Continuous-time filters - Continuous and Discrete Time Signals and Systems - page 354

From Fig. 7.12, it is clear that Eq. (7.68), or alternatively Eq. (7.67), represents
a highpass to lowpass transformation. We now exploit this transformation to
design a highpass filter.

Example 7.10
Design a highpass Butterworth filter with the following specifications:

stop band (0 ≤ξ ≤ 50 radians/s) − 1dB ≤ 20 log 10 H ( ξ ) ≤ 0;

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

Solution
Using Eq. (7.67) with ξ p

= 100 radians/s to transform the specifications from

the domain s = γ + j ξ of the highpass filter to the domain S = σ + j ω of the

lowpass filter, we obtain

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

stop band ( ω < 1 radian/s) 20 log 10 H ( ω ) ≤ 15 dB .

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

− 1 / 20 − 15 / 20

(1 − δ p ) = 10 = 0 . 8913 and δ s = 10 = 0 . 1778.

The gain terms G p and G s are given by

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( ξ p /ξ s )

= 1
2

ln(0 . 2588 / 30 . 6327)
ln(1 / 2)

N = 3 . 4435 .

We round off the order of the filter to the higher integer value as N = 4.
Using the pass-band constraint, Eq. (7.31), the cut-off frequency of the
required Butterworth filter is given by

ω s
( G s ) 1 / 2 N

2
(30 . 6327) 1 / 8

ω c = = = 1 . 3039 radians/s .