Chapter 8 The Complete Response of RL and RC Circuits
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- Cordelia Kelley
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1 Chaper 8 The Complee Response of RL and RC Circuis Seoul Naional Universiy Deparmen of Elecrical and Compuer Engineering
2 Wha is Firs Order Circuis? Circuis ha conain only one inducor or only one capacior can be represened by a firs-order differenial equaion. These circuis are called firs-order circuis (a) Firs, separae he energy sorage elemen from he res of he circui. (b) Nex, replace he circui conneced o a capacior by is Thevenin equivalen circui, or replace he circui conneced o an inducor by is Noron equivalen circui. (because he volage in capacior circui or he curren in inducor circui are o be coninuous.)
3 Response of he Firs Order Circuis Consider he firs-order circui shown in Figure The circui is a seady sae before he swich is closed v( ) = B cos( ), < 0 φ The swich closes a ime =0. v( 0) = B cos( φ), = 0 Afer he swich is closed, he capacior volage is v( ) = Ke / τ + M cos( δ ) (8.2-3) Transien response Seady-sae response
4 The responses are called: ransien par of he response ransien response seady-sae par of he response seady sae response The response, v(), given by Eq , is called he complee response complee response= ransien response+ seady sae response In general, he complee response of a firs-order circui can be represened as he sum of wo par, he naural response ( which is he ransien response) and he forced response (which is he seady sae response): complee response= naural response+ forced response Naural response: he general soluion of he (homogeneous) differenial equaion represening he firs-order circui, when he inpu is zero. Forced response: a paricular soluion of he differenial equaion represening he circui when here is non-zero inpu.
5 The names The naural response of a firs-order circui will be of he form naural response = Ke ( 0)/τ When 0 =0, hen naural response = Ke /τ The consan K in he naural response depends on he iniial condiion. For example, he capacior volage a ime 0
6 Special Inpus o he Firs Order Circuis In his chaper, we will consider hree cases. In heses cases he inpu o he circui afer he disurbance will be (1) a consan v ( ) = s V o (2) an exponenial v s ( ) = V 0 e /τ (3) a sinusoid v s ( ) = V0 cos( ω + θ ) These hree cases are special because he forced response will have he same form as he inpu.
7 Plans o find complee response Here is our plan for finding he complee response of firs-order circuis: Sep 1: Find he seady sae (forced) response before he disurbance. Evaluae his response a ime = 0 o obain he iniial condiion of he energy sorage elemen. (X(0): where i comes from.) Sep 2: Find he seady sae (forced) response afer he disurbance. (X( ): where i goes ulimaely.) Sep 3: Add he ransien (naural) response=ke -/τ o he seady sae (forced) response o ge he complee response. Use he iniial condiion o evaluae he consan K.
8 The Response of a Firs-Order Circui o a Consan Inpu FIGURE In Figure 8.3-1a we find a firs-order circui. X(0)= (R3/(R1+R2+R3)*Vs is he iniial Seady Sae response X( )= (R3/(R2+R3)*Vs is he final Seady Sae response. The ransien response can be obained using he Thevenin circui shown in Fig.8.3-1b. R RR Voc = V R s = R + R and R2 + R3 s 2 3
9 The Response of a Firs-Order Circui o a Consan Inpu The capacior curren is given by d i () = C v () d Apply KVL o Figure 8.3-1b o ge d Voc = Ri () + v () = R( C v ()) + v () d Therefore, d v () Voc v () + = (8.3-1) d RC RC FIGURE 8.3-1
10 Repea he same for he inducor circui using Noron eq. circui. FIGURE I sc = V R s 2 R = RR 2 3 R + R 2 3 s
11 Solving he inducor circui The inducor volage is given by d v () = L i () d Apply KCL o Figure 8.3-1b o ge d L i () v () Isc = + i () = d + i () R R Therefore, d R R i () + i () = I SC (8.3-2) d L L FIGURE 8.3-2
12 Now coming o a general form of soluion Equaion and have he same form. Tha is d d x () x () + = τ K (8.3-3) The parameer τ is called he ime consan. We will solve his differenial equaion by separaing he variables and inegraing. Then we will use he soluion of Eq o obain soluions of Eqs and We may rewrie Eq. (3) as Or, separaing he variables, dx Kτ x = d τ dx d = x kτ τ
13 Second page-- Forming he indefinie inegral, we have dx 1 = d + D x kτ τ where D is a consan of inegraion. Performing he inegraion, we have ln( x Kτ ) = + D τ Solving for x gives = + / x() Kτ Ae τ where A=e D, which is deermined from he iniial condiion, x(0).
