Digital Signal Processing Reference
In-Depth Information
2
2
c
k
¼½
sin
ða
k
Þ
sin h
ðbÞ
þ½
cos
ða
k
Þ
cos h
ðbÞ
(C.41)
where
a
k
¼ð
2
k þ
1
Þp=ð
2
nÞ
for
k ¼
0
;
1
;
/
; n=
2
1
(C.42)
p
1
þ
ε
2
For the unit passband gain and the filter order as an even number, we require that
B
n
ð
0
Þ¼
1
=
,
so that the maximum magnitude of the ripple on the passband equals 1. Then we have
n=
2
1
k ¼
0
c
k
=
p
1
þ
ε
Y
K ¼
2
(C.43)
b ¼
sin h
1
ð
1
=
ε
Þ=n
(C.44)
p
x
sin h
1
2
ðxÞ¼
ln
ðx þ
þ
1
Þ
(C.45)
Equations
(C.32)
to (C.45) are applied to compute the normalized Chebyshev transfer function. Now
let us look at the following illustrative examples.
EXAMPLE C.4
Compute the normalized Chebyshev transfer function for the following specifications:
Ripple ¼ 0.5 dB
n ¼ 2
Solution:
n=2 ¼ 1
Applying Equations
(C.39)
to (C.45), we obtain
a
0
¼ð2 0 þ 1Þp=ð2 2Þ¼0:25p
ε
2
¼ 10
0:10:5
1 ¼ 0:1220, 1=
ε
¼ 2:8630
b ¼ sin h
1
p
2:8630
2
þ 1
ð2:8630Þ=n ¼ lnð2:8630 þ
Þ=2 ¼ 0:8871
b
0
¼ 2 sin ð0:25pÞ sin hð0:8871Þ¼1:4256
c
0
¼½sin ð0:25pÞ sin hð0:8871Þ
2
2
þ½cos ð0:25pÞ cos hð0:8871Þ
¼ 1:5162
p
1 þ 0:1220
K ¼ 1:5162=
¼ 1:4314
Finally, the transfer function is derived as
1:4314
s
2
þ 1:4256s þ 1:5162
B
2
ðsÞ¼
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