Baixe sadiku 5ª edição ci... eletricos e extras - chapter 15 laplace transform e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! CHAPTER 15 - LAPLACE TRANSFORM List of topics for this chapter: Definition and Properties of the Laplace Transform Inverse Laplace Transform Application to Circuits Transfer Functions Convolution Integral Application to Integrodifferential Equations Applications (=, DEFINITION AND PROPERTIES OF THE LAPLACE TRANSFORM The Laplace transform is an integral transformation of a function f(t) from the time domain into the frequency domain, giving F(s). Problem 15.1 Find the Laplace transform for (a) (10-10e*) u(t) e 0) qdo-106D)u(t-1) (o) (101087) uft—1) 10 10 3 e) TS s+3 a nlré = 108% 10º Ê 1 ) ( 3 ) — =10e8--— |= - O) s s+3 CAs s+3 19e s(s+3) (e) Manipulate this to match a transform pair using the time-shifting property. (0-102%) u(t—1) = (10-10e0+D) u(t— 1) = (10-10€%(De?) u(t—1) The Laplace transform is 10e5 10e%e” s s+3 = oe (17en+3) s(s+3) 265 Problem 15.2 (a) 6) (e) (d) O) (b) (e) (d) Find the Laplace transform for [te + eTu(t) + +tJu() [t+PJu(t=1) [t+ret+(tt-Det+etTu(t) 1 2 (63 (612) í s 2, Ss s 6 sS!+28+6 + q=——— = ã Manipulate this to match a transform pair using the time-shifting property. [+ TU(=D) =[(t-+D+G-1+D?] uft-1) =[ED+HAGD +266-D+I]u(t=1) [D+ +2(-D+2)] u(t-1) =[(t-D? +36 -D+2]u(t-1) The Laplace transform is Ze” + 3e* + 3es Ss so [387 +3s+2] First, expand the third term. Then, combine like terms. ntret+t-Del + eu) =[tret+te! det tie ]u(t) =[t-3e!+tet+t2eJu(t) The Laplace transform is 13 14.2 sS (+) (SAD (+) + INVERSE LAPLACE TRANSFORM Finding the inverse Laplace transform of F(s) involves two steps: 1. Decomposing F(s) into simple terms using partial fraction expansion, and 2. Finding the inverse of each term by matching entries in a Laplace transform table. 266 Begin completing the square by letting s>+as+b=s +20us+0? +P? Also, let As+A,=A(s+0)+B,B vo Ailsto) B,B Then, "= rap Groap andtheinversetransformis (D=A e“ cosB)+B, e“ sin(Pt)+ E (t) +F(s) Problem 15.3 Find the inverse Laplace transform for 1 (a) 556 1 (o) s2+9 1 (9 s2+28+1 (a) ett (b) Manipulate this to match a transform pair. s-llzão) 82 +9 3As!+9 The inverse Laplace transform is(1/3) sin(3t) (c) Manipulato this to match a transform pair. 1 a s2+28 +17 (+? The inverse Laplace transform is te! Problém 15.4 Find the inverse Laplace transform for 1 a, (a s?+6s s+2 b pa 6) s2 +45+3 s2 +28+1 tc) si +82 +45+4 None of the above match a transform pair. Manipulate these problems, perform partial fraction expansion, and then use a table of Laplace transform pairs to find the inverse Laplace transforms. 269 Manipulate and perform partial fraction expansion to get the following. 1 Lo 6 -1/6 e) s2 +68 s(s+6) 5 *3+6 º s+2 s+2 y2 12 O raIS IDO E TOS s? +48+3 (s+IXs+3) s+1 s+3 ' seas (SHI s+1 ss ( 2 ) O asasa (+ 4 ra alga Now, the inverse Laplace transforms are 1 Lo. (a) ç u(t)— ç e 1 -t 1 3 1 t “dt (b) z€ +5e =ale +e*] (c) cos(2t) + 0.5sin(2t) Problem 15.5 Find the inverse Laplace transform for 82 +28+2 (a ses +4s+4 o ! 