Discrete and Digital Signals and Systems - Digital Signal Processing: An Introduction with MATLAB and Applications - page 112

The analog filter s-domain transfer function is H a (s) = LT{h a (t)}. Now the
poles of H a (s) are transformed to the poles of H(z) through the relation z ¼ e sT s :
The method can be summarized as follows (see, for example, A. V. Oppenheim
and R. Schafer, Discrete-Time Signal Processing, Prentice-Hall, 1989):
1. Expand the analog transfer function by partial fractions as P m ¼ 1

c m

s p m . Note that

p m 's can be non-distinct.
2. The required transfer function is

H ð z Þ¼ T s X

1 e p m T s z 1 ¼ T s X

M M

c m z

z z m

c m

m ¼ 1 m ¼ 1

where z m ¼ e p m T s

are the poles in the z-domain.

Note 1 If there is no aliasing, H ð e jX Þ¼ H a ð x Þ when p X p :
Note 2 Since the relation X ¼ xT s is applicable, the jx axis is transformed into
the circumference of the unit circle e jX ¼ e jxT s :
Note 3 Since the poles are transformed according to z ¼ e sT s , stability is
preserved.

Example Using the impulse invariance method, design a digital Butterworth LPF
with the following specifications:

1. T s = 0.01 sec (hence, f s = 100 Hz),
2. f c = 10 Hz (therefore x c = 20 p rad/sec),
3. G m = 1, gain B 0.1 (i.e., -20 dB) for 20 B f B f s /2 = 50 Hz).

Solution:
It is necessary to first find an analog Butterworth LPF with the above specifica-
tions. The gain should be less than -20 dB for a normalized frequency of f c C
20/10 = 2 (normalized w.r.t f c ). From the graph in Tables—Stopband gain for a
B-LPF, the filter order is found to be n = 4. From Tables—Denominator Poly-
nomial Coefficients for Normalized LPF's, the transfer function of this normalized
filter is seen to be:
H N ð s Þ¼

a o
1 þ 2 : 6131s N þ 3 : 4142s N þ 2 : 6131s N þ s N

Now H( N (s)| s=0 = a 0 = G dc = G m = 1, and s N = s/x c . Substituting these val-
ues for a o and s N into the above equation, yields:

1 : 55e7
1 : 55e7 þ 6 : 48e5s þ 1 : 34e4s 2 þ 164 : 1s 3 þ s 4 :

H a ð s Þ¼

H a (s) can then be expanded using partial fractions. This can be easily done in
MATLAB using B = [1 164.1 1.34e4 6.48e5 1.55e7],A = [1.55e7],
[R,P,K] = residue(A,B) where R gives the coefficients c m , P gives the poles