Digital Signal Processing Reference
In-Depth Information
Example 5.5 Complete Discrete-Time System Example
Problem:
Consider the difference equation for a discrete-time system shown
below. Determine the impulse response, transfer function, and frequency response
of the system.
y
(
n
)
=
2
⋅
y
(
n
−
1
−
1
81
⋅
y
(
n
−
2
+
0
68
⋅
y
(
n
−
3
+
1
⋅
x
(
n
)
+
3
⋅
x
(
n
−
1
+
3
⋅
x
(
n
−
2
+
1
⋅
x
(
n
−
3
Also, find the location of the poles and zeros for the system as well as draw the
system diagram. Finally, determine the output of the system if the input is the
analog signal
x
(
t
) shown below. Assume that the sampling frequency is 20,000
samples per second.
x
(
t
)
=
10
+
5
cos(
2
π
⋅
2
000
⋅
t
−
60
°
)
+
20
sin(
2
π
⋅
8
000
⋅
t
+
30
°
)
Solution:
We begin the solution of the problem by transforming the
difference equation to determine the transfer function for the system such that
−
1
−
2
−
3
Y
(
z
)
=
2
.
⋅
z
⋅
Y
(
z
)
−
1
.
81
⋅
z
⋅
Y
(
z
)
+
0
.
68
⋅
z
⋅
Y
(
z
)
+
1
.
⋅
X
(
z
)
−
1
−
2
−
3
+
3
0
⋅
z
⋅
X
(
z
)
+
3
.
⋅
z
⋅
X
(
z
)
+
1
.
⋅
z
⋅
X
(
z
)
or
−
1
−
2
−
3
Y
(
z
)
1
0
+
3
.
⋅
z
+
3
.
⋅
z
+
1
.
⋅
z
=
=
H
(
z
)
X
(
z
)
−
1
−
2
−
3
1
.
+
2
.
⋅
z
−
1
.
81
⋅
z
+
0
.
68
⋅
z
After we convert
H
(
z
) to positive powers of
z
, we factor the transfer function
in order to determine the pole and zero locations. From the expression, we see that
there are three zeros at
z
= −1.0 and three poles at
z
= 0.8 and 0.6 ±
j
0.7:
3
(
z
+
1
0
H
(
z
)
=
2
(
z
−
0
8
⋅
(
z
−
1
2
⋅
z
+
0
85
)
The frequency response can now be easily determined from the transfer
function, as indicated below:
j
Ω
3
(
e
+
1
.
0
)
j
Ω
H
(
e
)
=
j
Ω
j
2
Ω
j
Ω
(
e
−
0
.
8
⋅
(
e
−
1
.
2
⋅
e
+
0
.
85
)