14 Here we go. Noe he soluion a he boom of his page. To find A, le =0. Then 0/ τ x(0) = Kτ + Ae = Kτ + A or A= x(0) Kτ Therefore, we obain Since = τ + τ / x ( ) K [ x(0) K ] e τ x( ) = lim x ( ) = Kτ This can be wrien as / x ( ) = x( ) + [ x(0) x( )] e τ
15 The plo Figure shows a plo of x() versus. We can deermine he values of (1) he slope of he plo a ime =0 (2) he iniial value of x() (3) he final value of x() from his plo. FIGURE A graphical echnique for measuring he ime consan of a firs-order circui
16 Apply his o he capacior circui Nex, we apply hese resuls o he RC circui in Figure Comparing Eqs and 8.3-3, we see ha Voc x() = v(), τ = RC, and k = RC Making hese subsiuions in Eq gives v ( ) = V + ( v(0) V ) e oc oc /( RC) This is he seady-sae or forced response. The sum of he naural and forced responses is he complee response; complee reponse = v ( ), forced response = VOC naural reponse = ( v(0) V ) e /( RC) OC
17 Apply his o he inducor circui Nex, compare Eqs and o find he soluion of he RL circui in Figure We see ha L L x() = i(), τ =, and K = ISC R R Making hese subsiuions in Eq gives i ( ) = I + ( i(0) I ) e SC SC ( R/ L ) Again, he complee response is he sum of he forced(seady-sae) response and he ransien(naural) response: complee reponse = i ( ), forced response = ISC naural reponse = ( i(0) I ) e SC ( R/ L )
18 Example Firs-Order Circui wih a Capacior Find he capacior volage afer he swich opens in he circui shown in Figure 8.3-4a. Wha is he value of he capacior volage 50ms afer he swich opens?
19 Soluion Iniial condiion Figure 8.3-4b shows he circui afer he swich opens. R = 10kΩ and V = 8V v (0) = 2V Time consan τ = R C = = = OC (10 10 )(2 10 ) ms Capacior volage where has unis of ms. v = e / 20 () 8 6 V FIGURE 8.3-4
20 Soluion To find he volage 50ms afer he swich opens, le =50. Then v 50/ 20 (50) = 8 6e = 7.51V Figure 8.3-4c shows a plo of he capacior volage as a funcion of ime FIGURE 8.3-4
21 Example Firs-Order Circui wih an Inducor Find he inducor curren afer he swich closes in he circui shown in Figure 8.3-5a. How long will i ake for he inducor curren o reach 2mA? FIGURE 8.3-4
22 Soluion Figure 8.3-5b shows he circui afer he swich closes. R = 1000Ω and I = 4mA Time consan 3 L τ = = = 5 10 = 5µ s R 1000 Inducor curren /5 i() = 4 4e ma Find he ime when he curren reaches 2mA. /5 2= 4 4e ma 2 4 = 5 ln( ) = 3.47µ s 4 Figure8.3-5c shows a plo of he inducor curren as a funcion of ime SC
23 Example Firs-Order Circui The swich in Figure 8.3-6a has been open for a long ime, and he circui has reached seady sae before he swich closes a ime =0. Find he capacior volage for 0. FIGURE 8.3-6
24 Soluion Figure 8.3-6b shows he appropriae equivalen circui while he swich is open. Analyzing he circui in Figure 8.3-6b using volage division gives v(0) = 12 = 7.2V Figure 8.3-6c shows he appropriae equivalen circui afer he swich closes. Afer he swich is closed v OC = 12 = 8V R = = = 3 3 Deparmen 30 of Elecrical 10 + and 60Compuer 10 Engineering, SNU kΩ
25 Soluion The ime consan is τ = R C = = = Consequenly v = e where has unis of ms (20 10 ) (2 10 ) ms / 40 ( ) V
