Continuous-time filters - Continuous and Discrete Time Signals and Systems - page 333

in Step 2, and the cut-off frequency ω c , using the following transformation:

H ( s ) = H ( S ) S = s /ω c .

Note that the transformation S = s / ω c represents scaling in the Laplace domain.
It is therefore clear that the normalized cut-off frequency of 1 radian/s used in
the normalized Butterworth filter is transformed to a value of ω c as required in
Step 3.

Step 5 Sketch the magnitude spectrum from the transfer function H ( s ) deter-
mined in Step 4. Confirm that the transfer function satisfies the initial design
specifications.

Examples 7.5 and 7.6 illustrate the application of the design algorithm.

Example 7.5
Design a Butterworth lowpass filter with the following specifications:

pass band (0 ≤ω≤ 5 radians/s) 0 . 8 ≤ H ( ω ) ≤ 1;

stop band ( ω > 20 radians/s) H ( ω ) ≤ 0 . 20 .

Solution
Using Step 1 of Algorithm 7.3.1.1, the gain terms G p and G s are given by

1
(1 − δ p ) 2

1
0 . 8 2

= − 1 = − 1 = 0 . 5625

G p

and

1
( δ s ) 2

1
0 . 2 2

G s = − 1 = − 1 = 24 .

Using Eq. (7.29), the order of the Butterworth filter is given by

= 1
2

ln( G p / G s )
ln( ω p /ω s )

= 1
2

ln(0 . 5625 / 24)
ln(5 / 20)

N = 1 . 3538 .

We round off the order of the filter to the higher integer value as N = 2.
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
H ( S ) = 1

S 2 + 1 . 414 S + 1 .
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

5
(0 . 5625) 1 / 4

ω c = = = 5 . 7735 radians/s .