0 si +38 +35 +1 , The inverse Laplace transforms are 1 1 (a) di + cos(2t — 36.87º) (b) zé e!) + , APPLICATION TO CIRCUITS Begin by transforming the circuit from the time domain to the s-domain. Solve the circuit using nodal analysis, mesh analysis, source transformation, superposition, or any circuit analysis technique with which we are familiar. Take the inverse Laplace transform of the solution and thus obtain the solution in the time domain. Solving circuits with initial conditions is a straightforward process following the same basic approach as circuíts without initial conditions. Let us start with a capacitor and see how we actually solve such circuits. Start with the defining equation for a capacitor. ie) = cÊeD dt ] 270 Taking the Laplace transform gives Ic(9)=CLs Vols) — ve(0)]= Cs Ve(s)— Ce (O) or Vel) = CI) + ve(O)/s. We now can use the following circuit model for capacitors with initial conditions. 1/€s L é Veís) ve(0)/s Now let us look at the indutor. The defining equation for the inductor is di, (t) D=L—. vi(b dt Taking the Laplace transform gives Vi (s)= L[sl, (s) ir (0)]=Lsl, (9) Li (0) or LO=VO/Ls+i (0/8. We can use the following circuit model for inductors with initial conditions. K6) > + Vr(s) Ls s q ir(0)/s Problem 15,6 Solve a first-order capacitive circuit with an initial condition using Laplace transforms. In Figure 15.1, solve for vc(t) forall t> 0, The initial value of vo(t) = 50 volts, or Ve(0) = 50 volts, and the value of vs(t) = 20 u(t) volts. vs(t) + 59 E 01F >= vt) 100 - Figure 15.1 2n Problem 15.7 Given the circuit in Figure 15.2, solve for i, (t) where v(t) = (10€)u(t) volts and i, (0) = -2 amps. iL(t) 502 “O S sH Figure 15.2 We now convert the above circuit into its Laplace equivalent as shown below. TO) 5 > AM + v6) L(9) ) vos ê 5s (um) O) ir(0)/s Write a mesh equation, where 1, (0)=-2 amps. -V(O+SI (9 +Ss[L(s)-(-2/8)]=0 5(s+ DI, (s)= V(s)-10 “Y-10 LO = 21) Since v(t) = (10€? )u(t) volts, V(s)= 10/(s+2). Now, substitute V(s) into the equation. 10/(5+2)-10 LO)="ano 5(s+1) 2 -2 10 Gançes se Simplifying the first term using partial fraction expansion, 2 -2 -2 L(9)=-—— +—— + —— 19) s+1 s+2 s+l -2 [i(s)=—— 9) s+2 Taking the inverse Laplace transform of 1, (s) gives i(D= (208) u(t) A 274 The answer can be easily checked by again summing the voltages around the loop using this value of i, (t). > of voltages = O [ 100% + (0200) 4,5 42 dae, uD=0 €-100% + (106%) + ()[2X2)e vç) = 0 £-200%! +20€!) u(t) = 0 The sum does equal zero. Problem 15.8 Find the current, i, (t) for allt, given v(t)= (10-10€)u(t) volts for the circuit in Figure 15.3. Note that there are no initial conditions. v(t) O if) Figure 15.3 sq 5H This problem is most easily solved using Laplace transforms. The first thing to do is to write a mesh equation since we are looking for a current. -v(D)+Si, ()+5di, (1)/dt = 0 Sdiç(D/dt+ Si, (1) = v(t) where v(t) = (10- 107") u(t) volts Taking the Laplace transform of both sides, noting that E = =0, -10 6XsI ()-0+51, o or (DL (= 242 SDLO=-c+2 1 «(2 2 dos cs) O ts Asa) Ps +D EA DG+2) Now perform a partial fraction expansion of both terms. 