Continuous-time filters - Continuous and Discrete Time Signals and Systems - page 332

and ω = ω s . Using Eq. (7.15), the two gains are given by

1
1 + ( ω p /ω c ) 2 N

H ( ω p ) 2 = (1 − δ p ) 2 ;

pass-band corner frequency ( ω = ω p ) =

(7.25)

1
1 + ( ω s /ω c ) 2 N

H ( ω s ) 2 = ( δ s ) 2 .

stop-band corner frequency ( ω = ω s ) =

(7.26)

Equations (7.25) and (7.26) can alternatively be expressed as follows:

1
(1 − δ p ) 2

( ω p /ω c ) 2 N = − 1 (7.27)

and

1
( δ s ) 2

( ω s /ω c ) 2 N

= − 1 .

(7.28)

Dividing Eq. (7.27) by Eq. (7.28) and simplifying in terms of N , we obtain the
following expression:

ln( G p / G s )

= 1
2 ln( ω p /ω s ) ,

N (7.29)

where the gain terms are given by

1
(1 − δ p ) 2

1
( δ s ) 2

G p = − 1 and G s = − 1 . (7.30)

Step 2 Using Table 7.2 or otherwise determine the transfer function for the nor-
malized Butterworth filter of order N . The transfer function for the normalized
Butterworth filter is denoted by H ( S ) with the Laplace variable S capitalized
to indicate the normalized domain.

Step 3 Determine the cut-off frequency ω c of the Butterworth filter using either
of the following two relationships:

ω p
( G p ) 1 / 2 N ;

pass-band constraint ω c = (7.31)

ω s
( G s ) 1 / 2 N .

stop-band constraint ω c = (7.32)

If Eq. (7.31) is used to compute the cut-off frequency, then the Butterworth filter
will satisfy the pass-band constraint exactly. Similarly, the stop-band constraint
will be satisfied exactly if Eq. (7.32) is used to determine the cut-off frequency.

Step 4 Determine the transfer function H ( s ) of the required lowpass filter
from the transfer function for the normalized Butterworth filter H ( S ), obtained