Appendix o se up he sysem of equaions. Assume R = Ω, R = 5Ω, R = Ω, R = 5Ω and V = 5V. a b c d Answer: V = 3.3 V, V =.698V a b in b) Creae m-files o se up and solve marix equaions for he node values for he following circui. Use boh mehods from he Appendix o se up he sysem of equaions. Assume R = Ω, R = 5Ω, R = Ω, R = 5Ω, R = 5Ω, R = 5Ω and V = 5V. a b c d e f Answer: Va = 3.87 VV, b =.5739 VV, c =.57V Turn in your m-files and wrie he node volages obained on your code. in Copyrigh Rober D. Throne 83

84 c) Creae m-files o se up and solve marix equaions for he node values for he following circui. Use boh mehods from he Appendix o se up he sysem of equaions. Assume R = Ω, R = 5Ω, R = Ω, R = 5Ω, R = 5Ω, R = 5Ω and V = 5V. a b c d e f Turn in your m-files and wrie he node volages obained on your code. in Copyrigh Rober D. Throne 84

85 Appendix In addiion o working well wih vecors, Malab is also very good a marices. In his Appendix we will show you how o se up and solve marix equaions in Malab. Consider he following circui, wih wo nodes V a and V b. We can wrie node equaions a hese wo nodes as Va Vin Va Va Vb + + = R R R Rearranging hese we ge a c b Vb Va Vb + = R R b d + + V + V = V Ra Rb Rc Rb Ra a b in Va + + b Rb Rb R V = d We can hen wrie hese in sandard marix form, Ax = b, as + + Vin Ra Rb Rc R b Va = R a V b + Rb Rb R d Before we pu his ino Malab, we need o assign some values o he resisors. Le s assume we have he values Ra = Ω, Rb = 5Ω, Rc = Ω, Rd = 5Ω and Vin = 5V. We can now consruc our marices in wo differen ways in Malab, as shown by he following code segmen Copyrigh Rober D. Throne 85

86 % % circui parameers % Ra = ; Rb = 5; Rc=; Rd = 5; Re = 5; Rf = 5; Vin = 5; % % Mehod % % creae marices and vecors ha are all zeros, hen fill hem in % A = zeros(,); % consruc a x marix filled wih zeros b = zeros(,); % consruc a x vecor filled wih zeros % % fill in he values % A(,) = /Ra + /Rb + /Rc; A(,) = -/Rb; A(,) = -/Rb; A(,) = /Rb + /Rd; b() = (/Ra)*Vin; % % solving he marix equaion % x = A\b % % Mehod % % consruc he marices and vecors direcly % % noe ha he columns are separaed by spaces, and each row % ends wih a semicolon (;) % A = [(/Ra+/Rb+/Rc) -/Rb; -/Rb (/Rb + /Rd)]; b = [(/Ra)*Vin; ] % % solving he marix equaion % x = A\b % % check your answer % A*x b % % exrac he node volages % Va = x() Vb = x() Copyrigh Rober D. Throne 86

87 5. Impulse Response, Sep Response, and Convoluion In his chaper we confine ourselves o sysems ha can be modeled as linear and ime-invarian, or LTI sysems. For hese ypes of sysems, we can deermine he oupu of he sysem o any inpu in a very sysemaic way. We can also deermine a grea deal abou he sysem jus by looking a how i responds o various ypes of inpus. The mos fundamenal of hese inpus is he impulse response, or he response of a sysem a res o an impulse. However, he response of a sysem o a sep is much easier o deermine and can be used o deermine he impulse response of any LTI sysem. 5. Impulse or Dela Funcions An impulse, or dela funcion, δ (), is defined as a funcion ha is zero everywhere excep a one poin, and has an area of one. Mahemaically, we can wrie his as δ( λ) =, λ µ µ δ( λ) dλ =, µ > Noe ha we do no know he value of he dela funcion a zero, i is undefined! We can hink of, or model, dela funcions as funcions ha exis in some ype of limi. For example, he funcions displayed in Figures 5., 5., and 5.3 can be hough of as differen models for dela funcions, since he mee our wo (simplisic) requiremens above. Figure 5.. Recangular model of an impulse (dela) funcion. Copyrigh Rober D. Throne 87

88 Figure 5.. Triangular model of an impulse (dela) funcion. Figure 5.3. Gaussian model of an impulse (dela) funcion. Alhough dela funcions are really idealized funcions, hey form he basis for much or he sudy of sysems. Knowing how a sysem will respond o an impulse (an impulse response) ells us a grea deal abou a sysem, and les us deermine how he sysem will response o any arbirary inpu. Copyrigh Rober D. Throne 88

89 There following wo very imporan properies of dela funcions will be used exensively: Propery : φ( ) δ( ) = φ( ) δ( ) b Propery (Sifing Propery): φ() δ( ) d = φ( ) a < < b a The firs propery is prey easy o undersand if we hink abou he definiion of a dela funcion. A dela funcion is zero everywhere excep when is argumen is zero, so boh sides of he equaion are zero everywhere excep a, and hen a boh sides have he same value. The second propery follows direcly from he firs propery as follows: b b b φ( ) δ( ) d = φ( ) δ( ) d = φ( ) δ( ) d = φ( ) a < < b a a a I is very imporan ha he limis of he inegral are such ha he dela funcion is wihin he limis of he inegral, or else he inegral is zero. Example 5... You should undersand each of he following ideniies, and how o use he wo properies o arrive a he correc soluion. eδ( ) = eδ( ) 5. Uni Sep (Heaviside) Funcions δ( ) = 4 δ( ) δ ( )d = 4 e δ ( ) d = e e δ ( ) d = δ( ) δ( ) d = We will define he uni sep funcion as τ > u( τ ) = τ < We will no define u (), hough some exbooks define u () =. The argumen of he uni sep was deliberaely no wrien as, since his someimes leads o some confusion when solving problems. I is Copyrigh Rober D. Throne 89

90 generally beer o remember ha he uni sep is one whenever he argumen (τ ) is posiive, and hen ry and figure ou wha his migh mean in erms of. Example The following are some simple examples wih uni sep funcions: a) u ( ) = for > or > b) u( ) = for > or > c) u 4 = for 4 > or > 3 3 Uni sep funcions also show up in inegrals, and i is useful o be able o deal wih hem in ha conex. The usual procedure is o deermine when he uni sep funcion (or funcions) are one, and hen do he inegrals. If he uni sep funcions are no one, hen he inegral will be zero. When you are done wih he inegral, you may need o preserve he informaion indicaing ha he inegral is zero unless he uni sep funcions are on, and his is usually done by including uni sep funcions. A few examples will hopefully clear his up. Example 5... Simplify u( λ) u( λ ) dλ as much as possible. We need boh uni sep funcions o be one, or he inegral is zero. We need hen u ( λ) = for λ > or > λ u( λ ) = for λ > or λ > The inegral hen becomes ()() dλ =. However, we are no done ye. We need o be sure boh of he uni sep funcions are, which means we need > λ >, or >. So he answer is zero for < and for >. The way we can wrie his compacly is ( ) u ( ), which is he final answer. Example 5... Simplify + λ he inegral is zero. We need hen u( λ ) = for λ > or λ > e u( λ ) dλas much as possible. We need he sep funcion o be one, or The inegral becomes + λ ( + ) e () dλ = e e = e ( e ) However, he inegral will be zero unless + > λ > or >. The final answer is hen e ( e ) u ( ). Copyrigh Rober D. Throne 9

91 Example Simplify u ( λδ ) ( λ+ ) dλas much as possible. This inegral has boh an impulse and a uni sep funcion. While we migh be emped o use he uni sep funcion o se he limis of he inegral, he bes (and easies) hing o do is o jus use he sifing propery of impulse funcions. This gives he resul u ( λδ ) ( λ+ ) dλ= u ( + ) Example Simplify ( ) e λ δ ( λ ) d λ as much as possible. We can again use he sifing propery wih his inegral, bu we mus be careful. If we do no inegrae pas he impulse funcion, he inegral will be zero. Hence we have ( ) ( ) e e λ > δ( λ ) dλ = < ( ) which we can wrie in a more compressed form as e u( ). Example Simplify + e λ δ( λ+ ) dλ as much as possible. Using he sifing propery we have + e λ δ( λ+ ) dλ = which we can wrie more compacly as e u( ). e < > Finally, if we consider inegraing an impulse, δ( λ) dλ, we will eiher ge a one (if we inegrae pas he impulse) or a zero (if we do no). Thus we have or > δ( λ) dλ = < δ( λ) dλ = u () If we differeniae boh sides of his we ge du() = δ () d This relaionship is imporan o remember, bu when doing inegrals i is generally a beer idea o remember wha condiions you may need o impose in order o deermine if and wha uni sep funcions will be required. Copyrigh Rober D. Throne 9

92 5.3 Impulse Response The impulse response of an LTI sysem is he response of he sysem iniially a res (no iniial energy, all iniial condiions are zero) o an impulse a ime =. The mos common way o denoe he impulse response is by lower case leers h and g, hough ohers are used. Example Consider he circui shown in Figure 5.4. Deermine he impulse response of he sysem. The circui is a simple volage divider, so we have R () b y = x() Ra + Rb and he impulse response is R () b h = () Ra R δ + b Figure 5.4. Circui used for Example Example Consider he circui shown in Figure 5.5. Deermine he impulse response of he sysem. We have or Then x () y () dy () y () = C + R d R dy() + y () = x () d RC RC d / RC / RC ye = e x () () d RC Inegraing from up o, and assuming he sysem is iniially a res, we have ( )/ RC y () = λ e x( λ) dλ RC The impulse response is hen given by ( λ)/ RC / RC h () = e δ( λ) dλ = e u () RC RC Copyrigh Rober D. Throne 9

93 Figure 5.5. Circui used for Example Example Consider he sysem described by he mahemaical model The impulse response will be given by 5.4 Sep Response ( λ) y ( ) = x ( ) + e x( λ+ ) dλ ( λ) ( + ) h ( ) = δ( ) + e δ( λ+ ) dλ = δ( ) + e u ( + ) The sep response of an LTI sysem is he response of he sysem iniially a res (no iniial energy, all iniial condiions are zero) o a sep a ime =. There is no common mehod for denoing he sep response, bu we will someimes denoe he sep response as s (). If we know he sep response of an LTI sysem, we can deermine he impulse response of he sysem using he relaionship d h () = [ s ()] d Example Deermine he sep response and hen use i o deermine he impulse response for he sysem in Example From he example we have The sep response is hen given by or ( )/ RC y () = λ e x( λ) dλ RC ( )/ RC s () = λ e u( λ) dλ RC / RC λ/ RC / RC / RC / RC s () = e e d e [ e ] u () [ e ] u () RC λ = = The impulse response is hen given by Copyrigh Rober D. Throne 93

94 d d / RC / RC / RC / RC h () = s ( ) = [ e ] u () = e u () + [ e ] δ () = e u () d d RC RC which is he same answer we obained before Convoluion Derivaion of he Convoluion Inegral We will derive he convoluion inegral using wo differen, hough equivalen mehods. Consider an LTI sysem wih inpu x (). We can approximae x () as a piecewise consan funcion over inervals of lengh T, as shown in Figure 5.6. Thus we have he approximaion k= x ( ) x( k T) u k T u k + T k= Nex, we can wrie he sep response of he sysem as s (). Because he sysem is ime-invarian, he response of he sysem o he inpu u ( ( k ) T) is s ( ( k ) T), and he response of he sysem o u ( ( k+ ) T) is s ( ( k+ ) T). Because he sysem is boh linear and ime-invarian, he response of he sysem o inpu x () can hen be approximaed as k= y ( ) x( k T) s k T s k + T k= Now we can approximae he derivaive of he sep response as s k T s k T + h ( k ) T or h ( k ) T s k T s k+ T Thus he oupu can be approximaed as k= y () xk ( T) h ( k T) T k= If we define λ = k T, hen as T he sum becomes an inegral and we have y () = x( λ) h ( λ) dλ This is one form of he convoluion inegral, which ells us how o deermine he oupu of an LTI sysem if we know he impulse response of he sysem and he sysem inpu. We would wrie his as y () = x () h (), where represens he convoluion operaor. Copyrigh Rober D. Throne 94

95 An alernaive derivaion for convoluion would be o sar wih he same approximaion k= x ( ) x( k T) u k T u k + T k= We can hen approximae he impulse response as or u k T u k T + δ ( k T ) T δ ( k T) T u k T u k+ T Figure 5.6. Approximaing he coninuous funcion x () as a piecewise consan funcion for he derivaion of he convoluion inegral. We hen have k= x () x( k T) δ ( k T) T k= Copyrigh Rober D. Throne 95

96 If he inpu o he LTI sysem is δ ( k T) he oupu will be h ( k T), so we can wrie he oupu as k= y () xk ( T) h ( k T) T k= If we define λ = k T, hen as T he sum again becomes an inegral and we have y () = x( λ) h ( λ) dλ By a change of variable we can show ha convoluion has he commuaive propery, or y () = h () x () = x () h () This means ha we can compue he oupu in one of wo equivalen ways: y ( ) = x( λ) h( λ) dλ = x( λ) h( λ) dλ Now ha we know how o deermine he oupu of a sysem given he impulse response and he inpu, we need o deermine he bes way o compue his. There are wo general ways for compuing he convoluion, analyical and graphical. Boh mehods give he same resuls, bu usually he answers iniially look differen. Analyical evaluaion of he convoluion inegral is generally used for very simple problems, and becomes unwieldy for very complicaed problems. Graphical evaluaion of he convoluion inegral is usually used for more complicaed problems and is also useful for visualizing wha is happening o he signals during convoluion. Analyical Evaluaion of he Convoluion Inegral In analyical evaluaion, we basically jus evaluae he inegral. I is necessary o uilize any sep funcion in he impulse response or sysem inpu o change he limis of he inegral. In addiion, i is imporan o remember o include any necessary uni sep funcions on he oupu, since he oupu of one sysem may be he inpu o anoher sysem. Example Deermine he oupu of an LTI sysem wih impulse response h() = Ae u() o inpu x( ) = Bu( ) Bu( ). To solve his problem we mus firs choose he way we are going o perform he convoluion. We will use he form Subsiuing in our funcions we have y () = h ( λ) x( λ) dλ ( λ)/ τ y( ) = Ae u( λ)[ Bu( λ ) Bu( λ )] dλ / τ Copyrigh Rober D. Throne 96

97 or ( λ)/ τ ( λ)/ τ y( ) = ABe u ( λ) u( λ ) dλ ABe u ( λ) u( λ ) dλ Using he sep funcions o change he limis on he inegrals we have Finally we have ( λ)/ τ ( λ)/ τ / τ λτ / / τ λτ / y() = ABe dλ ABe dλ = ABe e dλ ABe e dλ / τ / τ / τ / τ / τ / τ y( ) = ABe τ[ e e ] u( ) ABe τ[ e e ] u( ) ( )/ τ ( )/ τ = τab[ e ] u( ) τab[ e ] u( ) Graphical Evaluaion of he Convoluion Inegral As wih analyical evaluaion, he firs hing o do is decide which of he wo forms of he convoluion inegral o use. Le s assume ha we are going o use he form y () = h ( λ) x( λ) dλ We need o keep in mind ha we wan he area under he produc of wo funcions, h ( λ) and x( λ ). In addiion, we need o remember ha we are inegraing wih respec o he dummy variable λ, no. This is imporan o undersand, since he funcion h ( λ) will be a differen places along he λ axis as he variable varies. In fac, he whole poin of doing graphical convoluion is o skech he funcion h ( λ) as a funcion of and x( λ ), deermine he overlap, and hen perform he inegraion. One simple mehod for being able o locae h ( λ) as a funcion of and λ is o look a h () and find suiable marker poins. Le s call wo such poins and. Then we can find where hese marker poins are on he λ axis as follows This will all make more sense wih a few examples. h ( ) = h ( λ) λ = h ( ) h ( ) = λ λ = Example Deermine he oupu of an LTI sysem wih impulse response h() = Ae u() o inpu x( ) = Bu( ) Bu( ). (This is he same problem as Example 5.5..) Le s use he inegral form y () = hλ ( ) x ( λ) dλ The op panel of Figure 5.7 displays he inpu signal. For his signal he mos convenien markers are x () and x (). If we can figure ou how hese poins move we can deermine he locaion of x ( λ) on he λ axis as a funcion of. For hese poins we have / τ Copyrigh Rober D. Throne 97

98 x() = x ( λ) λ = x() = x ( λ) λ = The boom panel of Figure 5.7 displays x ( λ) as a funcion of λ. From his figure we can deermine where his funcion is as varies. We nex need o graph picures of h( λ ) and x ( λ) and look for imes when he produc of he funcions is no zero. For hese funcions here are hree differen picures, corresponding o <,, and, as shown in Figure 5.8. For < here is no place he produc of he funcions is no zero, so he oupu y () =. For we have which agrees wih our previous answer. y() = Ae Bdλ = ABτ e λτ / ( )/ τ Figure 5.7. Inpu signal for Example The original signal x () is shown in he op panel, wih he wo convenien marker poins. The lower panel shows x ( λ) as a funcion of λ, and shows how hese marker poins move as he parameer is varied. Noe ha he funcion x () has been flipped (reversed) from is original orienaion. Copyrigh Rober D. Throne 98

99 Figure 5.8. Plos of h( λ ) and x ( λ) (dashed line) for Example 5.5. for represenaive values of. For hese graphs A=, B=3, and τ =.8. For we have y() = Ae Bdλ = ABτ e e λτ / ( )/ τ ( )/ τ ( )/ τ Our soluion is hen y() = ABτ e ( )/ τ ( )/ τ ABτ e e Our soluion should be coninuous, so we need o check he values a he boundaries. We have y() = / τ y() = ABτ [ e ] While his looks differen han our previous answer, i is really he same hing for his range of. To see his noe ha from before we had ( )/ τ ( )/ τ y( ) = τab[ e ] u( ) τab[ e ] u( ) Copyrigh Rober D. Throne 99

100 If boh uni sep funcions are one and we have The soluion is ploed in Figure 5.9. ( )/ τ ( )/ τ ( )/ τ ( )/ τ y( ) = τab[ e ] τab[ e ] = ABτ[ e e ] Figure 5.9. Resul (oupu y ()) for Example 5.5. assuming A=, B=3, and τ =.8. ( )/ τ Example Deermine he oupu of an LTI sysem wih impulse response h( ) = Ae u( ) o inpu x () = u () u ( ) 3 u ( 3). Le s use he inegral form y () = h ( λ) x( λ) dλ There is really only one marker poin of noe, ha of h (), which gives h() = h ( λ) λ =. There are four differen graphs we need for his example,, 3, 3 4, and 4. For he produc of he funcions h ( λ) and x( λ) is zero, so y () =. For 3 we have he siuaion shown in Figure 5.. Evaluaing he inegrals we have Copyrigh Rober D. Throne

101 ( λ )/ τ ( )/ τ λτ / ( )/ τ ( )/ τ ( )/ τ y() = e () dλ = e e dλ = τe e = τ e Figure 5.. Firs overlapping region for Example assumingτ =.5. This figure is valid for 3 For 3 4we have he siuaion shown in Figure 5.. Evaluaing he inegral we have ( λ )/ τ ( )/ τ λτ / ( )/ τ / τ ( 3)/ τ ( )/ τ y ( ) = e () dλ = e e dλ = τe e = τ e e Figure 5.. Second overlapping region for Example assumingτ =.5. This figure is valid for 3 4. Copyrigh Rober D. Throne

102 For 4 we have he siuaion shown in Figure 5.. For his siuaion we will need wo inegrals, In summary we have ( λ )/ τ ( λ )/ τ y () = e () dλ+ e ( 3) dλ = 3 ( )/ τ λτ / ( )/ τ λτ / = e e dλ 3e e dλ 3 ( )/ τ / τ ( )/ τ ( )/ τ 3/ τ = τe e 3τe e e ( 3)/ τ ( )/ τ ( 4)/ τ = τ e e 3τ e y () 3 ( )/ τ τ e = ( 3)/ τ ( )/ τ τ e e 3 4 ( 3)/ τ ( )/ τ ( 4)/ τ τ e e 3τ e 4 Figure 5.. Third overlapping region for Example assumingτ =.5. This figure is valid for 4 Checking he values of y () a each boundaries we have y() = / τ y(3) = τ [ e ] / τ 3/ τ y(4) = τ[ e e ] The final soluion is ploed in Figure 5.3. Copyrigh Rober D. Throne

103 Figure 5.3. Resul (oupu y ()) for Example assumingτ =.5. Example Deermine he oupu of an LTI sysem wih impulse response h ( ) = u [ ( + ) u ( )] o inpu x () = u () u ( ) + u ( ). Le s use he inegral form y () = h ( λ) x( λ) dλ There are wo marker poins of noe, ha of, h( ) which gives h( ) = h ( λ) λ = + h() = h ( λ) λ = There are six differen graphs we need for his example,,,,, 3 and 3. For he produc of he funcions h ( λ) and x( λ) is zero, so y () =. For we have he siuaion depiced in Figure 5.4, and + y ( ) = ( λ)() dλ = ( ) Copyrigh Rober D. Throne 3

104 Figure 5.4. Iniial overlapping region for Example This figure is valid for. For we have he siuaion depiced in Figure 5.5, and y ( ) = ( λ)() dλ = Figure 5.5. Second overlapping region for Example This figure is valid for. Copyrigh Rober D. Throne 4

105 For we have he siuaion depiced in Figure 5.6, and + y ( ) = ( λ)() dλ+ ( λ)() dλ = ( 6+ 6) Figure 5.6. Third overlapping region for Example This figure is valid for. For 3we have he siuaion depiced in Figure 5.7, and + y ( ) = ( λ)() dλ = 4+ 3 Figure 5.7. Fourh overlapping region for Example This figure is valid for 3. Copyrigh Rober D. Throne 5

106 For 3 we have he siuaion depiced in Figure 5.8, and + y ( ) = ( λ)() dλ = Figure 5.8. Final overlapping region for Example This figure is valid for 3. In summary we have y () = + ( ) ( 6 6) Noe ha alhough he inpu o he sysem sars a ime =, he oupu sars a ime =. Thus he sysem is noncausal. The sysem oupu is graphed in Figure 5.9. Copyrigh Rober D. Throne 6

107 .5 y() Time (sec) Figure 5.9 Resul (oupu y ()) for Example Noe ha he sysem is no causal since he inpu sars a = bu he oupu sars a = 5.6 Causaliy and BIBO Sabiliy Now ha we can wrie he oupu of an LTI sysem in erms of he convoluion of he inpu wih he impulse response, we can also deermine some fairly simple ess o deermine if an LTI sysem is BIBO sable or causal. We have If we know he inpu is bounded, x ( ) y () = hλ ( ) x ( λ) dλ N, hen we know y ( ) h( λ) x ( λ) dλ h( λ) x ( λ) dλ h( λ) Ndλ = N h( λ) dλ Thus an LTI sysem will be BIBO sable if h( λ) dλ < Copyrigh Rober D. Throne 7

