Normalized Butterworth and Chebyshev Functions - Digital Signal Processing: Fundamentals and Applications - page 812

EXAMPLE C.5
Compute the normalized Chebyshev transfer function for the following specifications:
Ripple ¼ 1dB
n ¼ 3
Solution:
ðn 1Þ=2 ¼ 1

Applying Equations (C.32) to (C.38) leads to
a 0 ¼ð2 0 þ 1Þp=ð2 3Þ¼p=6
ε 2 ¼ 10 0:11 1 ¼ 0:2589, 1= ε ¼ 1:9653
b ¼ sin h 1

p

1:9653 2 þ 1 ð1:9653Þ=n ¼ lnð1:9653 þ Þ=3 ¼ 0:4760

b 0 ¼ 2 sin ðp=6Þ sin hð0:4760Þ¼0:4942
c 0 ¼½sin ðp=6Þ sin hð0:4760Þ

2 2

þ½cos ðp=6Þ cos hð0:4760Þ ¼ 0:9942

sin hðbÞ¼sin hð0:4760Þ¼0:4942
K ¼ 0:4942 0:9942 ¼ 0:4913

We can derive the transfer function as

0:4913
ðs þ 0:4942Þðs 2 þ 0:4942s þ 0:9942Þ

B 3 ðsÞ¼

Finally, the unfactored form is found to be

0:4913
s 3 þ 0:9883s 2 þ 1:2384s þ 0:4913

B 3 ðsÞ¼