Infinite Impulse Response Filter Design - Digital Signal Processing: Fundamentals and Applications - page 336

1; 500 2p=8; 000
2

¼ 1:0691 10 4 rad=sec

u asl ¼ 16; 000 tan

Now, adjusting the unit gain for the center frequency of 2,500 Hz leads to the following:
Fixing u apl ¼ 2:2022 10 4 , we have u aph ¼ u 0

u apl ¼ ð2:3946 10 4 Þ 2

2:2022 10 4 ¼ 2:6038 10 4
and the passband bandwidth: W ¼ u aph u apl ¼ 4; 016
Fixing u asl ¼ 1:0691 10 4 , u ash ¼ u 0

u asl ¼ ð2:3946 10 4 Þ 2

2:10691 10 4 ¼ 5:3635 10 4
and the stopband bandwidth: W s ¼ u ash u asl ¼ 42; 944

u aph ¼ ð2:3946 10 4 Þ 2

u 0

2:6110 10 4 ¼ 2:1961 10 4

Again, fixing u aph ¼ 2:6110 10 4 , we have u apl ¼

and the passband bandwidth: W ¼ u aph u apl ¼ 4; 149

u ash ¼ ð2:3946 10 4 Þ 2

u 0

8:0437 10 4 ¼ 0:7137 10 4
and the stopband bandwidth: W s ¼ u ash u asl ¼ 73; 300
For an aggressive bandpass design, we select u apl ¼ 2:2022 10 4 , u aph ¼ 2:6038 10 4 ,
u asl ¼ 1:0691 10 4 , u ash ¼ 5:3635 10 4 and for a smaller bandwidth for passband.
Thus, we obtain the prototype specifications:

Fixing u ash ¼ 8:0437 10 4 , u asl ¼

v s ¼ð53; 635 10; 691Þ=ð26; 110 21; 961Þ¼10:6932

ε 2 ¼ 10 0:10:5 1 ¼ 0:1220
10 0:110 1 . 0:1220 ¼ 73:7705

cosh 1 h ð73:7705Þ 0:5 i

cosh 1 10:6932 ¼

ln 8:5890 þ

p

8:5890 2 1

¼ 0:9280

n ¼

p

10:6932 2 1

ln 10:6932 þ

Rounding up n leads to n ¼ 1.
Next, we apply the design steps:

1.

u aph ¼ 2:6038 10 4 rad=sec; u apl ¼ 2:2022 10 4 rad=sec

W ¼ 4; 016 rad=sec; u 0 ¼ 5:7341 10 8

2. Performing the prototype transformation (lowpass to bandpass), we obtain

H P ðsÞ¼ 2:8628
s þ 2:8628

and

HðsÞ¼H P ðsÞ

1:1878 10 4 s
s 2 þ 1:1878 10 4 s þ 5:7341 10 8

2:8628Ws
s 2 þ 2:8628Ws þ u 0 ¼

¼

s 2 þu 0
sW

s¼