108 Nex, le s assume we wan o find he oupu of an LTI sysem a he ime, so we have We can hen break he inegral ino wo pars, y ( ) = h ( λ) x( λ) dλ y ( ) = h ( λ) x( λ) dλ+ h ( λ) x( λ) dλ If he sysem is causal, hen he second inegral mus be zero, since i depends on fuure values of he inpu. In order of he second inegral o be zero we need h ( λ) = for λ (, ) Le s assume λ = +, >. Subsiuing his ino our expression for he impulse response we have h ( λ) = h ( [ + ]) = h( ) =, > or h ( ) =, < This means he impulse response mus be zero for any ime less han zero in order for he LTI sysem o be causal. In summary, an LTI sysem is BIBO sable if h( λ) dλ < and is causal if h ( ) =, < Noe ha hese are independen properies, a sysem can be sable and no causal, or causal and no sable. 5.7 Convoluion Properies and Inerconneced Sysems There are a number of useful and imporan properies of convoluion. Among he mos useful are he following: Commuaive Propery: y () = h () x () = x () h () Associaive Propery: h() [ h() x ()] = [ h() h()] x () Disribuive Propery: h () [ x() + x()] = h () x() + h () x() The commuaive propery means ha y ( ) = x( λ) h( λ) dλ = x( λ) h( λ) dλ Copyrigh Rober D. Throne 8

109 so ha we have wo equivalen ways of deermining he sysem oupu, y (). A convenien mehod of presening he relaionship beween he inpu, oupu, and impulse response of a sysem is depiced in Figure 5.7. Figure 5.. Inpu, impulse response, and oupu for an LTI sysem. If we have wo LTI sysems in series, as shown in Figure 5., hen we can relae he inpu o he oupu as follows: v () = x () h () Then y () = v () h() = x () h() h() Using he commuaive propery we can wrie his as [ ] y () = [ h() h()] x () The impulse response beween he inpu and oupu is hen h () = h() h() Figure 5.. Two LTI sysems conneced in series. If we have o LTI sysems in parallel, as shown in Figure 5., hen we can relae he inpu o he oupu as follows Combining we have v () = x () h () and w () = x () h () y () = v () + w () = x () h() = x () h() Using he associaive and disribuive properies we hen have y () = [ h() + h()] x () Copyrigh Rober D. Throne 9

110 Hence he sysem ransfer funcion is hen h () = h() + h() Figure 5.. Two LTI sysems conneced in parallel. Example Consider he sysem shown in Figure 5.3. In his figure we have negaive feedback. For his sysem we have Figure 5-3. Sysem for Example Combining we have e () = x () v () and v () = y () h () y () = e () h() = [ x () v ()] h() = ] x () y () h()] h() We can rearrange his as or y () + y () h() h () = x () h( ) y( ) [ δ () + h() h()] = x () [ h( ) ] This is an awkward expression, bu a his poin we canno simplify i anymore. Copyrigh Rober D. Throne

111 Example Consider he sysem shown in Figure 5.4. Again we have negaive feedback in his sysem. Here we have e () = x () h () v () v () = y () h4 () w() = e() h() + x() h5() y () = w () h() 3 Figure 5-4. Sysem for Example Saring wih our expression for y () and working backwards we have y () = w () h() Simplifying we ge 3 y() = [() e h () + x() h ()] h () 5 3 y () = [{ x () h() v ()} h() + x () h()] h() 5 3 y() = [{ x() h() ( y() h ())} h () + x() h ()] h () Finally y() = x() h() h () h () y() h () h () h () + x() h () h () y() [ δ () + h () h () h ()] = x() [ h() h () h () + h () h ()] Trying o deermine he sysem impulse response in his way is very difficul, if no impossible. In addiion, if we are rying o modify one of he subsysems sysem o change he behavior of he overall sysem, his mehod of relaing he inpu o he oupu does no lend iself o any inuiion or easy analysis. This is one of he primary reasons Laplace and Fourier ransforms were developed. Laplace ransforms are hence our nex opic. Copyrigh Rober D. Throne

112 Chaper 5 Problems ) Simplify he following expressions as much as possible 3 ( ) () λ δ( λ ) + ( ) () λ δ( λ+ ) a) g ( ) = e δ ( ) b) g = e dλ c) g = e dλ d) g) ( ) ( ) ( + ) g = e λ δ λ d λ e) 3 g () = e ( ) d g ( ( ) ) ( ) = e λ δ λ + d λ f) g ( ( ) ) e ( ) = λ δ λ d λ λ δ λ λ 4 h) g () = e λ u( λδ ) ( λ+ ) dλ i) g ( ) = eu λ ( λ ) δ( λ 3) dλ Scrambled Answers: ( ) ( ) 3 3 4( ) ( ) ( ) ( ) ( 4), + ( ),, ( 4), + ( ), e u e u e e u e u e u ( ), e + u ( 3), e u( ), e δ ( ) ) Simplify he following expressions as much as possible a) d) g) g = e δ b) ( ) ( ) ( ) 3 ( ) () e ( ) 3 g = λ δ λ dλ e) 5 ( ) ) ( ) ( ) ( ) () = λ δ( λ+ ) 3 ( ) g ( = e λ δ λ dλ c) g e dλ g ( ) = e λ δ( λ+ ) dλ f) g ( ) = e λ δ ( λ 3) d λ g () = e λ δ ( λ ) d λ 4λ h) g ( ) = e u( λ ) δ( λ+ ) dλ i) g ( ) = eu λ ( λ ) δ( λ 3) dλ Scrambled Answers: ( ) ( ) 3 3 4( ) ( 3) ( ) ( ) e u ( 7), e + u( ), e, e u ( 4), e + u ( ), e u ( 3), e + u ( + ), e u ( + ), eδ ( ) 3) Deermine he impulse response for each of he following sysems a) y () = x () + x ( ) b) ( λ ) () ( λ) λ y = e x dλ c) y () = e x ( λ) dλ d) y () y () = x () e) y () = y () + x ( + ) f) y ( ) = x ( ) ( + ) / Scrambled Answers: e u ( + ), e, e u ( ), e u ( ), u ( ), δ( ) + δ( ) 4) Deermine he impulse response for each of he following sysems a) ( λ ) ( ) ( ) ( λ) y = x + e x dλ b) ( λ ) ( ) = ( λ+ ) λ y e x dλ c) y () = e x ( + λ) dλ d) y () y () 6 x ( ) + = e) y () 3 y () = x ( + ) f) y ( ) = x ( 3) Scrambled Answers: e u ( + ), e u ( + ), u ( 3), e, δ ( ) + e u ( ), e u ( ) ( + ) 3( + ) ( ) Copyrigh Rober D. Throne

113 5) For he following wo sysems, ( λ ) λ a) y () = e x( λ+ ) dλ b) y ( ) = e u( λ) x( λ) dλ i) compue he impulse response direcly ii) compue he sep response direcly ii) show ha he derivaive of he sep response is he impulse response for hese wo sysems. You should ge wo erms for he derivaive, bu one of he erms is zero (you should be able o show why his is rue)! Noe ha his is a general propery of LTI sysems, and is imporan since i is much easier o deermine he sep respones of a sysem han i is o deermine he impulse response of an acual sysem. 6) For he following inpulse responses and inpus, deermine he sysem oupu using analyical evaluaion of he convoluion inegral. Be sure o include any required uni sep funcions in your answers. a) ( ) ( ) h = e u ( ), x) ( = u ( -) b) h ( ) = e u () + δ (), x() = e u ( -) c) h ( ) = δ( ) + δ( ), x() = e u( ) d) h ( ) = δ( ), x ( ) = δ ( + ) Scrambled Answers ( ) ( ) ( ) y () = δ ( + ), y () = [ e ] u ( ), y () = e u ( ), y () = e u () + e u ( ) : 7) For LTI sysems wih he following impulse responses, deermine if he sysem is BIBO sable. a) h ( ) = δ ( ) b) h () = u () c) h () = e u () d) h () = e u () Answers: 3 are sable, one is unsable 8) Consider and LTI sysem wih impulse response h (). We assume ha he impulse response begins a some ime h. Tha is, he impulse response is zero before his ime. Hence we can wrie h( ) = h ( ) u( h). Nex assume we have an inpu ha sars (is zero before) some ime x, so we can wrie x( ) = x ( ) u( x ). Noe ha we do no really care wha he funcions look like here, only heir iniial imes. In doing he graphical evaluaion, jus make hem blobs. a) Deermine, using graphical evaluaion, he firs possible ime a which he oupu of he sysem, y (), can be non-zero, and call his ime y. This ime should be a funcion of h and x b) Show ha, in order for he sysem o be causal, h. Tha is, he impulse response mus be zero for negaive imes. Copyrigh Rober D. Throne 3

114 9) For he following inerconneced sysems, i) deermine he overall impulse response (he impulse response beween inpu x() and oupu y()) and ii) deermine if he sysem is causal. a) h ( -), c) () = u h() = u( + ) b) h() = u ( -), h() = δ ( + ) h() = e u ( -), h () = δ () + u() ( ) Scrambled Answers: h () = u ( ), h () = u (), h () = u ( + ), h () = u ( + ) + u ( ), h = δ + u + e u h = u + δ + ( ) () () () ( ), () ( ) ( ) Three sysems no causal ) For he following inerconneced sysems, i) deermine he overall impulse response (he impulse response beween inpu x() and oupu y()) and ii) deermine if he sysem is causal. a) h ( -), b) () = u h() = u( + ) h() = e u ( -), h() = δ ( ) ( ) Copyrigh Rober D. Throne 4

115 Scrambled Answers: ( 3) h () = e u ( 3), h ( ) = u ( ) + u ( + ) ( ) h () u (), h () e u ( = = ) + δ ( ) One sysem is no causal ) For he following sysems, deermine he relaionship beween he inpu and he oupu Answers: y ()*[ δ () + h ()* h () + h ()* h ()] = x()* h()* h ()* h () [ ] y()*[ δ () + h()* h ()* h () + h ()* h ()] = x*[ h()* h () * h ( )] ) Consider a linear ime invarian sysem wih impulse response given by h ( ) = [ u ( + ) u ( )] and inpu x ( ) = u ( ) 3 u ( 3) + u ( 4), shown below h() x() Using graphical evaluaion, deermine he oupu y () = h () x (). Specifically, you mus Copyrigh Rober D. Throne 5

116 Flip and slide h (), NOT x () Show graphs displaying boh h ( λ) and x( λ) for each region of ineres Deermine he range of for which each par of your soluion is valid Se up any necessary inegrals o compue y (). Your inegrals mus be complee, in ha hey canno conain he symbols x( λ ) or h ( λ) bu mus conain he acual funcions. Your inegrals canno conain any uni sep funcions DO NOT EVALUATE THE INTEGRALS!! Answer: + ( λ)() dλ 3 + ( λ)() dλ+ ( λ)( ) dλ y () = ( λ)() dλ+ ( λ)( ) dλ ( λ)() dλ+ ( λ)( ) dλ ( λ)( ) dλ Copyrigh Rober D. Throne 6

117 3) Consider a noncausal linear ime invarian sysem wih impulse response given by h ( ) = [ ( ) ][ u ( + ) u ( )] The inpu o he sysem is given by x ( ) = u ( ) u ( 3) u ( 4) + u ( 6) h() Time (sec).5.5 x() Time (sec) Using graphical evaluaion, deermine he oupu y (). Specifically, you mus Flip and slide h (), NOT x () Show graphs displaying boh h ( λ) and x( λ) for each region of ineres Deermine he range of for which each par of your soluion is valid Se up any necessary inegrals o compue y (). Your inegrals mus be complee, in ha hey canno conain he symbols x( λ ) or h ( λ) bu mus conain he acual funcions. Your inegrals canno conain any uni sep funcions DO NOT EVALUATE THE INTEGRALS!! Copyrigh Rober D. Throne 7

118 Answer 3 + y () = [ ( λ ) ]() dλ+ [ ( λ ) ]( ) dλ [ ( λ ) ]() dλ 3 [ ( λ ) ]() dλ 3 6 [ ( λ ) ]( ) dλ [ ( λ ) ]( ) dλ 6 8 Copyrigh Rober D. Throne 8

119 4) Consider a noncausal linear ime invarian sysem wih impulse response given by h () = e [ u ( + ) u ( )] The inpu o he sysem is given by x ( ) = u ( ) 5 u ( ) + 3 u ( 3) 3.5 h() Time(sec) 4 3 x() Time(sec) Using graphical evaluaion, deermine he oupu y (). Specifically, you mus Flip and slide h (), NOT x () Show graphs displaying boh h ( λ) and x( λ) for each region of ineres Deermine he range of for which each par of your soluion is valid Se up any necessary inegrals o compue y (). Your inegrals mus be complee, in ha hey canno conain he symbols x( λ ) or h ( λ) bu mus conain he acual funcions. Your inegrals canno conain any uni sep funcions DO NOT EVALUATE THE INTEGRALS!! Copyrigh Rober D. Throne 9

120 Copyrigh Rober D. Throne Answer: ( ) ( ) ( ) 3 ( ) ( ) 3 ( ) ( ) 3 ( ) () () ( 3) () ( 3) 3 () ( 3) () ( 3) 5 5 e d e d e d e d e d y e d e d e d λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ + + = + + +

121 5) Consider a linear ime invarian sysem wih impulse response given by The inpu o he sysem is given by h ( ) = [ u ( + ) u( )] x ( ) = [ u ( + ) u ( )] + [ u ( 6) u ( 8)] 4 h() Time(sec) 4 x() Time(sec) Using graphical evaluaion, deermine he oupu y ().Specifically, you mus Flip and slide h (), NOT x () Show graphs displaying boh h ( λ) and x( λ) for each region of ineres Deermine he range of for which each par of your soluion is valid Se up any necessary inegrals o compue y (). Your inegrals mus be complee, in ha hey canno conain he symbols x( λ ) or h ( λ) bu mus conain he acual funcions. Your inegrals canno conain any uni sep funcions DO NOT EVALUATE THE INTEGRALS!! Copyrigh Rober D. Throne

122 Answer: 4 + ( λ) () dλ 4 + ( λ) () dλ ( λ) () dλ 3 y () = ( λ) () dλ ( λ) () dλ ( λ) () dλ ) Malab From he class websie download he files homework5.m and convolve.m. homework5.m is a scrip file ha ses up he ime arrays and funcions, and he invokes he funcion convolve.m o compue he convoluion of he wo funcions. homework5.m hen plos he wo funcions o be convolved, and hen he resuling convoluion of he wo funcions. Noe ha we are passing argumens o he funcion convolve and he funcion is reurning values o he calling program. In par c of his problem you will need o complee his code, don ry o use i unil hen. ( T) a) Show( by performing he convoluion) ha if h () = e u ( ), x ( ) = e u ( T), hen T y() = h() x() = e e ( T) u( T) b) Show (by performing he convoluion) ha if h() = e u (), x () = u () u( T), hen - -( T - ) y () = h () x () = [- e ] u ()-[- e ] u ( - T) c) Read he Appendix a he end of his homework and hen complee he code for he funcion convolve.m. d) Use he scrip homework5.m o compue and plo he convoluion for he funcions of par a wih T =.5 (i should defaul o his). If you have done everyhing correcly he analyical (rue) convoluion should be indisinguishable from he esimaed (numerical) convoluion. Copyrigh Rober D. Throne

123 e) Commen ou your funcions from par d, and hen modify he code in homework5.m o compue he convoluion for he funcions in par b, wih T =. Be sure o also plo he correc (rue) soluion o check for any errors. f) The funcions in pars b and e illusrae a capacior charging and hen discharging, somehing we have seen before. Change he widh of he pulse o he pulse is five ime consans long, and run he simulaion. Turn in work for pars a and b, and your plos for pars d and e, and your code. 7) An LTI sysem has inpu, x (), impulse response, h (), and oupu, y (). Use convoluion o deermine he ime parameers (a, b, c, d) and he ampliude, e. Noe ha hese figures are no drawn o scale and you mus show deailed work o jusify your answer. Answers:,4,5,7, Copyrigh Rober D. Throne 3

124 Appendix Alhough his is a coninuous ime course, and Malab works in discree-ime, we can use Malab o numerically do convoluions, under cerain resricions. The mos imporan resricion is ha he spaces beween he ime samples be he same for boh funcions. Anoher oher resricion is ha he funcions really need o reurn o zero (or very close o zero) wihin he ime frame we are examining hem. Finally, we need fine enough resoluion (he sampling inerval mus be sufficienly small) so ha our sampled signals are a good approximaion o he coninuous signal. Firs, we need o have wo funcions o convolve. Le s assume we wan o convolve he funcions ( T) h ( ) = e u ( ) and x () e = u ( T) Le s denoe he ime vecor ha goes wih h as h, and he ime vecor ha goes wih x as x. Then we creae he funcions wih somehing like % x = [-:.:8]; % ime from - o 8 wih an incremen of. seconds h = [-:.:8]; % h *(<)+exp(-).*(>=); T =.5; x *(<T)+exp(-(-T)).*(>=T) Nex we will need o deermine he ime inerval beween samples. We can deermine his as d = x()-x(); I doesn maer if we use x or h, since he sample inerval mus be he same. Now we can use Malabs conv funcion o do he convoluion. However, since we are rying o do coninuous ime convoluion we need o do some scaling. To undersand why, le s look again a convoluion: y () = x( λ) h ( λ) dλ If we were o ry and approximae his inegral using discree-ime samples, wih sampling inerval, we could wrie y () yk ( ) x( λ) xn ( ) h ( λ) h([ k n] ) We can hen approximae he inegral as n= n= y ( ) = x( λ) h ( λ) dλ yk ( ) = xn ( h ) ([ k n] ) = xn ( h ) ([ k n] ) n= n= The Malab funcion conv compues he sum xn ( h ) ([ k n] ), so o approximae he n= coninuous ime inegral we need o muliply (or scale) by y = d*conv(x,x);. Hence n= Copyrigh Rober D. Throne 4

125 Finally, we need o deermine a ime vecor ha corresponds o y. To do his we need o deermine where he saring poin should be. Le s consider he convoluion y () = x( λ) h ( λ) dλ Le s assume x () is zero unil ime, so we could wrie x () = xu () ( ) for some funcion x (). Similarly, le s assume h () is zero unil ime, so we could wrie h () = hu () ( ) for some funcion h (). The convoluion inegral is hen y ( ) = x ( λ) u( λ ) h ( λ) u ( λ ) dλ = x ( λ) h ( λ) dλ This inegral will be zero unless, or +. Hence we know ha y () is zero unil = +. This means for our convoluion, he iniial ime of he oupu is he sum of he iniial imes: iniial_ime = x()+h(); Every sample in y is separaed by ime inerval d. We need o deermine how long y is, and creae an indexed array of his lengh n = [:lengh(y)-]; Finally we can consruc he correc ime vecor (y) ha sars a he correc iniial ime and runs he correc lengh y = d*n + iniial_ime; Copyrigh Rober D. Throne 5

126 6. Laplace Transforms In many applicaions we have inerconnecions of LTI sysems. We can deermine he oupu of he sysem by using convoluion in he ime-domain, bu his ofen proves o be difficul when we have more han jus a few inerconnecing sysems. Someimes we don wan o jus compue he oupu, bu raher we wan o be able o deermine properies of he sysem in a simple way. We will uilize Laplace ransforms for his, hough in some applicaions he use of Fourier ransforms may be more appropriae. 6. Laplace Transform Definiions The wo-sided Laplace ransform of a signal x () is defined o be while he one-sided Laplace ransform is defined as s X ( s) = x() e d s X() s = xe () d The only difference beween he wo definiions is he lower limi, and one and wo sided refers o inegraing one on side of zero or boh sides of zero. In his course we will only use he one-sided ransform, and when we refer o he Laplace ransform we mean he one sided ransform. However, you should be aware he wo-sided ransform has some use when we are dealing wih noncausal sysems. Looking a he form of he inegral, since he exponen mus be dimensionless, we can conclude ha he variable s has he unis of /ime. There are a few convenions we need o know abou. Firs of all, he lower limi on he one sided Laplace ransform is generally wrien as, or saring a a ime jus before zero. This is paricularly useful when deermining he Laplace ransform of an impulse cenered a zero. Secondly, a very common convenion is o use lower case leers for ime-domain funcions, and capial leers for he corresponding ransform domain. We usually wrie x ( ) X( s) or { x ()} = X( s) o show ha x () and X() s are ransform pairs. We dill denoe he Laplace ransform operaor as. Finally, he complex variable s is someimes wrien in erms of is real and imaginary pars as s = σ + jω. This is paricularly useful in deermining if he inegral is finie (or can be made o be finie) or is infinie. 6. Basic Laplace Transforms Le s sar off by deermining he Laplace ransform of some basic signals. As you will see someime we need o pu condiions on σ o be sure he inegral converges. This condiion defines he region of convergence. Example 6... For x ( ) = δ ( ) we have sλ s ( ) = δ( λ) λ = δ( λ) λ = X s e d e d Copyrigh Rober D. Throne 6

127 since he dela funcion is conained in he region of inegraion. Hence we have he ransform pair x () = δ () X() s = Example 6... For x () = δ ( ), where, we have sλ s s X( s) = δ( λ ) e dλ = e δ( λ ) dλ = e since again he dela funcion is locaed in he region of inegraion. Hence we have he ransform pair s x () = δ ( ) X() s = e Noe ha if < he inegral will be zero. Example For x () = u () we have λ λ= s sλ sλ e X() s = u( λ) e dλ = e dλ = s λ= A his poin we canno really evaluae his inegral unless we pu some condiions on σ. Le s make he subsiuion s = σ + jω and we have e Using Euler s ideniy jωλ e = cos( ωλ) + j sin( ωλ) We can deermine he magniude of j as j e ωλ = cos ( ωλ) + sin ( ωλ) = ( σ + jω ) λ λ= σλ + jωλ λ= σλ jωλ λ= e e e e X() s = = = s s s λ= λ= λ= j Hence he erm e ωλ does no conribue o he convergence of he inegral. Tha means ha ω does no conribue o he convergence of he inegral. Tha leaves us wih σ. If σ >, hen he exponen in he exponenial is negaive, when we evaluae i a he limi of infiniy we ge zero. Hence we have he ransform pair x () = u () X() s =, σ > s Noe ha he condiion σ >, or he real par of s mus be posiive, defines he region of convergence. Hence he inegral will converge if R{s} > and we can rewrie he ransform pair as x () = u () X( s) =, R {} s > s Example For x () = u ( ) we have sλ = s sλ sλ e e λ λ λ s s λ= X() s = u( ) e d = e d = =, R {} s > λ Copyrigh Rober D. Throne 7

