LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS
[CHAP. 3
Fig. 3-15 Transform circuit, Writing the voltage law for the loop, we get
Solving for I(s), we have
v-u, 1 v-u, 1 -I(s)= s R + 1/Cs R s + l/RC Taking the inverse Laplace transform of I(s), we obtain
( b ) From Fig.3.15 we have
Substituting I ( s ) obtained in part (a) into the above equation, we get
Taking the inverse Laplace transform of V,(s),we have
3.41. In the circuit in Fig. 3-16(a) the switch is in the closed position for a long time before it is opened at t = 0. Find the inductor current i(t) for t 2 0. When the switch is in the closed position for a long time, the capacitor voltage is charged to 10 V and there is no current flowing in the capacitor. The inductor behaves as a short circuit, and the inductor current is = 2 A. Thus, when the switch is open, we have i ( O - ) = 2 and u,(O-) = 10; the input voltage is 10 V, and therefore it can be represented as lOu(t). Next, using Fig. 3-10, we construct the transform circuit as shown in Fig. 3-16(b).