Tutorials - Digital Signal Processing: An Introduction with MATLAB and Applications - page 279

X ð z Þ ¼ 1 þð b = 3 Þ z 1

) H ð z Þ¼ Y ð z Þ

1 ð b = 2 Þ z 1 ¼ z þð b = 3 Þ

ð 5 Þ

z ð b = 2 Þ

The system has a zero at z =-b/3 and a pole at z = b/2.
For stability we should have: b =j j \1 ! ) j b j \2 :

Q: From Eq. 5 above, find the impulse response h(n) and the difference equation
of the above system.

Tutorial 44

Q: Using the impulse invariance method, design a digital Chebychev LPF with
the following specifications:

1. T s = 0.01 s (hence, f s = 100 Hz),
2. f c = 10 Hz () x c ¼ 20p rad/s),
3. G m = 1, gain \ 0.1 (i.e., -20 dB) for 30 B f B f s /2 = 50 Hz),
4. 3 dB ripple is allowed in the passband.

Solution: We need an analog Chebychev filter with the above specifications,
with gain less than -20 dB for normalized frequency f n C 30/10 = 3 (normalized
w.r.t f c ). From Tables we find the filter order n = 2, with normalized transfer
function:

H N ð s Þ¼ a o
0 : 7079 þ 0 : 6449s N þ s N
Since n is even, we have G dc ¼ G m =

p

¼ 1 =

p

1 þ e 2

1 : 9953 ¼ 0 : 7079.

Hence, a o /0.7079 = 0.7079 ) a o = 0.5. The denormalized analog transfer

function is obtained by the substitution s N = s/x c = s/(20p) to get:

1974
2795 þ 40 : 5s þ s 2

H a ð s Þ¼

Using Partial Fraction Expansion (Tutorial 19), we can write H a (s) as follows:
H a ð s Þ¼ c 1
s p 1

þ c 2

s p 2 ;

w her e p 1 =-20.25 + 48.18i, p 2 ¼ p 1 ¼ 20 : 25 48 : 18i, c 1 ¼ 20 : 2i ;

c 2 ¼ c 1 ¼ 20 : 2i :
The z-domain poles are:

z 1 ¼ exp ð p 1 T s Þ¼ 0 : 68 þ 0 : 45i ; z 2 ¼ exp ð p 2 T s Þ¼ z 1 ¼ 0 : 68 0 : 45i :