128 The region of convergence is he same as for he non-delayed uni sep. So we have he ransform pair s e x () = u ( ) X() s =, R {} s > s a Example For x ( ) = e u ( ) we have ( s+ a) λ λ= aλ sλ ( s+ a) λ e X() s = e u( λ) e dλ = e dλ = ( s + a) λ= Now in order for he inegral o converge, we need R { s+ a} >.We can rewrie his as R {} s +R {} a > or R {} s > R {} a. Hence he region of convergence is defined as R {} s > R{} a and we have he Laplace ransform pair x () = e a u () X () s =, R {} s > R{} s + a a Example For x () = cos( ω u ) () we will need o use Euler s ideniy in he form Compuing he Laplace ransform we have e cos( ω ) = + e jω jω sλ jωλ jωλ sλ X( s) = cos( ) u( ) e d = e + e e d ωλ λ λ λ λ= λ= ( jω s) λ ( jω+ s) λ ( jω s) λ ( jω+ s) λ e e = e dλ+ e d λ = jω s ( jω + s) λ= λ= Boh inegrals will converges if R {} s >, and hen we have s+ jω + s jω s X( s) = + = = s jω s+ jω s + ω s + ω The Laplace ransform pair is hen s x () = cos( ω ) u () X( s) =, R {} s > s + ω Example Le s assume we wan o find he Laplace ransform of he derivaive of have dx() dx( λ) sλ = e dλ d dλ x (). Then we Copyrigh Rober D. Throne 8

129 In order o evaluae his we will need o use inegraion by pars. For wo funcionsu( λ ) and v( λ), we can wrie d( uv) = vdu + udv. Rearranging his we ge he usual form for inegraion by pars, udv = uv vdu. For our inegral we have dv( λ) dx( λ) = dλ dλ s so dv = dx or v( λ) = x( λ). We also have u( λ) = e λ, so du( λ) sλ = se dλ or s du = se λ dλ Combining hese we have dx() dx( λ) sλ sλ λ= sλ = e dλ = x( λ) e + s x( λ) e dλ = x( ) + sx( s) d dλ λ= So we have he ransform pair Table 6. summarizes some common Laplace ransforms. 6.3 Laplace Transforms of RLC Circuis dx() = sx ( s) x( ) d In his secion we will deermine he Laplace ransforms of resisors, capaciors, and inducors, and hen use hese relaionships in some examples o deermine he oupu of some RLC circuis in he Laplace domain. In he nex secions we will review he use of parial fracions o go back from he Laplace domain o he ime domain. Consider a resisor elemen shown in he lef panel of Figure 6.. If v () is he volage across he resisor, i () is he curren hrough he resisor, and R is he resisance, hen we have by Ohm s law v () = ir (). Taking Laplace ransforms of boh sides of his equaion we have { v ()} = { i ()R} or Vs () = IsR (). Thus, in he Laplace domain, he equivalen impedance is sill jus R and he resisive circui elemen in he lef panel of Figure 6.3. is replaced wih he circui elemen in he righ panel of Figure 6.. Copyrigh Rober D. Throne 9

130 { δ } () = { u ()} { u() } = s = s m u = () m ( m )! s a { e u ()} = s + a a { e u() } = ( s+ a) ( m ) a e u = () ( m )! ( s+ a) s = { cos( ω u ) ( )} s + ω { sin( ω u ) ( )} = s + ω α { e cos( ω u ) ( )} ω ( s + α) + ω α ω { e sin( ωu ) ()} = ( s + α) + ω { } a { } = dx() = sx ( s) x( ) d m s + α d x () = d as x ( a) = e X() s e x () = X( s+ a) s X ( s) sx( ) x ( ) x, a > = ax ( as) a Table 6.. Some common Laplace ransform pairs. Copyrigh Rober D. Throne 3

131 Figure 6.. Resisive elemen in he ime domain (lef) and Laplace domain (righ). Consider he capaciive circui elemen shown in he lef panel of Figure 6.. If v () is he volage across he capacior, i () is he curren hrough he capacior, and C is he capaciance, hen we have by Ohm s dv() law i () = C. Taking Laplace ransforms of boh sides of his equaion we have d or Rewriing his we have dv() dv() d d { i() } = I() s = C = C = C{ sv () s v() } I() s = CsV () s Cv() v( ) Vs () = Is () + Cs s If for now we ignore he iniial condiions we have Vs () = Is () Cs which means he capacior has an equivalen impedance of. If we make he subsiuion s = jω, we Cs have he equivalen impedance which is idenical o wha you used in sinusoidal (phasor) analysis. jωc Thus, in he Laplace domain, he capacior circui elemen in he lef panel of Figure 6. is replaced wih he wo circui elemens in he righ panel of Figure 6.. Copyrigh Rober D. Throne 3

132 Figure 6.. Capaciive elemen in he ime domain (lef) and Laplace domain (righ). Consider he inducive circui elemen shown in he lef panel of Figure 6.3. If v () is he volage across he inducor, i () is he curren hrough he inducor, and L is he inducance, hen we have by Ohm s di() law. v () = L Taking Laplace ransforms of boh sides of his equaion we have d di() di() { v() } = V () s = L = L = L{ si() s i() } d d or V () s = LsI() s Li() We can rearrange his equaion ino is more common form as i( ) Is () = Vs () + Ls s Again, if we ignore he iniial condiion erm we have Is () = Vs () which means he inducor has he Ls equivalen impedance sl. Making he subsiuion s = jω we have he equivalen impedance jωl which is idenical he wha you used in sinusoidal analysis. Thus, in he Laplace domain, he inducor circui elemen in he lef panel of Figure 6.3 is replaced wih he circui elemens in he righ panel of Figure 6.3. Figure 6.3. Inducive elemen in he ime domain (lef) and Laplace domain (righ). Copyrigh Rober D. Throne 3

133 Noe ha he subsiuion s = jω and ignoring he iniial condiions gives he same equivalen impedances for hese circui elemens as you used before in your sinusoidal seady sae analysis. We will have more o say abou his subsiuion when we alk abou frequency response. However, his is only one possible value of s. In he following examples we use Laplace ransforms of he circui elemens o wrie he oupu of he circui in erms of he inpu and iniial condiions Example Consider he RC circui shown in Figure 6.4. The oupu volage is he volage across he resisor, and he inpu is volage v () in. We assume i () is he curren flowing in he circui. We wan o wrie he oupu of he circui in erms of he inpu and iniial volage on he capacior, v( ), in he Laplace domain. The circui is redrawn in he Laplace domain in Figure 6.5. Going around he loop we have v( ) V() s IsR () Is () = Cs s in Solving for he curren we have v( ) Vin () s Is () = s R + Cs The sysem oupu is he volage across he resisor, so we have v( ) Vin () s R s Vou () s = IsR () = R + Cs Finally we can wrie he oupu as he sum of wo differen pars. The Zero Sae Response (ZSR) is he response of he sysem o he inpu alone, assuming no iniial condiions. The Zero Inpu Response (ZIR) is he response of he sysem o he iniial condiions alone, assuming here is no inpu. Hence our final soluion is v( ) R Vin () s R in() ( ) o u () s V s RCs v RC V s = = R R RCs + RCs Cs Cs ZSR ZR I Copyrigh Rober D. Throne 33

134 Figure 6.4. Circui for Example Figure 6.5. Circui from Example 6.3. in he Laplace domain. Example Consider he RL circui shown in Figure 6.6. We again assume he oupu of he sysem is he volage across he resisor, and wan o deermine he oupu of he sysem in erms of he inpu volage and he iniial curren in he inducor i( ) in he Laplace domain. In Figure 6.7 he circui has been redrawn in he Laplace domain. To analyze his circui, le s define he volage across he inducor as V () L s, so we have VL () s = Vin () s Vou () s. Equaing currens we hen have Rearranging we ge and finally Vou () s VL () s i( ) Vin () s Vou () s i( ) = + = + R Ls s Ls s R + Ls Vin () s i( ) Vou () s + = Vou () s = + R Ls RLs Ls s Copyrigh Rober D. Throne 34

135 V ou R RL () s = Vin () s i() R Ls + + R + Ls ZSR ZIR Figure 6.6. Circui for Example Figure 6.7. Circui from Example 6.3. in he Laplace domain. Copyrigh Rober D. Throne 35

136 Example Consider he RLC circui shown in Figure 6.8. We again assume he oupu of he sysem is he volage across he resisor, and wan o deermine he oupu of he sysem in erms of he inpu curren and boh he iniial volage across he capacior v( ) and he iniial curren in he inducor i( ) in he Laplace domain. In Figure 6.9 he circui has been redrawn in he Laplace domain. To analyze his circui we need o equae all of he currens, as follows: v( ) Vou () s () s Vou s Vou () s i( ) Iin() s = R Ls s Cs Combining erms we have i( ) Iin() s = Vou () s + Cs + Cv() + R Ls s or RLCs + Ls + R i( ) Iin() s = Vou () s Cv() RLs + s Finally we ge Iin () s RLs RLi( ) + RLCsv( ) Vou () s = + RLCs + Ls + R RLCs + Ls + R ZSR ZIR Figure 6.8. Circui for Example Figure 6.9. Circui from Example in he Laplace domain. Copyrigh Rober D. Throne 36

137 6.4 Transfer Funcions and he Impulse Response Assume we have and LTI sysem ha is iniially a res (all iniial condiions are zero). We define he ransfer funcion as he raio of he Laplace ransform of he oupu divided by he Laplace ransform of he inpu, or { oupu( ) } H ( s) = { inpu( ) } Here we have denoed he ransfer funcion as H() s, hough obviously oher leers are possible. For example if we have and LTI sysem wih inpu x () and oupu and have he Laplace ransform pairs x () X() s and y () Y() s, hen he ransfer funcion is defined as Y() s H() s = X() s We can rearrange his relaionship o be Ys () = HsX () () s Example In Example 6.3. he ransfer funcion is Vou () s RCs H() s = = V () s RCs + Example In Example 6.3. he ransfer funcion is Example In Example he ransfer funcion is in Vou ( s) R H( s) = = V () s R + Ls V () s RLs H() s = = in ou Iin() s RLCs + Ls + R Noe ha he ransfer funcion comes from he Zero Sae Response (ZSR), since he Zero Inpu Response (ZIR) includes he iniial condiions, and we assume all iniial condiions are zero when deermining he ransfer funcion. If we know a sysem is LTI, hen we know ha he oupu is he convoluion of he inpu wih he impulse response, y () = h () x () = h( λ) x ( λ) dλ Nex, le s assume he inpu is causal and he sysem is causal, so boh are zero for <. Then we have y () = h () x () = h( λ) x ( λ) dλ = h( λ) x ( λ) dλ Since his is an equaliy, we can ake he Laplace ransform of each side of he equaion, Copyrigh Rober D. Throne 37

138 s s Y() s = { y ()} = Y() s = ye () d = { h( ) x() } = h( λ) x( λ) dλ e d Rearranging he order of inegraion we have ( ) ( ) s s h λ x λ dλ e d = h( λ) x( λ) e d dλ Nex, le s le σ = λ in he innermos inegral. As far as his inegral is concerned, λ is jus a consan parameer so dσ = d. The inegral hen becomes h λ x λ e d dλ = h λ x σ e σ dλ = h λ e dλ x σ e dσ = H s X s s s( σ+ λ) sλ sσ ( ) ( ) ( ) ( ) d ( ) ( ) ( ) ( ) Hence we have derived he convoluion propery of Laplace ransforms, y () = h () x () Ys () = HsXs () () We have also derived an imporan relaionship beween he impulse response funcion H() s. These are Laplace ransform pairs, h () and he ransfer h () H( s) 6.5 Poles, Zeros, and Pole-Zero Plos In mos insances, he ransfer funcion is of he form Ns ( ) As ( z)( s z)...( s zm) H() s = = Ds ( ) ( s p)( s p)...( s pn ) where Ns () and Ds () are polynomials in s. The zeros of he ransfer funcion, z, z,..., zm, are he roos of values Ns (), i.e., he values of s ha make Ns () zero. The poles of he ransfer funcion, p, p,..., p, are he roos of Ds (), i.e., he values of s ha make Ds () n zero. If he degree of he numeraor is less han he degree of he denominaor, or m< n, hen he ransfer funcion is sricly proper, while if he degree of he numeraor is less han or equal o he degree of he denominaor, or m n, he ransfer funcion is proper. Clearly a ransfer funcion ha is sricly proper is also proper. Noe also ha if he ransfer funcion has real valued coefficiens, hen he poles or zeros of he ransfer funcion mus occur as complex conjugae pairs. Someimes i is useful o plo he poles and zeros of a ransfer funcion. This pole-zero plo is an alernaive way of presening he informaion conained in he algebraic represenaion of H() s. As you will see laer, he pole-zero plo allows us o easily visually deermine he response of an LTI sysem o a Copyrigh Rober D. Throne 38

139 sinusoid of differen frequencies. The pole-zero plo is jus a plo of poles (represened by x s) and zeros (represened by o s) in he complex plane, where he horizonal axis represens real values, he verical axis represens imaginary values. Thus any complex number can be represened in his plane. Example Deermine he pole-zero plo for ( s )( s+ + j)( s+ j) H() s = ss ( + )( s+ 4 + j)( s+ 4 j) Noe ha his is a sricly proper ransfer funcion. The zeros of he ransfer funcion are a, --j, and - +j, while he poles are a, -, -4-j, and -4+j. The pole-zero plo for his ransfer funcion is shown in Figure Parial Fracions for Compuing Inverse Laplace Transforms The firs hing we need o do before using he parial fracion echnique o find he inverse Laplace ransform is o be sure he ransfer funcion is a raio of polynomials. If here is a ime delay ( e s ) we remove his and accoun for i when we are done. In addiion, we need a sricly proper raio of polynomials. This means ha he order of he numeraor polynomial mus be less han he order of he denominaor polynomial. If his is no he case, hen we use long division and use parial fracions on he remainder. Finally, we need o be sure he denominaor polynomial is monic. This means he leading coefficien in he denominaor polynomial is a one. Copyrigh Rober D. Throne 39

140 Figure 6.. Pole-zero plo for ransfer funcion of Example Example Prepare he ransfer funcion s e ( s+ ) H() s = ( s+ 3)( s+ 4) so we can use parial fracions o deermine he inverse Laplace ransform. The firs hing we need o do is o remove he ime delay erm, and wrie s H() s = e Gs ( ) where s + Gs () = ( s+ 3)( s+ 4) Since Gs () is a sricly proper raio of polynomials and he denominaor polynomial is monic, we are ready for parial fracions now. Example Prepare he ransfer funcion s + H() s = ( s + ) for parial fracion expansion. Here here is no ime-delay, bu he ransfer funcion is no sricly proper so we need o do some long division. We have hen s + s + 4s+ 3 Hs () = = = = Gs () ( s+ ) s + 4s+ 4 s + 4s+ 4 where 4s + 3 Gs () = ( s + ) Since Gs () is a sricly proper raio of polynomials and he denominaor polynomial is monic, we are ready for parial fracions now. Example Prepare he ransfer funcion s e ( s + s+ ) H() s = (s+ )( s+ ) for parial fracion expansion. Here we have a ime delay, he raio of polynomials is no sricly proper, and he denominaor is no monic. Firs we pull ou he ime-delay par and do he long division, wih he resul s e ( s + s+ ) s s + s+ s.5s H() s = = e = e.5 (s+ )( s+ ) s + 5s+ s + 5s+ Nex we scale he denominaor so i is monic Copyrigh Rober D. Throne 4

141 Finally we have where s.5s.5 s.5s.5 H( s) = e.5 = e.5 s +.5s+ ( s+.5)( s+ ) [ ] s Hs ( ) = e.5 Gs ( ).5s.5 Gs () = ( s+.5)( s+ ) Parial Fracions wih Disinc Poles Le s assume we have a sricly proper ransfer funcion Ns () K( s z)( s z) ( s zm) H() s = = D() s ( s p )( s p ) ( s p ) n The poles of he sysem are a p, p,..., pn and he zeros of he sysem are a. z, z,..., zm Since we have disinc poles, we know ha pi p j for i j. We also assume ha Ns () and Ds () have no common facors so here is no pole/zero cancellaion. We would like o find he corresponding impulse response h (). To do his we assume H( s) = a + a an s p s p s p If we can find he a hen i will be easy o deermine h () i since we know p i () s p e u i To find a we firs muliply by s p, ) ) ( ) ( ) ( s p s p H s a a... a ( s p = n s p s p Nex ake he limi as s p. Since all of he poles are disinc all of he erms on he righ hand side of he equaion are zero, excep for he firs erm. Hence we have n n Similarly we ge And in general a = lim( s p ) H ( s) a s p = lim ( s p ) H ( s) s p a = lim( s p ) H ( s) i s pi i Copyrigh Rober D. Throne 4

142 Example For he ransfer funcion s + H() s = ( s+ )( s+ 3) deermine he corresponding impulse response. We have H() s = a + a s+ s+ 3 Then ( s+ )( s+ ) ( s+ ) a = lim( s+ ) H( s) = lim = lim = s s ( s+ )( s+ 3) s ( s+ 3) So and ( s+ 3)( s+ ) ( s+ ) a = lim( s+ 3) H( s) = lim = lim = s 3 s 3( s+ )( s+ 3) s 3( s+ ) H() s = + s + s + 3 h e u e u 3 () = () + () I is ofen unnecessary o wrie ou all of hese seps, since we know in advance here will be a cancellaion wih one of he poles. In paricular, when we wan o find we know we will have a cancellaion beween he s p i erm in he numeraor and he s pi erm in he denominaor. In fac, when we o find ai, we can jus ignore (or cover up) he s pi erm in he denominaor. This is usually jus called he cover up mehod. For our previous example, we have a a = = lim s ( s + ) ( s + ) = ( s + 3) ( s + ) + + lim s 3 ( s ) ( s 3) = a i This can ofen jus be done in your head! Example Consider he ransfer funcion from Example We have s e ( s+ ) s H() s = = e Gs ( ) ( s+ 3)( s+ 4) where ( s + ) Gs () = ( s+ 3)( s+ 4) We now do parial fracions for Gs () and use he fac ha he erms corresponds o a ime delay of unis. We have hen e s Copyrigh Rober D. Throne 4

143 s + G() s = = a + a ( s+ 3)( s+ 4) s+ 3 s + 4 ( s + ) a = lim = s 3 ( s + 3) ( s + 4) lim s 4 ( 3) ( 4) Then 3 4 g () e = u () + 3 e u () Finally we ake he ime delay ino accoun o ge he final answer a = ( s + ) s+ s+ = 3 h g e u e u 3( ) 4( ) ( ) = ( ) = ( ) + 3 ( ) Example Consider he ransfer funcion from Example We have where We hen have s e ( s + s+ ) s.5s.5 s H() s = = e.5 = e.5 ( ) (s+ )( s+ ) ( s+.5)( s+ ).5s.5 Gs () = ( s+.5)( s+ ) [ Gs].5s.5 Gs () = = a + a ( s+.5)( s+ ) s+.5 s+ So If we hen define hen.5s.5 5 a = = slim.5 ( s +.5) ( s + ).5s.5 a = lim = s ( s+.5) ( s+ ) 3 5 g = e u + e u 3.5 () () () Fs ( ) =.5 Gs ( ) 5 f = g = + e u e u 3 Finally we ake he ime delay ino accoun, s H( s) = e Fs ( ).5 ( ).5 ( ).5 ( ) ( ) Copyrigh Rober D. Throne 43

144 5 h = f = + e u e u 3.5( ) ( ) ( ) ( ).5 ( ) ( ) Example Consider he circui in Example Here we have he relaionship Vin () s RCs v( ) RC Vou () s = RCs + RCs + Insead of deermining he impulse response, le s deermine he sysem oupu for a sep inpu of A ampliude A. Hence we have vin ( ) = Au(), or Vin () s =. Then we can wrie he oupu as s ARCs v( ) RC Vou () s = s( RCs + ) RCs + Nex we need o make he denominaor polynomials monic, so we have As v( ) A v( ) Vou () s = = s s+ s+ s+ s+ RC RC RC RC This problem is already in he correc for, and we have / RC / RC / RC vou () = Ae u () v() e u () = A v() e u () If we look a he circui we have modeling (in Figure 6.4), his answer makes sense. The iniial volage across he resisor is he difference beween he iniial applied volage and he iniial volage across he capacior. In addiion, he volage across he resisor should approach zero as ime increases, since evenually he volage on he capacior will reach he applied volage. Finally, i is clear ha he ime consan for his simple circui is τ = RC and his is he ime consan of our resuls. Example Consider he circui in Example Here we have he relaionship R RL Vou () s = Vin () s + i() R + Ls R + Ls Again le s deermine he oupu when he inpu is a sep of ampliude A. Then we have R A RL Vou ( s) = + i( ) R + Ls s R + Ls We need o make he denominaors monic, so we rewrie his as AR ( ) L R Vou s = + i( ) R R s s+ s + L L Applying parial fracion o he firs erm we have Copyrigh Rober D. Throne 44

145 In he ime-domain we hen have A A R Vou ( s) = + i( ) s R R s+ s+ L L L L vou () = A( e ) u () + Ri() e u () Parial Fracions wih Repeaed Poles If here are repeaed poles wih no oher poles, hen he inverse Laplace ransform is very sraighforward, using he formula m p e u () m ( m )! ( s+ p) If insead of isolaed repeaed poles and nonrepeaed poles, hen we need o use a differen form for he parial fracions for he repeaed poles. If we assume we have a pole a p ha is of order m, hen for his pole we will use an expansion of he form a + a + + a m m ( s+ p) ( s+ p) ( s+ p) We will also need a new approach for deermining some of he expansion coefficiens for he repeaed poles. There are wo common approaches Muliply boh sides of he equaion by s and aking he limi as s Selec convenien values for s and evaluae boh sides of he equaion for hese values of s These echniques are probably mos easily explained by he use of examples. In each of hese examples you should noe ha here are always as many unknowns as here are poles! Example Find he impulse response ha corresponds o he ransfer funcion R H () s = ( s + )( s + ) To do his we look for a parial fracion expansion of he form H() s = a + a + a3 s+ s+ ( s+ ) For he disinc pole we have a = lim = s ( ) s + ( s + ) Nex we deermine he coefficien for he highes power of he repeaed roo. In his case we muliply boh sides of our parial fracion expansion by ( s + ) and hen ake he limi as s a3 = lim = s ( ) ( ) s+ s+ Finally, o ge he value of a we mus resor o a differen mehod. If we muliply boh sides by s and ake he limi as s we have R Copyrigh Rober D. Throne 45

