Definition: The z-transform X(z) of a causal discrete – time signal x(n) is defined as
X(z)=∑n=0∞x(n)z−nX(z)=∑n=0∞x(n)z−n size 12{X \( z \) = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { size 11{x \( n \) }} z rSup { size 8{ - n} } } {}
(1)
z is a complex variable of the transform domain and can be considered as the complex frequency. Remember index n can be time or space or some other thing, but is usually taken as time. As defined above ,
X(z)X(z) size 12{X \( z \) } {} is an integer power series of
z−1z−1 size 12{z rSup { size 8{ - 1} } } {} with corresponding
x(n)x(n) size 12{x \( n \) } {} as coefficients. Let’s expand
X(z)X(z) size 12{X \( z \) } {}:
X(z)=∑n=−∞∞x(n)z−n=x(0)+x(1)z−1+x(2)z−2+...X(z)=∑n=−∞∞x(n)z−n=x(0)+x(1)z−1+x(2)z−2+... size 12{X \( z \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x \( n \) z rSup { size 8{ - n} } =x \( 0 \) } +x \( 1 \) z rSup { size 8{ - 1} } +x \( 2 \) z rSup { size 8{ - 2} } + "." "." "." } {}
(2)
In general we write
X(z)=Z[x(n)]X(z)=Z[x(n)] size 12{X \( z \) =Z \[ x \( n \) \] } {}
(3)
In
Equation 1 the summation is taken from
n=0n=0 size 12{n=0} {} to
∞∞ size 12{ infinity } {}, ie ,
X(z)X(z) size 12{X \( z \) } {} is not at all related to the past history of
x(n)x(n) size 12{x \( n \) } {} . This is
one–sided or
unilateral z-transform . Sometime the one–sided z-transform has to take into account the initial conditions of
x(n)x(n) size 12{x \( n \) } {} (see section 4.7).
In general , signals exist at all time , and the two-sided or bilateral z–transform is defined as
H(z)=∑n=−∞∞h(n)z−n=...x(−2)z2+x(−1)z+x(0)+x(1)z−1+x(2)z−2+...H(z)=∑n=−∞∞h(n)z−n=...x(−2)z2+x(−1)z+x(0)+x(1)z−1+x(2)z−2+...alignl { stack {
size 12{H \( z \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h \( n \) z rSup { size 8{ - n} } } } {} #
matrix {
{} # {} # {}
} = "." "." "." x \( - 2 \) z rSup { size 8{2} } +x \( - 1 \) z+x \( 0 \) +x \( 1 \) z rSup { size 8{ - 1} } +x \( 2 \) z rSup { size 8{ - 2} } + "." "." "." {}
} } {}
(4)
Because
X(z)X(z) size 12{X \( z \) } {} is an infinite power series of
z−1z−1 size 12{z rSup { size 8{ - 1} } } {} , the transform only exists at values where the series converges (i.e. goes to zero as
n→∞n→∞ size 12{n rightarrow infinity } {} or -
∞∞ size 12{ infinity } {}). Thus the z-transform is accompanied with its region of convergence (ROC) where it is finite (see section 4.4).
A number of authors denote
X+(z)X+(z) size 12{X rSup { size 8{+{}} } \( z \) } {} for one-side z-transform.
Example 1 Find the z–transform of the two signals of
Figure 1
Solution
(a) Notice the signal is causal and monotically decreasing and its value is just
0.8n0.8n size 12{0 "." 8 rSup { size 8{n} } } {}for
n≥0n≥0 size 12{n >= 0} {}. So we write
x
(
n
)
=
0
.
8
n
u
(
n
)
x
(
n
)
=
0
.
8
n
u
(
n
)
size 12{x \( n \) =0 "." 8 rSup { size 8{n} } u \( n \) } {}
and use the transform (4.1)
X
(
z
)
=
∑
n
=
0
∞
x
(
n
)
z
−
n
=
1
+
0
.
8z
−
1
+
0
.
64
z
−
2
+
0
.
512
z
−
3
+
.
.
.
=
1
+
(
0
.
8z
−
1
)
+
(
0
.
8z
−
1
)
2
+
(
0
.
8z
−
1
)
3
+
.
.
.
X
(
z
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=
∑
n
=
0
∞
x
(
n
)
z
−
n
=
1
+
0
.
8z
−
1
+
0
.
64
z
−
2
+
0
.
512
z
−
3
+
.
.
.
=
1
+
(
0
.
8z
−
1
)
+
(
0
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8z
−
1
)
2
+
(
0
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8z
−
1
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3
+
.
.
.
alignl { stack {
size 12{X \( z \) = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {x \( n \) z rSup { size 8{ - n} } } } {} #
matrix {
{} # {} # {}
} =1+0 "." 8z rSup { size 8{ - 1} } +0 "." "64"z rSup { size 8{ - 2} } +0 "." "512"z rSup { size 8{ - 3} } + "." "." "." {} #
matrix {
{} # {} # {}
} =1+ \( 0 "." 8z rSup { size 8{ - 1} } \) + \( 0 "." 8z rSup { size 8{ - 1} } \) rSup { size 8{2} } + \( 0 "." 8z rSup { size 8{ - 1} } \) rSup { size 8{3} } + "." "." "." {}
} } {}
Applying the formula of infinite geometric series which is repeated here
1+a+a2+a3+...=∑n=0∞an=11−a∣a∣<11+a+a2+a3+...=∑n=0∞an=11−a∣a∣<1 size 12{1+a+a rSup { size 8{2} } +a rSup { size 8{3} } + "." "." "." = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {a rSup { size 8{n} } } = { {1} over {1 - a} } matrix {
{} # {} # {}
} \lline a \lline <1} {}
(5)
to obtain
X
(
z
)
=
1
1
−
0
.