1 1 1 1 no) 275 1. -2 1 rob) º Now to find i, (t) all we have to do is to take the inverse transform of L, (5). i(O=241-2e! +e”'u(t) A It is interesting to check and see if our answer is correct. To do that, we need to place this into the original equation. Sdi, (D/dt+ Si (1) = vt) where Sdi, ()/dt=S(Q)0-(D2e)]-2e*3 u(t) = (208! — 20?) u(t) Sic (t) = (NB 28! +? u(t)= (10-208" +10€23 u(t) Clearly, (20e! — 207) u(1) + [10- 208! +10679 u(t) = [10-10e7) u(t) = v(t) and our answer checks. Problem 15.9 Given the circuit in Figure 15.4, R, =100, R, =100, C=1/10F,and v(t) = 10u(t) volts. Calculate ip, (t). R inf) vo) c s R Figure 15.4 Ri ira(t) Erx(s) ' —ANAr > > v(t) > Re $ R, = Ve) C 78 Cs s R; Time domain Frequency domain = Ryld/Cs Orar, ao O co)= R;/Cs RR; where RIWC)=7 qe” RCs+] e 276 5, -5 5. -10 5 V()=Vi(g)=|>+—— |[+|D4+4-—— + —— e 69) 265) (à S) (2 s+1 =) 10-10 V()=Vt)=+—— s s+l Therefore, vi(t) = (10 — 10e'3 u(t) V and vo(t)= (10-10) u(t) V Problem 15.11 Develop the matrix representation for the circuit shown in Figure 15.6, using nodal analysis with three nodes. 119 15F so 15H vt) > $»o uno $%9 io Figure 15.6 Transforming Figure 15.6 to the frequency domain, e 0 yo Ss yo 5 15 y AMA E MA DODON va s 20 E 10 ER) Q Hs) Using nodal analysis, v, 2V69),N-0 MN =V, =0 10 20 5/s MM (MO, WoVi q 5/s 10 5+15s VV, + V,-0 “I69)=0 S415s 20 TS) Simplifying these equations, 2W,-2V(9)+V,+48V, -48V, = 0 ICO + DIV, — VD) + Gs +DV, +(D(V, — V))=O e (IV, — Vi) + (Bs +1)V, — (20)/3s + DI(s) = O 279 Combining like terms, Q+1+48)V, —4sV, =2V(s) 0 [C2)Bs + DIV, + [OB +D+0s+D+2]V,; -2V, = 0 -4V, +[4+0Gs +DIV, = (2035 + DI(s) Simplifying further, (48 +3)V, -4sV, = 2. V(s) (68? 28) V, + (68? +55 +3)V, —-2V, =0 -4V, +08+5)V; = (20X3s + 1) I(s) Therefore, the matrix equation is | 45+3 -4s 0 T A | 2V(s) 1 -68º-2s. 682 +58+3 -2 | V, |= o | 0 -4 3s+5 l V; | (20)(3s + 1) K(s) Problem 15.12 Given the circuit in Figure 15.7, write the s-domain equations for V,, V,, V,, and V,. DO NOT SOLVE. it 10 q w 0.1F v 0.1F vs 0,1 F v. a Je c 4 E NR W ) vt) nos 19 $ 100 s 100 s BD sidg + Figure 15.7 Converting this circuit to the s-domain yields, : ÁS 10 10/s 10/s 10/s ! (s) Vi Va Va o Ma Lo IV IV ves) 10 s 10 É 10 s 10 E ) SI) — Using nodal analysis, Vim Vo- - Atnode 1: MV, Nc0, MN Vo 10 10 10/s = 280 V,— — — Atnode 2: 2 Mi N200 Mo Vig 0 10/s 10 10/s Vi Vi, 5-0, Mi Atnode 3: = 10/s 10 10/s Vaca M4-O V(S)-=V Atnode 4: 105 + 10 -51,(8)=0 where L)=o Simplifying, (s+DVi-sV, = V(s) -SV, +08 +DV, -sV, =0 -sV, +(28+D)V,-sV,=0 SV-sV,+(s+DV, =5V(s) Finally, collect the equations and place them into a matrix form. s+2 "5 0 0 V Vís) -s 2841 -s 0 A o O cs 241 slvyi lo 5 0 -s s+1][ Vi 5Vís) e Problem 15.13 [15.49] | For the circuit in Figure 15.8 find v,(t) for t>0. 