Digital Signal Processing Reference
In-Depth Information
Multiplication By a Ramp
z
dX(z)
dz
Z(nx
[
n
]
)
=−
;
ROC: ROC(
x
[
n
]
)
Example 2.12.
Determine the
z
-transform of the sequence
nu
[
n
]
.
We get initially
z
d
1
z
d
z
−
1
]
−
1
)
X(z)
=−
dz
(
z
−
1
)
=−
dz
(
[
1
−
1
−
z
−
1
0
z
0
+
z
−
1
−
z
−
1
)
−
2
(z
−
2
))
=
=−
z(
−
(
1
=
+
z
−
2
We can verify this using the following code, which yields the sequence
nu
(
1
−
z
−
1
)
2
1
−
2
z
−
1
[
n
]
i.e., [0,1,2...].
y = filter([0 1],[1,-2,1],[1,zeros(1,20)])
2.3.7 COMMON Z-TRANSFORMS
Sequence
z
-Transform
ROC
δ
[
n
]
1
∀
z
z
−
1
)
u
[
n
]
1
/(
1
−
|
z
|
>
1
z
−
1
)
−
u
[−
n
−
1
]
1
/(
1
−
|
z
|
<
1
a
n
u
[
n
]
−
az
−
1
)
1
/(
1
|
z
|
>
|
a
|
b
n
u
bz
−
1
)
−
[−
−
]
−
|
|
|
|
n
1
1
/(
1
z
<
b
(a
sin
ω
o
)z
−
1
[
a
n
sin
ω
o
n
]
u
[
n
]
|
z
|
>
|
a
|
(
2
a
cos
ω
o
)z
−
1
a
2
z
−
2
1
−
+
(a
cos
ω
o
)z
−
1
[
a
n
cos
ω
o
n
]
u
[
n
]
|
z
|
>
|
a
|
(
2
a
cos
ω
o
)z
−
1
a
2
z
−
2
1
−
+
az
−
1
(
1
−
az
−
1
)
2
na
n
u
[
n
]
|
z
|
>
|
a
|
bz
−
1
nb
n
u
−
[−
−
]
|
z
|
<
|
b
|
n
1
bz
−
1
)
2
(
1
−
Example 2.13.
with a
single-pole filter having its pole at 0
.
9. Use the representation to determine the time domain convolution.
Represent in the
z
-domain the time domain convolution of the sequence
[
1,-0
.
9
]
The convolution in the time domain of the two sequences can be equivalently achieved in the
z
-domain by multiplying the
z
-transforms of each time domain sequence, i.e., in this case
0
.
9
z
−
1
)
(
1
−
0
.
9
z
−
1
)
=
1
−
(
1
Looking at the table of common
z
-transforms, we see that the unit impulse is the time domain
signal that has as its
z
-transform the value 1. This can be confirmed computationally (for the first 20
samples) using the following code:
