Digital Signal Processing Reference
In-Depth Information
Example 5.3 Determining the Transfer Function
Problem:
Determine the transfer function of the system described in Example
5.1, which has an impulse response of
n
⋅
h
(
n
)
=
b
u
(
n
)
1
Find the location of the poles and zeros of the transfer function as well as the
difference equation of the system from the transfer function.
Solution:
Using the fourth entry in the table of
z
-transform pairs, we can
determine that the transfer function of the system described is
Y
(
z
)
z
1
H
(
z
)
=
=
=
−
1
(
z
−
b
)
X
(
z
)
(
−
b
⋅
z
)
1
1
We see that there is a single pole (denominator root) at
z
=
b
1
and a single
zero (numerator root) at
z
= 0. By cross-multiplication, the following equation
results:
−
1
Y
(
z
)
−
b
⋅
z
⋅
Y
(
z
)
=
X
(
z
)
1
By taking the inverse
z
-transform of this equation and applying the time shift
property of the
z
-transform, we can determine the same difference equation, as in
Example 5.1:
y
(
n
)
=
b
⋅
y
(
n
−
1
+
x
(
n
)
1
Another way that a discrete-time system can be described is by drawing a
system diagram to represent a general difference equation. Figure 5.4 shows the
system diagram for (5.21). In the figure, delays are represented by
z
-1
and
multiplication by triangular symbols. Of course, if a nonrecursive filter was being
represented, the lower half of the system diagram would be eliminated since the
output of such a system does not depend on past values of the output.
In this system diagram, we see that each time the signal moves through a
delay element the output is delayed by one sample interval. Although the system
diagram is correct and will produce the correct output relationship, there is a more
efficient description of the system. First, if we transform the general difference
equation of (5.22), we have
⎛
N
⎞
M
∑
∑
⎜
⎝
−
k
⎟
⎠
−
k
Y
(
z
)
⋅
1
−
b
⋅
z
=
X
(
z
)
⋅
a
⋅
z
k
k
k
=
1
k
=
0
(5.27)
