δ
[
n
]
=
u
[
n
]
-
u
[
n
-
1
]
size 12{δ \[ n \] " ="u \[ n \]"- "u \[ n "- "1 \] } {}
i.e., the unit sample is the first difference of the unit step function.
u
[
n
]
=
∑
n
=
−
∞
∞
δ
[
m
]
size 12{u \[ n \] = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \[ m \]} } {}
i.e., the unit step is a running sum of the unit sample.
The Z-transform of the unit step is
Z
{
u
[
n
]
}
=
∑
n
=
−
∞
∞
u
[
n
]
z
−
n
=
∑
n
=
0
∞
z
−
n
=
∑
n
=
0
∞
z
−
1
n
size 12{Z lbrace u \[ n \] rbrace = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {u \[ n \]z rSup { size 8{ - n} } ={}} Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {z rSup { size 8{ - n} } ={}} Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { left [z rSup { size 8{ - 1} } right ] rSup { size 8{n} } } } {}
This is a sum of a geometric series of the form
∑
n
=
0
∞
α
n
=
1
1
−
α
for
∣
α
∣
<
1
size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {α rSup { size 8{n} } } = { {1} over {1 - α} } matrix {
{} # {}} ital "for" matrix {
{} # {}} \lline α \lline<1} {}
Therefore, the geometric series for Z{u[n]} converges provided that |z−1|<1 or |z|>1. Therefore,
Z
{
u
[
n
]
}
=
1
1
−
z
−
1
=
z
z
−
1
for
∣
z
>
1
∣
size 12{Z lbrace u \[ n \] rbrace = { {1} over {1 - z rSup { size 8{ - 1} } } } = { {z} over {z - 1} } matrix {{} # {}
} ital "for"matrix {
{} # {}} \lline z>1 \lline } {}
The region of convergence is outside a circle of radius one called the unit circle.
On the unit circle,
z
=
e
j
Ω
size 12{"z "=" e" rSup { size 8{j %OMEGA } } } {} . As we shall see, the unit circle in the z-plane plays a role analogous to the jω-axis in the s-plane.
3/ Causal exponential DT time function
If
x
[
n
]
=
a
n
u
[
n
]
size 12{x \[ n \] =" a" rSup { size 8{n} } u \[ n \]} {} then the Z-transform is
X
~
(
z
)
=
1
1
−
az
−
1
for
∣
z
∣
>
∣
a
∣
size 12{ {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix {
{} # {}} ital "for" matrix {
{} # {}} \lline z \lline>\lline a \lline } {}
The sum converges provided |az−1|<1 or |z|>|a| so that
X
~
(
z
)
=
∑
n
=
−
∞
∞
a
n
z
−
n
u
[
n
]
=
∑
n
=
0
∞
az
−
1
n
=
1
1
−
az
−
1
size 12{ {X} cSup { size 8{ "~" } } \( z \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {a rSup { size 8{n} } z rSup { size 8{ - n} } u \[ n \] } = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { left [ ital "az" rSup { size 8{ - 1} } right ]rSup { size 8{n} } } = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } } {}
The region of convergence (ROC) of
X
~
(
z
)
size 12{ {X} cSup { size 8{ "~" } } \( z \) } {} is |z|>|a|, i.e., outside a circle of radius a.
Thus we have
x
[
n
]
=
a
n
u
[
n
]
⇔
Z
X
~
(
z
)
=
1
1
−
az
−
1
for
∣
z
∣
>
∣
a
∣
size 12{x \[ n \] =a rSup { size 8{n} } u \[ n \]{ dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix {{} # {}
} ital "for" matrix {{} # {}
} \lline z \lline>\lline a \lline } {}
What happens if a<0?
x
[
n
]
=
a
n
u
[
n
]
⇔
Z
X
~
(
z
)
=
1
1
−
az
−
1
for
∣
z
∣
>
∣
a
∣
size 12{x \[ n \] =a rSup { size 8{n} } u \[ n \]{ dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix {
{} # {}} ital "for" matrix {
{} # {}} \lline z \lline>\lline a \lline } {}
4/ Relation between causal DT time functions and pole-zero diagrams
Demo of relation of pole-zero diagram to time function.
