IIR filter design - Continuous and Discrete Time Signals and Systems - page 731

Step 2 Using the analog filter design techniques, design an analog filter H ( s )
based on the transformed specifications obtained in step 1.

Step 3 Using the bilinear transformation s = ( z − 1) / ( z + 1) (obtained by rear-
ranging Eq. (16.25) to express z in terms of s ), derive the z-transfer function
H ( z ) from the s-transfer function H ( s ).

Step 4 Confirm that the z-transfer function H ( z ) obtained in step 3 satisfies the
design specifications by plotting the magnitude spectrum H ( Ω ) . If the design
specifications are not satisfied, increase the order N of the analog filter designed
in step 2 and repeat from step 2.

We now illustrate the application of the above algorithm in Example 16.4.

Example 16.4
Repeat Example 16.3 using the bilinear transformation.

Solution
Choosing k = 1 (sampling interval T = 2), step 1 transforms the pass-band and
stop-band corner frequencies into the CT frequency domain:

pass-band corner frequency ω p = tan(0 . 5 Ω p ) = tan(0 . 5 0 . 25 π )

= 0 . 4142 radians / s;

stop-band corner frequency ω s = tan(0 . 5 Ω s ) = tan(0 . 5 0 . 75 π )

= 2 . 4142 radians / s .

The transformed specifications of the CT filter are given by

pass-band (0 ≤ω≤ 0 . 4142 radians/s) 0 . 8 ≤ H ( ω ) ≤ 1;

stop-band ( ω > 2 . 4142 radians/s) H ( ω ) ≤ 0 . 20 .

Step 2 designs the analog filter based on the transformed specifications. As in
Example 16.3, we use the Butterworth filter. The gain terms for the filter stay
the same as in Example 16.3:

1
(1 − δ p ) 2

G p =

− 1 = 0 . 5625

and

1
( δ s ) 2

G s = − 1 = 24

The order N of the filter is given by

1
2

ln( G p / G s )
ln( ω p /ω s )

1
2

ln(0 . 5625 / 24)
ln(0 . 4142 / 2 . 4142)

N = = = 1 . 0646 ,