Signals and Systems - Electrical Engineering

338 CHAPTER 5: Frequency Analysis: The Fourier Transform

So that for 0≤ 0 <∞, if we compute the length and the angle ofZE( 0 )andPE( 0 ), the ratio of these

lengths gives the magnitude response and the difference of their angles gives the phase response.

For a filter with a transfer function

H(s)=

∏

∏i(s−zi)

k(s−pk)

wherezi,pkare zeros and poles of H(s) with vectorsZEi()=j−ziandPEk()=j−pk, going from each of

the zeros and poles to the frequency at which we are computing the magnitude and phase response in thej

axis, gives

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

∏

∏iZEi()

kEPk()

=

∏

∏i|ZEi()|

︸k|PE︷︷k()|︸

|H(j)|

e

j

[∑

i∠(EZi())−

∑

k∠(PEk()

]

︸ ︷︷ ︸

ej∠H(j)

(5.25)

nExample 5.17

Consider series RC circuit with a voltage sourcevi(t). Choose the output to obtain low-pass and

high-pass filters and use the poles and zeros of the transfer functions to determine their frequency

responses. LetR= 1 ,C=1 F, and the initial conditions be zero.

Solution

n Low-pass filter:Let the output be the voltage across the capacitor. By voltage division, we obtain

that the transfer function of the filter is

H(s)=

VC(s)

Vi(s)

=

1 /Cs

R+ 1 /Cs

For dc frequency, the capacitor behaves as an open circuit so that the output voltage equals the

input voltage, and for very high frequencies the impedance of the capacitor tends to zero so

that the voltage across the capacitor also goes to zero. This is a low-pass filter.

LetR= 1 andC=1 F, so

H(j)=

1

1 +j

=

1

PE()

Drawing a vector from the poles=−1 to any point on thejaxis givesPE(), and for different

frequencies we get

= 0 PE( 0 )= 1 ej^0

= 1 PE( 1 )=

√

2 ejπ/^4

=∞ PE(∞)=∞ejπ/^2