Digital Signal Processing Reference
In-Depth Information
j
/
2
j
/
2
H
(
z
)
=
−
−
(
α
+
j
β
)
T
−
1
−
(
α
−
j
β
)
T
−
1
1
−
e
⋅
z
1
−
e
⋅
z
(6.4)
( )
−
(
α
+
j
β
)
T
−
(
α
−
j
β
)
T
−
1
j
/
2
⋅
(
e
−
e
)
⋅
z
H
(
z
)
=
(6.5)
−
(
α
+
j
β
)
T
−
1
−
(
α
−
j
β
)
T
−
1
(
−
e
⋅
z
)
⋅
(
−
e
⋅
z
)
−
α
T
−
1
e
⋅
sin(
β
T
)
⋅
z
H
(
z
)
=
−
α
T
−
1
−
2
α
T
−
2
1
−
2
⋅
e
⋅
cos(
β
T
)
⋅
z
+
e
⋅
z
(6.6)
Example 6.2 Butterworth Impulse Invariant Filter Design
Problem:
Determine the impulse invariant digital filter for a second-order
Butterworth approximation function, as shown below. Notice that
H
(
s
) is
normalized and therefore has a passband edge frequency of 1 rad/sec or (1/2π Hz).
Determine the differences that result from choosing sampling periods of
T
= 1.0
sec and
T
= 0.1 sec.
1
H
(
s
)
=
2
s
+
1
4142
⋅
s
+
1
Solution:
We can first factor the analog transfer function and use partial
fraction expansion to determine
j
0
.
7071
j
0
7071
H
(
s
)
=
−
s
+
0
.
7071
+
j
0
7071
s
+
0
.
7071
−
j
0
7071
The digital transfer function can then be determined by using the results indicated
in (6.3) to (6.6):
−
0
.
7071
⋅
T
−
1
1
.
4142
⋅
e
⋅
sin(
0
.
7071
⋅
T
)
⋅
z
H
(
z
)
=
−
0
.
7071
⋅
T
−
1
−
1
.
4142
⋅
T
−
2
1
−
2
⋅
e
⋅
cos(
0
.
7071
⋅
T
)
⋅
z
+
e
⋅
z
We can now make the substitution of the different sampling periods in the
general form to find the two distinct transfer functions:
−
1
.
45300
⋅
z
H
(
z
)
=
−
1
−
2
T
=
1
.
0
1
−
0
.
74971
⋅
z
+
0
.
24312
⋅
z
