Signal Processing and Linear Systems-B.P.Lathi copy

# 6 11 i nitial condition generators for a c apacitor a

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Unformatted text preview: a re i llustrated in Figs. 6.14b a nd 6.14c. For o ur s ituation, t he c ircuit in Fig. 6.14c applies. Fig. 6.14d shows t he t ransformed v ersion o f t he c ircuit in Fig. 6.14a a fter switching. N ote t hat t he i nductors L l + M , L2 + M , a nd - M a re 3 ,4, a nd - 1 h enries w ith i mpedances 3s, 4s, a nd - 8 respectively. T he i nitial condition voltages in t he t hree b ranches a re ( Ll + M)Yl(O) = 6, (L2 + M)Y2(O) = 4, a nd - M[Yl(O) - Y2(0)] = - 1, respectively. T he two loop e quations of t he c ircuit a ret 10 F ig. 6 .14 T he L aplace t ransform a nal sis of a c . . by t he t ransformed c ircuit m ethod. y o upled m ductlve n etwork ( Example 6.15) t The time domain equations (loop equations) are dYl Ll d t dYl (2s dY2 + ( Rl + R2)Yl(t) - R2Y2(t) + M dt = 10u(t) M dt - R2Yl (t) dY2 + L2 d t + (R2 + R3)Y2(t) = T he Laplace transform of these equations yields E q. (6.61). (d) S + 3 )Yl(S) + (s (s - I)Yl(S) - I)Y2(s) + (3s + 2)Y2(S) =~ +5 s =5 or 2S 0 [ +3 s- 1 S3s 1] +2 [ Yl(S)] [5S~1O] Y2(s) 5 (6.61) 6 Continuous-Time System Analysis Using t he Laplace Transform 406 6.4 Analysis of Electrical Networks: T he T ransformed Network a nd +:1 ( I) 28 2 + 98 + 4 Y l(8) = 8(82 + 38 + 1) 4 8 Therefore Y l(t) vI ( I) + &quot;2 ( I) 1 8 + 0.382 = (4 - b&gt;----: + 8 + 2.618 - . + (a) e-O.3B2t _ e-2.61Bt) u (t) 4 07 - v ( I) 2 -A&quot;I ( I) (b) Similarly 2 8 + 28 + 2 Y2(8) = 8(82 + 38 + 1) 2 8 1.618 8 + 0.382 0.618 8 + 2.618 =-----+--a nd Y2(t) = (2 _1.618e-o.3B2t + 0.618e-2.61Bt) u (t) (c) T he o utput v oltage v o(t) F ig. 6 .16 O perational a mplifier a nd i ts e quivalent circuit. = Y2(t) = (2 - 1.618e-o.3B2t + 0.618e-2.61Bt) u (t) • E xercise E 6.8 For t he R LC c ircuit in Fig. 6.15, t he i nput is s witched o n a t t = O. T he i nitial conditions a re y (O-) = 2 a mps a nd vc(O) = 50 volts. F ind t he loop current y(t) a nd t he c apacitor voltage f :, v c(t) for t 2: o. Answer: y (t) = l OV2e- t c os (2t + Bl.BO)u(t), 2T 4r t =O v c(t) = [24 + 3 1.62e- t cos (2t - 2Q 34.7°)]u(t) \ l IH Q Y:; + &quot;c ( I) F ig. 6 .15 C ircuit for Exercise E6.8. 6 .4-1 (d) Analysis o f Active Circuits A lthough we have considered examples of only passive networks so far, the circuit analysis procedure using t he Laplace transform is also applicable t o active circuits. All t hat is needed is t o replace t he active elements with their mathematical models (or equivalent circuits) a nd proceed as before. T he o perational amplifier (depicted by t he t riangular symbol in Fig. 6.16a) is a well-known element in m odern electronic circuits. T he t erminals with t he p ositive a nd t he negative signs correspond t o noninverting a nd inverting terminals, respectively. This means t hat t he p olarity of t he o utput voltage V 2 is t he s ame as t hat o fthe i nput voltage a t t he t erminal marked by t he positive sign (noninverting). T he o pposite is t rue for t he inverting terminal, marked by t he negative sign. Figure 6.16b shows t he model (equivalent circuit) of t he o perational amplifier (op amp) in Fig. 6...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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