Signals and Systems - Electrical Engineering

5.7 Convolution and Filtering 339

Since there are no zeros, the frequency response of this filter depends inversely on the behavior

of the pole vectorPE(). The frequency responses for these three frequencies are:

H(j 0 )= 1 ej^0

H(j 1 )=0.707e−jπ/^4

H(j∞)= 0 e−jπ/^2

Thus, the magnitude response is unity at=0 and it decays as the frequency increases. The

phase is zero at=0,−π/4 at=1, and−π/2 at→∞. The magnitude response is even

and the phase response is odd.

n High-pass filter:Consider then the output being the voltage across the resistor. Again by voltage

division we obtain the transfer function of this circuit as

H(s)=

Vr(s)

Vs(s)

=

CRs

CRs+ 1

Again, letC=R=1, so the frequency response is

H(j)=

j

1 +j

=

ZE()

PE()

The vectorZE()goes from zero at the origins=0 tojin thejaxis, and the vectorPE()

goes from the poles=−1 tojin thejaxis. The vectors and the frequency response, at three

different frequencies, are given by

= 0 PE( 0 )= 1 ej^0 ZE( 0 )= 0 ejπ/^2 H(j 0 )=

ZE( 0 )

PE( 0 )

= 0 ejπ/^2

= 1 PE( 1 )=

√

2 ejπ/^4 ZE( 1 )= 1 ejπ/^2 H(j 1 )=

ZE( 1 )

PE( 1 )

=0.707ejπ/^4

=∞ EP(∞)=∞ejπ/^2 ZE(∞)=∞ejπ/^2 H(j∞)=

ZE(∞)

PE(∞)

= 1 ej^0

Thus, the magnitude response is zero at=0 (this is due to the zero ats=0, makingZE( 0 )= 0

as it is right on top of the zero), and it grows to unity as the frequency increases (at very high

frequency, the lengths of the pole and the zero vectors are alike and so the magnitude response

is unity and the phase response is zero). n

Remarks

n Poles create “hills” at frequencies in the jaxis in front of the poles imaginary parts. The closer the pole
is to the jaxis, the narrower and higher the hill. If, for instance, the poles are on the jaxis (this would
correspond to an unstable and useless filter) the frequency response at the frequency of the poles will be
infinity.