Infinite Impulse Response Filter Design - Digital Signal Processing: Fundamentals and Applications - page 329

Solution:
a. From the specifications, the digital frequencies are
u dp ¼ 2pf ¼ 2pð3; 000Þ¼6; 000p rad/sec
u ds ¼ 2pf ¼ 2pð1; 000Þ¼2; 000p rad/sec
and T ¼ 1=f s ¼ 1=8; 000 sec

Using the design procedure, it follows that

u dp T
2

6; 000p=8; 000
2

u ap ¼ T tan ¼ 3:8627 10 4 rad=sec

¼ 16; 000 tan

u ds T
2

2; 000p=8; 000
2

¼ 6:6274 10 3 rad=sec

u as ¼ 16; 000 tan ¼ 16; 000 tan

We find the lowpass prototype specification as follows:

v s ¼ u ps =u sp ¼ 3:8627 10 4 =6:6274 10 3 ¼ 5:8284 rad=s and A s ¼ 25 dB

Then the filter order is computed as

ε 2 ¼ 10 0:10:5 1 ¼ 0:1220

10 0:125 1 . 0:1220 ¼ 2; 583:8341

cosh 1 h ð2; 583:8341Þ 0:5 i

cosh 1 5:8284

p

50:8314 2 1

ln 50:8314 þ

ln 5:8284 þ

¼ 1:8875

n ¼ ¼

p

5:8284 2 1

We select n ¼ 2 for the lowpass prototype function. Following the steps of the design procedure, it follows that

1.

u p ¼ 3:8627 10 4 rad=sec

2. Performing the prototype transformation (lowpass to lowpass) using the prototype filter in Table 8.4 , we have

1:4314
s 2 þ 1:4256s þ 1:5162 and

H P ðsÞ¼

0:9441s 2
s 2 þ 3:6319 10 4 s þ 9:8407 10 8

1:4314

HðsÞ¼H P ðsÞj s

u a ¼

¼

u p
s

2

þ1:4256 u p
s

þ 1:5162

3. Applying the BLT, we convert the analog filter to the digital filter as follows:

s¼16;000ðz1Þ=ðzþ1Þ

0:9441s 2
s 2 þ 3:6319 10 4 s þ 9:8407 10 8

HðzÞ¼