146 s s s s lim sh( s) = lim = lim a + a + a3 s s ( s+ )( s+ ) s s+ s+ ( s + ) Simplifying his we have = a+ a or a = a = Hence we have H() s = s+ s+ ( s+ ) wih corresponding impulse response h () = e u () e u () e u () As an alernaive mehod for deermining, we sar wih This expression mus be rue for all values of s as long as boh sides remain finie. Le s choose a convenien value of s, like s =. Then we have H() = = + a 4 4 or =a which again give us a =. Example Deermine he impulse response ha corresponds o he ransfer funcion s + H() s = s ( s+ )( s+ 3) The parial fracion expansion we need is of he form a H() s = = + a ( s+ )( s+ ) s+ s+ ( s+ ) H() s = a + a + a 3 + a4 s s s+ s+ 3 Firs we find he coefficiens ha correspond o he disinc poles and he coefficien ha goes wih he highes power of he repeaed pole. a = lim s s s + = ( s+ )( s+ 3) 6 Copyrigh Rober D. Throne 46

147 A his poin we have s + a3 = lim = s s ( s+ ) ( s + 3) 4 s + a4 = lim = s 3 s ( s+ ) ( s+ 3) 9 s + H () s = = a + + s ( s+ )( s+ 3) s 6 s 4 s+ 9 s+ 3 To find he unknown coefficien, we will muliply by s and le s, ss ( + ) s s s s lim sh ( s) = lim = lim a + + s s s ( s+ ) ( s+ 3) s s 6 s 4 s + 9 s + 3 or This simplifies o Finally we have = a a = 36 s + H() s = = + + s ( s+ )( s+ 3) 36 s 6 s 4 s+ 9 s+ 3 which corresponds o he impulse response 3 h () u () u () e = + u () + e u () As an alernaive o aking limis in he expression s + H () s = = a + + s ( s+ )( s+ 3) s 6 s 4 s+ 9 s+ 3 we can make he subsiuion s = in boh sides of he expression (noe ha we canno use s =, s =, or s = 3, since hese subsiuions make he funcion infinie). The we have which again yields = a+ + = a+ = a a = 36 Example Deermine he impulse response for he ransfer funcion Copyrigh Rober D. Throne 47

148 s + 3 H() s = ss ( + ) ( s+ ) The appropriae parial fracion expansion is of he form s + 3 H() s = = a + a + a + a + a ss ( + ) ( s+ ) s s+ ( s+ ) s+ ( s+ ) Again we firs find he easy coefficiens, a, a3and a5. s a = lim = s s ( s+ ) ( s+ ) 4 s + 3 a3 = lim = s ( ) s s+ ( s + ) s + 3 a5 = lim = s ( ) ( ) ss+ s+ Nex we use limis ss ( + 3) 3 s s s s lim sh ( s) = lim = lim + a + a 4 s s ss ( + ) ( s+ ) s 4 s s ( s+ ) s ( s+ ) + + or We need anoher equaion, so le s le = a + a s = 3, in he expression which yields s H() s = = + a + a ss ( + ) ( s+ ) 4 s s+ ( s+ ) s+ ( s+ ) 4 = a a4 4 or a a4 = Solving hese wo equaions yields a = and a 4 =. Finally, we have 4 s H() s = = + ss ( + ) ( s+ ) 4 s s+ ( s+ ) 4 s+ ( s+ ) which corresponds o he impulse response Copyrigh Rober D. Throne 48

149 3 7 h = u + e u e u e u e u 4 4 () () () () () () Parial Fracions wih Complex Conjugae Poles The ransform pairs we are primarily going o use wih complex conjugae poles are a s+ a e cos( b) u( ) ( s + a) + b a b e sin( b) u( ) ( s + a) + b Noe ha complex conjugae poles always resul in sines and cosines (or a single sine/cosine wih a phase angle). We will ry and make erms wih complex conjugae poles look like hese erms by compleing he square in he denominaor. Tha is, we need o be able o wrie he denominaor as Ds () = ( s+ a) + b Noe ha if we canno wrie he denominaor in his form, he poles are no complex conjugaes! To deermine he correc values of a and b, use he fac ha he coefficien of s should be a. Once we find a, he value for b is easy o find. A few examples will make his clear. Example Assume Ds ( ) = s + s + and we wan o wrie his in he correc form. Firs we recognize ha he coefficien of s is, so ha a =, or a =. We hen have D() s = s + s+ = ( s+ ) + b = s + s+ + b 4 So hen = and we can deermine. Thus we have 4 + b 7 b = 7 Ds ( ) s s ( s ) = + + = + + Example Assume Ds () = s + 3s+ 5and we wan o wrie his in he correc form. The 3 coefficien of s is 3, so we have a = 3 and a =. We hen have 3 9 Ds () = s + 3s+ 5 = ( s+ ) + b = s + 3s+ + b 4 Copyrigh Rober D. Throne 49

150 Then 5 9 = + b and we can deermine b =. 4 4 Now ha we know how o complee he square we will look a wo simple examples of complex conjugae poles, hen a more complicaed examples. Example Assuming H() s =, deermine he corresponding impulse response h (). s + s+ From our previous example we know H() s = 7 ( s + ) + This almos has he form we wan, which is b a e sin( b) u() ( s+ a) + b However, o use his form we need a b in he numeraor. To achieve his we will muliply and divide by 7 b =, 7 7 H() s = = ( s+ ) + ( s+ ) + and we can deermine 7 h () = e sin () 7 u s Example Assuming H() s =, deermine he corresponding impulse response h (). s + 3s+ 5 From our previous example we know s H() s = 3 s + + This is almos he form we wan, which is s+ a H() s = ( s+ a) + b However, o use his form, we will add and subrac 3/ in our ransfer funcion, Copyrigh Rober D. Throne 5

151 s+ s+ H() s = = s+ + s+ + s+ + The firs erm is now wha we wan, bu we need o scale he second erm, 3 3 s+ s+ 3 3 H() s = = s+ + s+ + s+ + s+ + So we finally have he impulse response h( ) = e cos( u ) ( ) e sin( u ) () The nex wo examples are much more involved, bu hey have he same general approach. In general, if here are complex conjugae poles we will look for parial fracions of he form cs ( + a) db ( ) + ( s+ a) + b ( s+ a) + b where c and d are he parameers o be deermined. Example Assume Y() s = s + s + s + Use parial fracions o deermine he corresponding ime domain funcion. We will have 3 cs ( + ) d A Y() s = = = + + We s+ s + s+ 3 s ( s+ ) s+ + s + + s + + need o deermine he hree coefficiens A, c, and d. To deermine A we use he cover-up mehod as before A = lim = = s s + s + s To deermine c, muliply boh sides by s and le s, Copyrigh Rober D. Throne 5

152 3 sc( s ) sd s + sa lim lim s ( )( ) = + + s+ s + s+ s+ 3 3 s+ + s+ + s or = A+ c= + c 3 So we have c =. Finally, we choose a convenien value for s, and evaluae boh sides. Le s choose 3 s =, so we have 3 cs ( ) d + A lim Y( s) = lim = lim + + s s ( s+ )( s + s+ ) s s s s or We can simplify his o which yields d = A 3 3 = + c ( ) + d ( ) = + d ( ) 6 6. Finally he ime-domain resul is y () = e u () e cos( u ) () + e sin( u ) () Example Find he sep response of he sysem wih ransfer funcion H ( s) = s + s + The sep response for his sysem will be given by A cs ( + ) d Y( s) = H( s) = = + + s ss ( + s+ ) s ( s + ) + ( s + ) + Using he cover-up mehod we ge A =. To ge c, muliply boh sides by s and le s, s sa sc( s + ) sd lim sy ( s) = = lim + + s ss ( + s+ ) s s ( s+ ) ( s+ ) + + or Copyrigh Rober D. Throne 5

153 = A+ c= + c which gives us c =. Finally, we se s o a convenien value and equae boh sides. In his case s = is a good choice. So we can conclude ha lim Y() s = lim ss ( s s s d = and he complee answer is 6.7 Block Diagrams, Signal Flow Graphs, and Mason s Rule For an LTI sysem, we know ha we can relae he inpu and oupu using convoluion and he impulse response, y () = h ()* x (). Using Laplace ransforms we have he equivalen relaionship Ys () = HsX () () s. Here H() s is called he ransfer funcion, and is he Laplace ransform of he impulse response h (). We can represen inerconneced sysems in he ime domain, as shown in Figure 6., where we have he inpu/oupu relaionships y ( ) = h()* h()* x () y () = [ h() + h()* ] x () y ()*[ δ () + g()* g()* h()] = x ()*[ g ()* g ()* g ( )] As expeced, hese relaionships are in erms of convoluion. A cs ( + ) d = lim s+ ) s ( s+ ) + ( s+ ) + = A+ d = + d y () = u () e cosu () () e sin() u () Copyrigh Rober D. Throne 53

154 Figure 6.. Time domain inerconnecion of sysems. Insead of dealing wih ime-domain funcions and convoluions, we can deal wih ransfer funcions and muliplicaions. The ransfer funcion represenaion of he sysems in Figure 6. can be represened in erms of ransfer funcions as shown in Figure 6.. In he ransfer funcion domain we have Ys () = H() sh() sxs () Y() s = [ H() s + H()] s X() s G() sg() sg() s Y() s = + G() sg() sh() s Now we have replaced convoluion wih muliplicaion and division, which is easier o work wih. As you become more familiar wih ransfer funcions, you will realize ha you can learn a grea deal abou he ime-domain performance wihou acually performing he inverse Laplace ransforms. Copyrigh Rober D. Throne 54

155 Figure 6.. Transform domain represenaion of sysems from Figure 6.. As an alernaive o block diagrams, we can represen he inerconnecion beween subsysems uilizing signal flow graphs. A signal flow graph is composed of direced branches and nodes. Jus like block diagrams, hese graphs show he flow of signals hroughou an inerconneced sysem bu raher han shown ransfer funcions in blocks, he ransfer funcions (or ransmiances) are wrien over he direced branches. Finally, in a signal flow graph all of he branches ino a node are summed. Hence, if you wan negaive feedback you need o include he negaive sign in one of he ransfer funcions. Before geing ino oo much deail, you should examine he following examples of block diagrams and he equivalen signal flow diagrams. In he following signal flow graphs, a indicaes he ransfer funcion is jus uniy. Copyrigh Rober D. Throne 55

156 Figure 6.3. Signal flow graph of a simple feedback sysem. Figure 6.4. Second example of a block diagram and he equivalen signal flow graph. Copyrigh Rober D. Throne 56

157 Figure 6.5. Third example of a block digraph and he equivalen signal flow graph. We would like o be able o deermine he overall sysem ransfer funcion form he inpu o he oupu of a sysem represened by a signal flow diagram. In order o do his we will need o inroduce some definiions and apply hem o an example problem. I is usually helpful o label some of he pahs and loops on he signal flow graph o keep rack of hem. As you are going hrough he following definiions and seps, ry o label hese on he diagram. Le s assume we have he (fairly complicaed) signal flow graph shown in Figure 6.6. Figure 6.6. Example signal flow graph used in he definiions example. Copyrigh Rober D. Throne 57

158 Definiion: A pah is any succession of branches, from inpu o oupu, in he direcion of he arrows, which does no pass any node more han once. A pah gain is he produc of he ransfer funcions (ransmiances) of he branches comprising he pah. For he signal flow graph in Figure 6.6, he pah gains are (arbirarily numbered): P = HHHH H, P = HHH H Definiion: A loop is any closed succession of branches, in he direcion of he arrows, which does no pass any node more han once. The loop gain is he produc of he ransfer funcions of he branches comprising he loops. For he signal flow graph in Figure 6.6, he loop gains are (arbirarily numbered): L = HH, L = HH, L = H H, L = H, L = H H H H H Definiion: To loops are ouching if hey have any node in common. A pah and a loop are ouching if hey have any node in common. Definiion: The deerminan of a signal flow graph is denoed by, and is compued by he formula = (sum of all loop gains) + (sum of all producs of gains of all combinaions of non ouching loops) (sum of producs of gains of all combinaions of 3 nonouching loops) + For he signal flow graph in Figure 6.6, he deerminan is = ( L+ L + L3+ L4 + L5) + ( LL + LL + LL + L L + L L ) ( LLL + LLL ) Definiion: The cofacor of a pah is he deerminan of he signal flow graph formed by deleing all loops ouching he pah. To ge he cofacor for each pah, you wrie ou he deerminan and cross off all loops ouching ha paricular pah. For he signal flow graph in Figure 6.6, pah P, = L4 pah P, = ( L + L ) + ( LL ) 4 4 Definiion: The ransfer funcion of he signal flow graph is given by he formula H sysem = P + P + P3 3 + Copyrigh Rober D. Throne 58

159 For he signal flow graph in Figure 6.6, he sysem ransfer funcion is hen P + P H sysem = In mos cases our sysems are no as complicaed as his one, bu his mehod will work for any sysem provided we are sysemaic. Le s do some examples now. You should ry hese and hen check your answers. Example For he signal flow graph in Figure 6.3 we have P = H, L = H( H) = - HH, = + HH, = H H sysem= + H H Example For he signal flow graph in Figure 6.4 we have P = HH 5 3, P = HHH 3, L = HHH 3 4, = L, =, = HH HHH 3 H sysem= + H H H 3 4 Example For he signal flow graph shown in Figure 6.5, we have P = HHH 3, L = HH 3, L = HH 3 4, = L L, = HHH 3 H sysem = + HH3 + HH 3 4. Example For he block diagram and corresponding signal flow graph from Figure 6.7, he ransfer funcion is H+ H3 H sysem= + HH Example For he block diagram and corresponding signal flow graph from Figure 6.8 he ransfer funcion is H3( + HH ) + H H sysem = + H H Example For he block diagram and corresponding signal flow graph from Figure 6.9, he ransfer funcion is H sysem HH 4 + HHH 3 = HHH HHH HH + HHH + HH Copyrigh Rober D. Throne 59

160 Figure 6.7. Block diagram and signal flow graph for Example Figure 6.8. Block diagram and signal flow graph for Example Copyrigh Rober D. Throne 6

161 Figure 6.9. Block diagram and signal flow graph for Example Copyrigh Rober D. Throne 6

162 Chaper 6 Problems ) In his problem we will derive some useful properies of Laplace ransforms saring from he basic relaionship s X() s = xe () d a) Le s assume x () is a causal signal (i is zero for < ). We can hen wrie x () = xu () () o emphasize he fac ha x () is zero before ime zero. If here is a delay in he signal and i sars a ime, hen we can wrie he signal as x ( ) = x ( ) u ( ) o emphasize he fac ha he signal is zero before ime. Using he definiion of he Laplace ransform and a simple change of variable in he inegral, show ha x( ) u( ) = X( s) e s { } 3s e b) Using he resuls from par a, deermine he inverse Laplace ransform of X() s = ( s+ )( s+ 4) ( 3) 4( 3) Answer: x( ) = ( 3) e e u dx () s c) Saring from he definiion of he Laplace ransform, show ha { x() } =. ds d) Using he resul from par c, and he ransform pair x() = e a u () X () s =, and some simple s + a calculus, show ha a a 3 a 6 { e u ()} =, { e u ()} =, 3 { e u ()} = 4 ( s+ a) ( s+ a) ( s+ a) Copyrigh Rober D. Throne 6

163 ) For he following circuis, deermine he ransfer funcion and he corresponding impulse responses. Scrambled Answers: Ra+ Rb R / / / b Rb L RaC RaC CRaRb () = bδ() e u (), () = δ() e u (), h ( = e u h = δ() e ] L RC a RC a RC a h R h ) (), () [ u () 3) For he following impulse responses and inpus, compue he sysem oupu using ransfer funcions. a) h ( ) = e u( ), x ( ) = u() b) ( ) ( ) h = e u (), x () = δ ( ) c) ( ) h = e u ( ), x ( ) = e u ( ) d) h ( ) = e u (), x () = ( ) u ( ) e) h f) h e u x e u ( ) 3 () = ( ), () = () () = e u (), x () = u( ) u( ) Scrambled Answers : ( ) ( ) y () = () ( ), () ( ) (), () ( ) ( ) e u e u y = e u y = e u ( ) y () = e u (), y () = [ + ( ) + e ] u ( ), y= e e e u ( ) 3( ) 3( ) ( ) [ ( ) ] ( ) Copyrigh Rober D. Throne 63

164 4) For he following circuis, deermine and expression for he oupu Vou () s in erms of he ZSR and ZIR. Do no assume he iniial condiions are zero. Also deermine he sysem ransfer funcion. Answers: R b RRC a b Vou () s = Vin () s + v() RaRbCs + Ra + Rb RaRbCs + Ra + Rb Ra( Rb + Ls) RaRbL Vou () s = Vin () s + i() ( Ra + Rb) Ls + RaRb ( Ra + Rb) Ls + RaRb 5) For he following ransfer funcions 3 5 H() s = H() s = H() s = s + s+ s + 4s+ 6 s + 6s + 4 H() s = H() s = s 4s+ 7 s + 4 By compuing he inverse Laplace ransform show ha he sep responses are given by y () = e cos( ) e sin( ) u () y() = e sin( ) e cos( ) ( ) u y () = e sin() e cos() u () y () = + e sin( 3) e cos( 3) u() 7 7 y() = cos( ) u ( ) 4 4 Copyrigh Rober D. Throne 64

165 6) For he following ransfer funcions, deermine boh he impulse response and he uni sep response. Scrambled Answers: s H() s = H() s = s+ s+ s+ s+ ( )( ) ( )(3 ) ) 8 5 () s + H ( s = H s = s + s+ s + s / /3 h() = e sin(3) u (), h () = e u () + e u () + e u (), h () = e u () e u (), 3 / 3 /3 h () = e cos( 3) u () + e sin( 3) u (), y () = u () e u () + e u (), 3 y () = u () + e sin( 3) u () e cos( 3) u( ), y () = e u () e u () e u (), y() = u() e sin(3) u () e cos(3) u() Copyrigh Rober D. Throne 65

166 7) For he following signal flow diagrams deermine he sysem ransfer funcion. You may use Maple. \ Answers: Y GGG 9( GG 6 7) + GGG 5 6 8( G) 3 = X G G GG G + GGGG + GGG + GGG GGGGG Y GGGGG GGGG 4 6 = X GG GG GG GGG G + GGGG + GGG + GGG + GGGG Y GGGG 9 G3 + GGGGG 6 8 G3( G) + GGGGG G3( G) = X G GG G GG + GGGG + GGG Copyrigh Rober D. Throne 66

167 8) We can also use Mason s rule for sysems wih muliple inpus and muliple oupus. To do his, we use superposiion and assume only one inpu is non-zero a a ime. The only hings ha changes are he pahs (which depends on he inpu and he oupu) and he cofacors (which depends on he pah). The deerminan does no change, since i is inrinsic o he sysem. For he following sysems, deermine he ransfer funcions from all inpus (R) o all oupus (Y). You may use Maple. Copyrigh Rober D. Throne 67

168 Answers: Y GGG R GG GGG GG GGGG 3 = Y GGG 3 5 GG 3 6 = R GG GGG GG + GGGG Y GG R GG GGG GG GGGG 3 = Y G4( GG 3 6) = R GG GGG GG + GGGG ,, Y GGG 3( + G5H) + GGGGG =, R + GH GGG + HG + GGHH HGGGG Y GGGG ( + G H ) R GH GGG HG GGHH HGGGG = Y GGGG + GGGGG, R GH GGG HG GGHH HGGGG Y R = G G G( + HG) + GH GGG + HG + GGHH HGGGG = Copyrigh Rober D. Throne 68

169 Copyrigh Rober D. Throne 69

170 7. Laplace Transform Applicaions In his chaper we will examine many applicaions of Laplace ransforms. While i is possible o go back o he ime-domain o deermine properies of a sysem, i is ofen more convenien o be able o deermine hese properies in he s-domain direcly. 7. Characerisic Polynomial, Characerisic Modes, and he Impulse Response Consider a ransfer funcion of he form H ( s) = Ns ( ) D( s) where Ns () and Ds () are polynomials in s wih no common facors. Ds () is called he characerisic polynomial of he sysem. The poles of he sysem are deermined from Ds () and hese give us mos of he informaion we need o compleely characerize he sysem. The ime-domain funcions ha correspond o he poles of he ransfer funcion are called he characerisic modes of he sysem. To deermine he characerisic modes of a sysem i is ofen easies o hink of doing a parial fracion expansion and deermining he resuling ime funcions. Finally, he impulse response is a linear combinaion of characerisic modes. A few examples will help. Example 7... Consider he ransfer funcion s + H () s = = a + a + a 3 + a s 4 s ( s+ )( s+ 3) s s s+ s+ 3 The characerisic polynomial is Ds ( ) = s ( s + )( s + 3) and he characerisic modes are ( ), ( ), 3 u u e u ( ), and e u (). The impulse response is a linear combinaion of hese characerisic 3 modes, h () = au () + a u () + ae u() + a e u(). 3 4 Example 7... Consider he ransfer funcion s 3 H( s) = = a + a + a3 + a 4 ss ( + ) ( s+ 3) s s+ ( s+ ) s+ 3 The characerisic polynomial is u ( ), e u ( ), e u ( ), and 3 e Ds ss s ( ) = ( + ) ( + 3) 3 modes, h () = au() + ae u () + ae u () + ae u (). and he characerisic modes are u (). The impulse response is a linear combinaion of hese characerisic 3 4 Copyrigh Rober D. Throne 7

171 Example Consider he ransfer funcion 3 s + H() s = = = a + a s + s s+ + s+ + s+ + The characerisic polynomial is Ds s s () = + + and he characerisic modes are 3 and e sin u (). The impulse response is he linear combinaion 3 3 h( ) = ae cos u( ) + ae sin u(). 3 e cos u () There are a few hings o keep in mind when finding he characerisic modes of a sysem: There are as many characerisic modes as here are poles in he ransfer funcion For complex conjugae poles of he form σ ± jωd he characerisic modes will be of he form σ σ e cos( ωdu ) ( ) and e sin( ωdu ) ( ). Noe ha we could combine hese ino he form e σ cos( ω + θ) u ( ) d Example If a ransfer funcion has poles a,, ± 3 j, and 5 ± j he characerisic modes will be e u ( ), e u ( ), e cos(3 u ) ( ), e sin(3 u ) ( ), 5 e cos( ) ( ) u, and 5 e sin( u ) ( ). 7. Asympoic Sabiliy We have previously inroduced he concep of Bounded Inpu Bounded Oupu, or BIBO, sabiliy. As we have seen, an LTI sysem is BIBO sable if h( λ) dλ < Anoher useful definiion of sabiliy, which is used ofen in conrol sysems, is ha of asympoic sabiliy. A sysem is defined o be asympoically sable if all of is characerisic modes go o zero as, or equivalenly, if lim h ( ) =. A sysem is defines o be asympoically marginally sable is all of is characerisic modes are bounded as, or equivalenly, if lim h ( ) M for some consan M. If a sysem is neiher asympoically sable or asympoically marginally sable, he sysem is asympoically unsable. In deermining asympoic sabiliy, he following mahemaical ruhs should be remembered: Copyrigh Rober D. Throne 7