26 Example Firs-Order Circui The swich in Figure 8.3-7a has been open for a long ime, and he circui has reached seady sae before he swich closes a ime =0. Find he inducor curren for 0. FIGURE 8.3-7
27 Soluion 1. Figure 8.3-7b shows he appropriae equivalen circui while he swich is open. The iniial inducor curren can be calculaed using Ohm s law: 12 i (0) = = 40mA Figure 8.3-7c shows he appropriae equivalen circui afer he swich closes. Afer he swich is closed I SC 3. The ime consan is 3 L τ = = = = 25µ s R Consequenly, where has of microseconds. 12 = = 60mA and R = 200Ω 200 i e / 25 ( ) = ma
28 Example Firs-Order Circui The circui in Figure 8.3-8a is a seady sae before he swich opens. Find he curren i() for >0. FIGURE 8.3-8
29 Soluion 1. We find he capacior volage. Before he swich opens, he capacior volage is equal o he volage of he 2-vol source. The iniial condiion is v (0) = 2V 2. Figure 8.3-8b shows he circui as i will be afer he swich is opened. The par of he circui conneced o he capacior has been replaced by is Thévenin equivalen circui in Figure 8.3-8c. The parameers of he Thévenin equivalen circui are v OC R = 8 = 4V = = = 60kΩ
30 Soluion 3. The ime consan is τ = R C = (60 10 ) (2 10 ) = = 120ms herefore, /120 v () = 4 2e V where has unis of ms 4. The node volage, v a () in Figure 8.3-8b va() 8 va() va() v() + + = /120 va() 8 va() va() (4 2 e ) + + = Solving for va(), we ge / (4 2 e ) /120 va () = = 4 e V 4 5. Finally, we calculae I() using Ohm s law: /120 va () 4 e /120 i ( ) = = = e µ A
31 Example Firs-Order Circui wih 0 0 Find he capacior volage afer he swich opens in he circui shown in Figure 8.3-9a. Wha is he value of he capacior volage 50ms afer he swich opens? FIGURE 8.3-9
32 Soluion 1. The 2-vol volage source forces he capacior volage o be 2 vols unil he swich opens. Consequenly, v( ) = 2V for 0.05s 2. In paricular, he in iniial condiion is v (0.05) = 2V 3. Figure 8.3-8b shows he circui afer he swich opens. We see ha R = 10kΩ and V = 8V oc 4. The ime consan for his firs-order circui conaining a capacior is τ = RC = 0.020s 5. Consequenly, he volage of he capacior is given by ( 50)/ 20 () 8 6 V v = e
33 Soluion 6. To find he volage 50ms afer he swich opens, le =100ms. Then v e (100 50)/ 20 (100) = 8 6 =7.51V 7. Figure 8.3-9c shows a plo of he capacior volage as a funcion of ime.
34 Example Firs-Order Circui wih 0 0 Find he inducor curren afer he swich closes in he circui shown in Figure a. How long will i ake for he inducor curren o reach 2mA? FIGURE
35 Soluion 1. The inducor curren will be 0A unil he swich closes. There, he iniial condiion is i(10 µ s) = 0A 2. Figure b shows he circui afer he swich closes. We see ha R = 1000Ω and I = SC 4mA 3. The ime consan for his firs-order circui conaining an inducor is 3 L 5 10 τ = = = = R μs 4. Consequenly, he curren of he inducor is given by ( 10)/5 ( ) 4 4 ma i = e
36 Soluion 6. To find he ime when he curren reaches 2mA, subsiue i()=2ma. Then ( 10)/5 2 = 4 4e ma Solving for gives 2 4 = 5 ln + 10 = 13.47μs 4 7. Figure c shows a plo of he inducor curren as a funcion of ime.
