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Characteristics of Prototype Analog Filters 407
where the operation ⌈x⌉ means “choose the smallest integer larger than
x”—for example, ⌈4.5⌉ =5. Since the actual N chosen is larger than required,
specifications can be either met or exceeded either at Ω p or at Ω s .
To satisfy the specifications exactly at Ω p ,
Ω c =
2N
Ω p
√ (10
R p/10 − 1 ) (8.50)
or, to satisfy the specifications exactly at Ω s ,
Ω c =
2N
Ω s
√ (10
A s/10
− 1 ) (8.51)
□ EXAMPLE 8.3 Design a lowpass Butterworth filter to satisfy
Passband cutoff: Ω p =0.2π ;Passband ripple: R p = 7dB
Stopband cutoff: Ω s =0.3π ; Stopband ripple: A s = 16dB
Solution From (8.49)
⌈ [(
log 10 10 0.7 − 1 ) / ( 10 1.6 − 1 )] ⌉
N =
= ⌈2.79⌉ =3
2 log 10 (0.2π/0.3π)
To satisfy the specifications exactly at Ω p, from (8.50) we obtain
Ω c =
0.2π
6 √ (10 0.7 − 1) =0.4985
To satisfy specifications exactly at Ω s, from (8.51) we obtain
Ω c =
0.3π
6 √ (10 1.6 − 1) =0.5122
Now we can choose any Ω c between the above two numbers. Let us choose
Ω c =0.5. We have to design a Butterworth filter with N =3and Ω c =0.5,
which we did in Example 8.1. Hence
H a(jΩ) =
0.125
(s +0.5) (s 2 +0.5s +0.25)
□
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