172 Example 7... Assume a sysem has poles a -,. and -. Is he sysem asympoically sable? The characerisic modes for his sysem are e u (), u () and e u (). Boh e u () and e u () go o zero as. u () does no go o zero as, bu i is bounded. Hence he sysem is asympoically marginally sable. Example 7... Assume a sysem has poles a -,, and ± 3j. Is he sysem asympoically sable? The characerisic modes for his sysem are e u (), e u (), e cos(3 u ) ( ), and e sin(3 u ) ( ). All of he modes go o zero as excep for e u (), which goes o infiniy. Hence he sysem is asympoically unsable. Example Assume a sysem has poles a -, -, ± j. Is he sysem asympoically sable? The characerisic modes for his sysem are e u (), e u(), e cos() () u, and e sin() u (). All of he characerisic modes of he sysem go o zero as, so he sysem is asympoically sable. From hese examples, if should be clear ha a sysem will be asympoically sable if all of he poles of he sysem are in he lef half plane (all of he poles have negaive real pars). This is a very easy es o remember. 7.3 Seling Time and Dominan Poles n a lim e = for all n > and a > a lim e cos( ω + θ) = for all a > d u() cos( ωd + θ) sin( ω + θ) d is bounded is bounded is bounded Once we hink abou represening he impulse response as a linear combinaion of characerisic modes, we can define asympoic sabiliy in erms of he way hese modes behave as. Anoher benefi of his approach is ha we can hink of he seling ime of a sysem, i.e., he ime he sysem akes o reach % of is final value, in erms of he seling ime of each of is characerisic modes. When we alk abou he seling ime of a sysem, we assume he sysem is asympoically sable he poles of he sysem are disinc (no repeaed poles) he inpu o he sysem is a sep Le s assume our sysem has ransfer funcion H() s wih corresponding impulse response h () = aφ() + aφ() + + anφn() Here he a i are he coefficiens we deermine using he parial fracion expansion, and he φ i () are he characerisic modes, i.e., ai a φi () s+ p i Copyrigh Rober D. Throne 7

173 Now le s assume he inpu o our sysem is a sep of ampliude A and we wan o use parial fracions o deermine he oupu, A Y() s = H() s = B + b + b + + bn s s s+ p s+ p s+ p In he ime-domain his will have he form y() = Bu() + bφ() + bφ() + + bnφn() The primary difference beween his and he impulse response of he sysem is he erm Bu(), which represens he final value of he sysem due o he sep, and a differen linear combinaion of he characerisic modes (he a are now b, bu he φ i ( ) remain he same.) i i / τi Recall ha he % seling ime for an exponenial funcion φi () = e u () is equivalen o four ime consans, Ts = 4τ. In order o deermine he seling of a sysem wih muliple characerisic modes, we deermine he seling ime corresponding o each characerisic mode. The longes seling ime deermines he seling ime of he sysem. Noe ha his is no an exac formula, since he acual seling ime of he sysem is also a funcion of he coefficiens. However, for mos insances his gives a reasonably good firs esimae. The pole ha produces he longes seling ime is called he dominan pole. If a sysem has wo complex conjugae poles ha produce he longes seling ime, hose poles are he dominan poles. Example Consider a sysem wih ransfer funcion H( s) = ( s+ )( s+ 5)( s + ) and corresponding impulse response h () = e u () e u () + e u () 9 9 The uni sep response of he sysem is 5 5 y () = u () e u () + e u () e u () Noe ha excep for he firs erm, which comes from he inpu, each of he oher ime funcions is he same as for he impulse response. The ime consan for each of he characerisic modes is,., and. wih corresponding seling imes of 4,.8, and.4 seconds. Hence we esimae he seling ime of he sysem o be he larges of hese, or Ts 4 seconds. Figure 7. plos he sysem oupu and each of he characerisic modes. As he plo indicaes, he seling ime is approximaely 4 seconds, hough his is no exac. n Copyrigh Rober D. Throne 73

174 .9.8 y() e - e -5 e Figure 7.. Response of sysem from Example Example Consider a sysem wih ransfer funcion and corresponding impulse response H() s = ( s ) ( s ) 9 h ( ) = e u ( ) e cos(3 u ( ) e ) u( ) sin(3 ) The uni sep response of he sysem is given by ( ).3846 ( ).6849 ( ).363 y = u e u e cos(3 ) u ( ).439 5e sin( 3 ) u () The ime consan for each of he characerisic modes is.,.5, and.5 wih corresponding seling imes of.4,., and. seconds. Hence we esimae he seling ime of he sysem o be he larges of hese, or Ts seconds. Figure 7. plos he sysem oupu and each of he characerisic modes. As he plo indicaes, he seling ime is approximaely seconds, hough his is no exac. Copyrigh Rober D. Throne 74

175 . y() e - e - cos(3) e - sin(3) Figure 7.. Response of sysem from Example Example Consider a sysem wih a ransfer funcion wih poles a -, -4, and -6. Esimae he seling ime of he sysem. We do no have o acually deermine he characerisic funcions of he impulse response o do his problem. The corresponding ime consans will be ½. ¼, and /6. The esimaed seling imes corresponding o each of hese ime consans is,, and.667 seconds. The larges seling ime is seconds, so ha is our esimae of he seling ime of he sysem. In his case, he dominan pole is he pole a -, since i leads o he larges seling ime. Example Consider a sysem wih a ransfer funcion wih poles a 4, 6, and ± 5j. Esimae he seling ime of he sysem. The ime consans ha correspond o hese pole locaions are ¼, /6, and ½ seconds. The seling ime associae wih each of hese ime consans is,.667, and seconds. Hence he esimaed seling ime is again seconds, and he dominan poles are a ± 5j. Example Deermine he esimaed seling ime for a ransfer funcion wih poles a 4 ± 5 j, 8 ± j, and -. The corresponding ime consans are.5,.5, and. seconds. The associaed seling imes are,.5, and.4 seconds. Hence he esimaed seling ime is second and he dominan poles are a 4 ± 5 j. I should be obvious by now ha he dominan poles are hose poles wih he real par closes o he axis. jω Copyrigh Rober D. Throne 75

176 7.4 Iniial and Final Value Theorems I is ofen necessary o deermine he iniial ( = ) or final ( ) value of a funcion represened in he s-domain. While i is possible o perform parial fracions and deermine he ime domain represenaion, his is ofen edious and we would like o be able o perform his compuaion direcly in he s-domain. Insead we generally use he iniial and final value Theorems, which are saed below: Iniial Value Theorem: If x () X() s and X() s is asympoically sable, hen lim x( ) = lim sx ( s) + s Final Value Theorem: If x () X() s and X() s is asympoically sable, hen lim x( ) = lim sx ( s) Noe ha hese final value heorems are also valid if here is a single pole a he origin. s Example Consider he ransform pair 3 X () s = x () = e u () e u () ( s+ )( s+ 3) Clearly all of he poles are in he lef half plane, so we can use he iniial value heorem, lim x( ) = = + lim sx ( s) = s Example Consider he ransform pair s X( s) = x () = e u () + 3 e u () ( s+ 3)( s+ 4) Clearly all of he poles are in he lef half plane, so we can use he iniial value heorem, lm i x() = 3= + lim sx ( s) = s Example Consider he ransform pair 3s + 6s X( s) = x () = e u () + e cos(5 ) u () ( s+ ) ( s+ 3) + 5 Clearly all of he poles are in he lef half plane, so we can use he iniial value heorem, Copyrigh Rober D. Throne 76

177 Example Consider he ransform pair Clearly all of he poles are in he lef half plane, wih he excepion of a single pole a he origin, so we can use he final value heorem, lim x ( ) = Clearly all of he poles are in he lef half plane, wih he excepion of a single pole a he origin, so we can use he final value heorem, lim x( ) = 3 lim sx ( s) = s 3 Example Consider he ransform pair s X ( s) = x () = u () 4 e cos( ) u () s ( s + 3) + 4 Clearly all of he poles are in he lef half plane, wih he excepion of a single pole a he origin, so we can use he final value heorem, As you will see as we go hrough his chaper, he iniial value heorem is ofen used o deermine he iniial amoun of effor needed for a sysem, while he final value heorem is used for deermining he final value of a sysem and he seady sae error. 7.5 Saic Gain lm i x() = + = 3 + lim sx ( s) = 3 s X ( s) = x( ) = u( ) s lim sx ( s) = Example Consider he ransform pair X ( s) = x = u e u s( s + 3) 3 3 s lm i x ( ) = lim sx ( s) = s 3 () () () The saic gain of a sysem is basically he seady sae gain of a sysem when he inpu is a sep funcion. Obviously he concep of he saic gain of a sysem only makes sense for sysems ha are asympoically sable. Since he saic gain of he sysem is measured during seady sae, all of he ransiens have died ou. Copyrigh Rober D. Throne 77

178 The final value heorem is generally used o find he saic gain of a sysem. Assume we have an asympoically sable sysem wih ransfer funcion H() s and he inpu x () o our sysem is a sep of ampliude A, x() = Au(). Then we can deermine he oupu of he sysem as A Y() s = H() sx() s = Hs () s Using he final value heorem, we can hen deermine he seady sae value of he oupu as A lim y( ) = lim sy ( s) = lim sh ( s) = lim H ( s) A = H () A s s s s The gain of he sysem is he oupu ampliude divided by he inpu ampliude, so we can compue he saic gain K as Oupu Ampliude H() A K = Inpu Ampliude = = H () A So we have he resul (for asympoically sable sysems) ha K = H () and lim y( ) = KA for a sep inpu of ampliude A. Example Deermine he saic gain and seady sae value of he oupu for he sysem s + H() s = s +.4s+ 3 for a sep inpu of ampliude 5. Here K = and lim y( ) = KA = 5 =. The response of his sysem is displayed in Figure 7.3. Example Deermine he saic gain and seady sae value of he oupu for he sysem H ( s ) = s + s + for a sep inpu of ampliude 5. Here K = and. lim y( ) = KA = 5 = The response of his sysem is displayed in Figure 7.4. Copyrigh Rober D. Throne 78

179 Figure 7.3. Response of sysem for Example 7.5. o a sep of ampliude 5. The saic gain of his sysem is /3 and he final value of he oupu is hen /3 = 3.3, as he figure shows. Figure 7.4. Response of sysem for Example 7.5. o a sep of ampliude 5. The saic gain of his sysem is and he final value of he oupu is hen, as he figure shows. Copyrigh Rober D. Throne 79

180 7.6 Ideal Second Order Sysems (Again) Recall ha he differenial equaion ha describes an ideal second order sysem is of he form d y () dy( ) + ζω n + ωny ( ) = Kωnx () d d where ζ is he damping raio, ω is he naural frequency, and K n is he saic gain of he sysem. To deermine he ransfer funcion we assume all of he iniial condiions are zero and ake he Laplace ransform of each erm, Combining hese we ge, d y () () = sy s d dy() ζωn = ζωnsy () s d { ωny} = { Kωn } () ω Y() s n x () = Kω X() s n s + ζωns + ω n Y( s) = Kω n X () s and ransfer funcion Kωn K H() s = = s + ζω ns+ ω ζ n s + s + ωn ωn The characerisic equaion for his sysem is s + ζωns + ωn =, and for he under damped case ( < ζ < )he poles of he sysem are given by s = ζω ± jω ζ = ζω ± jω = σ ± jω n n n d d Figure 7.5 displays he relaionship beween hese parameers in he complex plane. We have previously deermined ha he response of he sysem o a sep inpu of ampliude A is given by ζωn y() = KA + ce sin( ω + φ ) where he consans c and φ are deermined by he iniial condiions. Wha we would like o examine now are how we can specify sysem pole locaions o achieve a desired seling ime, percen overshoo, and ime o peak. d Copyrigh Rober D. Throne 8

181 + + Figure 7.5. Relaionship beween damping raio ( ζ ), naural frequency ( ωn ), and damped frequency ( ω ) in he complex plane for an under damped sysem ( < ζ < d ). The poles of he sysem are locaed a s = σ ± jω. Seling Time We have approximaed he seling ime for a sysem wih disinc poles o be he ime i akes for he slowes characerisic mode o reach four ime consans. The ime consan for each characerisic mode is he reciprocal of he magniude of he real par of he pole. For example, for a sysem wih complex σ poles a s = σ ± jω. The corresponding characerisic modes are of he form σ e cos( ω), e sin( ω), max and he ime consan is τ = / σ. Thus if we wan our sysem o have a seling ime of T s, hen we have 4 max Ts = 4τ = < Ts σ or d 4 max T s < σ Copyrigh Rober D. Throne 8

182 This means he magniude of he real par of he pole mus be greaer han poles of he sysem mus be o he lef of 4 max T s, or equivalenly, all. Alhough his relaionship was derived for an ideal second order sysem, i is generally a good iniial approximaion for any sysem wihou repeaed poles. 4 max T s Example Deermine he allowed pole locaions in he s-plane ha correspond o a seling ime of max less han or equal o.5 seconds. Here T s =.5 and σ = 4 = = max T s Hence all poles mus be o he lef of -.6. The allowable pole locaions are shown as he shaded region in Figure 7.6. Figure 7.6. The pole locaions corresponding o a seling ime less ha.5 seconds are shown in he shaded region, o he lef of -.6 on he real axis. Noe ha he seling ime consrain only affecs he real par of he pole, no he imaginary par of he poles. Example Deermine he allowed pole locaions in he s-plane ha correspond o a seling ime of max less han or equal o.5 seconds. Here T s =.5 and σ = 4 = = max T s Copyrigh Rober D. Throne 8

183 Hence all poles mus be o he lef of -8. The allowable pole locaions are shown as he shaded region in Figure 7.7. Figure 7.7. The pole locaions corresponding o a seling ime less ha.5 seconds are shown in he shaded region, o he lef of -8 on he real axis. Noe ha he seling ime consrain only affecs he real par of he pole, no he imaginary par of he poles. Percen Overshoo From our previous analysis, he percen overshoo (PO) is compued as PO = e In order o deermine he pole locaions in he s-plane ha produce an accepable percen overshoo, we need o do some simple algebra. Le s define max max PO O = max Here O is he maximum overshoo, no expressed as a percenage. To se an upper bound on he allowed percen overshoo we have ζπ ζ % PO PO max Copyrigh Rober D. Throne 83

184 or Nex we need o solve for ζ, e e ζπ ζ ζπ ζ ζπ ζ max PO = O O max max ln( O ) ζ ln( O ζ π max max ) Squaring boh sides and expanding we have max ζ ln( O ) ζ π max ln( O ) ζ ζ π max ln( O ) ln( O ) ζ + π π max ln( O ) ζ π max ln( O ) + π max Finally, we use he relaionship depiced in Figure 7.5, cos( θ) = ζ, or cos ( ζ) = θ, where he angle θ is measured from he negaive real axis. Noe ha since we usually specify a maximum allowed percen overshoo we have deermined he maximum allowed value of he damping raio, ζ. Since he damping raio increases (and he damping decreases) as he poles ge closer o he real axis, specifying he maximum allowed percen overshoo deermines he maximum angle allowed. Thus he allowable pole locaions o achieve a maximum percen overshoo will have he shape of a wedge. Finally, noe ha his relaionship is only valid for ideal second order sysems. Example Deermine he allowable pole locaions so an ideal second order sysem will have a maximum percen overshoo of 5%. We have and hen max 5 O = =.5 Copyrigh Rober D. Throne 84

185 ln(.5) ζ π =.44 ln(.5) + π cos ( ) = cos (.44) = 6. ζ 6 o This means he maximum angle wih he real axis is 66. degrees. The allowable pole locaions are shown as he shaded region in Figure 7.8. Figure 7.8. The pole locaions corresponding o a maximum percen overshoo of 5% are shown in he shaded region, wihin he wedge.. Noe ha he percen overshoo affecs boh he real and imaginary pars of he poles. Example Deermine he allowable pole locaions so an ideal second order sysem will have a maximum percen overshoo of %. We have and hen max O = =. Copyrigh Rober D. Throne 85

186 ln(.) ζ π =.59 ln(.) + π ζ = = cos ( ) cos (.59) o This means he maximum angle wih he real axis is 53.8 degrees. The allowable pole locaions are shown as he shaded region in Figure 7.9. Figure 7.9. The pole locaions corresponding o a maximum percen overshoo of % are shown in he shaded region, wihin he wedge. Noe ha he percen overshoo affecs boh he real and imaginary pars of he poles. Noe also ha he angle of his wedge is narrower han ha in Figure 7.8 since he allowed percen overshoo in ha case is smaller (% compared o 5% in Figure 7.8). Example Deermine he allowable pole locaions for an ideal second order sysem so he response o a sep will have a percen overshoo less han 5% and a seling ime of less han seconds. To solve his problem we need o deermine he accepable regions for each consrain, and hen deermine if here is any overlapping region so ha boh consrains will be saisfied. To mee he percen overshoo requiremen we have Copyrigh Rober D. Throne 86

187 max 5 O = =.5 ln(.5) ζ π =.69 ln(.5) + π cos ( ζ ) = cos (.69) = 464. o max To mee he seling ime requiremen we have T s =. and σ = 4 = 4. = max T s Hence he region of he s-plane ha mees boh consrains is o he lef of - and wihin a wedge wih an angle of 46.4 degrees. Each of hese individual regions, and he overlapping region are displayed in Figure 7.. Figure 7.. The pole locaions corresponding o a maximum percen overshoo of 5% and a seling ime of seconds are shown in he shaded regions. The overlapping region shows he locaion in he s- plane where boh condiions are me. Copyrigh Rober D. Throne 87

188 Time o Peak From our previous analysis, he ime o peak for an under damped ideal second order sysem is given by π Tp = ωd From Figure 7.5 we can easily see ha ω d, he damped frequency, is jus he imaginary par of he poles. Hence his consrain will only consrain he imaginary pars of he poles. If we define he max maximum allowable ime o peak as T p, hen we have π max Tp = TP ωd or π ω max d T p max which indicaes he imaginary par of he pole mus be larger han π /. Example Deermine he pole locaions for an ideal second order sysem ha correspond o a ime max o peak of less han.5 seconds for an ideal second order sysem. We have T p =.5 and hen π π = = 6.83 < ω max d T p.5 This means he imaginary par of he poles mus be larger han The accepable pole locaions are shown as he shaded region in Figure 7.. Example Deermine he pole locaions for an ideal second order sysem ha correspond o a maximum ime o peak of less han or equal o.5 seconds and a seling ime of less han or equal o max seconds. We have T p =.5 and hen π π = =.9 < ω max d T p.5 max To mee he seling ime requiremen we have T s =. and σ = 4 = 4 4. = max T s Hence o mee boh requiremens we need he real pars of he poles o he lef of -4, and he imaginary pars greaer han.9 (or less han -.9). The accepable pole locaions are shown in Figure 7.. Example Deermine he pole locaions for an ideal second order sysem ha corresponds o a ime o peak of less han or equal o 3 seconds, a seling ime of less han seconds, and a percen overshoo max of less han %. We have T p = 3. and hen T p Copyrigh Rober D. Throne 88

189 π max T p π = = 3..5 < ωd To mee he seling ime requiremen we have max T s =. and σ = 4 4. max T = = s max Finally o mee he percen overshoo requiremen we have O = =. ln(.) ζ π =.456 ln(.) + π cos ( ζ ) = cos (.456) = 6. 9 o Figure 7.3 displays he accepable regions for each of he hree requiremens, and he overlapping region where all hree requiremens are me. and Copyrigh Rober D. Throne 89

190 Figure 7.. The pole locaions corresponding o a maximum ime o peak of.5 seconds. This corresponds o he imaginary par of he poles being larger han 6.8. Noe ha his consrain affecs only he imaginary pars of he pole. Figure 7.. The pole locaions corresponding o a maximum ime o peak of.5 seconds and a seling ime less han seconds. The seling ime consrain means he real par of he poles mus be le han -4, and he peak ime consrain means he absolue value of he imaginary par of he pole mus be greaer han.9. The pole locaions ha mee boh of hese consrains is he overlapping regions, labeled as Accepable pole locaions. Copyrigh Rober D. Throne 9

191 Figure 7.3. The accepable pole locaions for an ideal second order sysem ha corresponds o a ime o peak of less han or equal o 3 seconds, a seling ime of less han seconds, and a percen overshoo of less han %. The pole locaions ha mee all of hese consrains is he overlapping regions, labeled as Accepable pole locaions. 7.7 Block Diagrams We ofen need o analyze and design inerconneced sysems. When we inroduced convoluion for LTI sysems, we demonsraed some simple inerconneced sysems. However, using convoluion echniques for hese sysems is ofen difficul. Insead we uilize ransfer funcions relaing he oupu of one sysem o he inpu of anoher sysem. We hen use he fac ha he ime domain relaionship y ( ) = h() h () h () x( ) 3 is equivalen o he s-domain algebraic relaionship Ys () = H() sh() sh() sxs () 3 Copyrigh Rober D. Throne 9

192 Example Consider he op-amp circui shown in Figure 7.4. The inpu o he circui is v () in, he oupu is v (), and v () is he volage depiced in he figure as he oupu of he firs op amp. ou m Figure 7.4. Proporional gain circui for Example A he negaive erminal of he firs op amp we have or Vin () s Vm () s + = R R m() s = Vin () s R So he ransfer funcion beween inpuv s and oupu V () s is V in () R m R H() s = R Similarly we have R V s V s 4 ou () = m() R3 So he ransfer funcion beween inpu () R4 Vm s and oupu Vou () s is H () s =. We can depic hese R3 relaionships graphically as Copyrigh Rober D. Throne 9

193 We can hen combine hese blocks as follows: This block diagram indicaes graphically ha R R4 RR 4 V m() s = Vin () s, Vou () s = Vm () s, and Vou () s = V in() s R R3 RR 3 Finally, we can wrie his as where k p is a proporionaliy consan. V () s = k V () s ou p in Example Consider he op-amp sysem shown in Figure 7.5. Figure 7.5. Inegral and gain op amp circui for Example A he negaive node of he firs op amp we have Vin () s Vm () s + =, or Vm () s = Vin () s R RC s Cs A he negaive node of he second op amp we again have We can depic hese relaionships graphically as R V s V s 4 ou () = m() R3 Copyrigh Rober D. Throne 93