37 Example Exponenial Response of a Firs-Order Circui Figure a shows a plo of he volage across he inducor in Figure b. a. Deermine he equaion ha represens he inducor volage as a funcion of ime. b. Deermine he value of he resisance R. c. Deermine he equaion ha represens he inducor curren as a funcion of ime. FIGURE
38 Soluion(a) 1. The inducor volage is represened by an equaion of he form The consans D, E, and F are described by From he plo, we see ha D for < 0 v () = a E + Fe for 0 D = v() when < 0, E = lim v(), and E + F = lim v() Consequenly, 0 for < 0 v () = a 4e for 0 2. One such poin is labeled on he plo in Figure b. We see v(0.14)=2v; a(0.14) ln(0.5) 2= 4e => a= = Consequenly, 0 for < 0 v () = 5 4e for 0 0+ D = 0, E = 0, and E + F = 4V
39 Soluion(b) 1. Figure a shows he circui immediaely afer he swich opens.in Figure b, he par of he circui conneced o he inducor has been replaced by is Thévenin equivalen circui. 2. The ime consan of he circui is given by L τ = = R 4 R+ 5 L τ = = R 4 R+ 5 Also, he ime consan is relaed o he exponen in v() by 1 R = = => R = 15Ω τ 4 5 =. Consequenly τ
40 Soluion(c) 1. The inducor curren is relaed o he inducor volage by 1 i ( ) = v( ) d i(0) L τ τ Figure show he circui before he swich opens. The inducor curren is given by 6 i ( ) = = 0.4A 15 In paricular, i(0-)=0.4a. The curren in an inducor is coninuous, so i(0+)=i(0-). Consequenly, i (0) = 0.4A 3. Reurning o he equaion for he inducor curren, afer he swich opens we have 1 5τ i ( ) = 4e dτ 0.4 ( e 1) e 4 + = + = In summary, i () = 0.4 for < e for 0 FIGURE
41 Sequenial Swiching Sequenial swiching occurs when a circui conains wo or more swiches ha change sae a differen insans. Figure 8.4-1a is an example of sequenial swiching. FIGURE 8.4-1
42 Sequenial Swiching (con d) Figure 8.4-1b shows he equivalen circui ha is appropriae for <0. i ( ) = 10[ A] < 0 Before he swich changes sae a ime =0. FIGURE i(0 ) = 10A [ ] Afer he swich changes sae a ime =0. + i(0 ) = 10A [ ] This is he iniial condiion a =0.
43 Sequenial Swiching (con d) FIGURE Figure 8.4-1c shows he equivalen circui a 0< <1 ms I = 0A and R = 2Ω sc [ ] [ ] Time consan 3 L 2 10 τ = = = = R 2 Inducor curren ms / ( ) (0) τ i = i e = 10e A for 0 < < 1ms [ ] Immediaely before =1ms 1 i(1 ) = 10e = 3.68A Immediaely afer =1ms + i(1 ) = 3.68A [ ] [ ] This is he iniial condiion a ime =1ms.
44 Sequenial Swiching (con d) FIGURE Figure 8.4-1d shows he appropriae equivalen circui. I = 0[ A] and R = 1Ω sc Time consan 3 L 2 10 τ = = = R 1 = Inducor curren [ ] ms = = > ( 0 ( ) ( ) )/ ( i i e τ 3.68e 1) / 2 A for 1ms 0 [ ] 0 denoes he ime when he swich changes sae 1ms in his example. FIGURE Figure shows a plo of he inducor curren.
45 Sequenial Swiching (con d) In some applicaions, swiching occurs a prescribed volage values raher han a prescribed imes. Figure a device, called comparaor, ha can be used o accomplish his kind of swiching. v () o VH if v > v = VL if v < v + + FIGURE 8.4-3
46 Sequenial Swiching (con d) In figure 8.4-4, a comparaor is used o compare he capacior volage o a hreshold volage V T, Suppose VA > VT > v (0) c The inpu volages of he comparaor are v+ = vc() and v = vt so he oupu volage of he comparaor is v () o V if v () > v = V if v () < v H c T L c T FIGURE 8.4-4
47 Sequenial Swiching (con d) We know ha he capacior volage of his firs-order circui will be Le 1 denoe he ime when he comparaor oupu volage swiches from V L o V H. Then v c ( 1 )=V T, so Solving for 1 gives v ( ) = V + ( v (0) V ) e c A c A V = V + [ v (0) V ] e T A c A /( RC ) 1 /( RC ) 1 = vc(0) VA RC ln( ) V V T A