1H 16 Oy MAN e + Guto (O 2H Sim w $20 e - Figure 15.8 Consider the circuit shown below. 281 Using the time das rastion property, -8)-1 80 = 5 aco) ree-eo) 8 gD= te vê -98" Hence, 8 8 4 Cudrtet-Set teto v(D= Su(y+ qto 9º 271 = 4 Ben, Lis v(D= 3 u(t) se + 27tº Problem 15.17 [15.59] Refer to the network in Figure 15.11. Find the following transfer functions: 0 H(O= VV) O o HO=VO/4) O HO=LO/E) (d) H(S9=I,(s)/V,(s) is 19 1H io 0000 Figure 15.11 (a) Consider the following circuit. I, 1 Vi s Vo IL MA “BUIO + Vs 1/s 2% 1/s 2% 1 Vo Atnode 1, 4 =sy 4 oiro ly | =|l+s+-|V/-— s=[lesto)Mi= ço O 284 Atnode o, e VN, =sV+V=(s+DV, Vi=(S +s+DV, (2) Substituting (2) into (1) Vo=(s+1+/9S2 +s+DV, 1/8 V, V,= (84282 +38+2)V, V 1 H(9)=L=———— 1) Vo si+28+35+2 (b) L=V-V=(8 +28 +38+DV, —(s + s+DV, L=(s'+s"+28+DV, ame 269) = IL si+s!+28+1 L=Yo O L=5 “CO ses! 42841 1 Lv (a H6s)= = “HO = Gras 43512 + 4 CONVOLUTION INTEGRAL The convolution of two signals consists of time-reversing one signal, shifting it, multiplying it point by point with the second signal, and integrating the product. Steps to evaluate the convolution integral : 1. Folding: Take the mirror image of h(A) about the ordinate axis to obtain h(-A). 2. Displacement : Shift or delay h(-À) by tto obtain h(t—A). 3. Miultiplication : Find the product of h(t - À) and x(A). 4. Integration : For a given time t, calculate the arca under the product h(t — A) x(A) for o O<A<ttoget y(t) att. 285 Problem 15.18 [1569] Giventhat R()=E,(9)=s/(8º +), tind LH[E (E, 69] by 0: convolution. : ED=1,()=cos(y) VIBE f cos(A) cos(t - à) da. cos(A) cos(B) = =lcosta +B)+cos(A-B)] Lil OE] 1 5! costt- 22) - cos(t) da t [EE o]-Fes. n+5 1 See, vino ol]- ia teos(t) + yRsiu(o º + + APPLICATION TO INTEGRODIFFERENTIAL EQUATIONS e Problem 15,19 Given the following is a matrix representation of a circuit in the frequency domain. Determine the value of v,(t) and v, (t). [ -s Tv()] [8/8] -s sslvo Lo | [ s | [s+2 5 ] O s se2llgs] Ls 28h] VAO] raras 0) (as) [o Hence, [ (s+2) Ss | 9e+2 ] [4 2] [vis | | E+D (g+D [8/8] | (asks+D | | ste | vol +) [o | 8 2| Lea es Lage! La | Therefore, v,(t) = [4- 2e'Ju(t) V and v(D=42e!'u(t)V. 286 Thus, Also, Finally, YR =1/5 or R=59 s(0.2+0)=s(1/4) or €=025-02=0,05F 2 Cs+ 1/10+1/Ls= +28 td 20s 0.05s +0.1+1/Ls = 0.055 +0.1+1/5s or L=5H + APPLICATIONS Problem 15.22 Given the op-amp circuit in Figure 15.13, determine the value of v. (t), where the value of vc (t) is equal to 2 voltsatt=0. In other words, ve (0) = 2 volts. 01 F Jc Ro volt so c(t) Ay - ——s ut Se” uít) Vou pos Figure 15.13 Begin by transforming the circuit from the time domain to the frequency domain. 105 28 fr o +40 ND 5 Vols) v Va ANN vol, Vo t + sHs+1) Vau E > 289 Writing a node equation at a, MV V-[Q/)+Va(s)] 5 10/s =0 which is only one equation with two unknowns. We need a constraint equation. V.