194 Again, we can combine hese as This block diagram indicaes graphically ha R 4 R 4 ki Vou () s = Vin () s = Vin () s = Vin () s RCs R3 RRCs 3 s Here we have wrien he proporionaliy consan as k i = R4 RRC 3 Example Consider he differenial amplifier circui shown in Figure 7.6. We assume v () = v () v (). A he negaive inpu erminal we have ou a b Vf () s V () s Va () s V () s + = R Rg and a he posiive erminal we have + + Vr () s V () s Vb () s V () s + = R R + Since under he ideal op amp assumpion V () s = V () s, we can rearrange hese as Vf () s Va() s + Vr () s Vb() s + = V () s + = V () s + = + R Rg R Rg R Rg R Rg Simplifying his we ge Rg Va () s Vb () s = Vou () s = Vr () s Vf () s R or Vou () s = k Vr () s Vf () s This relaionship is depiced graphically in Figure 7.7 and is commonly used in feedback sysems. Noe ha if R = R hen k =. g g Copyrigh Rober D. Throne 94

195 Figure 7.6. Differenial amplifier circui used in Example Figure 7.7. Block diagram for he differenial amplifier circui. This configuraion is he basis for feedback. Example Consider he model of an armaure conrolled DC moor shown in Figure 7.8. Te armaure (he par ha does he work) is locaed on he roor, while he field (he par ha creaes he magneic field) is locaed on he saor. The field source is consan and hence he srengh of he field does no vary, his means he consans Ke and K do no vary. The sysem inpu is he applied volage v () a. The developed moor orque, Tm (), is proporional o he curren flowing in he loop, ia (), so Tm() = Kia(). The moor also develops a back emf, eb (), which is proporional o he speed of he moor, eb() = Keω(). The moor is used o drive a load wih momen of ineria J, damping B, and a load orque T (). l Copyrigh Rober D. Throne 95

196 To deermine he curren we have or The orque developed by he moor is hen Figure 7.8. Model of an armaure conrolled DC moor. V () s I () s [ R + L s] = E () s a a a a b I V () s E () s = R + Ls a b a () s a a Tm() s = KI a() s This orque is hen used o spin he load and overcome fricional forces and any exernal applied loads. The free body diagram for he load is shown below. Conservaion of angular momenum gives J θ = Tm B θ TL In he Laplace domain we ge s JΘ () s + BsΘ () s = T () s T () s or Tm() s TL() s sθ () s =Ω () s = sj + B m L Copyrigh Rober D. Throne 96

197 Finally, we have E () s = K Ω() s b e The block diagram for his sysem is shown in Figure 7.9. Noe ha his sysem includes feedback. Figure 7.9. Block diagram represenaion for he armaure conrolled DC moor from Example Alhough block diagrams and ransfer funcions are widely used in various engineering disciplines for showing he inerconnecion of subsysems, one needs o be careful o avoid loading effecs when represening a sysem wih a block diagram. Consider he simple resisive circui shown in Figure 7. wih inpu v and oupu v (). in () ou Figure 7.. Circui used o demonsrae loading. A he node wih volage v () m we can wrie Vin () s Vm () s Vm () s Vm () s = + R R R + R a b c d This can be simplified o be Rb( Rc + R ) d Vm ( s) = Vin () s ( Ra + Rb)( Rc + Rd) + RaRb and finally we ge RbR d Vou () s = Vin () s ( Ra + Rb)( Rc + Rd) + RR a b Now suppose we decided o wrie he circui as wo inerconneced subsysems, as shown in Figure 7.. Copyrigh Rober D. Throne 97

198 Figure 7.. Circui from Figure 7. wrien (incorrecly) as wo subsysems. The ransfer funcion for he firs sysem is clearly Vm() s Rb = V () s R + R and he ransfer funcion for he second subsysem is Vou () s Rd = Vm() s Rc + Rd From his we can deermine he sysem ransfer funcion o be Vou () s Rb Rd = Vin () s ( Ra + Rb) ( Rc + Rd ) and he oupu is RbR d Vou () s = Vin () s ( Ra + Rb )( Rc + Rd) This is clearly he wrong answer, so wha when wrong? In general, if a ransfer funcion changes afer a sysem is conneced o i, he connecions is said o have a loading effec. For our circui he relaionship beween he sysem inpu v () in and he oupu vm () changes when he second half of he circui is added o he sysem. For elecronic sysem we can ofen inser an isolaing amplifier o remove loading beween subsysems. However, loading can occur for non-elecrical sysems also, so you need o be aware of i. 7.8 Feedback Sysems in a b Assume we have a ransfer funcion, G () p s, ha represens a sysem, and we wan o make his sysem behave in a cerain manner. Typically we call any sysem we are rying o conrol a plan. For example, assume we have he mass-spring-damper sysem shown in Figure 7.. In his sysem, he inpu is he volage applied o a moor (which is modeled as a simple gain) and he moor oupu is a force applied o he car. The sysem oupu is he displacemen of he car from equilibrium. Copyrigh Rober D. Throne 98

199 Figure 7.. Spring-mass-damper sysem. The moor is modeled as a simple gain. The inpu o our plan is he conrol signal u () and he oupu is he displacemen of he car x (). Noe ha in his conex u () is no necessarily a uni sep. I can be any allowable inpu. However, i is convenional in conrol sysems o label he inpu o a plan as u (). A free body diagram of our sysem is as follows: Applying conservaion of linear momenum we ge he equaion of moion, mx () = ku m () kx () bx () which gives us he ransfer funcion for he plan as km Gp () s = s m + sb+ k There are wo general mehods for rying o conrol he behavior of a plan, open loop conrol and closed loop conrol. These wo mehods are displayed in Figure 7.3. For he open loop conrol sysem, we have he sysem ransfer funcion Y() s G () () () () = G o s = c sgp s Rs For he closed loop sysem we need o do a bi more work. In analyzing a closed loop sysem we usually look a an inermediae signal ha relaes inpu and oupu and hen ry o eliminae any inermediae signals. So for his sysem we have Es () = Rs () YsHs () () We can hen wrie he oupu in erms of he error signal as Y() s = EsG () sg() s c p Copyrigh Rober D. Throne 99

200 Figure 7.3. Open loop and closed loop conrol of a plan. Here r () is he reference inpu, u () is he conrol effor or conrol signal, y () is he sysem oupu, and e () is he error signal. Ofen he ransfer funcion in he feedback loop, H() s, is some ype of ransducer which convers he oupu o he same form a he inpu. Finally we need o remove he error signal from hese wo equaions, Y() s = [ R() s Y() s H( s) ] Gc() s Gp( s) Rearranging hese we ge he closed loop ransfer funcion. Y() s Gc() sgp() s Go () s = R() s = + H() sgc() sgp() s A his poin, i should be obvious o you ha using his ransfer funcion o deermine properies of he sysem is much easier han he equivalen ime-domain convoluion based represenaion, y() [ δ () + h() g () g ()] = r() g () g () Mos conrol sysems are closed loop conrol sysems. As you will see, a closed loop sysem has he abiliy o correc for errors in modeling he plan, or if he plan changes over ime as componens age. 7.9 Seady Sae Errors c p c p Ofen we design a conrol sysem o rack, or follow, he reference inpu. How well he sysem racks he reference inpu is usually hen divided ino wo pars: he ransien (ime-varying) response, and he seady sae response. The mos common reference inpu is a sep inpu, and we can use our previously defined measures of seling ime, percen overshoo, and rise ime o measure how well our sysem racks he sep inpu during he ransien ime. Once he sysem has reached seady sae we ofen wan o use he seady sae error as a measure of how well our sysem racks he inpu. Copyrigh Rober D. Throne

201 The seady sae error, e ss, is usually defined as he difference beween he reference inpu, r (), and oupu of he sysem, y (), in seady sae, or While we can use parial fracions and inverse Laplace ransforms o compue his, i is ofen easier o do his compuaion in he s-domain using he final value Theorem. If we assume he sysem is asympoically sable, hen we have If we assume our sysem has ransfer funcion ss s, hen we have Finally, if we assume our inpu is a sep of ampliude A, r() = Au(), hen and Clearly for a seady sae error of zero, we wan gain of he sysem o be one. [ ] e = lim r () y () ss [ ] [ ] e = lim r ( ) y ( ) = lim s Rs ( ) Y( s) ss G () s Example Assume we have he sysem ransfer funcion. Noe also ha his means we wan he saic G () s = ( s+ )( s+ ) and he inpu o he sysem is a sep of ampliude 3. Deermine he seady sae error in he ime and s- domain. In he ime-domain we can use parial fracions, s [ ] = im [ () sr() ] = lim () [ s] e = lim srs () Ys () l srs G() s srs G() s s A Rs () = s s [ ()] e = lim A G s ss G () = or Then Ys () = G () srs () = = 3 + ( s+ )( s+ ) s s s+ s+ 3 3 y = u e u + e u () () 3 () () ess = lim[3 u () y ()] = lim[3 u () u () + 3 e u () e u ()] = In he s-domain we have Copyrigh Rober D. Throne

202 Example For he sysem depiced in Figure 7.4, deermine he value of he prefiler gain,, so he seady sae error for a sep is zero. For his sysem we have he closed loop ransfer funcion 3 Gpf 3G G () s = s+ s+ = pf ( s+ )( s+ ) + 5 s+ s+ For zero seady sae error we need 3 ess = lim A[ G( s) ] = 3[ G()] = s () 3 G pf G = = 7 G pf So we need G pf = 7 / 3. Noe ha in his case we do no really need o simplify he ransfer funcion, we can direcly evaluae he ransfer funcion a s =, 3 3 Gpf Gpf 3 G () s = = = G pf Clearly in his example we wan he prefiler gain o be 7/3. Figure 7.4. Block diagram for Example For a uni sep inpu, he prefiler should be 7/3 o produce zero seady sae error. Example For he sysem depiced in Figure 7.5, deermine he value of he prefiler so he seady sae error for a sep is zero. For his sysem we have he closed loop ransfer funcion G pf G () s = ss + s+ + ss + s+ Copyrigh Rober D. Throne

203 Noe ha we canno immediaely se s = in his form. We could muliply he ransfer funcion ou, bu i is easier o jus muliply he op and boom by s, G pf G () s = s + s+ s + s + s+ Now we can se s = o ge G () = Gpf, so for a zero seady sae error we need he prefiler o be. Figure 7.5. Block diagram for Example For a uni sep inpu, he prefiler should be one o produce zero seady sae error. Example For he sysem depiced in Figure 7.6, deermine he value of he parameer k so he seady sae error for a uni sep is less han or equal o., ess.. For his sysem we have he closed loop ransfer funcion 3s + k G 4 5 () s = s + s + 3s + + k s + s and Then we wan or This means we need k 4.5. e ss k G () = + k = = k = G () + k + k + k Copyrigh Rober D. Throne 3

204 Figure 7.6. Block diagram for Example For a uni sep inpu, if we wan he seady sae error less han or equal o., we need k > Iniial Conrol Effor Alhough i is ofen sraighforward o design a conrol sysem o produce a given seady sae error or an accepable ransien response, someimes hese conrollers require a conrol effor ha is no possible o produce. In many, hough no all, insances, he iniial conrol effor is he larges conrol effor when he inpu is a sep. In order o quickly deermine he iniial conrol effor we use he iniial value Theorem. Recall ha he iniial value Theorem saed ha if x () X( s) and X() s is asympoically sable, hen lim x( ) = lim sx ( s) + s For our sandard closed loop sysem in Figure 7.7, he conrol effor is denoed by U() s for his as follows:. We can solve and Hence Combining hese we have Es () = Rs () H() sy() s Y() s = EsG () sg() s U() s = EsG () () c s Y() s = U() sg U( s) E( s) = G ( s) c c p p () s U() s = R() s H() s Gp () s U() s G () s c U() s = RsG () () s H() sg() sg() su() s c c p This yields he following expression for he conrol effor Gc () srs () U() s = + H () sg () sg () s c p Copyrigh Rober D. Throne 4

205 Finally, o deermine he iniial conrol effor, we have + sgc () s R() s u( ) = lim su ( s) = lim s s + H () sgc () sgp () s If we assume he inpu is a sep of ampliude A hen we have + Gc () sa u( ) = lim s + H () sg () sg () s As an example, le s consider a plan wih he ransfer funcion 5 G () s = p s + s + We will assume he closed loop conrol configuraion shown in Figure 7.7, and look a he resuls using hree differen conrollers. We will assume our reference inpu is a uni sep for all hree examples. c p 5 Figure 7.7. Block diagram used for conrolling he plan G () s =. p s + s + In he firs case, we will assume we have a proporional (P) conroller, where he conrol effor is proporional o he error signal. Here we will have G ( s ) = k, k = c p p where we have assigned k p =. The seady sae error for his sysem can easily be deermined o be e ss =.. The closed loop poles are a ± j.5 which gives an approximae seling ime of 4 seconds. In he second case, we will assume we have a proporional plus inegral (PI) conroller. Here he conrol effor is made up of wo componens, one is proporional o he error signal, and one is proporional o he inegral of he error signal. Here we will have ki Gc( s) = kp +, kp =.4, ki =. s where we have assigned k p =.4 and k i =.. The seady sae error for his sysem is zero, e ss =. The closed loop poles are a (approximaely).6 ± j.94 and.8, which gives an approximae seling ime of 6.7 seconds. Copyrigh Rober D. Throne 5

206 In he hird case, we will assume we have a proporional plus derivaive (PD) conroller. Here he conrol effor is again made up of wo componens, one is again proporional o he error signal and he oher is proporional o he derivaive of he error signal. This conroller will have he form G( s) = k + ksk, = 8, k =.4 c p d p d where we have assumed k p = 8 and k d =.4. The seady sae error for his sysem is approximaely.5. The closed loop poles are a (approximaely). ± j6. 6 which gives an approximae seling ime of seconds. The response of he plan o each of hese conrollers is shown in Figure 7.8. Nex le s look a he iniial conrol effor for each of hese conrollers. For he proporional conroller we have k + p u( ) = lim = kp = s 5 + ()( k p ) s + s+ For he proporional plus inegral conroller we have ki k p + + u( ) = lim s = kp =.4 s ki 5 + () k p + s s + s+ Finally, for he proporional plus derivaive conroller we have kp + kds + u( ) = lim = s 5 + ()( kp + kds) s + s + For hese hree conrollers, he PI conroller requires he leas iniial conrol effor. While he conrol effor for he P conroller is finie, i may be more difficul o implemen his conroller using op amps wih fixed volage sources. Finally, he iniial conrol effor is infinie for he PD conroller. However, ofen his means he source will jus saurae and no reach infiniy. However, i is somehing you will need o be aware of. Copyrigh Rober D. Throne 6

207 5 Figure 7.8. The response of he plan G () s = o he proporional (P) conroller, p s s G + + ( c s ) =. he proporional plus inegral (PI) conroller G ( c s ) =.4 +, and he proporional plus derivaive s conroller G ( c s ) = s. The inpu was a uni sep. Copyrigh Rober D. Throne 7

208 Chaper 7 Problems ) For he following ransfer funcions, deermine he characerisic polynomial he characerisic modes if he sysem is (asympoically) sable, unsable, or marginally sable s ss ( ) a) H() s = b) H() s = c) H() s = ss ( + )( s+ ) ( s+ ) ( s + s+ ) s ( s+ ) s d) H() s = e) H() s = ( s )( s+ )( s + ) ( s + )( s+ ) Parial Answer: sable, unsable,m 3 marginally sable ) For a sysem wih he following pole locaions, esimae he seling ime and deermine he dominan poles a) -,-,-4,-5 b) -4, -6, -7, -8 c) -+j, --j, -, -3 d) -3-j, -3+j, -4+j, -4-j Scrambled Answers: 4/3, 4, 4, 3) Deermine he saic gain for he sysems represened by he following ransfer funcions, and hen he seady sae oupu for an inpu sep of ampliude 3: s s 4 H( s) = +, H() s =, H() s = s + s+ s + 4s+ 4 s + s + Answers:,.5, -4, 6,.75, - (his should be very easy) Copyrigh Rober D. Throne 8

209 4) Consider he following simple feedback conrol block diagram. The plan, he hing we wan o 3 conrol, has he ransfer funcion Gp () s = and he conroller is a proporional conroller, so s + G () s = k. c p a) Deermine he seling ime of he plan alone (assuming here is no feedback) b) Deermine he closed loop ransfer funcion, k p G () s c) Deermine he value of so he seling ime of he sysem is.5 seconds. d) If he inpu o he sysem is a uni sep, deermine he oupu of he sysem. e) The seady sae error is he difference beween he inpu and he oupu as. Deermine he seady sae error for his sysem. 3 8 Parial Answer: y() = e u ( ), ess=.5 4 5) Show ha he following circui can be used o implemen he PI conroller U() s RR 4 R4 Gc() s = = kp + ki = + E() s s RR RRC s 3 3 Copyrigh Rober D. Throne 9

210 6) The following figure shows hree differen circuis, which are subsysems for a larger sysem. We can wrie he ransfer funcions for hese sysems as G K ω low low a () s = s + ωlow Khighs Gb () s = Gc () s = K s + ω high ap Deermine he parameers Klow, ωlow, Khigh, ωhigh, and K ap in erms of he parameers given (he resisors and capaciors). Answers: K Ra R Rc =, ω =, K =, ω =, K = R R C R RC R low low high high ap a b b Copyrigh Rober D. Throne

211 7) Consider he following simple feedback conrol block diagram. The plan, he hing we wan o 3 G conrol, has he ransfer funcion p () s = s and he conroller is an inegral conroller, so () k G i + c s = s. a) Deermine he closed loop ransfer funcion, G () s b) Deermine he poles of value of G () s and show hey are only real if < k i <. Noe ha he bes 3 possible seing ime is 4 seconds. Use k i = for he remainder of his problem. 3 c) If he inpu o he sysem is a uni sep, deermine he oupu of he sysem. d) The seady sae error is he difference beween he inpu and he oupu as. Deermine he seady sae error for his sysem. Parial Answers: ( ) y = e e u( ), e ss = 8) Consider he following simple feedback conrol block diagram. The plan is Gp () s =. The inpu s + 4 is a uni sep. a) Deermine he seling ime and seady sae error of he plan alone (assuming here is no feedback) b) Assuming a proporional conroller, G () s = k, deermine he closed loop ransfer funcion, G () s c) Assuming a proporional conroller, G () s k, deermine he value of k so he seady sae error for a uni sep is., and he corresponding seling ime for he sysem. d) Assuming a proporional conroller, G () s k,deermine he value of k so he seling ime is.5 seconds, and he corresponding seady sae error. e) Assuming an inegral conroller, G () s = k / s, deermine closed loop ransfer funcion, G () s f) Assuming an inegral conroller, Gc() s = ki / s, deermine he value of ki so he seady sae error for a uni sep is less han. and he sysem is sable. Parial Answers: T c c c c p = p p i = p p =, e =.5, k = 8, k =, T =., e =.5, k > s ss p p s ss i Copyrigh Rober D. Throne

212 9) (Model Maching) Consider he following closed loop sysem, wih plan G () p s and conroller G () c s. One way o choose he conroller is o ry and make your closed loop sysem mach a ransfer funcion ha you choose (hence he name model maching). Le s assume ha our desired closed loop ransfer funcion, G () o s, our plan can be wrien in erms of numeraors and denominaors as N () () o s Np s Go() s = Gp() s = D () s D () s No() s Dp () s Show ha our conroller is hen Gc () s = N () s D () s N () s p o [ ] o Noe ha here are some resricions here, in ha for implemenaion purposes he conroller mus be sable, and i mus be proper. ) For he following sysem, wih plan Gp () s =, and conroller G s + () c s o p a) Using he resuls from problem 9, deermine he conroller so ha he closed loop sysem maches a second order ITAE (Inegral of Time and Absolue Error) opimal sysem, i.e., so ha he closed loop ransfer funcion is ω G () s = s +.4ωs+ ω ω ( s + ) Anwes. Gc () s =, noe ha here is a pole/zero cancellaion beween he conroller and he ss ( +.4 ω) plan and here is a pole a zero in he conroller. b) Show ha he damping raio for his sysem is.7, he closed loop poles of his sysem are a.7ω ± j.74ω. For faser response should ω be large or small? c) Wha is he seady sae error for his sysem if he inpu is a uni sep? d) Deermine he conroller so ha he closed loop sysem maches a hird order deadbea sysem, i.e., so ha he closed loop ransfer funcion is 3 ω G () s = 3 3 s +.9ωs +.ωs+ ω 3 ω ( s + ) Ans. Gc () s =, noe ha here is a pole/zero cancellaion beween he conroller ss ( +.9ωs+. ω) and he plan and here is a pole a zero in he conroller. e) Wha is he seady sae error of his sysem for a uni sep inpu? Copyrigh Rober D. Throne

213 ) When implemening a conrol sysem, we need o be careful he conrol effor does no ge oo large. Someimes his is overcome by changing he feedback srucure. Consider he following wo feedback conrol srucures, For he firs srucure, he iniial conrol effor is given by + sr() s Gc () s u( ) = lim su ( s) = lim s s + G () sg () s For he second srucure he iniial conrol effor is given by + sr() s Ga () s u( ) = lim su ( s) = lim s s + G () s + G () s G () s Le s assume we can break up he conroller so ha we can wrie he ransfer funcions as a) Show ha if we can pariion he conroller so ha Then we can wrie he closed loop ransfer funcions as Hence, if D () a s is a consan, he wo srucures produce he same closed loop poles. 3 b) Assume we have he plan Gp () s =, and assume we are rying o use a PD conroller. So s + Show ha if he inpu is a sep of ampliude A, hen for he wo conrol srucures we have c [ ] Np () s Nc() s Na() s Nb() s G () s =, G () s =, G () s =, G () s = D () s D () s D () s D () s p c a b p c a b G () s + G () s = G () s a b c Nc() snp() s Dc() sna() snp() s G o() s =, Go() s = Dc() sdp() s+ Nc() snp() s Da() s Dc() sdp() s+ Nc() snp( s) G () s = k + k s, G () s = k, G () s = k s c p d a p b d a b p p Copyrigh Rober D. Throne 3