48 Example) Comparaor Circui Consider he circui shown in Figure The iniial value of he capacior volage is v c (0)=1.667 vols. Wha value of resisance, R, is required if he comparaor is o swich from V L =0 o V H =5 vols a ime 1 =1 ms FIGURE 8.4-5
49 Soluion Figure shows a specific example of he circui in Figure We ge = R R 10 = Then, solving for R: (1 10 ) ln (1 10 ) ln(2) R = = 1.44kΩ 6 ln(2) 10
50 Example) Comparaor Circui In Figure 8.4-6, a comparaor is used o compare he resisor volage, v R (), o a hreshold volage, V T. Suppose VA > V (0) T > RiL Deermine he ime 1 when he comparaor oupu volage swiches from V L o V H FIGURE 8.4-6
51 Soluion 1. The resisor curren is equal o he inducor curren, so 2. The comparaor does no disurb he firs-order circui consising of he volage source, resisor, and inducor. The inducor curren is 3. Nex, 1 is he ime when Ri L ( 1 )=V T, so 4. Solving for 1 gives v () = Ri () R VA VA il( ) = + ( il(0) ) e R R V = V + ( Ri (0) V ) e T A L A L ( R/ L)/ ( R/ L)/ 1 1 = L RiL(0) VA ln( ) R V V T A
52 Sabiliy of Firs-Order Circuis Complee response The circui is sable When τ>0, he naural response vanishes as ->0. The circui is unsable When τ<0, he naural response grows wihou bound as ->0. In mos applicaions, he behavior of unsable circuis is undesirable and is o be avoided. How can we design firs-order circuis o be sable? x () = xn() + xf () / τ x ( ) = Ke ( naural response) x n f ( ) (forced response) R >0 is required o make a firs-order circui be sable. ( τ = R C or τ = L / R )
53 Example Response of an Unsable Firs- Order Circui The Firs-order circui shown in Figure 8.5-1a is a seady sae before he swich closes a =0. This circui conains a dependen source and so may be unsable. Find he capacior volage, v(), for >0. FIGURE 8.5-1
54 Soluion 1. We calculae he iniial condiion from he circui in Figure 8.5-2b. 1. Apply KCL o he op node of he dependen curren source 2. Consequenly, There is no volage drop across he resisor and v (0) = 12V 2. Calculae he open-circui volage using he circui in Figure 8.5-1c. 1. Wriing a KVL equaion, we ge 2. We find + i 2i = 0 i = = (5 10 ) i+ (10 10 ) ( i 2 i) i = 2.4mA 3. Applying Ohm s law o he 10-kΩ resisor, we ge V = i i = OC 3 (10 10 ) ( 2 ) 24V
55 Soluion 3. Calculae he Thévenin resisance using he circui shown in Figure 8.5-1d. 1. Apply KVL o he loop consising of he wo resisors o ge 2. Solving for he curren, 3. Applying Ohm s law o he 10-kΩ resisor, we ge 4. The Thévenin resisance is given by 5. The ime consan is 4. The complee response is v ( ) = 24 12e ( ) ( T ) = (5 10 ) i I i i i = 2I T ( ) V = I + i 2i = I 3 3 T T T R VT = = 10kΩ I T τ = RC= 20ms / 20
56 Example Designing Firs-Order Circuis o Be Sable The circui considered in Example has been redrawn in Figure 8.5-2a, wih he gain of he dependen source represened by he variable B. Wha resricions mus be placed on he gain of he dependen source o ensure ha i is sable? Design his circui o have a ime consan of +20ms. FIGURE 8.5-2
57 Soluion Figure 8.5-2b he circui used o calculae R 1. Applying KVL o he loop consising of he wo resisors i+ V T = 0 2. Solving for he curren gives i = Applying KCL o he op node of he dependen source, we ge VT + i Bi + I 0 3 T = V T 4. Combining hese equaions, we ge 1 B VT IT =
58 Soluion 5. The Thevenin resisance is given by R 3 VT = = I 2B 3 T B<3/2 is required o ensure ha R is posiive and he circui is sable. 6. To obain a ime consan of +20ms requires 7. which in urn requires = 2B 3 Therefore B=1. This suggess ha we can fix he unsable circui by decreasing he gain of he dependen source from 2A/A o 1 A/A
