Normalized Butterworth and Chebyshev Functions - Digital Signal Processing: Fundamentals and Applications - page 807

Setting P n ð 0 Þ¼ 1 for the unit passband gain, we have
K ¼ r n ¼ 1 = ε

Let us examine the following examples.

EXAMPLE C.1
Compute the normalized Butterworth transfer function for the following specifications:
Ripple ¼ 3dB
n ¼ 2
Solution:
n=2 ¼ 1
q k ¼ð2p 0 þ pÞ=ð2 2Þ¼0:25p
ε 2 ¼ 10 0:13 1
r ¼ 1 and K ¼ 1

Applying Equation (C.11) leads to

1
s 2 þ 2 1 cos ð0:25pÞs þ 1 2 ¼

1
s 2 þ 1:4141s þ 1

P 2 ðsÞ¼

EXAMPLE C.2
Compute the normalized Butterworth transfer function for the following specifications:
Ripple ¼ 3dB
n ¼ 3
Solution:
ðn 1Þ=2 ¼ 1
ε 2 ¼ 10 0:13 1
r ¼ 1 and K ¼ 1
q k ¼ð2p 1Þ=ð2 3Þ¼p=3
From Equation (C.9) , we have

1
ðs þ 1Þðs 2 þ 2 1 cos ðp=3Þs þ 1 2 Þ

P 3 ðsÞ¼

1
ðs þ 1Þðs 2 þ s þ 1Þ

¼

For the unfactored form, we get

1
s 3 þ 2s 2 þ 2s þ 1

P 3 ðsÞ¼