214 Ak + A + p A u( ) =, ess=, u( ) =, ess= + 3k + 3k + 3k p d p This second conrol srucure is ofen used wih PID conrollers, as I-PD and PI-D conrollers. ) (Malab Problem) Consider he following Sallen-Key filer, shown in he figure below: We are going o solve for he oupu of his circui for a range of frequencies using Malab. a) Download he m-file homework9.m from he class websie. You will need o modify his code in his problem. b) Wrie node equaions for his sysem in erms of he nodes Va, Vb, and Vou and he inpu volage Vin. Noe ha you will no be able o wrie a node equaion a he oupu of he op amp, bu will have o relae Vb and Vou using he ideal op-amp assumpion. The impedance for a capacior is Z( jω) = jω C. In Malab, we can use j as he square roo of - (hough you may have o muliply by j explicily). The capacior is microfarad, he resisors are ohms, and he parameer A is.5. These parameers have already been enered in he program. The frequency will be changing however. c) In he Malab program, wrie he node equaions in erms of he marix M, and he inpu vecor b. Assume we wrie he marix equaion as Mx = b, or Va Vin [ M] V b = V ou d) Use Malab o solve he marix equaion for he nodal volages (x). In his problem we only care abou he oupu volage, he hird elemen of x. Assume he inpu volage is vol wih a phase of zero degrees. Copyrigh Rober D. Throne 4

215 e) Use he Malab commands abs and angle o ge he magniude and phase of he oupu signal. The command angle gives he phase in radians, and you will need o conver his o degrees. Your final plo should look like ha in he following figure. Turn in your code and your plo. 5 4 Vou (vols) Frequency (rad/sec) Vou (degrees) Frequency (rad/sec) : Copyrigh Rober D. Throne 5

216 Copyrigh Rober D. Throne 6

217 8. Seady Sae Frequency Response Consider he response of wo LTI sysem wih ransfer funcions 5 H() s = s + 8 H() s = s +.s+ o he inpus x () = sin( ωu ) () forω = 5,, and 5 radians/sec. Figure 8. displays he response of he firs sysem o he hree inpu sinusoids, while Figure 8. displays he response of he second sysem o he hree inpu sinusoids. In boh figures, he inpu sinusoid is displayed as a dashed line and he oupu is a coninuous line. Boh figures include heavy solid lines ha bound he ampliude of he oupu sinusoid in seady sae. As he figures demonsrae, boh of he sysems go hrough some iniial ransiens, and hen reach a seady sae response. The firs sysem has a pole a - and he second sysem has is poles a approximaely.6 ± 3. j, so he % seing imes for he wo sysems are esimaed o be 4 and 6.7 seconds which corresponds prey well wih he resuls in he figures. Once he sysems come ino seady sae, he oupu of he sysem has a consan ampliude, and here is a consan relaionship beween he inpu signal and he oupu signal. I is imporan o noe ha he seling ime for he sysem is no a funcion of he frequency of he inpu, bu is a propery of he sysem! A his poin we know how o quickly esimae he seling ime of a sysem based on he poles of he sysem, and we nex wan o be able o quickly esimae he seady sae oupu of an asympoically sable sysem wih a sinusoidal inpu. In order o do his we firs need o review Euler s ideniies and wrie hem in a differen form han you are used o seeing. 8. Euler s Ideniy and Oher Useful Relaionships The usual form of Euler s ideniy is j e ω = ω + ω cos( ) jsin( ) We can also wrie his as cos( ) jsin( ) j e ω = ω ω If we add and subrac hese wo expressions we ge j j e ω + e ω = cos( ω ) jω jω e e = jsin( ω) Finally, we ge our alernae forms of Euler s ideniy jω jω e + e cos( ω) = jω jω e e sin( ω) = j This form of Euler s ideniy is very useful in boh his and in laer courses. Copyrigh Rober D. Throne 7

218 y () y () y 3 () Time (sec) Figure 8.. Response of he firs sysem o sinusoids of 5,, and 5 radians/sec. The esimaed seling ime of he sysem is 4 seconds. The inpu signal is he dashed line and he oupu signal is he solid line. The hick solid line bounds he seady sae ampliude. Nex, le s assume our usual case of a proper ransfer funcion ha is he raio of wo polynomials, m m bms + bm s + + bs + b H() s = n n s + an s + + as + a Since his is proper we have m n. Le s also assume ha we have a real valued sysem, so ha if he inpu is a real valued funcion he oupu will be a real valued funcion. This means ha all of he coefficiens mus be real values. To undersand his, remember ha his ransfer funcion means he inpu and oupu are relaed by he differenial equaion d n y n m m () () () x() x() () a d y n a dy a () b d d + y dx m m () n = + b b bx n m m d d d d d d Hence, if he inpu is real and we wan he oupu o be real all of he coefficiens mus also be real. Copyrigh Rober D. Throne 8

219 5 y () y () y 3 () Time (sec) Figure 8.. Response of he second sysem o sinusoids of 5,, and 5 radians/sec. The esimaed seling ime of he sysem is 6.7 seconds. The inpu signal is he dashed line and he oupu signal is he solid line. The hick solid line bounds he seady sae ampliude. Nex, le s assume we wan o evaluae he ransfer funcion a s = jω and also a s = jω and if we can relae he wo. Remember ha as we raise j o increasing powers we cycle hrough he same four values, i.e., j = j, j =, j = j, j =, and j = j so we are righ back where we sared. To deermine wha is going on wih ransfer funcions we will look a low order sysems and build are way up. Finally, remember ha o deermine he complex conjugae of a number, you replace j wih j. Copyrigh Rober D. Throne 9

220 Firs Order Sysem: bs+ b jbω + b jbω + b * H( s) =, H( jω) =, H( jω) = = H ( jω) s+ a jω + a jω + a Second Order Sysem: bs + bs+ b bω + jbω + b H() s =, H( jω ) =, s + a s + a ω + ja ω + a b ω jbω + b H( jω ) = H ( j ) * = ω ω jaω + a Third Order Sysem: 3 3 b3s + bs + bs+ b jb3ω bω + jbω + b H() s =, H( jω 3 ) =, 3 s + a s + a s + a jω a ω + ja ω + a jb ω b ω jbω + b H( jω = H j 3 3 * ) = ( ω 3 jω aω jaω + a Fourh Order Sysem: b4s + b3s + bs + bs+ b b4ω jb3ω bω + jbω + b H() s =, H( jω 4 3 ) =, 4 3 s + a s + a s + a s + a ω ja ω a ω + ja ω + a 4 3 b ω + jb ω b ω jbω + b H( jω ) = H ( j ) * = ω 4 3 ω + ja3ω aω jaω + a ) A his poin i is prey sraighforward o generalize he relaionship H * ( jω) = H( jω). The las hing we need o do is look a represening he ransfer funcion in polar form o deermine wo imporan properies of ransfer funcions for real-valued sysems. Assume ha we have he complex funcion wrien in recangular form as z( ω) = a( ω) + jb( ω) where a( ω ) and b( ω ) are real valued funcions. We can wrie his in polar form as ( ) ( ) ( ) ( ) jd ω z c e z( ) e j z ω ω = ω = ω where c( ω) = a( ω) + b( ω) = z( ω) d ω b( ω) a( ω) ω ( ) = an = z( ) Now le s assume we have a complex valued ransfer funcion H( jω ). We can wrie his in polar form as j H( jω ) H( jω) = H( jω) e Nex, le s look a H * ( jω ) and H( jω) o derive some useful relaionships. Copyrigh Rober D. Throne

221 H ( jω) = H ( jω) e = H( jω) e * * j H( jω) j H( jω) H( jω) = H( jω) e = H( jω) e j H( jω) j H( jω) Now since we know ha for a real valued sysem H * ( jω) = H( jω) we can conclude he following useful informaion: H( jω) = H ( jω) (he magniude is an even funcion of frequency) H( jω) = H( jω) ( he phase is an odd funcion of frequency) Alhough we have only shown his o be rue for ransfer funcions ha are raios of polynomials, i is rue in general for any ransfer funcion ha corresponds o a real valued sysem. I may also seem a bi odd o alk abou negaive frequencies, bu his is necessary for he mahemaics o work ou. Ofen when we plo he frequency response of a sysem we will plo boh negaive and posiive frequencies, even if we can only measure he posiive frequencies. Wih his background, we can now solve our problem. 8. Seady Sae Response o a Periodic Inpu Le s assume we have sricly proper raional ransfer funcion for a real valued asympoically sable sysem which has n disinc poles. Then we can represen he ransfer funcion as b s + b s + + bs+ b b s + b s + + bs+ b H() s = = m m m m m m m m n n s + an s + + as + a ( z p)( z p) ( z pn) Nex, assume he inpu o our sysem is a cosine wih ampliude A, x () = Acos( ωu ) (). We can hen use Laplace ransforms o deermine he oupu of he sysem, As As Y() s = H() s X() s = H() s = H() s s + ω ( s jω )( s+ jω ) Using parial fracion expansion we will have As B B Bn C C Y() s = H() s = ( s jω )( s+ jω ) s p s p s+ pn s jω s+ jω In he ime domain we can represen his as jω jω y() = Bφ () + Bφ () + + Bφ () + Ce + C e n n where φ i () is he i h characerisic mode. Now, since we have assumed our sysem is asympoically sable, in seady sae, as, all of he characerisic modes will go o zero and we will have jω jω yss () = Ce + Ce Using he parial fracion expansion we have Copyrigh Rober D. Throne

222 jω A C = AH ( jω ) = H( jω ) C jω + jω jω A = AH ( jω ) = H( jω ) jω jω so we have A jω A jω yss () = H( jω) e + H( jω) e Now we need he relaionships we derived in he las secion. We can wrie j H( jω ) H( jω) = H( jω) e j H( jω ) H ( jω) = H( jω) e Insering his ino our seady sae response, and combining exponens and facoring ou common erms we have j( ω+ H( jω)) j( ω+ H( jω)) e + e yss ( ) = A H( jω ) Using Euler s ideniy we finally have yss ( ) = A H( jω) cos( ω+ H( jω)) This is a very imporan resul ha you will use repeaedly in he fuure. Alhough we assumed our sysem had disinc poles, i should be obvious ha ha is no necessary. Also, alhough we assumed he inpu conained no phase, ha was o keep he mahemaics o a simpler level. If he inpu has a phase, i is jus added o he oupu. In summary if we have a proper, raional ransfer funcion for a real valued sysem ha is asympoically sable, hen we can compue he seady sae oupu of he sysem as follows x () A + H s yss = A H j H j = cos( ω θ) ( ) ( ) ( ω ) cos( ω + ( ω ) + θ) Noe ha his resul is only valid afer he ransiens have died ou, usually afer approximaely four ime consans (i.e., he sysem has passed he seling ime). Example 8... Deermine he seady sae oupu for he sysem represened by he ransfer funcion H() s = s + 3 o if he inpu o he sysem is x ( ) = 4cos(3+ 45 ). Firs, we mus recognize ω = 3, θ = 45 o, and A = 4. Nex we compue he magniude and phase of he ransfer funcion, Combining hese we ge H( j3) =, H( j3) =.75, H( j3) = 45 j3+ 3 o o y ss ( ) = (4)(.75) cos( ) = 4.3 cos(3 ) o Copyrigh Rober D. Throne

223 Example 8... Deermine he seady sae oupu for he sysem represened by he ransfer funcion s H() s = ( s+ )( s+ ) o if he inpu o he sysem is x ( ) = 3sin(+ 3 ). Firs we mus conver he sine o a cosine, x ( ) = 3sin(+ 3 o ) = 3cos(+ 3 o 9 o ) = 3cos( 6 o ) Nex we recognizeω =, θ = 6 o, and A = 3. Evaluaing he ransfer funcion a he correc frequency we have j o H( j) =, H( j).36, H( j) ( j )( j ) = + + = Finally we have o o o yss ( ) = (3)(.36) cos( ) =.9487 cos( ) =.9487 sin( ) Noe ha we subraced 9 degrees o conver from he cosine o he sine, and hen added 9 degrees when we were done o conver back o he sine. Since he iniial phase is jus carried hrough, we really did no have o conver from he cosine o he sine o do his problem. In shor, we could have used he relaionship x = Asin( ω + θ) H( s) y ( ) = A H( jω ) sin( ω + H ( jω ) + θ) () ss Finally, noe ha he noaion H( jω ) usually refers o evaluaing he Laplace ransform of an asympoically sable sysem a he frequency ω, H ( jω ) () = H s s= j ω 8.3 Compuaion of H( jω ) Using The Pole-Zero Diagram In addiion o being able o compue H ( jω ) analyically, i is someimes useful o uilize he pole-zero diagram o deermine he magniude and phase of he ransfer funcion. Wih some pracice, his mehod allows one o ge a feeling for how he magniude and phase of he ransfer funcion changes as he frequency ω is changed. Le s assume we have a sricly proper ransfer funcion wrien in erms of poles and zeros as H() s = K( s z)( s z) ( s- zm) ( s p )( s p ) ( s p ) n We need o evaluae his a s = jω, so we have K( jω z)( jω z) ( jω - zm) H( jω ) = ( jω p )( jω p ) ( jω p ) n Copyrigh Rober D. Throne 3

224 Nex, we need o represen each of hese erms in polar form, raher han in recangular form. To sandardize his, le s wrie jφ ( ) i ω Im zi jω zi = βie, βi = ( Re( zi) + ( ω Im( zi)), φi = an Re( zi ) jθ ( ) i ω Im pi jω pi = αie, αi = ( Re( pi) + ( ω Im( pi)), θi = an Re( pi ) β i and α i represen he disances beween zero z i and pole p i from he poin jω, respecively. φ i represens he angle (measured from he posiive real axis) beween zero z i and jω, while θ i represens he angle beween pole p i and jω. While hese formulas look prey messy, you will see hey are acually fairly easy o undersand graphically. Finally we have This means φ Kβe βe... β e Kββ... βm H( jω ) = = e α α... α αα... α jφ jφ j m m jθ jθ jθ n e e ne n φ φ φ θ θ θ j( m n) Kββ... βm H( jω ) = αα... αn H( jω) = φ+ φ + + φm θ θ θn To see how o use hese formulas, le s do a few examples. In words, he firs formula means ha he magniude is equal o K imes he produc of disances from he zeros o jω divided by he produc of disances from he poles o jω. The second formula indicaes ha he phase of he ransfer funcion is he sum of he phases beween he zeros and he poin jω minus he sum of he phases beween he poles and he poin jω. A few examples will help clear his up, and you will see i is quie easy o view his graphically. Example 8.3. For he sysem wih ransfer funcion 4 H( s) = s + 4 use he pole zero plo o deermine he frequency response a frequencies ω =, 5, and 8 radians/sec. We will firs look a he magniude and hen he phase of H ( jω ) for he hree frequencies. Figures shows he corresponding pole-zero plos and he disances we need o compue. A a frequency of radians/second (Figure 8.3), we firs need o compue he disance beween he pole a -4 and he poin j, so we have and hen α = ( 4 ) + ( ) = Copyrigh Rober D. Throne 4

225 K 4 H( j) = = =.894 α We nex need o deermine he angle beween he pole a -4 and he poin j, which we can compue as o θ = an = Hence he phase is 6.6 o. Thus we have H( j ) = o. 4 Figure 8.3. Pole-zero plo of H() s = evaluaed a radians/second. The disance beween he pole s + 4 and he poin j is used in he magniude compuaion, while he angle beween he pole and he poin j is used in he phase compuaion. A a frequency of 5 radians/second (Figure 8.4), we need o compue he disance beween he pole a -4 and he poin j 5 so we have α = ( 4 ) + ( 5) = 4 and hen K 4 H( j5) = = =. 65 α 4 Similarly, o compue he phase we need he angle beween he pole a -4 and he poin j 5, so we have Copyrigh Rober D. Throne 5

226 and hence H( j 5) = o. 5 θ = = 4 o an Figure 8.4. Pole-zero plo of H() s = evaluaed a 5 radians/second. The disance beween he pole s + 4 and he poin j 5 is used in he magniude compuaion, while he angle beween he pole and he poin j 5 is used in he phase compuaion. A a frequency of 8 radians/second (Figure 8.5), we need o compue he disance beween he pole a -4 and he poin j 8, so we have and hen The phase is given by and we have H( j 8) = o. α = + = ( 4 ) ( 8) 8 K 4 H( j8) = = =. 447 α 8 θ = an 8 4 = 63.4 o Copyrigh Rober D. Throne 6

227 4 Figure 8.5. Pole-zero plo of H() s = evaluaed a 8 radians/second. The disance beween he pole s + 4 and he poin j8is used in he magniude compuaion, while he angle beween he pole and he poin j8is used in he phase compuaion. Figure 8.6 displays he magniude and phase as a coninuous funcion of frequency, wih he poins we jus calculaed shown wih a sar. Noe ha he magniude is an even funcion of he frequency, and he phase is an odd funcion of he frequency. Copyrigh Rober D. Throne 7

228 Figure 8.6. Magniude and phase plos of we evaluaed in Example H() s 4 =. The sars (*) are locaed a he discree poins s + 4 Example Consider he sysem wih ransfer funcion H ( s) = s + s+ s+ + j s+ = ( 6 8)( ) ( 8 4 )( 8 4 )( ( )( j) s + s+ s+ s+ + j s+ j s+ ) Use he pole-zero diagram o deermine H( j ). The appropriae pole-zero diagram is shown in Figure 8.7. Compuing he disances and angles beween he zeros and he poin j we have 8 o β = ( ) + ( ) = 64 =.86, φ = an = β = ( ) + ( ) = 44 = 5.6, φ = an = 5.9 Compuing he disances and angles beween he poles and he poin j we have o Copyrigh Rober D. Throne 8

229 We can hen compue α = ( ) + ( ) = 4 =.98, θ = an =.3 o α = ( 8 ) + (4 ) = 68 = 8.46, θ = an = α3 = ( 8 ) + ( 4 ) = =., θ3 = an = ββ (.86)(5.6) H( j) = =.38 ααα (.98)(8.46)() 3 H ( j) = φ + φ θ θ θ = o o o o o o o o A plo of he magniude and phase of his ransfer funcion as a coninuous funcion of frequency is shown in Figure 8.8. In his figure, he poin on he plo we jus compued is idenified wih a sar (*). While compuing he magniude and phase of he ransfer funcion his way can become quie edious if we need o compue i for many frequencies, i does help o visualize he frequency response. For example, if a sysem has a single real pole a p, hen he smalles he disance beween his poin and any poin on he jω axis occurs whenω =. This means he maximum of H( jω) occurs when ω = and will be monoonically decreasing in magniude as he frequency increase. If a sysem has wo complex conjugae poles, such as a σ ± jω, hen a he frequency ω = ω he denominaor ends o have is smalles value, and he magniude of he ransfer funcion is larges value. This is no always rue, bu is a general rend especially when he imaginary par of he pole is large compared o he real par of he pole ( ω >> σ ). Figure 8-9 demonsraes his ype of behavior. In his figure, he imaginary par of he pole is fixed a ± 6 as he real par varies. The magniude of each plo is scaled o a maximum of one. As he figure shows, as he magniude of he imaginary par of he pole becomes larger han he magniude of he real par of he pole, we see peaks or resonances a he frequency corresponding o he imaginary par of he poles. Copyrigh Rober D. Throne 9

230 Figure 8.7. Pole-zero diagram for Example We are evaluaing he ransfer funcion ( s+ + j)( s+ j) H () s = a he frequency radians/sec. ( s j)( s+ 8 4 j)( s+ ) Copyrigh Rober D. Throne 3

231 ( s+ + j)( s+ j) Figure 8.8. Plo of H () s = as a coninuous funcion of frequency. This is ( s j)( s+ 8 4 j)( s+ ) he ransfer funcion from Example 8.3., and he sars (*) denoe he poins we compued in ha example. Copyrigh Rober D. Throne 3

232 Figure 8.9. This figure shows he magniude of he frequency response of various sysems ha have wo complex conjugae poles. All of hese sysems have poles of he form σ ± 6 where he real par of he pole varies from -3 o -. The magniudes for hese sysems have been scaled o have a maximum value of one. Copyrigh Rober D. Throne 3

233 8.4 Using Decibels Before he adven of calculaors and compuers, people needed an easy way o skech frequency response plos, and i was deermined ha he use of a logarihmic scale was he bes choice. This is primarily because of he following relaionships log( ab) = log( a) + log( b) log( a/ b) = log( a) log( b) log( a N ) = Nlog( a) In many engineering sysems, we are ineresed in he relaionship beween he inpu and he oupu. One very useful measure for many applicaions is o measure he raio of he oupu power o he inpu power, so his idea is also included in our measuremen scale. We hen define he power gain G in decibels (or dbs) as G db P ou log = Pin Noe ha we are using a base en logarihmic scale. If we hink abou a circui, we ofen wan o measure he raio of he oupu volage (or curren) o he inpu volage (or curren). I is cusomary o assume we measure he volage across (or hrough) a common resisance (ypically we assume ohm), so we have G db V / log ou R V ou V ou = log = log = Vin / R Vin Vin Now noe ha if we have a ransfer funcion relaing he inpu and oupu, hen we have and G db V ou () s = H() s V () s in Vou ( s) = H( s) V ( s) in V ou = log = log H Vin We can also measure he power wih respec o a known reference power. The reference power is measured as he rms value of he signal across (hrough) a ohm resisor. However, since we are assuming he common one ohm resisor, we jus reference i o an rms volage. For example, o measure a signal in dbv, we reference he signal o a vol rms signal, Vrms GdBV = log vol ( rms) Copyrigh Rober D. Throne 33

234 To measure power in dbmv, we reference our signal o a mv rms signal, and compue i as Vrms GdBmV = log millivol ( rms) In wha follows, we will jus use he convenion above, ha 8.5 Bode Plos GdB = log H( jω) A very common mehod of represening he frequency response of a ransfer funcion is wih a Bode plo. This form of represenaion was very imporan before he adven of compuers. However, i is sill very useful o be able o quickly skech Bode plos o esimae he frequency response of a sysem, and also o be able o deermine how he frequency response may change as poles and zeros are added o a ransfer funcion. The Bode plo is based on knowledge of how o consruc he Bode plo for basic building blocks ha make up our ransfer funcions and hen combining he Bode plos for hese basic building blocks ino he Bode plo for he enire ransfer funcion. The abiliy o easily combine he Bode plos of he building blocks is based on he properies of he polar form of wriing complex funcions of frequency and properies of he logarihm,. The building blocks ha make up he ransfer funcions we have been sudying are he following: Consan gains, K Inegraors or differeniaors, Simple poles or zeros, ( τ s + ) n s n ζ Complex conjugae poles or zeros, s + s+ ωn ωn Basically, any ransfer funcion we have sudied (excep for hose wih delays) can be wrien as a combinaion of hese building blocks. Nex, le s assume we have wo complex funcions, A( jω) and B( jω ). We can represen hese in polar form as A( jω) = A( jω) e B( jω) = B( jω) e n j A(jω ) j B(jω ) Le s now look a wo new funcions of hese basic funcions, Z ( jω) = A( jω) B( jω) A( jω) Z( jω) = B( jω) Copyrigh Rober D. Throne 34

