Chapter 8 The Complete Response of RL and RC Circuits

Size: px
Start display at page:

Download "Chapter 8 The Complete Response of RL and RC Circuits"

Transcription

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.

Homework-8(1) P8.3-1, 3, 8, 10, 17, 21, 24, 28,29 P8.4-1, 2, 5

Homework-8(1) P8.3-1, 3, 8, 10, 17, 21, 24, 28,29 P8.4-1, 2, 5 Homework-8() P8.3-, 3, 8, 0, 7, 2, 24, 28,29 P8.4-, 2, 5 Secion 8.3: The Response of a Firs Order Circui o a Consan Inpu P 8.3- The circui shown in Figure P 8.3- is a seady sae before he swich closes a

More information

(b) (a) (d) (c) (e) Figure 10-N1. (f) Solution:

(b) (a) (d) (c) (e) Figure 10-N1. (f) Solution: Example: The inpu o each of he circuis shown in Figure 10-N1 is he volage source volage. The oupu of each circui is he curren i( ). Deermine he oupu of each of he circuis. (a) (b) (c) (d) (e) Figure 10-N1

More information

8. Basic RL and RC Circuits

8. Basic RL and RC Circuits 8. Basic L and C Circuis This chaper deals wih he soluions of he responses of L and C circuis The analysis of C and L circuis leads o a linear differenial equaion This chaper covers he following opics

More information

Chapter 7 Response of First-order RL and RC Circuits

Chapter 7 Response of First-order RL and RC Circuits Chaper 7 Response of Firs-order RL and RC Circuis 7.- The Naural Response of RL and RC Circuis 7.3 The Sep Response of RL and RC Circuis 7.4 A General Soluion for Sep and Naural Responses 7.5 Sequenial

More information

Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits

Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits Lecure 13 RC/RL Circuis, Time Dependen Op Amp Circuis RL Circuis The seps involved in solving simple circuis conaining dc sources, resisances, and one energy-sorage elemen (inducance or capaciance) are:

More information

ES 250 Practice Final Exam

ES 250 Practice Final Exam ES 50 Pracice Final Exam. Given ha v 8 V, a Deermine he values of v o : 0 Ω, v o. V 0 Firs, v o 8. V 0 + 0 Nex, 8 40 40 0 40 0 400 400 ib i 0 40 + 40 + 40 40 40 + + ( ) 480 + 5 + 40 + 8 400 400( 0) 000

More information

CHAPTER 12 DIRECT CURRENT CIRCUITS

CHAPTER 12 DIRECT CURRENT CIRCUITS CHAPTER 12 DIRECT CURRENT CIUITS DIRECT CURRENT CIUITS 257 12.1 RESISTORS IN SERIES AND IN PARALLEL When wo resisors are conneced ogeher as shown in Figure 12.1 we said ha hey are conneced in series. As

More information

INDEX. Transient analysis 1 Initial Conditions 1

INDEX. Transient analysis 1 Initial Conditions 1 INDEX Secion Page Transien analysis 1 Iniial Condiions 1 Please inform me of your opinion of he relaive emphasis of he review maerial by simply making commens on his page and sending i o me a: Frank Mera

More information

First Order RC and RL Transient Circuits

First Order RC and RL Transient Circuits Firs Order R and RL Transien ircuis Objecives To inroduce he ransiens phenomena. To analyze sep and naural responses of firs order R circuis. To analyze sep and naural responses of firs order RL circuis.

More information

ECE 2100 Circuit Analysis

ECE 2100 Circuit Analysis ECE 1 Circui Analysis Lesson 35 Chaper 8: Second Order Circuis Daniel M. Liynski, Ph.D. ECE 1 Circui Analysis Lesson 3-34 Chaper 7: Firs Order Circuis (Naural response RC & RL circuis, Singulariy funcions,

More information

CHAPTER 6: FIRST-ORDER CIRCUITS

CHAPTER 6: FIRST-ORDER CIRCUITS EEE5: CI CUI T THEOY CHAPTE 6: FIST-ODE CICUITS 6. Inroducion This chaper considers L and C circuis. Applying he Kirshoff s law o C and L circuis produces differenial equaions. The differenial equaions

More information

Direct Current Circuits. February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 1

Direct Current Circuits. February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 1 Direc Curren Circuis February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 1 Ammeers and Volmeers! A device used o measure curren is called an ammeer! A device used o measure poenial difference

More information

Voltage/current relationship Stored Energy. RL / RC circuits Steady State / Transient response Natural / Step response

Voltage/current relationship Stored Energy. RL / RC circuits Steady State / Transient response Natural / Step response Review Capaciors/Inducors Volage/curren relaionship Sored Energy s Order Circuis RL / RC circuis Seady Sae / Transien response Naural / Sep response EE4 Summer 5: Lecure 5 Insrucor: Ocavian Florescu Lecure

More information

EEEB113 CIRCUIT ANALYSIS I

EEEB113 CIRCUIT ANALYSIS I 9/14/29 1 EEEB113 CICUIT ANALYSIS I Chaper 7 Firs-Order Circuis Maerials from Fundamenals of Elecric Circuis 4e, Alexander Sadiku, McGraw-Hill Companies, Inc. 2 Firs-Order Circuis -Chaper 7 7.2 The Source-Free

More information

EECE251. Circuit Analysis I. Set 4: Capacitors, Inductors, and First-Order Linear Circuits

EECE251. Circuit Analysis I. Set 4: Capacitors, Inductors, and First-Order Linear Circuits EEE25 ircui Analysis I Se 4: apaciors, Inducors, and Firs-Order inear ircuis Shahriar Mirabbasi Deparmen of Elecrical and ompuer Engineering Universiy of Briish olumbia shahriar@ece.ubc.ca Overview Passive

More information

EE202 Circuit Theory II , Spring. Dr. Yılmaz KALKAN & Dr. Atilla DÖNÜK

EE202 Circuit Theory II , Spring. Dr. Yılmaz KALKAN & Dr. Atilla DÖNÜK EE202 Circui Theory II 2018 2019, Spring Dr. Yılmaz KALKAN & Dr. Ailla DÖNÜK 1. Basic Conceps (Chaper 1 of Nilsson - 3 Hrs.) Inroducion, Curren and Volage, Power and Energy 2. Basic Laws (Chaper 2&3 of

More information

( ) ( ) if t = t. It must satisfy the identity. So, bulkiness of the unit impulse (hyper)function is equal to 1. The defining characteristic is

( ) ( ) if t = t. It must satisfy the identity. So, bulkiness of the unit impulse (hyper)function is equal to 1. The defining characteristic is UNIT IMPULSE RESPONSE, UNIT STEP RESPONSE, STABILITY. Uni impulse funcion (Dirac dela funcion, dela funcion) rigorously defined is no sricly a funcion, bu disribuion (or measure), precise reamen requires

More information

RC, RL and RLC circuits

RC, RL and RLC circuits Name Dae Time o Complee h m Parner Course/ Secion / Grade RC, RL and RLC circuis Inroducion In his experimen we will invesigae he behavior of circuis conaining combinaions of resisors, capaciors, and inducors.

More information

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4.

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4. PHY1 Elecriciy Topic 7 (Lecures 1 & 11) Elecric Circuis n his opic, we will cover: 1) Elecromoive Force (EMF) ) Series and parallel resisor combinaions 3) Kirchhoff s rules for circuis 4) Time dependence

More information

Signal and System (Chapter 3. Continuous-Time Systems)

Signal and System (Chapter 3. Continuous-Time Systems) Signal and Sysem (Chaper 3. Coninuous-Time Sysems) Prof. Kwang-Chun Ho kwangho@hansung.ac.kr Tel: 0-760-453 Fax:0-760-4435 1 Dep. Elecronics and Informaion Eng. 1 Nodes, Branches, Loops A nework wih b

More information

Basic Circuit Elements Professor J R Lucas November 2001

Basic Circuit Elements Professor J R Lucas November 2001 Basic Circui Elemens - J ucas An elecrical circui is an inerconnecion of circui elemens. These circui elemens can be caegorised ino wo ypes, namely acive and passive elemens. Some Definiions/explanaions

More information

ECE 2100 Circuit Analysis

ECE 2100 Circuit Analysis ECE 1 Circui Analysis Lesson 37 Chaper 8: Second Order Circuis Discuss Exam Daniel M. Liynski, Ph.D. Exam CH 1-4: On Exam 1; Basis for work CH 5: Operaional Amplifiers CH 6: Capaciors and Inducor CH 7-8:

More information

Chapter 9 Sinusoidal Steady State Analysis

Chapter 9 Sinusoidal Steady State Analysis Chaper 9 Sinusoidal Seady Sae Analysis 9.-9. The Sinusoidal Source and Response 9.3 The Phasor 9.4 pedances of Passive Eleens 9.5-9.9 Circui Analysis Techniques in he Frequency Doain 9.0-9. The Transforer

More information

Lab 10: RC, RL, and RLC Circuits

Lab 10: RC, RL, and RLC Circuits Lab 10: RC, RL, and RLC Circuis In his experimen, we will invesigae he behavior of circuis conaining combinaions of resisors, capaciors, and inducors. We will sudy he way volages and currens change in

More information

Solutions to Assignment 1

Solutions to Assignment 1 MA 2326 Differenial Equaions Insrucor: Peronela Radu Friday, February 8, 203 Soluions o Assignmen. Find he general soluions of he following ODEs: (a) 2 x = an x Soluion: I is a separable equaion as we

More information

Electrical Circuits. 1. Circuit Laws. Tools Used in Lab 13 Series Circuits Damped Vibrations: Energy Van der Pol Circuit

Electrical Circuits. 1. Circuit Laws. Tools Used in Lab 13 Series Circuits Damped Vibrations: Energy Van der Pol Circuit V() R L C 513 Elecrical Circuis Tools Used in Lab 13 Series Circuis Damped Vibraions: Energy Van der Pol Circui A series circui wih an inducor, resisor, and capacior can be represened by Lq + Rq + 1, a

More information

Chapter 8 The Complete Response of RL and RC Circuits

Chapter 8 The Complete Response of RL and RC Circuits Chaper 8 he Complee Response of R and RC Ciruis Exerises Ex 8.3-1 Before he swih loses: Afer he swih loses: 2 = = 8 Ω so = 8 0.05 = 0.4 s. 0.25 herefore R ( ) Finally, 2.5 ( ) = o + ( (0) o ) = 2 + V for

More information

University of Cyprus Biomedical Imaging and Applied Optics. Appendix. DC Circuits Capacitors and Inductors AC Circuits Operational Amplifiers

University of Cyprus Biomedical Imaging and Applied Optics. Appendix. DC Circuits Capacitors and Inductors AC Circuits Operational Amplifiers Universiy of Cyprus Biomedical Imaging and Applied Opics Appendix DC Circuis Capaciors and Inducors AC Circuis Operaional Amplifiers Circui Elemens An elecrical circui consiss of circui elemens such as

More information

R.#W.#Erickson# Department#of#Electrical,#Computer,#and#Energy#Engineering# University#of#Colorado,#Boulder#

R.#W.#Erickson# Department#of#Electrical,#Computer,#and#Energy#Engineering# University#of#Colorado,#Boulder# .#W.#Erickson# Deparmen#of#Elecrical,#Compuer,#and#Energy#Engineering# Universiy#of#Colorado,#Boulder# Chaper 2 Principles of Seady-Sae Converer Analysis 2.1. Inroducion 2.2. Inducor vol-second balance,

More information

ln 2 1 ln y x c y C x

ln 2 1 ln y x c y C x Lecure 14 Appendi B: Some sample problems from Boas Here are some soluions o he sample problems assigned for Chaper 8 8: 6 Soluion: We wan o find he soluion o he following firs order equaion using separaion

More information

Chapter 10 INDUCTANCE Recommended Problems:

Chapter 10 INDUCTANCE Recommended Problems: Chaper 0 NDUCTANCE Recommended Problems: 3,5,7,9,5,6,7,8,9,,,3,6,7,9,3,35,47,48,5,5,69, 7,7. Self nducance Consider he circui shown in he Figure. When he swich is closed, he curren, and so he magneic field,

More information

d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3

d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3 and d = c b - b c c d = c b - b c c This process is coninued unil he nh row has been compleed. The complee array of coefficiens is riangular. Noe ha in developing he array an enire row may be divided or

More information

23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes

23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes Represening Periodic Funcions by Fourier Series 3. Inroducion In his Secion we show how a periodic funcion can be expressed as a series of sines and cosines. We begin by obaining some sandard inegrals

More information

Phys1112: DC and RC circuits

Phys1112: DC and RC circuits Name: Group Members: Dae: TA s Name: Phys1112: DC and RC circuis Objecives: 1. To undersand curren and volage characerisics of a DC RC discharging circui. 2. To undersand he effec of he RC ime consan.