8z
−
1
=
z
z
−
0
.
8
X
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z
)
=
1
1
−
0
.
8z
−
1
=
z
z
−
0
.
8
size 12{X \( z \) = { {1} over {1 - 0 "." 8z rSup { size 8{ - 1} } } } = { {z} over {z - 0 "." 8} } } {}
The result can be left in either of the two forms .
(b) The signal is alternatively positive and negative with increasing value .The signal is divergent . We can put the signal in the form
x
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n
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=
(
−
1
.
2
)
n
−
1
u
(
n
−
1
)
x
(
n
)
=
(
−
1
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2
)
n
−
1
u
(
n
−
1
)
size 12{x \( n \) = \( - 1 "." 2 \) rSup { size 8{n - 1} } u \( n - 1 \) } {}
which is
(−1.2)nu(n)(−1.2)nu(n) size 12{ \( - 1 "." 2 \) rSup { size 8{n} } u \( n \) } {} delayed one index(sample) . Let’s use the transform (4.1)
X
(
z
)
=
∑
n
=
0
∞
x
(
n
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z
−
n
=
0
+
1
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0
(
z
−
1
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−
1
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2
(
z
−
1
)
2
+
1
.
44
(
z
−
1
)
3
−
1
.
718
(
z
−
1
)
4
+
.
.
.
=
z
−
1
[
1
+
(
−
1
.
2z
−
1
)
+
(
−
1
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2z
−
1
)
2
+
(
−
1
.
2z
−
1
)
3
+
.
.
.
]
=
z
−
1
1
1
+
1
.
2z
−
1
=
z
−
1
1
+
1
.
2z
−
1
=
1
z
+
1
.
2
X
(
z
)
=
∑
n
=
0
∞
x
(
n
)
z
−
n
=
0
+
1
.
0
(
z
−
1
)
−
1
.
2
(
z
−
1
)
2
+
1
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44
(
z
−
1
)
3
−
1
.
718
(
z
−
1
)
4
+
.
.
.
=
z
−
1
[
1
+
(
−
1
.
2z
−
1
)
+
(
−
1
.
2z
−
1
)
2
+
(
−
1
.
2z
−
1
)
3
+
.
.
.
]
=
z
−
1
1
1
+
1
.
2z
−
1
=
z
−
1
1
+
1
.
2z
−
1
=
1
z
+
1
.
2
alignl { stack {
size 12{X \( z \) = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {x \( n \) z rSup { size 8{ - n} } } } {} #
matrix {
{} # {} # {}
} =0+1 "." 0 \( z rSup { size 8{ - 1} } \) - 1 "." 2 \( z rSup { size 8{ - 1} } \) rSup { size 8{2} } +1 "." "44" \( z rSup { size 8{ - 1} } \) rSup { size 8{3} } - 1 "." "718" \( z rSup { size 8{ - 1} } \) rSup { size 8{4} } + "." "." "." {} #
matrix {
{} # {} # {}
} =z rSup { size 8{ - 1} } \[ 1+ \( - 1 "." 2z rSup { size 8{ - 1} } \) + \( - 1 "." 2z rSup { size 8{ - 1} } \) rSup { size 8{2} } + \( - 1 "." 2z rSup { size 8{ - 1} } \) rSup { size 8{3} } + "." "." "." \] {} #
matrix {
{} # {} # {}
} =z rSup { size 8{ - 1} } { {1} over {1+1 "." 2z rSup { size 8{ - 1} } } } = { {z rSup { size 8{ - 1} } } over {1+1 "." 2z rSup { size 8{ - 1} } } } = { {1} over {z+1 "." 2} } {}
} } {}
The inverse z-transform
The inverse z-transform is denoted by
Z−1Z−1 size 12{Z rSup { size 8{ - 1} } } {}:
x(n)=Z−1[X(z)]x(n)=Z−1[X(z)] size 12{x \( n \) =Z rSup { size 8{ - 1} } \[ X \( z \) \] } {}
(6)
The signal
x(n)x(n) size 12{x \( n \) } {}and its transform constitutes a transform pair
x(n)→Z(z)x(n)→Z(z) size 12{x \( n \) widevec {} Z \( z \) } {}
(7)
One way to find the inverse transform , whenever possible , is to utilize just the z-transform definition. General methods of the inverse z-transform are discursed in section 4.5 and 4.6
Example 2 Find the inverse z-transform of the following
-
X
(
z
)
=
z
z
−
0
.
8
X
(
z
)
=
z
z
−
0
.
8
size 12{X \( z \) = { {z} over {z - 0 "." 8} } } {}
-
X
(
z
)
=
1
z
+
1
.
2
X
(
z
)
=
1
z
+
1
.
2
size 12{X \( z \) = { {1} over {z+1 "." 2} } } {}
Solution
(a) Let’s write
X
(
z
)
=
z
z-
0
.
8
=
1
1
−
0
.
8z
−
1
=
1
+
(
0
.
8z
−
1
)
+
(
0
.
8z
−
1
)
2
+
(
0
.
8z
−
1
)
3
+
.
.
.
=
1
+
0
.
8z
−
1
+
0
.
64
z
−
2
+
0
.
512
z
−
3
+
.
.
.
X
(
z
)
=
z
z-
0
.
8
=
1
1
−
0
.
8z
−
1
