In going from
∑
k
=
0
N
a
k
y
[
n
-
k
]
=
∑
k
=
0
m
b
k
x
[
n
-
k
]
∑
k
=
0
N
a
k
y
[
n
-
k
]
=
∑
k
=
0
m
b
k
x
[
n
-
k
]
(1)to
H
(
z
)
=
Y
(
z
)
X
(
z
)
H
(
z
)
=
Y
(
z
)
X
(
z
)
(2)we did not specify an ROC. If we factor H(z)H(z), we can plot the poles and zeros in the zz-plane as below.
Several ROCs may be possible. Each ROC corresponds to a different impulse
response, so which one should we choose? In general, there is no “right”
choice, however, there are some choices that make sense in practice.
In particular, if h[n]h[n] is causal, i.e., if h[n]=0h[n]=0, n<0n<0, then the ROC extends
outward from the outermost pole. This can be seen in the examples up to this point. Moreover, recall
that a system is BIBO stable if the impulse response h∈ℓ1(Z)h∈ℓ1(Z). In this case,
H
(
z
)
=
∑
n
=
-
∞
∞
h
[
n
]
z
-
n
≤
∑
n
=
-
∞
∞
h
[
n
]
z
-
n
H
(
z
)
=
∑
n
=
-
∞
∞
h
[
n
]
z
-
n
≤
∑
n
=
-
∞
∞
h
[
n
]
z
-
n
(3)
Consider the unit circle z=ejωz=ejω. In this case we have |z-n|=|e-jωn|=1|z-n|=|e-jωn|=1, so that
H
(
e
j
ω
)
≤
∑
n
=
-
∞
∞
h
[
n
]
<
∞
H
(
e
j
ω
)
≤
∑
n
=
-
∞
∞
h
[
n
]
<
∞
(4)for all ωω. Thus, if a system is BIBO stable, the ROC of H(z)H(z) must include the unit
circle. In general, any ROC containing the unit circle will be BIBO stable.
This leads to a key question – are stability and causality always compatible? The answer is no. For example,
consider
H
(
z
)
=
z
2
(
z
-
2
)
(
z
+
1
2
)
=
4
5
z
z
-
2
+
1
5
z
z
+
1
2
H
(
z
)
=
z
2
(
z
-
2
)
(
z
+
1
2
)
=
4
5
z
z
-
2
+
1
5
z
z
+
1
2
(5)and its various ROC's and corresponding inverses. If the ROC contains the unit-circle (so that the corresponding system is stable) and is not to contain any poles, then it must extend inward towards the origin, and hence it cannot be causal. Alternatively, if the ROC is to extend outward, it will not contain the unit-circle so that the corresponding system will not be BIBO stable.
![Graph with horizontal axis Re[z] and vertical axis Im[z]. There are three X marks and four circle marks on the graph.](15_1.png)