More information

Section 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients

Section 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients Secion 3.5 Nonhomogeneous Equaions; Mehod of Undeermined Coefficiens Key Terms/Ideas: Linear Differenial operaor Nonlinear operaor Second order homogeneous DE Second order nonhomogeneous DE Soluion o homogeneous

More information

LAPLACE TRANSFORM AND TRANSFER FUNCTION

LAPLACE TRANSFORM AND TRANSFER FUNCTION CHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION Professor Dae Ryook Yang Spring 2018 Dep. of Chemical and Biological Engineering 5-1 Road Map of he Lecure V Laplace Transform and Transfer funcions

More information

Math 333 Problem Set #2 Solution 14 February 2003

Math 333 Problem Set #2 Solution 14 February 2003 Mah 333 Problem Se #2 Soluion 14 February 2003 A1. Solve he iniial value problem dy dx = x2 + e 3x ; 2y 4 y(0) = 1. Soluion: This is separable; we wrie 2y 4 dy = x 2 + e x dx and inegrae o ge The iniial

More information

EE 301 Lab 2 Convolution

EE 301 Lab 2 Convolution EE 301 Lab 2 Convoluion 1 Inroducion In his lab we will gain some more experience wih he convoluion inegral and creae a scrip ha shows he graphical mehod of convoluion. 2 Wha you will learn This lab will

More information

2.4 Cuk converter example

2.4 Cuk converter example 2.4 Cuk converer example C 1 Cuk converer, wih ideal swich i 1 i v 1 2 1 2 C 2 v 2 Cuk converer: pracical realizaion using MOSFET and diode C 1 i 1 i v 1 2 Q 1 D 1 C 2 v 2 28 Analysis sraegy This converer

More information

LabQuest 24. Capacitors

LabQuest 24. Capacitors Capaciors LabQues 24 The charge q on a capacior s plae is proporional o he poenial difference V across he capacior. We express his wih q V = C where C is a proporionaliy consan known as he capaciance.

More information

( ) = Q 0. ( ) R = R dq. ( t) = I t

( ) = Q 0. ( ) R = R dq. ( t) = I t ircuis onceps The addiion of a simple capacior o a circui of resisors allows wo relaed phenomena o occur The observaion ha he ime-dependence of a complex waveform is alered by he circui is referred o as

More information

dv 7. Voltage-current relationship can be obtained by integrating both sides of i = C :

dv 7. Voltage-current relationship can be obtained by integrating both sides of i = C : EECE202 NETWORK ANALYSIS I Dr. Charles J. Kim Class Noe 22: Capaciors, Inducors, and Op Amp Circuis A. Capaciors. A capacior is a passive elemen designed o sored energy in is elecric field. 2. A capacior

More information

SOLUTIONS TO ECE 3084

SOLUTIONS TO ECE 3084 SOLUTIONS TO ECE 384 PROBLEM 2.. For each sysem below, specify wheher or no i is: (i) memoryless; (ii) causal; (iii) inverible; (iv) linear; (v) ime invarian; Explain your reasoning. If he propery is no

More information

i L = VT L (16.34) 918a i D v OUT i L v C V - S 1 FIGURE A switched power supply circuit with diode and a switch.

i L = VT L (16.34) 918a i D v OUT i L v C V - S 1 FIGURE A switched power supply circuit with diode and a switch. 16.4.3 A SWITHED POWER SUPPY USINGA DIODE In his example, we will analyze he behavior of he diodebased swiched power supply circui shown in Figure 16.15. Noice ha his circui is similar o ha in Figure 12.41,

More information

non-linear oscillators

non-linear oscillators non-linear oscillaors The invering comparaor operaion can be summarized as When he inpu is low, he oupu is high. When he inpu is high, he oupu is low. R b V REF R a and are given by he expressions derived

More information

Chapter 2. First Order Scalar Equations

Chapter 2. First Order Scalar Equations Chaper. Firs Order Scalar Equaions We sar our sudy of differenial equaions in he same way he pioneers in his field did. We show paricular echniques o solve paricular ypes of firs order differenial equaions.

More information

2.7. Some common engineering functions. Introduction. Prerequisites. Learning Outcomes

2.7. Some common engineering functions. Introduction. Prerequisites. Learning Outcomes Some common engineering funcions 2.7 Inroducion This secion provides a caalogue of some common funcions ofen used in Science and Engineering. These include polynomials, raional funcions, he modulus funcion

More information

Chapter 2: Principles of steady-state converter analysis

Chapter 2: Principles of steady-state converter analysis Chaper 2 Principles of Seady-Sae Converer Analysis 2.1. Inroducion 2.2. Inducor vol-second balance, capacior charge balance, and he small ripple approximaion 2.3. Boos converer example 2.4. Cuk converer

More information

V L. DT s D T s t. Figure 1: Buck-boost converter: inductor current i(t) in the continuous conduction mode.