59 The Uni Sep Source The Uni sep forcing funcion as a funcion of ime ha is zero for < 0, and uniy for > 0. u ( ) = < > Applicaion of a consan-volage source a = 0 using wo swiches boh acing a = v () = Vu ( ) 0 0 Single-swich equivalen circui for he sep volage source Symbol for he sep volage source
60 The Uni Sep Source A pulse signal has a consan nonzero value for a ime duraion of Δ= 1-0 Pulse source v () = Vu ( ) Vu ( ) < 0 = V < < 0 1 < Two-sep volage sources
61 The Uni Sep Source Le us consider he applicaion of a pulse o an RL circui as shown in Figure Here we le 0 =0. The pulse is applied o he RL circui when i(0)=0. FIGURE Since he circui is linear, we may use he principle of superposiion, so ha i = i + i 1 2 where i 1 is he response o V 0 u() and i 2 is he response o V 0 u(- 1 ) The response of an RL circui o a consan forcing funcion applied a = n is where τ=l/r. V 0 ( )/ i (1 e n τ = ) when > R n
62 The Uni Sep Source The wo soluions o he wo-sep sources are V0 / τ i1 = (1 e ) when 0 R V0 ( 1 )/ τ i2 = (1 e ) when > 1 R Adding he responses, we have V0 / τ (1 e ) 0 < 1 i = R V0 e / τ / ( e τ 1) > R 1 The response a =1 is V0 1 / τ i ( 1) = (1 e ) R If 1 is greaer han τ, he response will approach V 0 /R before saring is decline, as shown in Figure The response a =2 1 is V 0 2( 1 / τ ) 1 / τ V 0 1 / τ 2( 1 i(2 / τ ) 1) = e ( e 1) = ( e e ) R R FIGURE 8.6-8
63 Example Firs-Order Circui Figure shows a firs-order circui. The inpu o he circui is he volage of he volage source, v s (). The oupu is he curren of he inducor, i 0 (). Deermine he oupu of his circui when he inpu is v s ()=4-8u() [V]. FIGURE 8.6-8
64 Soluion 1. The response of he firs-order circui will be a i () = A + Be for > 0 o 2. Circuis used o calculae he seady-sae reponse (a)before =0 (b) afer =0 (0) 0 = + = + a(0) i A Be A B a( ) i = A + Be = A 0 ( ) A+ B= 0.2A A= 0.2A B = 0.4A 3. The value of he consan a is deermined from he ime consan τ. 1 a = τ = L R
65 Soluion 4. Figure shows he circui used o calculae R. Therefore, R = 20Ω 20 1 a = = 2 10 s 5. Subsiuing he values of A, B and a gives FIGURE [ A ] for 0 i0 () = e [ A] for 0
66 Example Firs-Order Circui Figure shows a firs-order circui. The inpu o he circui is he volage of he volage source, v s (). The oupu is he volage across he capacior, v o (). Deermine he oupu of his circui when he inpu is v s ()=7-14u()V. FIGURE
67 Soluion 1. The response of he firs-order circui will be a v () = A + Be for > 0 o 2. Circuis used o calculae he seady-sae reponse (a)before =0 (b) afer =0 (0) 0 = + = + a(0) v A Be A B 5 A+ B= 7 = 4.38V 3+ 5 a( ) v = A + Be = A 0( ) 5 A= ( 7) = 4.38V 3+5 B = 8.76V 3. The value of he consan a is deermined from he ime consan τ. 1 a = τ = RC
68 Soluion 4. Figure shows he circui used o calculae R. (5)(3) R = = 1.875Ω 5+ 3 Therefore, a 1 1 = = (1.875)( ) s 5. Subsiuing he values of A, B and a gives FIGURE V for 0 v0 () = e V for 0
69 The Response of a Firs-Order Circui o a Nonconsan Source The differenial equaion an RL or RC circui is represened by he general form dx() + ax() = y() (8.7-1) d Consider he derivaive of a produc of wo erms such ha d a dx a a dx a ( xe ) = e + axe = ( + ax) e (8.7-2) d d d The erm wihin he parenheses on he righ-hand side of Eq is exacly he form on he lef-hand side of Eq Therefore, dx a a d a a ( + ax) e = ye or ( xe ) = ye d d Inegraing boh sides of he second equaion, we have a a xe = ye d + K
70 The Response of a Firs-Order Circui o a Nonconsan Source Therefore, x = e ye d + Ke a a a (8.7-1) For he case where he source is a consan so ha y()=m, we have a a a a x = e M e d + Ke = + Ke = x f + xn naural response : a forced response : x = M / a x n f M a = ke (8.7-2) Consider he case where y(), he forcing funcion, is no a consan. naural response : forced response : x n = ke a b a b a a ( a+ b) 1 a ( a+ b) e x f = e e e d = e e d = e e = a+ b a+ b