235 If we wrie hese new funcions in polar form we will have Z ( jω) = A( jω) e B( jω) e = A( jω) B( jω) e j A( jω) j B( jω) j( A( jω) + B( jω)) j A( jω ) A( jω) e A( jω) ( ω) = = j B( jω ) Z j e B( jω) e B( jω) j( A( jω) B( jω)) Then we can wrie he magniudes and phases as, Z ( jω) = A( jω) B( jω) Z ( jω) = A( jω) + B( jω) A( jω) Z( jω) = Z( jω) = A( jω) B( jω) B( jω) Represen he magniudes in erms of dbs, hen we have [ ] Z ( jω) = log A( jω) B( jω) = log A( jω) + log B( jω) = A( jω) + B( jω) db db db A( jω) Z ( jω) = log = log A( jω) log B( jω) = A( jω) B( jω) B( jω) db db db Finally, represening he magniude in db and he phase in eiher degrees or radians we have Z ( jω) = A( jω) + B( jω) Z ( jω) = A( jω) + B( jω) db db db Z ( jω) = A( jω) B( jω) Z ( jω) = A( jω) + B( jω) db db db I is hese simple relaionships ha form he basis of consrucing he Bode plo. We basically need o figure ou how o plo he magniude or phase of a building block funcion, and hen add he responses. Noe ha a Bode plo really refers o wo differen plos, a magniude plo and a phase plo. Finally, i will be convenien o plo he Bode plos as semi-log plos, where he frequency axis is ploed on a logarihmic scale and he magniude (in db) or phase is ploed on a linear scale Consan Terms The firs building block is he consan erm in he ransfer funcion, H( jω ) = K. We can represen his j K in erms of magniude and phase as H( jω ) = K e. I is imporan o noe ha he magniude mus be zero or posiive. Hence if he gain is negaive, hen his will be included in he phase. For example, we o j8 can wrie 8= 8e. Nex, when we look a he magniude we have H( jω ) db = log K We can break his ino hree regions < K < log K = log K > log K < K = K > Copyrigh Rober D. Throne 35

236 In summary, he magniude plo of a consan will be a fla line, and he phase plo will also be a fla line Inegraors and Differeniaors n The ransfer funcion for hese building blocks are of he form H() s = s where n can be posiive (differeniaors) or negaive (inegraors). To deermine he frequency response we look a o n n 9 ( e ) e o n j9 jn H( jω) = ( jω) = ω = ω The phase of his building block is hen easy o deermine as n 9 o, which is jus a consan. To deermine he magniude par, we have n log ω = nlog ω A his poin, i is imporan o noe ha when ω = we have log ω = log =, so he poin (,) will be on he Bode plo for his building block. Nex, le s see how much he magniude changes as he frequency changes by a facor of (a decade). To do his, le s assume ha a some frequencyω we haveω = a, and hen we have anoher frequencyω a decade laer, ω = a+. Looking a he magniudes for hese frequencies we have log ω = n log ω = n log = na lo n a n a g ω = nlo ω log + = g = n n ( a+ ) Thus in a decade (facor of ) change in frequency, he magniude changes by n db. This is corresponds o a slope of n db/decade. Example Skech he Bode plo for he ransfer funcion Gs () =. The magniude of his ransfer s funcion goes hrough he poin ( rad / sec, db) and has a slope of - db/decade. The phase is a consan 9 o. The Bode plo for his sysem is shown in Figure 8.. Example Skech he Bode plo for he ransfer funcion Gs () = s. The magniude of his ransfer funcion goes hrough he poin ( rad / sec, db) and has a slope of +4 db/decade. The phase is a consan 8 o. The Bode plo for his sysem is shown in Figure 8.. Example Skech he Bode plo for he ransfer funcion Gs ( ) =. Noe ha here are wo s building blocks for his example, Gs () = = G() s G() s. We will consruc he Bode plo for he s individual componens, and hen add hem. For G ( s ) = we have log = 4dB and he phase is o. For G () s = he magniude goes hrough he poin ( rad / sec, db) wih a slope of s db/decade, and he phase is a consan 9 o. The Bode plos for G and G are shown in Figure 8. as Copyrigh Rober D. Throne 36

237 dashed lines. The Bode plo for Gsis () hen deermined by adding hese componens, and is displayed as a solid line in Figure Magniude (db) Phase (deg) Frequency (rad/sec) Figure 8.. Bode plo of he ransfer funcion ( rad / sec, db). Gs () = s. Noe ha ha a diamond delineaes he poin Copyrigh Rober D. Throne 37

238 8 6 Magniude (db) Phase (deg) Figure 8.. Bode plo of he ransfer funcion ( rad / sec, db). Frequency (rad/sec) Gs () = s. Noe ha ha a diamond delineaes he poin Copyrigh Rober D. Throne 38

239 6 4 Magniude (db) Phase (deg) Frequency (rad/sec) Figure 8.. Bode plo of he ransfer funcion Gs ( ) =. The Bode plos of he wo subcomponens s are shown as dashed lines, he final Bode plo is shown as a solid line. In he phase diagram, he subcomponen a -9 degrees overlaps wih he final phase of -9 degrees (since he oher subcomponen s phase is zero). Noe ha he sum of any wo values (a a given frequency) on he dashed lines equals he value on he solid line a he same frequency. Copyrigh Rober D. Throne 39

240 Simple Poles and Zeros This building block has he form H ( s) = ( τ s + ) n where again n is negaive (for a pole) and posiive (for a zero). To ge he frequency response, we look a n τω an n τω j jn an n H ( jω) = ( τ jω + ) = ( τω) + e = ( τω) + e The phase response is ωτ H( jω) = n an Raher han rying o exacly deermine wha his looks like, le s jus look a a few poins. For small frequencies, ω, we have H( jω) n an. For very large frequencies, ω, we have o H( jω) nan n9. Finally, when ω = we have o H( jω) nan n45 τ. A more precise formula is as follows: o. H ( jω) ω = one decade before τ τ o H( jω) n45 ω = τ o H ( jω) n9 ω = one decade afer τ τ. When we skech he phase as follows: for ω < he phase is zero, and for ω > he phase is n 9 o. τ τ We hen connec hese poins wih a sraigh line. n The magniude response is H( jω) = ( τω) +, and again we will look a hree poins. For small frequencies, ω, we have H( jω) db log( ). For large very frequencies, ω, we have n og H ( jω) db log ( τω) = n log ( τω) = n log τ + n log ω nl ω This means ha for large frequencies we will have a slope of n db/decade. Finally, when ω = we τ H( jω ) = log = nlog ( ) = 3n. Summarizing hese resuls we have have [ ] db n Copyrigh Rober D. Throne 4

241 Magniude Phase o. ω db H ( jω) a ω = one decade before τ τ o ω = 3 n db H ( jω) n45 τ o ω slope n db / decade H ( jω) n9 a ω = one decade afer τ τ Example Skech he Bode plo for he ransfer funcion Gs () =.s +. Here we have τ =, and his is referred o as he break or corner frequency. The magniude is hen zero unil ω = =. A τ his poin he magniude decreases linearly wih a slope of - db/decade. Boh he approximae (sraigh line) approximaion and rue magniude porion of he Bode plo is shown in he op panel of Figure 8.3. The approximae magniude is shown in he dashed line while he rue magniude is shown as a solid line. Noe ha a he corner frequency here is a difference of approximaely 3 db beween he esimae and rue magniude plo. The phase plo is zero unil. =. = and is 9o for τ frequencies above = =. Beween hese poins we draw a sraigh line. The esimaed phase τ plo (dashed line) and rue phase plos are shown in he lower panel of Figure 8.3. Example Skech he Bode plo for he ransfer funcion Gs () = s +. This ransfer funcion is again made up of wo building blocks, G () s = and G () s = s+. The magniude of he firs building block is jus he consan log. ( ) = 4 db and he phase is zero. The corner frequency for he second building block is rad/sec. The magniude of he esimae will be zero before his corner poin, and will have a posiive slope of 4 db/decade for frequencies above his. The phase of his componen will be zero unil rad/sec, and will be 8 o for frequencies above rad/sec. The esimaed phase will be linear beween hese poins. The esimaed magniude and phase plos, as well as he correc magniude and phase plos, are shown in Figure 8.4. In his plo, he building block esimaes are dashed lines, he sum of he building blocks are doed lines, and he correc plos are shown as solid lines. Example Skech he Bode plo approximaion for he ransfer funcion need o pu his ino he correc form, Gs ( ) = = = ss ( + ) s (.s+ ) s(.s+ ) We will hen have hree componens, or building blocks, Gs () = ss ( + ). Firs we Copyrigh Rober D. Throne 4

242 G () s = G () s = s G3 () s = s+ The sraigh line approximaions for he building blocks, and he final Bode plo are shown in Figure 8.5. Example Skech he Bode plo for he ransfer funcion need o pu his ino he correc form, s + Gs () =. Again, we firs ( s+ )( s+ ) 3 s+ s+ ( s+ ) Gs () = = = ( s+ )( s+ ) (.s+ ) (.s+ ) (.s+ )(.s+ ) The building blocks are hen 3 G ( s) = G ( s) = ( s+ ) G ( s) = (.s+ ) 3 4 G ( s) = (.s + ) The sraigh line approximaion for he building blocks, and he final Bode plo are shown in Figure 8.6. s Example Skech he Bode plo for he ransfer funcion Gs () = ( s + ) funcion ino he correc form and idenifying he building blocks we have. Puing his ransfer 3 s s s Gs () = = = ( s+ ) (.s+ ) ( (.s + ) ) 3 G ( s) = G () s = s G ( s) = (.s+ ) 3 The sraigh line approximaion for he building blocks, and he final Bode plo are shown in Figure 8.7. Copyrigh Rober D. Throne 4

243 Magniude (db) Frequency (rad/sec) - Phase (deg) Frequency (rad/sec) Figure 8.3. Bode plo for he ransfer funcion Gs () =.s +. Noe he 3 db discrepancy a he corner frequency in he magniude plo. The Bode plo of he sraigh line approximaion is shown as dashed lines while he exac Bode plo is shown as a solid line. Copyrigh Rober D. Throne 43

244 4 Magniude (db) Frequency (rad/sec) 5 Phase (deg) Frequency (rad/sec) Figure 8.4. Bode plo for he ransfer funcion Gs () = s +. Noe he 6 db discrepancy a he corner frequency in he magniude plo. The sraigh line approximaions of he individual componens are shown as dashed lines, he acual Bode plo is shown as a coninuous line. Noe ha for any given frequency, if you sum he values of he sraigh line approximaions you ge reasonable approximaions o he rue (coninuous curve) Bode plo. Copyrigh Rober D. Throne 44

245 4 Magniude (db) Frequency (rad/sec) 5 Phase (deg) Frequency (rad/sec) Figure 8.5. Bode plo for he ransfer funcion Gs ( ) = =. The sraigh line ss ( + ) s(.s+ ) approximaions of he individual componens are shown as dashed lines, he acual Bode plo is shown as a coninuous line. Noe ha for any given frequency, if you sum he values of he sraigh line approximaions you ge reasonable approximaions o he rue (coninuous curve) Bode plo. Copyrigh Rober D. Throne 45

246 8 6 Magniude (db) Frequency (rad/sec) 5 Phase (deg) Frequency (rad/sec) 3 ( s + ) Figure 8.6. Bode plo for he ransfer funcion Gs () =. The sraigh line (.s+ )(.s+ ) approximaions of he individual componens are shown as dashed lines, he acual Bode plo is shown as a coninuous line. Noe ha for any given frequency, if you sum he values of he sraigh line approximaions you ge reasonable approximaions o he rue (coninuous curve) Bode plo. Copyrigh Rober D. Throne 46

247 Magniude (db) Frequency (rad/sec) 5 Phase (deg) Frequency (rad/sec) 3 s Figure 8.7. Bode plo for he ransfer funcion. Gs () = (.s + ) The sraigh line approximaions of he individual componens are shown as dashed lines, he acual Bode plo is shown as a coninuous line. Noe ha for any given frequency, if you sum he values of he sraigh line approximaions you ge reasonable approximaions o he rue (coninuous curve) Bode plo Complex Conjugae Poles and Zeros The basic building block for a sysem wih complex conjugae poles or zeros can be pu ino he sandard form n ζ Gs () = s + s+ ωn ωn Evaluaing his a s = jω we ge Copyrigh Rober D. Throne 47

248 where we have defined n ω ω G( jω) = + jζ = u + jζu ωn ω n n ( ) ω u =. Compuing he magniude and phase we have ω n G( jω) = ( u ) + ( ζu), G( jω) = n a n Le s firs look a he magniude in decibels, n ζ u u n + ] n [( ) + ( ) G( jω) = log [( u ) ( ζu) = log u ζu ] db For ω we have u, and G( jω), so G( jω). 4n ω For larger frequencies, ω >> ωn, we have G( jω) u =, so in decibels we have ωn 4 G( jω) nlog [ u ] = 4n log u = 4n log ω 4nlog ω 4nlog ω db So for large frequencies we have a slope of 4n db/decade. db n For ω ω we have u. Then we have G( jω) ( ζ) n, and in decibels, n G( jω) db nlog Now if we have wo regions o consider, 4n ( ζ ) < ζ <.5 log( ζ) <.5 < ζ < log( ζ) > Figure 8.8 displays he magniude plo of a he second order sysem Kωn G() s = s + ζωns+ ωn for K =, ω =, and ζ =.,.,.5.5,.75,.99. Noe ha here is a sharp peak a he resonan n frequency, which is given by ωr = ωn ζ. No ha for ζ >.5 =.77 here is no resonance. As his figure shows, he smaller he damping raio, he higher he peak frequency will be. Now we need o look a he phase of he response of his sysem. For low frequencies, u we have G( jω) nan ( ζu) = o ζ. For large frequencies, ω >> ωn we have G( jω) nan n8 o. ζ o Finally, forω = ωn ( u = ) we have G( jω) = nan = n9. Figure 8.9 shows he phase plo of Copyrigh Rober D. Throne 48

249 he second order sysem Kωn G() s = for K =, ω n =, and s + ζω s+ ω n n ζ =.,.,.5.5,.75,.99. Noice ha as he damping raio is smaller, he phase ransiion is much sharper. Also noice ha all of he phase plos pass hrough 9 o when ω = ω n = rad/sec. We can hen summarize our rules as ω db o ωn H ( jω) a ω = one decade before ωn ω = ω depends on ζ o H ( jω) n45 n Magniude Phase ( ) ( ) o ω slope 4 n db / decade H ( jω) n8 a ω = ω one decade afer ω n n 8.6 Bandwidh, Filer Types, and Qualiy Facor How a sysem affecs an inpu signal is ofen used o classify he sysem ype in erms of is filering characerisics. Someimes, when we are designing a conrol sysem for example, we are primarily concerned wih making he oupu mach he inpu wih a reasonable ransien response, and are no primarily concerned wih he frequency response of our sysem. In oher siuaions, such as designing a circui for a music sysem, we are designing a sysem o enhance or remove frequencies from he inpu signal. In hese cases we are ineresed in alking abou hese filering characerisics ha are mos relevan in hose applicaions. Three of he mos useful sysem characerisics in filering applicaions are he bandwidh of he filer, he filer ype, and he qualiy of he filer. Copyrigh Rober D. Throne 49

250 Magniude (db) Frequency (rad/sec) Figure 8.8. Magniude of he frequency response for he ransfer funcion K =, ω =, and ζ =.,.,.5.5,.75,.99 n Kωn G() s = s + ζω s+ ω n n for Copyrigh Rober D. Throne 5

251 5. Phase (deg) Frequency (rad/sec) Figure 8.9. Magniude of he frequency response for he ransfer funcion K =, ω =, and ζ =.,.,.5.5,.75,.99 n Kωn G() s = s + ζω s+ ω n n for 8.6. Sysem Bandwidh The sysem bandwidh is usually defined as he frequency range for which he power of he oupu is wihin one half of he peak oupu power. Thus we deermine he bandwidh as he difference beween he minimum and maximum frequencies as which max Pou ( ω ) = Pou In erms of db s we have Copyrigh Rober D. Throne 5

252 log P ( ) = log P = log + log P [ ω ] max max ou ou ou Then we have log 3 db Hence we deermine he bandwidh as where Bandwidh = ω ω high low This form is useful for reading informaion off of Bode plos. Someimes we don wan o have o conver o Bode plos. Then we can jus use Example Consider he RC circui shown in Figure 8., The ransfer funcion for his circui is clearly H() s = RCs + In erms of frequency response we have H( jω) = jωrc + The magniude of he ransfer funcion is hen H( jω) = ( RCω) + Clearly he maximum of his ransfer funcion occurs when ω =, so for his sysem We nex need o find he frequency a which max log [ P ( ω) ] log [ P ] 3 db ou ωlow ω ωmax max Pou ( ω) Pou ωlow ω ωmax In erms of ransfer funcions, we have max H( jω) H ( jω) ωlow ω ωmax or max log H( jω) log H ( jω) 3 db [ ] ωlow ω ωmax max H ( jω ) = ou max H ( jω) H ( jω) = Copyrigh Rober D. Throne 5

253 Since our ransfer funcion is a decreasing funcion of frequency, we can jus find he larges frequency for which his is rue, H( jω) = ( RCω ) + = which leads o RCω =, or ω =. Thus he bandwidh for his sysem is RC ωhigh ωlow = = RC RC A Bode plo of he magniude of his ransfer funcion for R= kω, C = µ f is shown in Figure 8.. We expec he bandwidh o be rad/sec. The maximum value of he ransfer funcion is (or RC = db), and he bandwidh is hen he frequency a which he magniude has dropped 3 db. This Bode plo has a dashed line a he -3 db poin, and you can see i inersecs he magniude of he ransfer funcion a rad/sec. Figure 8.. RC circui for Example Figure 8. displays hree differen bandpass sysems wih bandwidhs of 4,, and 5 radians/sec. In hese figures, he -3 db line is shown as a dashed line, and is measured from he peak ampliude. Copyrigh Rober D. Throne 53

254 -5 Magniude (db) Frequency (rad/sec) Figure 8.. Bode plo of he magniude of he ransfer funcion for Example The maximum value of he ransfer funcion is ( db), so he bandwidh will be deermined by he poin where he magniude falls o 3dB. This occurs a rad/sec. Copyrigh Rober D. Throne 54

255 Magniude (db) Magniude (db) Frequency (rad/sec) Frequency (rad/sec) 45 Magniude (db) Frequency (rad/sec) Figure 8.. Three differen bandpass sysems wih bandwidhs of 4,, and 5 radians/sec. The -3 db line is shown as a dashed line, and is measured from he peak ampliude Filer Types In discussing filer ypes, i is good o remember he basic relaionship ha if he inpu signal o a sable sysem is x ( ) = A cos( ω + φ o ), hen he seady sae oupu of he sysem will be yss ( ) = A H( jω ) cos( ωo + φ+ H( jωo )). The filer ype is deermined by he magniude of he ransfer funcion a various frequencies, since his direcly affecs he oupu signal. Specifically, if he magniude of he ransfer funcion as a specific frequency is zero, he oupu signal is zero. Alernaively, if he magniude of he ransfer funcion a a specific frequency is one, hen he inpu Copyrigh Rober D. Throne 55

256 signal passes wihou change. In alking abou ideal filer ypes, his is generally wha we are looking a: does he signal in a range of inpu frequencies pass or is he signal removed from he oupu. The four major filer ypes are displayed in Figure 8.3, as ideal filers. Noe ha he magniude of he ransfer funcion is symmeric abou he real axis, and we are ploing he magniude response for boh posiive and negaive frequencies. Alhough he phase of he filer is imporan in some applicaions, i is no imporan in classifying he filer ype. The firs characerizaion is he lowpass filer, and his filer passes signals wih low frequencies and removes high frequencies from he inpu signal. Noe ha he lowpass filer is cenered a zero frequency, bu he bandwidh is only measured from o he edge of he passpband. Nex, a highpass filer removes all low frequencies and only allows high frequencies o pass. The bandwidh of a high pass filer is no really meaningful. A bandpass filer allows frequencies only in a range of frequencies o pass. Finally, a bandrejec or noch filer removes only very specific frequencies. Noch filers are ofen used o remove 6 Hz noise from AC power sysems. Figure 8.3. Ideal lowpass, highpass, bandpass, and bandrejec filers. Copyrigh Rober D. Throne 56

257 Qualiy Facor of a Filer The qualiy facor, or Q, of a bandpass filer is simply defined o be he raio of he cener frequency o he bandwidh of he filer. Cener Frequency Q = Band Widh Noe ha he Q of a filer is dimensionless. The qualiy facor gives a measure of he widh of he pass band, independen of he locaion of filer on he frequency axis. Figure 8.4 shows hree filers wih he same cener frequency and differen bandwidhs, and hence differen Q s. Noe ha he narrower he filer, he higher he Q (since he numeraor is he same for all filers in his case). Figure 8.5 gives an example of hree filers wih he same Q, bu a hree differen frequencies. Noe ha in his case, as he cener frequency is increased, he bandwidh mus also increase o mainain he same Q. Copyrigh Rober D. Throne 57

258 Figure 8.4. These hree filers have he same cener frequency bu differen qualiy (Q) facors. The filer qualiy facor decreases from he op filer o he boom filer in his figure. Figure 8.5. These hree filers have he same qualiy facors. 8.7 Gain and Phase Margins We now wan o reurn o he concep of asympoic sabiliy, and deermine if we can use he frequency response of a sysem o help us. Le s consider a closed loop sysem wih ransfer funcion given by G() s G () s = + H () sgs () We know he sysem will be asympoically sable if all of he poles of he closed loop sysem are in he (open) lef half plane (he real pars of he poles are negaive). When he sysem is asympoically marginally sable, he poles are on he jω axis. We can view his as he poin a which he sysem is abou o become unsable. If we have poles on he jω axis, hen we have + H( jω) G( jω ) = We can rewrie his as H( jω) G( jω ) = We can hen break his ino wo differen condiions, a magniude condiion and a phase condiion: magniude condion : H ( jω) G( jω) = phase condiion : H ( jω) G( jω) = ± 8 When boh of hese condiions are me, hen he sysem is asympoically marginally sable. However, we can also used hese condiions o deermine how close o be unsable is our sysem. The addiional ampliude of G( jω) H( jω ) and he addiional phase of G( jω) H( jω) which resul in purely imaginary poles of G () s are measures of he allowable olerances or changes in G( jω) H( jω) for sysem sabiliy. This leads o he ideas of gain margins and phase margins. o Copyrigh Rober D. Throne 58