V L. DT s D T s t. Figure 1: Buck-boost converter: inductor current i(t) in the continuous conduction mode. ECE 445 Analysis and Design of Power Elecronic Circuis Problem Se 7 Soluions Problem PS7.1 Erickson, Problem 5.1 Soluion (a) Firs, recall he operaion of he buck-boos converer in he coninuous conducion

More information

Chapter 1 Fundamental Concepts

Chapter 1 Fundamental Concepts Chaper 1 Fundamenal Conceps 1 Signals A signal is a paern of variaion of a physical quaniy, ofen as a funcion of ime (bu also space, disance, posiion, ec). These quaniies are usually he independen variables

More information

Simulation-Solving Dynamic Models ABE 5646 Week 2, Spring 2010

Simulation-Solving Dynamic Models ABE 5646 Week 2, Spring 2010 Simulaion-Solving Dynamic Models ABE 5646 Week 2, Spring 2010 Week Descripion Reading Maerial 2 Compuer Simulaion of Dynamic Models Finie Difference, coninuous saes, discree ime Simple Mehods Euler Trapezoid

More information

Some Basic Information about M-S-D Systems

Some Basic Information about M-S-D Systems Some Basic Informaion abou M-S-D Sysems 1 Inroducion We wan o give some summary of he facs concerning unforced (homogeneous) and forced (non-homogeneous) models for linear oscillaors governed by second-order,

More information

u(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x

u(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x . 1 Mah 211 Homework #3 February 2, 2001 2.4.3. y + (2/x)y = (cos x)/x 2 Answer: Compare y + (2/x) y = (cos x)/x 2 wih y = a(x)x + f(x)and noe ha a(x) = 2/x. Consequenly, an inegraing facor is found wih

More information

MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions

MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions MA 14 Calculus IV (Spring 016) Secion Homework Assignmen 1 Soluions 1 Boyce and DiPrima, p 40, Problem 10 (c) Soluion: In sandard form he given firs-order linear ODE is: An inegraing facor is given by

More information

Inductor Energy Storage

Inductor Energy Storage School of Compuer Science and Elecrical Engineering 5/5/ nducor Energy Sorage Boh capaciors and inducors are energy sorage devices They do no dissipae energy like a resisor, bu sore and reurn i o he circui

More information

Designing Information Devices and Systems I Spring 2019 Lecture Notes Note 17

Designing Information Devices and Systems I Spring 2019 Lecture Notes Note 17 EES 16A Designing Informaion Devices and Sysems I Spring 019 Lecure Noes Noe 17 17.1 apaciive ouchscreen In he las noe, we saw ha a capacior consiss of wo pieces on conducive maerial separaed by a nonconducive

More information

10. State Space Methods

10. State Space Methods . Sae Space Mehods. Inroducion Sae space modelling was briefly inroduced in chaper. Here more coverage is provided of sae space mehods before some of heir uses in conrol sysem design are covered in he

More information

Problem Set #1. i z. the complex propagation constant. For the characteristic impedance:

Problem Set #1. i z. the complex propagation constant. For the characteristic impedance: Problem Se # Problem : a) Using phasor noaion, calculae he volage and curren waves on a ransmission line by solving he wave equaion Assume ha R, L,, G are all non-zero and independen of frequency From

More information

6.2 Transforms of Derivatives and Integrals.

6.2 Transforms of Derivatives and Integrals. SEC. 6.2 Transforms of Derivaives and Inegrals. ODEs 2 3 33 39 23. Change of scale. If l( f ()) F(s) and c is any 33 45 APPLICATION OF s-shifting posiive consan, show ha l( f (c)) F(s>c)>c (Hin: In Probs.

More information

Continuous Time. Time-Domain System Analysis. Impulse Response. Impulse Response. Impulse Response. Impulse Response. ( t) + b 0.

Continuous Time. Time-Domain System Analysis. Impulse Response. Impulse Response. Impulse Response. Impulse Response. ( t) + b 0. Time-Domain Sysem Analysis Coninuous Time. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 1. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 2 Le a sysem be described by a 2 y ( ) + a 1

More information

Predator - Prey Model Trajectories and the nonlinear conservation law

Predator - Prey Model Trajectories and the nonlinear conservation law Predaor - Prey Model Trajecories and he nonlinear conservaion law James K. Peerson Deparmen of Biological Sciences and Deparmen of Mahemaical Sciences Clemson Universiy Ocober 28, 213 Ouline Drawing Trajecories

More information

EE202 Circuit Theory II

EE202 Circuit Theory II EE202 Circui Theory II 2017-2018, Spring Dr. Yılmaz KALKAN I. Inroducion & eview of Fir Order Circui (Chaper 7 of Nilon - 3 Hr. Inroducion, C and L Circui, Naural and Sep epone of Serie and Parallel L/C

