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

Moreover t he s implification techniques t hat have

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Unformatted text preview: tage o f 10 volts is r epresented a ccording t o t he a rrangement i n F ig. 6.11b w ith a s eries v oltage s ource v (0) / s = 10/ s . N ote t hat t he i mpedance o f t he i nductor is s a nd t hat o f t he c apacitor is 5 /s. T he i nput o f l Ou(t) is r epresented b y i ts L aplace t ransform 1 0/s. T he t otal v oltage i n t he l oop is +2= 2 , a nd t he l oop i mpedance is .!f E xample 6 .14 T he switch in t he circuit of Fig. 6.13a is in the closed position for a long time before t == 0, when it is opened instantaneously. Find the currents YI(t) a nd Y2(t) for t:::: O. Inspection of this circuit shows t hat when t he switch is closed and the steady-state conditions are reached, t he c apacitor voltage V c == 16 volts, and t he i nductor current Y2 == 4 amps. Therefore, when t he switch is opened ( at t == 0 ), t he initial conditions are v c(O-) == 16 a nd Y2(0-) == 4. Figure 6.13b shows t he t ransformed version of the circuit in Fig. 6.13a. We have used equivalent sources to account for t he initial conditions. T he initial capacitor voltage of 16 volts is represented by a series voltage of 1 6/s a nd the initial inductor current of 4 amps is represented by a source of value LY2(0) == 2. From Fig. 6.13b, t he loop equations can be written directly in t he frequency domain as ~ .!f (s + 2 + ~). T herefore [ 2 Y (s)=-- .... s+2+~ 2s S2 + 2s + 5 w hich c onfirms o ur e arlier r esult i n E xample 6.10. +1 i 6- i s 1 1 [~l [ YI (S) +2 Y2(S) Application of Cramer's rule t o this equation yields -5 YI(S) == 5 == 2 24(s + 2) s 2+7s+12 (s - 24 48 24(s + 2) ==--+-+ 3)(8 + 4) 8+3 8+4 ~==aw:E!&. 404 6 C ontinuous-Time S ystem A nalysis U sing t he L aplace T ransform 6 .4 A nalysis o f E lectrical N etworks' T he T rans' dN . l orme e twork 405 a nd Y l(t) = ( _24e- 3t + 4 8e- 4t ) u (t) R1 =2, 4(s + 7) s 2+7s+12 16 12 = s+3-s+4 Y2(S) = R2=R3=1 M= I. S imilarly, we o btain £1 =2. £;=: + 5V (a) v 0 ( I) a nd Y2(t) = ( 16e- 3t - 12e- 4t ) u (t) We also c ould h ave c omputed Yl(S) a nd Y2(S) u sing T Mvenin's t heorem b y replacing t he c ircuit t o t he r ight of t he c apacitor ( right o f t erminals ab) w ith i ts T Mvenin e quivalent, as shown in Fig. 6.13c. F igure 6.13b shows t hat t he T Mvenin i mpedance Z (8) a nd t he T Mvenin s ource V (s) are: Z (s)= V (s) = M t(~+l) =~ t + ~ + 1 5s + 12 -! 5 i +} V £1 -4 2=-+1 5s + 12 A ccording t o F ig. 6.13c, t he c urrent Y l(s) is given by 1 - V(s) Yl(S) = ~ L 1-M l..;, L I+M ~ == ( b) + Z (s) l..;,+M (e) 24(s + 2) S2 + 7s + 12 a c onclusion w hich confirms t he e arlier result. manner. • We m ay d etermine Y 2 (s) i n a similar 4s • E xample 6 .15 T he s witch in t he c ircuit in Fig. 6.14a is a t p osition a for a long t ime before t = 0, w hen i t is moved i nstantaneously t o p osition b. D etermine t he c urrent Y l(t) a nd t he o utput voltage vo(t) for t ~ O. J ust b efore switching, t he values of t he l oop c urrents a re 2 a nd 1, respectively, t hat is; Yl(O-) = 2 a nd Y 2(0-) = 1. T he e quivalent c ircuits for two t ypes o f i nductive couplings...
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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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