=V,-0. Then, VOO QI) + Vo] À 5 10/s 0 -2V(9)-[+sVa(1=0 v g=HO,2 (2/5 ),2. 10 -2 ms s Us Asa ts SGD Partial fraction expansion of the first term leads to 10 ,-2 -12,10 Vas(g) = 2 + s s+1.5 s s+1 Taking the inverse Laplace transform of both sides produces Vou (1) = 4-12+10€) u(t) V Problem 15.23 In Figure 15.14, find v,(t) and i,(t), given that v(t) = (10e9u(t) volts. mo 10H v(t) Figure 15.14 290 Begin by transforming the time domain circuit to the frequency domain. 20 AMA 10s va “UU0O - Vo t T(s) g + vs O) Vols) 1 Atnodea: Mr V6) Vac Vols) 10s 2 O 0 We cannot help ourselves by writing any more node equations. So we have one equation and two unknowns. We need another equation without generating any additional unknowns. So we go to the constraint equation. V4=V,=0 So, nów our node equation becomes YO, MO q 10s 20 Vols) YO 20 10s which leads to vi) Êve Taking the Laplace transform of v(t) gives vg= s+1 Substituting for V(s) yields -2 - vo) s As+1V s(s+1) After partial fraction expansion, vg) = 120,20 s s+l Taking the inverse Laplace transform produces Vo (O) = (-20+20€9 u(t) V 291 Voa(S) = Va(8) Cs R, Therefore, R,Cs Vou (8) = Vin (8) Problem 15.26 Given R, =100 kQ. What value of €, in Figure 15.16, yields Voa (8) == [Vit dt Taking the Laplace transform, -1 Vol) = Vis(S) From Problem 15.25, we know that -1 Vor(8) = Rs Vin(S) Thus, R,C=1 cd Lo 1x10s=10uF = or =D o - , R, 100x10º 7” MO Problem 15.27 Given R| =10k2, R, =1000, C=50F,and V, ()=10€% u(t). Calculate the total energy that v,, delivers to the circuit shown in Figure 15.16. Also, find tho total energy delivered to R, . wít) = foco dt=1/R foz (D dt To find the total energy that v,, delivers to the circuit, wa (O=R, fi (o dt we, (D=1/10,000 [acesa 0e 2) dr =100/10,000 fes dr = (1/1000-1/40*| 9) =(1/100X-1/4e +1/4) = (1/400)1 - e) 294 To find the total encrgy delivered to R,, va (D= UR, eo at First, find vo«(t). From Problem 15.25, we know that Vai) = reg Vols) where RC = (10x 10º)%50x 10%) = 500x10º = 0,5 Then, Vols) = vit) = "2 v.69) out R, Cs in s in Taking the Laplace transform of v; (t), 10 Val) =——— s+2 Substitute V, (s) into the equation for Voy(S). 10 )- -20 = 2y Voa (8) ="5 Via(8) = (2 s e s66+2) Take the partial fraction expansion of V.y(s)- Vai (S) = 10 + Jo out s s+2 Hence, Vil) = (10+10€%) u(t) Now, Vea (D= UR feio de =1/100 feio +10€2)(10+10672) dr =1/100 [aooes — 200º +100) dt = fes dr— a fo dt+ ; ldr =[(-1/9e 4 +1/48]-(CUCI/De? +1/2]+ = (Act D+ (et A D+t 295 t Problem 15.28 [15.89] A gyrator is a device for simulating an inductor in a network, A 0 basic gyrator is shown in Figure 15.17. By finding V, (s)/I, (), show that the inductance produced by the gyratoris L= CR? R c R My Rm Wi tam + L Figure 15.17 The gyrator is equivalent to two cascaded inverting amplifiers. Let V, be the voltage at the output of the first op amp. -R = V,.=V V= q n= 4, o -1/sC 1 V=0R Vo SeR Yi I Vo Vo “O RO SRIC v TS SRÊC V, 1 =sL, whenL=R?C o 296