More information

h[n] is the impulse response of the discrete-time system:

h[n] is the impulse response of the discrete-time system: Definiion Examples Properies Memory Inveribiliy Causaliy Sabiliy Time Invariance Lineariy Sysems Fundamenals Overview Definiion of a Sysem x() h() y() x[n] h[n] Sysem: a process in which inpu signals are

More information

EE100 Lab 3 Experiment Guide: RC Circuits

EE100 Lab 3 Experiment Guide: RC Circuits I. Inroducion EE100 Lab 3 Experimen Guide: A. apaciors A capacior is a passive elecronic componen ha sores energy in he form of an elecrosaic field. The uni of capaciance is he farad (coulomb/vol). Pracical

More information

Section 2.2 Charge and Current 2.6 b) The current direction is designated as the direction of the movement of positive charges.

Section 2.2 Charge and Current 2.6 b) The current direction is designated as the direction of the movement of positive charges. Chaper Soluions Secion. Inroducion. Curren source. Volage source. esisor.4 Capacior.5 Inducor Secion. Charge and Curren.6 b) The curren direcion is designaed as he direcion of he movemen of posiive charges..7

More information

4.6 One Dimensional Kinematics and Integration

4.6 One Dimensional Kinematics and Integration 4.6 One Dimensional Kinemaics and Inegraion When he acceleraion a( of an objec is a non-consan funcion of ime, we would like o deermine he ime dependence of he posiion funcion x( and he x -componen of

More information

Finish reading Chapter 2 of Spivak, rereading earlier sections as necessary. handout and fill in some missing details!

Finish reading Chapter 2 of Spivak, rereading earlier sections as necessary. handout and fill in some missing details! MAT 257, Handou 6: Ocober 7-2, 20. I. Assignmen. Finish reading Chaper 2 of Spiva, rereading earlier secions as necessary. handou and fill in some missing deails! II. Higher derivaives. Also, read his

More information

Università degli Studi di Roma Tor Vergata Dipartimento di Ingegneria Elettronica. Analogue Electronics. Paolo Colantonio A.A.

Università degli Studi di Roma Tor Vergata Dipartimento di Ingegneria Elettronica. Analogue Electronics. Paolo Colantonio A.A. Universià degli Sudi di Roma Tor Vergaa Diparimeno di Ingegneria Eleronica Analogue Elecronics Paolo Colanonio A.A. 2015-16 Diode circui analysis The non linearbehaviorofdiodesmakesanalysisdifficul consider

More information

ECE-205 Dynamical Systems

ECE-205 Dynamical Systems ECE-5 Dynamical Sysems Course Noes Spring Bob Throne Copyrigh Rober D. Throne Copyrigh Rober D. Throne . Elecrical Sysems The ypes of dynamical sysems we will be sudying can be modeled in erms of algebraic

More information

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k Challenge Problems DIS 03 and 0 March 6, 05 Choose one of he following problems, and work on i in your group. Your goal is o convince me ha your answer is correc. Even if your answer isn compleely correc,

More information

HOMEWORK # 2: MATH 211, SPRING Note: This is the last solution set where I will describe the MATLAB I used to make my pictures.

HOMEWORK # 2: MATH 211, SPRING Note: This is the last solution set where I will describe the MATLAB I used to make my pictures. HOMEWORK # 2: MATH 2, SPRING 25 TJ HITCHMAN Noe: This is he las soluion se where I will describe he MATLAB I used o make my picures.. Exercises from he ex.. Chaper 2.. Problem 6. We are o show ha y() =

More information

Laplace Transforms. Examples. Is this equation differential? y 2 2y + 1 = 0, y 2 2y + 1 = 0, (y ) 2 2y + 1 = cos x,

Laplace Transforms. Examples. Is this equation differential? y 2 2y + 1 = 0, y 2 2y + 1 = 0, (y ) 2 2y + 1 = cos x, Laplace Transforms Definiion. An ordinary differenial equaion is an equaion ha conains one or several derivaives of an unknown funcion which we call y and which we wan o deermine from he equaion. The equaion

More information

Two Coupled Oscillators / Normal Modes

Two Coupled Oscillators / Normal Modes Lecure 3 Phys 3750 Two Coupled Oscillaors / Normal Modes Overview and Moivaion: Today we ake a small, bu significan, sep owards wave moion. We will no ye observe waves, bu his sep is imporan in is own