Conclusions on relation between causal DT time functions and pole-zero diagrams
5/ Anti-causal exponential DT time function
If
x
[
n
]
=
-a
n
u
[
-n
-
1
]
size 12{x \[ n \] =" -a" rSup { size 8{n} } u \[ "-n" "- "1 \]} {} then the Z-transform is
X
~
(
z
)
=
1
1
−
az
−
1
for
∣
z
∣
<
∣
a
∣
size 12{ {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix {
{} # {}} ital "for" matrix {
{} # {}} \lline z \lline<\lline a \lline } {}
The sum converges provided
∣
a
−
1
z
∣
<
1
size 12{ \lline a rSup { size 8{ - 1} } z \lline<1} {} or |z|<|a| so that
X
~
(
z
)
=
−
∑
n
=
−
∞
∞
a
n
u
[
−
n
−
1
]
z
−
n
=
−
∑
n
=
−
∞
−
1
az
−
1
n
=
−
∑
n
=
1
∞
a
−
1
z
n
=
−
a
−
1
z
1
−
a
−
1
z
=
1
1
−
a
−
1
z
alignl { stack {
size 12{ {X} cSup { size 8{ "~" } } \( z \) = - Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {a rSup { size 8{n} } u \[ - n - 1 \]z rSup { size 8{ - n} } } = - Sum cSub { size 8{n= - infinity } } cSup { size 8{ - 1} } { left [ ital "az" rSup { size 8{ - 1} } right ] rSup { size 8{n} } } } {} #matrix {
{} # {}} = - Sum cSub { size 8{n=1} } cSup { size 8{ infinity } } { left [a rSup { size 8{ - 1} } z right ] rSup { size 8{n} } } = - { {a rSup { size 8{ - 1} } z} over {1 - a rSup { size 8{ - 1} } z} } = { {1} over {1 - a rSup { size 8{ - 1} } z} } {}} } {}
The region of convergence (ROC) of
X
~
(
z
)
size 12{ {X} cSup { size 8{ "~" } } \( z \) } {} is |z|<|a|, i.e., inside a circle of radius a.
Thus we have
x
[
n
]
=
−
a
n
u
[
−
n
−
1
]
⇔
Z
X
~
(
z
)
=
1
1
−
az
−
1
for
∣
z
∣
<
∣
a
∣
size 12{x \[ n \] = - a rSup { size 8{n} } u \[ - n - 1 \]{ dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix {
{} # {}} ital "for" matrix {
{} # {}} \lline z \lline<\lline a \lline } {}
6/ Summary of causal and anti-causal geometric sequences
Causal and anti-causal sequences can have the same
X
~
(
z
)
size 12{ {X} cSup { size 8{ "~" } } \( z \) } {} formula but different ROCs
x
[
n
]
=
a
n
u
[
n
]
⇔
Z
X
~
(
z
)
=
1
1
−
az
−
1
for
∣
z
∣
>
∣
a
∣
x
[
n
]
=
−
a
n
u
[
−
n
−
1
]
⇔
Z
X
~
(
z
)
=
1
1
−
az
−
1
for
∣
z
∣
<
∣
a
∣
alignl { stack {
size 12{x \[ n \]=a rSup { size 8{n} } u \[ n \] { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix {{} # {}
} ital "for" matrix {{} # {}
} \lline z \lline>\lline a \lline } {} #
x \[ n \]= - a rSup { size 8{n} } u \[ - n - 1 \] { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix {{} # {}
} ital "for" matrix {{} # {}
} \lline z \lline<\lline a \lline {}
} } {}
Causal sequences have ROCs outside a circle that intersects the pole.
Anti-causal sequences have ROCs inside a circle that intersects the pole.
Bounded sequences have ROCs that include the unit circle.
Unbounded sequences have ROCs that do not include the unit circle.
