Signals and Systems - Electrical Engineering

244 C H A P T E R 4: Frequency Analysis: The Fourier Series

circuit shown in Figure 4.1 where the input is

vs(t)= 1 +cos(10,000t)

with components of low frequency,=0, and of large frequency,=10,000 rad/sec. The output

vc(t)is the voltage across the capacitor in steady state. We wish to find the frequency response of

this circuit to verify that it is alow-pass filter(it allows low-frequency components to go through,

but filters out high-frequency components).

Solution

Using Kirchhoff’s voltage law, this circuit is represented by a first-order differential equation,

vs(t)=vc(t)+

dvc(t)

dt

Now, if the input isvs(t)=ejt, for a generic frequency, then the output isvc(t)=ejtH(j).

Replacing these in the differential equation, we have

ejt=ejtH(j)+

dejtH(j)

dt

=ejtH(j)+jejtH(j)

so that

H(j)=

1

1 +j

or the frequency response of the filter for any frequency. The magnitude ofH(j)is

|H(j)|=

1

√

1 +^2

which is close to one for small values of the frequency, and tends to zero when the frequency

values are large—the characteristics of a low-pass filter.

For the input

vs(t)= 1 +cos(10,000t)=cos( 0 t)+cos(10,000t)

(i.e., it has a zero frequency component and a 10,000-rad/sec frequency component) using Euler’s

identity, we have that

vs(t)= 1 +0.5

(

ej10,000t+e−j10,000t

)

and the steady-state output of the circuit is

vc(t)= 1 H(j 0 )+0.5H(j10,000)ej10,000t+0.5H(−j10,000)e−j10,000t

≈ 1 +

1

10,000

cos(10,000t−π/ 2 )≈ 1