More information

2.9 Modeling: Electric Circuits

2.9 Modeling: Electric Circuits SE. 2.9 Modeling: Elecric ircuis 93 2.9 Modeling: Elecric ircuis Designing good models is a ask he compuer canno do. Hence seing up models has become an imporan ask in modern applied mahemaics. The bes

More information

7. Capacitors and Inductors

7. Capacitors and Inductors 7. Capaciors and Inducors 7. The Capacior The ideal capacior is a passive elemen wih circui symbol The curren-volage relaion is i=c dv where v and i saisfy he convenions for a passive elemen The capacior

More information

5.2. The Natural Logarithm. Solution

5.2. The Natural Logarithm. Solution 5.2 The Naural Logarihm The number e is an irraional number, similar in naure o π. Is non-erminaing, non-repeaing value is e 2.718 281 828 59. Like π, e also occurs frequenly in naural phenomena. In fac,

More information

Chapter #1 EEE8013 EEE3001. Linear Controller Design and State Space Analysis

Chapter #1 EEE8013 EEE3001. Linear Controller Design and State Space Analysis Chaper EEE83 EEE3 Chaper # EEE83 EEE3 Linear Conroller Design and Sae Space Analysis Ordinary Differenial Equaions.... Inroducion.... Firs Order ODEs... 3. Second Order ODEs... 7 3. General Maerial...

More information

Name: Total Points: Multiple choice questions [120 points]

Name: Total Points: Multiple choice questions [120 points] Name: Toal Poins: (Las) (Firs) Muliple choice quesions [1 poins] Answer all of he following quesions. Read each quesion carefully. Fill he correc bubble on your scanron shee. Each correc answer is worh

More information

1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.

1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx. . Use Simpson s rule wih n 4 o esimae an () +. Soluion: Since we are using 4 seps, 4 Thus we have [ ( ) f() + 4f + f() + 4f 3 [ + 4 4 6 5 + + 4 4 3 + ] 5 [ + 6 6 5 + + 6 3 + ]. 5. Our funcion is f() +.

More information

Chapter 3 Boundary Value Problem

Chapter 3 Boundary Value Problem Chaper 3 Boundary Value Problem A boundary value problem (BVP) is a problem, ypically an ODE or a PDE, which has values assigned on he physical boundary of he domain in which he problem is specified. Le

More information

4.1 - Logarithms and Their Properties

4.1 - Logarithms and Their Properties Chaper 4 Logarihmic Funcions 4.1 - Logarihms and Their Properies Wha is a Logarihm? We define he common logarihm funcion, simply he log funcion, wrien log 10 x log x, as follows: If x is a posiive number,

More information

15. Vector Valued Functions

15. Vector Valued Functions 1. Vecor Valued Funcions Up o his poin, we have presened vecors wih consan componens, for example, 1, and,,4. However, we can allow he componens of a vecor o be funcions of a common variable. For example,

More information

Computer-Aided Analysis of Electronic Circuits Course Notes 3

Computer-Aided Analysis of Electronic Circuits Course Notes 3 Gheorghe Asachi Technical Universiy of Iasi Faculy of Elecronics, Telecommunicaions and Informaion Technologies Compuer-Aided Analysis of Elecronic Circuis Course Noes 3 Bachelor: Telecommunicaion Technologies

More information

5.1 - Logarithms and Their Properties

5.1 - Logarithms and Their Properties Chaper 5 Logarihmic Funcions 5.1 - Logarihms and Their Properies Suppose ha a populaion grows according o he formula P 10, where P is he colony size a ime, in hours. When will he populaion be 2500? We

More information

Chapter 4 The Fourier Series and Fourier Transform

Chapter 4 The Fourier Series and Fourier Transform Represenaion of Signals in Terms of Frequency Componens Chaper 4 The Fourier Series and Fourier Transform Consider he CT signal defined by x () = Acos( ω + θ ), = The frequencies `presen in he signal are

More information

Experimental Buck Converter

Experimental Buck Converter Experimenal Buck Converer Inpu Filer Cap MOSFET Schoky Diode Inducor Conroller Block Proecion Conroller ASIC Experimenal Synchronous Buck Converer SoC Buck Converer Basic Sysem S 1 u D 1 r r C C R R X

More information

Chapter 6. Laplace Transforms

Chapter 6. Laplace Transforms 6- Chaper 6. Laplace Tranform 6.4 Shor Impule. Dirac Dela Funcion. Parial Fracion 6.5 Convoluion. Inegral Equaion 6.6 Differeniaion and Inegraion of Tranform 6.7 Syem of ODE 6.4 Shor Impule. Dirac Dela