7/ Two-sided DT time function
Next we consider the two-sided sequence
x
[
n
]
=
a
∣
n
∣
=
a
−
n
u
[
−
n
−
1
]
+
a
n
u
[
n
]
size 12{x \[ n \] =a rSup { size 8{ \lline n \lline } } =a rSup { size 8{ - n} } u \[ - n - 1 \]+a rSup { size 8{n} } u \[ n \] } {}
where we assume 0<a<1. We use previous results to obtain the Z-transform
X
~
(
z
)
=
1
1
−
a
−
1
z
−
1
+
1
1
−
az
−
1
size 12{ {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - a rSup { size 8{ - 1} } z rSup { size 8{ - 1} } } } + { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } } {}
Therefore, we have
X
~
(
z
)
=
1
1
−
a
−
1
z
−
1
+
1
1
−
az
−
1
for
a
<
∣
z
∣
<
1
/
a
size 12{ {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - a rSup { size 8{ - 1} } z rSup { size 8{ - 1} } } } + { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix {
{} # {}} ital "for" matrix {
{} # {}} a<\lline z \lline<1/a} {}
which can be written as
X
~
(
z
)
=
z
−
1
(
a
−
a
−
1
)
(
1
−
a
−
1
z
−
1
)
(
1
−
az
−
1
)
=
z
(
a
−
a
−
1
)
(
z
−
a
−
1
)
(
z
−
a
)
x
[
n
]
=
a
∣
n
∣
⇔
Z
X
~
(
z
)
=
z
(
a
−
a
−
1
)
(
z
−
a
−
1
)
(
z
−
a
)
alignl { stack {
size 12{ {X} cSup { size 8{ "~" } } \( z \) = { {z rSup { size 8{ - 1} } \( a - a rSup { size 8{ - 1} } \) } over { \( 1 - a rSup { size 8{ - 1} } z rSup { size 8{ - 1} } \) \( 1 - ital "az" rSup { size 8{ - 1} } \) } } = { {z \( a - a rSup { size 8{ - 1} } \) } over { \( z - a rSup { size 8{ - 1} } \) \( z - a \) } } } {} #x \[ n \] =a rSup { size 8{ \lline n \lline } } { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {z \( a - a rSup { size 8{ - 1} } \) } over { \( z - a rSup { size 8{ - 1} } \) \( z - a \) } } {}} } {}
8/ Conclusions on relation between DT time functions and pole-zero diagrams
More rapidly changing exponential time functions, both growing and decaying, have poles that lie further from the point z = 1 in the z-plane.
Two-minute miniquiz problem
Problem 11-1
Find the Z-transform including the ROC for
x
[
n
]
=
(
0
.
5
)
n
−
4
u
[
n
−
4
]
size 12{x \[ n \] = \( 0 "." 5 \) rSup { size 8{n - 4} } u \[ n - 4 \]} {}
Solution
We use the Z-transform of the causal exponential time function and time delay property to solve this problem.
(
0
.
5
)
n
u
[
n
]
⇔
Z
1
1
−
0
.
5z
−
1
for
∣
z
∣
>
0
.
5
(
0
.
5
)
n
−
4
u
[
n
−
4
]
⇔
Z
z
−
4
1
−
0
.
5z
−
1
for
∣
z
∣
>
0
.
5
alignl { stack {
size 12{ \( 0 "." 5 \) rSup { size 8{n} } u \[ n \]{ matrix {
{} # {}} { dlrarrow } cSup { size 8{Z} } } cSup {} matrix {
{} # {}} { {1} over {1 - 0 "." 5z rSup { size 8{ - 1} } } } matrix {
{} # {}} ital "for" matrix {
{} # {}} \lline z \lline>0 "." 5} {} #
\( 0 "." 5 \) rSup { size 8{n - 4} } u \[ n - 4 \]{ matrix {
{} # {}} { dlrarrow } cSup { size 8{Z} } } cSup {} matrix {
{} # {}} { {z rSup { size 8{ - 4} } } over {1 - 0 "." 5z rSup { size 8{ - 1} } } } matrix {
{} # {}} ital "for" matrix {
{} # {}} \lline z \lline>0 "." 5 {}
} } {}
Note that, the delayed sequence has the same ROC as the original sequence.
III. CONCLUSIONS
The Z-transform is capable of representing a rich class of DT time functions. Z-transform pairs can be obtained by combining
Z-transform properties.
the Z-transforms of elementary time functions.
Exercises .
Solutions of Exercises.