Digital Signal Processing Reference
In-Depth Information
imp = 0.9.ˆ(0:1:100); sig = [1,-0.9]; td = conv(imp,sig)
which returns the value of
td
as a unit impulse sequence. Note that even though the single pole generates
an infinitely long impulse response, this particular input sequence results in a response which is identically
zero after the first sample of output.
Example 2.14.
The sequence randn(1,8) is processed by an LTI system represented by the difference
equation below; specify another difference equation which will, when fed the output of the first LTI
system, return the original sequence.
y
[
n
]=
x
[
n
]+
x
[
n
−
2
]+
1
.
2
y
[
n
−
1
]−
0
.
81
y
[
n
−
2
]
We rewrite the difference equation as
y
[
n
]−
1
.
2
y
[
n
−
1
]+
0
.
81
y
[
n
−
2
]=
x
[
n
]+
x
[
n
−
2
]
and then write the
z
-transform as
1
.
2
z
−
1
0
.
81
z
−
2
)
z
−
2
)
Y(z)(
1
−
+
=
X(z)(
1
+
which yields
z
−
2
1
+
Y (z)/X(z)
=
0
.
81
z
−
2
The
b
and
a
coefficients for this system are
b
= [1,1] and
a
= [1,-1.2,0.81]. To obtain the inverse system,
simply exchange the values for
b
and
a
, i.e., obtain the reciprocal of the
z
-transform. To check the answer,
we will first process the input sequence with the original difference equation, then process the result with
the inverse difference equation:
x = randn(1,8), y = filter([1,1],[1,-1.2,0.81],x);
ans = filter([1,-1.2, 0.81],[1,1],y)
1
−
1
.
2
z
−
1
+
2.3.8 TRANSFER FUNCTIONS, POLES, AND ZEROS
LTI System Representation
If an LTI system is represented by the difference equation
N
−
1
M
−
1
y
[
n
]+
a
k
y
[
n
−
k
]=
b
m
x
[
n
−
m
]
k
=
1
m
=
0
Taking the
z
-transform and using properties such as the shifting property, we get
N
−
1
M
−
1
a
k
z
−
k
)
b
m
z
−
m
Y(z)(
1
+
=
X(z)
k
=
1
m
=
0
