Digital Signal Processing Reference
In-Depth Information
2.6.6 OFF-UNIT-CIRCLE ZEROS AND DECAYING SIGNALS
When transfer function zeros are off the unit circle in the
z
-plane, their effect on the transfer function
(for steady-state complex exponentials) is diminished. Such “off-unit-circle zeros” do not drive the output
of the system to zero for a steady-state, unity-magnitude complex exponential, but rather for one which
has a decaying (for zeros inside the unit circle) or growing (for zeros outside the unit circle) magnitude
with time.
Example 2.42.
Determine, for an FIR having the following transfer function
0
.
9
z
−
1
1
−
(2.22)
what input signal will result in a zero-valued output signal.
We can set expression (2.22) to zero, i.e.,
0
.
9
z
−
1
1
−
=
0
(2.23)
and solving for
z
we get
z
= 0.9.
To see this in the time domain, consider the decaying exponential series
x
[
n
]
= 1, 0.9, 0.81, ..., i.e.,
0
.
9
n
x
[
n
]=
where 0
. Let's compute the first few terms of the response of the single zero system whose
z
-transform is shown at (2.22) and whose difference equation is
≤
n
≤∞
[
]=
[
]−
[
−
]
(2.24)
Since Eq. (2.24) represents an FIR, the impulse response is identical to the FIR coefficients, which
are [1, -0.9]. We can then convolve
x
y
n
x
n
0
.
9
x
n
1
[
]
n
with the system impulse response to obtain the system output
y
[
n
]
:
n = 0:1:20; x = 0.9.ˆn;
y = conv([1, -0.9],x)
stem(y)
Note that the first and last samples of the above computation are not zero-why?
Example 2.43.
Compute the output for an FIR with one zero at 0.9 and a constant amplitude signal
having the same frequency as that of the zero, i.e., DC.
We can use the same call as above, but with the signal amplitude constant:
n = 0:1:20; x = 1.ˆn;
y = conv([1, -0.9],x)
stem(y)
or, alternatively,
stem(filter([1,-0.9],[1],[1.ˆ(0:1:50)]))
The output sequence is 1 followed by the steady-state value of 0.1 and the final transient (non-
steady-state) value as the end of
x
[
n
]
is reached.
