0195136047.pdf

3.1 SINUSOIDAL STEADY-STATE PHASOR ANALYSIS 105

Y(s). Note that both the impedance and the admittance are in general functions of the variable
s, and they are reciprocal of each other. Such expressions as Equations (3.1.11) through (3.1.16)
relate the amplitudes of the exponential voltages and currents, and are thefrequency-domain
representationsof the elements. Networks drawn using impedance or admittance symbols are
known astransformed networks,which play a significant role in finding the network response, as
shown in the following examples.

EXAMPLE 3.1.1

Consider anRLCseries circuit excited byv(t)=Vestin the time domain. Assume no initial
capacitor voltage or inductive current att=0. Draw the transformed network in thes-domain
and solve for the frequency-domain forced response of the resultant current. Then find thet-
domain forced responsei(t).

Solution

The forced response is produced by the particular excitation applied. The KVL equation for the
circuit of Figure E3.1.1 (a) is

−

• v(t) i(t) C

R L

(a)

−

+

V I^1

sL

sC

R

(b)

Figure E3.1.1RLCseries circuit withv(t)=Vest.(a)Time domain.(b)Transformed network ins-domain.

v(t)=Ri(t)+L

di(t)

dt

+

1

C

∫t

−∞

i(τ)dτ

The corresponding transformed network in thes-domain is shown in Figure E3.3.1(b), for which
the following KVL relation holds. (Note that the initial capacitor voltage att=0 is assumed to
be zero.)

V=RI+LsI+

1

Cs

I

Solving forI, one gets

I=

V

R+Ls+( 1 /Cs)

=

V

Z(s)

whereZ(s) can be seen to be the addition of each series impedance of the elements. The time
function corresponding to the frequency-domain response is given by

i(t)=Iest=

V

R+Ls+( 1 /Cs)

est

which is also an exponential with the same exponent contained inv(t).