More information

8.022 (E&M) Lecture 9

8.022 (E&M) Lecture 9 8.0 (E&M) Lecure 9 Topics: circuis Thevenin s heorem Las ime Elecromoive force: How does a baery work and is inernal resisance How o solve simple circuis: Kirchhoff s firs rule: a any node, sum of he currens

More information

Final Spring 2007

Final Spring 2007 .615 Final Spring 7 Overview The purpose of he final exam is o calculae he MHD β limi in a high-bea oroidal okamak agains he dangerous n = 1 exernal ballooning-kink mode. Effecively, his corresponds o

More information

Sub Module 2.6. Measurement of transient temperature

Sub Module 2.6. Measurement of transient temperature Mechanical Measuremens Prof. S.P.Venkaeshan Sub Module 2.6 Measuremen of ransien emperaure Many processes of engineering relevance involve variaions wih respec o ime. The sysem properies like emperaure,

More information

MEMS 0031 Electric Circuits

MEMS 0031 Electric Circuits MEMS 0031 Elecric Circuis Chaper 1 Circui variables Chaper/Lecure Learning Objecives A he end of his lecure and chaper, you should able o: Represen he curren and volage of an elecric circui elemen, paying

More information

6.01: Introduction to EECS I Lecture 8 March 29, 2011

6.01: Introduction to EECS I Lecture 8 March 29, 2011 6.01: Inroducion o EES I Lecure 8 March 29, 2011 6.01: Inroducion o EES I Op-Amps Las Time: The ircui Absracion ircuis represen sysems as connecions of elemens hrough which currens (hrough variables) flow

More information

EECE 301 Signals & Systems Prof. Mark Fowler

EECE 301 Signals & Systems Prof. Mark Fowler EECE 3 Signals & Sysems Prof. Mark Fowler Noe Se #2 Wha are Coninuous-Time Signals??? Reading Assignmen: Secion. of Kamen and Heck /22 Course Flow Diagram The arrows here show concepual flow beween ideas.

More information

Math 36. Rumbos Spring Solutions to Assignment #6. 1. Suppose the growth of a population is governed by the differential equation.

Math 36. Rumbos Spring Solutions to Assignment #6. 1. Suppose the growth of a population is governed by the differential equation. Mah 36. Rumbos Spring 1 1 Soluions o Assignmen #6 1. Suppose he growh of a populaion is governed by he differenial equaion where k is a posiive consan. d d = k (a Explain why his model predics ha he populaion

More information

Chapter 1 Electric Circuit Variables

Chapter 1 Electric Circuit Variables Chaper 1 Elecric Circui Variables Exercises Exercise 1.2-1 Find he charge ha has enered an elemen by ime when i = 8 2 4 A, 0. Assume q() = 0 for < 0. 8 3 2 Answer: q () = 2 C 3 () 2 i = 8 4 A 2 8 3 2 8

More information

Electrical and current self-induction

Electrical and current self-induction Elecrical and curren self-inducion F. F. Mende hp://fmnauka.narod.ru/works.hml mende_fedor@mail.ru Absrac The aricle considers he self-inducance of reacive elemens. Elecrical self-inducion To he laws of

More information

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3. Mah Rahman Exam Review Soluions () Consider he IVP: ( 4)y 3y + 4y = ; y(3) = 0, y (3) =. (a) Please deermine he longes inerval for which he IVP is guaraneed o have a unique soluion. Soluion: The disconinuiies

More information

MAE143A Signals & Systems - Homework 2, Winter 2014 due by the end of class Thursday January 23, 2014.

MAE143A Signals & Systems - Homework 2, Winter 2014 due by the end of class Thursday January 23, 2014. MAE43A Signals & Sysems - Homework, Winer 4 due by he end of class Thursday January 3, 4. Quesion Zener diode malab [Chaparro Quesion.] A zener diode circui is such ha an oupu corresponding o an inpu v

More information

dy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page

dy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page Assignmen 1 MATH 2270 SOLUTION Please wrie ou complee soluions for each of he following 6 problems (one more will sill be added). You may, of course, consul wih your classmaes, he exbook or oher resources,

More information

ENGI 9420 Engineering Analysis Assignment 2 Solutions

ENGI 9420 Engineering Analysis Assignment 2 Solutions ENGI 940 Engineering Analysis Assignmen Soluions 0 Fall [Second order ODEs, Laplace ransforms; Secions.0-.09]. Use Laplace ransforms o solve he iniial value problem [0] dy y, y( 0) 4 d + [This was Quesion

More information