Continuous-time filters - Continuous and Discrete Time Signals and Systems - page 335

Example 7.6
Design a lowpass Butterworth filter with the following specifications:

pass band (0 ≤ω≤ 50 radians/s)

− 1dB ≤ 20 log 10 H ( ω ) ≤ 0;

stop band ( ω > 100 radians/s) 20 log 10 H ( ω ) ≤− 15 dB .

Solution
Expressed on a linear scale, the pass-band gain is given by (1 − δ p ) = 10

− 1 / 20

=

− 15 / 20 = 0.1778.
Using Step 1 of Algorithm 7.3.1.1, the gain terms G p and G s are given by

0 . 8913. Similarly, the stop-band gain is given by δ s = 10

1
(1 − δ p ) 2

1
0 . 8913 2

G p = − 1 = − 1 = 0 . 2588

and

1
( δ s ) 2

1
0 . 1778 2

G s = − 1 = − 1 = 30 . 6327 .

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

= 1
2

ln( G p / G s )
ln( ω p /ω s )

= 1
2

ln(0 . 2588 / 30 . 6327)
ln(50 / 100)

N = 3 . 4435 .

We round off the order of the filter to the higher integer value as N = 4.
Using Step 2 of Algorithm 7.3.1.1, the transfer function H ( S ) of the normal-
ized Butterworth filter with a cut-off frequency of 1 radian/s is given by

1
( S 2 + 0 . 7654 S + 1)( S 2 + 1 . 8478 S + 1) .

H ( S ) =

Using the pass-band constraint, Eq. (7.31), in Step 3 of Algorithm 7.3.1.1, the
cut-off frequency of the required Butterworth filter is given by

ω p
( G p ) 1 / 2 N

50
(0 . 2588) 1 / 8

ω c = = = 59 . 2038 radians/s .

Using Step 4 of Algorithm 7.3.1.1, the transfer function H ( s ) of the required
Butterworth filter is obtained by the following transformation:

1
( S 2 + 0 . 7654 S + 1)( S 2 + 1 . 8478 S + 1)

H ( s ) = H ( S ) S = s /ω c = ,

S = s / 59 . 2038

which simplifies to

(3 . 5051 10 3 ) 2
( s 2 + 45 . 3146 s + 3 . 5051 10 3 )( s 2 + 109 . 396 s + 3 . 5051 10 3 )

H ( s ) =

or

1.2286 10 7
s 4 + 154.7106 s 3 + 1 . 1976 10 4 s 2 + 5.4228 10 5 s + 1.2286 10 7 .

H ( s ) =