Infinite Impulse Response Digital Filter Design - Practical Analog and Digital Filter Design - page 149

Example 6.4 Butterworth Step Invariant Filter Design

Problem: Determine the step invariant digital filter for a second-order
normalized Butterworth approximation function, as shown below. Determine the
differences that result for sampling periods of T = 1.0 sec and T = 0.1 sec.

1

H ( s ) =

2

s + 1 4142 ⋅ s + 1

Solution: Again, we first determine the output of the analog system to an
input step function and then use partial fraction expansion to determine the
individual terms.

1 0 + j 0 0 − j 0

G ( s ) = − −

s s + 0 7071 + j 0 7071 s + 0 7071 − j 0 7071

The step response can then be determined by finding the inverse Laplace
transform of G ( s ), as indicated below:

− ( 0 . 7071 + j 0 . 7071 ) t − ( 0 . 7071 − j 0 . 7071 ) t

g ( t ) = 1 . 0 − ( 0 . 5 + j 0 . 5 ⋅ e − ( 0 . 5 − j 0 . 5 ⋅ e

Then, after sampling at intervals of T , the discrete-time step response is
determined to be

− ( 0 . 7071 + j 0 . 7071 ) nT − ( 0 . 7071 − j 0 . 7071 ) nT

g ( nT ) = 1 . 0 − ( 0 . 5 + j 0 . 5 ⋅ e − ( 0 . 5 − j 0 . 5 ⋅ e

The discrete-time response to the step input can then be determined by using
the z -transform.

1 0 + j 0 0 − j 0

G ( z ) = − −

− 1 ( 0 7071 + j 0 7071 ) T − 1 ( 0 7071 − j 0 7071 ) T − 1

1 − z 1 − e z 1 − e z

Now, by combining these terms over a common denominator (and performing
a considerable amount of complex algebra), we have the system response to a step
input:

− 1 − 0 707 T − 1

z + e [(sin( 0 707 ⋅ T ) − cos( 0 707 ⋅ T )] ⋅ z

G ( z ) =

− 1 − 0 707 ⋅ T − 1 − 1 414 ⋅ T − 2

( − z ) ⋅ [ − 2 ⋅ e cos( 0 707 T ) ⋅ z + e z ]

− 1 414 T − 2 − 0 707 T − 2

e z − e [sin( 0 707 ⋅ T ) + cos( 0 707 ⋅ T )] ⋅ z

+

− 1 − 0 707 ⋅ T − 1 − 1 414 ⋅ T − 2

( − z ) ⋅ [ − 2 ⋅ e cos( 0 707 ⋅ T ) ⋅ z + e z ]