Infinite Impulse Response Filter Design - Digital Signal Processing: Fundamentals and Applications - page 351

After algebra simplification, we obtain the second-order digital filter as

0:1 0:09767z 1
1 1:7735z 1 þ 0:8187z 2

HðzÞ¼

b. The magnitude and phase frequency responses are shown in Figure 8.30 and MATLAB Program 8.13 is
given. The passband gain of the digital filter is higher than that of the analog filter, but their shapes are
same.

Program 8.13. MATLAB program for Example 8.16.

% Example 8.16
% Plot the magnitude responses j H(s) j and j H(z) j

% For the Laplace transfer function H(s)
f ¼ 0:0.1:5;T ¼ 0.1; % Initialize analog frequency range in Hz and sampling interval
w ¼ 2*pi*f; % Convert the frequency range to radians/second
hs ¼ freqs([1 0], [1 2 5],w); % Calculate analog filter frequency responses
phis ¼ 180*angle(hs)/pi;
% For the z-transfer function H(z)
% Calculate digital filter frequency responses
hz ¼ freqz([0.1 -0.09766],[1 -1.7735 0.8187],length(w));
phiz ¼ 180*angle(hz)/pi;
% Plot magnitude and phase responses
subplot(2,1,1), plot(f,abs(hs), ' x ' ,f, abs(hz), ' - ' ),grid;
xlabel( Frequency (Hz) ); ylabel( Magnitude Responses )

' ' ' '

subplot(2,1,2), plot(f,phis, x ,f, phiz, - ); grid;

' ' ' '

xlabel( Frequency (Hz) ); ylabel( Phases (degrees) )

' ' ' '

8.7 POLE-ZERO PLACEMENT METHOD FOR SIMPLE INFINITE IMPULSE
RESPONSE FILTERS
This section introduces a pole-zero placement method for a simple IIR filter design. Let us first
examine the effects of the pole-zero placement on the magnitude response in the z-plane ( Figure 8.31 ) .
In the z-plane, whenwe place a pair of complex conjugate zeros at a given point on the unit circlewith
an angle q (usually we do), wewill have a numerator factor of ðz e jq Þðz e jq Þ in the transfer function.
Its magnitude contribution to the frequency response at z ¼ e j U is ðe j U e jq Þðe j U e jq Þ . When
U ¼ q , the magnitude will reach zero, since the first factor ðe jq e jq Þ¼ 0 contributes zero magnitude.
When a pair of complex conjugate poles are placed at a given point within the unit cycle, we have
a denominator factor of ðz re jq Þðz re jq Þ , where r is the radius chosen to be less than and close to 1 to
place the poles inside the unit circle. The magnitude contribution to the frequency response at U ¼ q
will rise to a large magnitude, since the first factor ðe jq re jq Þ¼ð 1 rÞe jq gives a small magnitude
of 1 r in the denominator. This small magnitude (1 r) is the length between the pole location and the
unit circle at the angle U ¼ q as shown in Figure 8.31. Note that the magnitude of e jq is 1.
Therefore, we can reduce the magnitude response using zero placement, while we increase the
magnitude response using pole placement. Placing a combination of poles and zeros will result in