Signals and Systems - Electrical Engineering

4.2 Eigenfunctions Revisited 241

frequency, with time generates a cosine of the indicated frequency, amplitude, and phase. The transfer

function is computed at s=j 0 or

H(s)|s=j 0 =H(j 0 )=

Y

X

(ratio of the phasors corresponding to the output Y and the input X). The phasor for the output

is thus

Y=H(j 0 )X=|Y|ej∠Y

Such a phasor is then converted into the sinusoid (which equals Eq. 4.6):

yss(t)=Re[Yej^0 t]=|Y|cos( 0 t+∠Y)

n A very important application of LTI systems isfiltering, where one is interested in preserving desired
frequency components of a signal and getting rid of less-desirable components. That an LTI system can be
used for filtering is seen in Equations (4.3) and (4.4). In the case of a periodic signal, the magnitude
|H(jk)|can be set ideally to one for those components we wish to keep and to zero for those we wish
to get rid of. Likewise, for an aperiodic signal, the magnitude|H(j)|could be set ideally to one for
those components we wish to keep and zero for those components we wish to get rid of. Depending on
the filtering application, an LTI system with the appropriate characteristics can be designed, obtaining the
desired transfer function H(s).

For a stable LTI with transfer functionH(s)if the input is

x(t)=Re[Aej(^0 t+θ)]=Acos( 0 t+θ)

the steady-state output is given by

y(t)=Re[AH(j 0 )ej(^0 t+θ)]

=A|H(j 0 )|cos( 0 t+θ+∠H(j 0 )) (4.7)

where

H(j 0 )=H(s)|s=j 0

nExample 4.1

Consider the RC circuit shown in Figure 4.1. Let the voltage source bevs(t)=4 cos(t+π/ 4 )volts.

the resistor beR= 1 , and the capacitorC=1 F. Find the steady-state voltage across the capacitor.

Solution

This problem can be approached in two ways.

n Phasor approach.From the phasor circuit in Figure 4.1, by voltage division we have the follow-

ing phasor ratio, whereVsis the phasor corresponding to the sourcevs(t)andVcthe phasor