71 Example Firs-Order Circui wih Nonconsan Source Find he curren i for he circui of Figure 8.7-1a for >0 when 2 10 ( )V v = e u Assume he circui is in seady sae a =0 - s FIGURE 8.7-1
72 Soluion 1. We expec i f o be 2. Wriing KVL arround he righ-hand mesh, we have di di L + Ri = vs or + 4i = 10e d d 2 3. Subsiuing i = Be, we have Hence, B=5 and f i f = Be 2 2Be + 4Be = 10 e or ( 2B + 4 B) e = 10e if = 5 2 e 4. The naural response can be obained by considering he circui shown in Figure 8.7-1b. This is he equivalen circui afer he swich opens. The naural response is i = Ae = Ae n ( R/ L ) 4 2
73 Soluion 5. The complee response is 4 2 i in i f Ae = + = + 5e 6. The consan A can be deermined from he value of he inducor curren a ime =0. The iniial inducor curren, i(0), can be obained by considering he circui shown in Figure 8.7-1c. This is he equivalen circui ha is appropriae before he swich opens. 7. From Figure 8.7-1c 8. Therefore, a =0 9. Therefore, 10 i(0) = = 2A 5 i = Ae + e = A (0) 5 5 2= A + 5 A = 3 i = e + e 4 2 ( 3 5 )A >0
74 Differenial Operaors We can define a differenial operaors such a 2 dx 2 d x sx = and s x = 2 d d Use of he s operaor is paricularly aracive when higher-order differenial equaions are involved. Then we use he s operaor, so ha n n d x s x = for n 0 n d We assume ha n=0 represens no differeniaion, so ha which implies s 0 x=x. 0 s =1 Because inegraion is he inverse of differeniaion, we define 1 x = xd s τ The operaor 1/s mus be shown o saisfy he usual rules of algebraic manipulaions. Of hese rules, he commuaive muliplicaion propery presens he only difficuly. Thus, we require 1 1 s s 1 s = s =
75 Differenial Operaors Firs, we examine Eq Muliplying Eq by s yields 1 d s x = x d or x x s d τ = We ry he reverse order by muliplying sx by he inegraion operaor o obain 1 dx sx = d x() x( ) s τ = d Therefore 1 sx = x only when x( ) = 0 s From a physical poin of view, we require ha all capacior volages and inducor currens be zero a =. Then he operaor 1/s can be said o saisfy Eq and can be manipulaed as an ordinary algebraic quaniy. Differenial operaors can be used o find he naural soluion of a differenial equaion. For example, consider he firs-order differenial equaion d x () + ax () = by () d
76 Differenial Operaors The naural soluion of his differenial equaion is s x () = Ke The homogeneous form of his equaion is d x () + ax () = 0 d To see ha x n () is a soluion of he homogeneous form of he differenial equaion, d ( Ke s s s s ) + a ( Ke ) = ske + ake = 0 d To obain he parameers s, we use he soluion s=-a. Consequenly, n sx + ax = ( s + a) x = 0 x () = Ke n a
77 Differenial Operaors As a second applicaion of differenial operaors, consider using he compuer program MATLAB o find he complee response of a firs-order circui. Differenial operaors are used o describe differenial equaions o MATLAB. The naural soluion of his differenial equaion is To represen his circui by a differenial equaion, apply KVL o ge d v + v = d d or 0.01 v ( ) + v ( ) = 4cos(100 ) d ( ) ( ) 4cos(100 ) 0 In he synax used by MATLAB, he differenial operaor is represened by D insead of s. Replace by he differenial operaor D o ge 0.01Dv + v = 4 cos(100 ) Enering he MATLAB commands v = dsolve( 0.01*Dv + v = 4*cos(100*), v(0)=-8 ) ezplo(v, [0, 2])
78 Differenial Operaors MATLAB responds by providing he complee soluion of he differenial equaion v = 2.*cos(100*)+2.*sin(100*)-10.*exp(-100.*) The plo of v() versus shown in Figure 8.8-1b.
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