Continuous-time filters - Continuous and Discrete Time Signals and Systems - page 358

Example 7.11
Design a bandpass Butterworth filter with the following specifications:

stop band I (0 ≤ξ ≤ 50 radians/s) 20 log 10 H ( ξ ) ≤− 20 dB;

pass band (100 ≤ξ ≤ 200 radians/s) − 2dB ≤ 20 log 10 H ( ξ ) ≤ 0;

stop band II ( ξ ≥ 380 radians/s) 20 log 10 H ( ξ ) ≤− 20 dB .

Solution
For ξ p1 = 100 radians/s and ξ p2 = 200 radians/s, Eq. (7.70) becomes

ω = 2 10 4 − ξ 2
100 ξ

,

to transform the specifications from the domain s = γ + j ξ of the bandpass
filter to the domain S = σ + j ω of the lowpass filter. The specifications for the
normalized lowpass filter are given by

pass band (0 ≤ω < 1 radian/s) − 2 ≤ 20 log 10 H ( ω ) ≤ 0;

stop band ( ω≥ min(3 . 2737 , 3 . 5) radians/s 20 log 10 H ( ω ) ≤− 20 .

The above specifications are used to design a normalized lowpass Butterworth
filter. Expressed on a linear scale, the pass-band and stop-band gains are given
by

− 2 / 20 − 20 / 20

(1 − δ p ) = 10 = 0 . 7943 and δ s = 10 = 0 . 1 .

The gain terms G p and G s are given by

1
(1 − δ p ) 2

1
0 . 7943 2

G p = − 1 = − 1 = 0 . 5850

and

1
( δ s ) 2

1
0 . 1778 2

G s = − 1 = − 1 = 99 .

The order N of the Butterworth filter is obtained using Eq. (7.29) as follows:

ln( G p / G s )
ln( ξ p /ξ s )

= 1
2

= 1
2

ln(0 . 5850 / 99)
ln(1 / 3 . 2737)

N = 2 . 1232 .

We round off the order of the filter to the higher integer value as N = 3.
Using the stop-band constraint, Eq. (7.31), the cut-off frequency of the low-
pass Butterworth filter is given by

ω s
( G s ) 1 / 2 N

= 3 . 2737
(99) 1 / 6

ω c = = 1 . 5221 radians/s .

The poles of the lowpass filter are located at

j π
2

+ j (2 n − 1) π
6

S = ω c exp