the inverse z-Transform
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Transcript the inverse z-Transform
The Inverse
z-Transform
The inverse z transformation can be performed by
using a few results from complex variable theory.
Suppose that we integrate one (1) from z=0 to z=1:
1
0
1
(1)dz z 1 0 1.
0
The integrals become more involved when the path
of integration is in the complex plane.
For example, suppose that we wish to integrate
around a unit circle in the (complex) z-plane. To do
the integration we make a substitution of variables:
z=ejq. We have dz = j ejq dq.
Im{z}
q p/2
qp
ze
q
radius = 1
q 3p/2
q0
jq
Re{z}
Suppose, first, that we wish integrate one (1) around
a semicircle C around the origin from 0 to p.
Im{z}
z e jq
C
qp
q
q0
Re{z}
p
(
1
)
dz
je
d
q
jq
0
C
p
j cosq j sin q dq
0
j 0 j 2 2.
Suppose that we integrate one (1) around a closed
unit circle C around the origin (0 to 2p).
Im{z}
C
z e jq
q
q0
q 2p
Re{z}
We have,
(
1
)
dz
2p
0
jq
je dq 0.
C
(The areas under the sine and cosine sum to zero
over a full period.)
Suppose that we integrate z around a closed unit
circle C around the origin (0 to 2p).
2p
z
dz
e
je
d
q
jq
jq
0
C
2p
j e
0
2 jq
dq 0.
Similarly, if we integrate z2 , z3 or zk (k 0) around
the unit circle we get zero.
What about integrating z-1?
2p
z
dz
e
1
0
jq
jq
je dq
C
j
2p
0
dq 2p j.
We do not get zero when we integrate z-1 around
the unit circle.
What about integrating z-2?
2p
z
dz
e
2
j 2q
0
jq
je dq
C
2p
j e
0
jq
dq 0.
Similarly, if we integrate z-3 , z-4 or z-k (k>1) around
the unit circle we get zero.
So integrating z-1 is unique, and
z
dz
2
p
j
.
1
C
So if we were to integrate a z-transform
X ( z)
n
x
[
n
]
z
,
n
we get
X ( z)dz
C
n
x[n] z n dz 2p jx[1].
C
(Only z-1 does not integrate to zero.)
In other words the integration “picks-out” the sample
x1. We can “pick-out” another sample by adjusting
the power of z:
X ( z) zdz
C
n
x[n] z n1 2p jx[2].
C
(The integral is zero except when –n+1 = -1, or n=2.)
Similarly,
2
X
(
z
)
z
dz
C
n
xk z n2 2p jx[3],
C
and
m1
X
(
z
)
z
dz
C
n
xk z nm1 2p jx[m].
C
Thus, we have found an expression for the inverse
z-transform:
x[n]
1
X
(
z
)
z
dz
.
n 1
2p j C
The path of the integral must be carried-out over a
region of convergence of X(z). [CROC.]
Example: Find the inverse z-transform of X(z)=1
using the complex inversion integral.
Solution: Plugging-in X(z)=1, we have
x[n]
1
z
2p j
C
n 1
dz.
We know that the integral is equal to zero except
when n=0. When n=0, the integral is equal to 2pj.
Thus, the final inverse transform is equal to
x[n]
1
2p j
2p j 1
when n=0. Hence we have x[n]=d[n].
Example: Find the inverse z-transform of X(z)=z-k
using the complex inversion integral.
Solution: Plugging-in X(z)=z-k, we have
x[n]
1
z
2p j
C
k
n 1
z dz.
We know that the integral is equal to zero except
when n=0. When n-k=0, or when n=k. Hence we
have x[n]=d[n-k].
Example: Find the inverse z-transform of
1
X ( z)
1
1 z
using the complex inversion integral.
Solution: Let us expand 1/(1-z-1) using the
geometric series:
1
1
2
1 z z
1
1 z
Solution: Plugging this into the complex inversion
integral we have
x[n]
1 z
2p j
1
1
2
n 1
z z dz.
C
For n=0, the only term in 1+z-1+z-2+… which
“survives” a non-zero integration is 1.
For n=1, the only term in 1+z-1+z-2 which “survives”
is z-1.
We see that there is exactly one term that “survives”
when n0. The resultant integral in each case is 2pj,
and the corresponding value for x[n] in each case is
one. Thus,
x[n] u[n].
Example: Find the inverse z-transform of
1
X ( z)
1 2
(1 z )
using the complex inversion integral.
Solution: Using the expansion for 1/(1-z-1), we have
2
2
1
1
2
1
z
z
1
1 z
1
2
1 2 z 3z
Solution: Plugging this into the complex inversion
integral we have
x[n]
1 2z
2p j
1
1
2
n 1
3z z dz.
C
For n=0, the only term in 1+z-1+z-2+… which
“survives” a non-zero integration is 1.
For n=1, the only term in 1+z-1+z-2 which “survives”
is 2z-1.
The resultant inverse z-transform is
x[n] (n 1)u[n 1] (n 1)u[n].
Example: Find the inverse z-transform of
1
X ( z)
1
1 az
using the complex inversion integral.
Solution: Expanding 1/(1-az-1) as a geometric
series:
1
1
2 2
1 az a z
1
1 az
Plugging this into the complex inversion integral we
have
x[n]
1 az
2p j
1
1
2 2
n 1
a z z dz.
C
For n=0, the only term in 1+az-1+a2z-2+… which
“survives” a non-zero integration is 1.
For n=1, the only term in 1+az-1+a2z-2 which
“survives” is az-1.
The resultant integral in the first case is 2pj, the
resultant integral in the second case is 2pja, etc.
The inverse z-transform is
x[n] a nu[n].
Example: Find the inverse z-transform of
1
1
X ( z)
1
1
1 az 1 bz
using the complex inversion integral.
Solution: We expand both 1/(1-az-1) and 1/(1-bz-1):
1
1
2 2
1 az a z
1
1 az
1
1
2 2
1 bz b z
1
1 bz
The product is
1
1
1
1
1 az 1 bz
1
2 2
1
2 2
1 az a z 1 bz b z
1
2
1 (a b) z (a ab b ) z
2
2
Plugging this into the complex inversion integral, we
wind-up with
1
a b
x[n] 2
2
a
ab
b
...
(n 0),
( n 1),
(n 2),
This same problem can be solved in a slightly
different manner:
1
X ( z)
1
1
(1 az )(1 bz )
2
z
( z a )(z b)
z
z
( z a )(z b)
B
A
z
z
a
z
b
A
B
,
1
1
1 az
1 bz
where
z
1
A
1
z b z a 1 bz
,
z a
and
z
B
za
z b
1
1
1 az
.
z b
The inverse z-transform is
x[n] Aa Bb u[n].
n
n
It can be shown algebraically (exercise) that this
expression is equivalent to the previous expression
for x[n].
In general, to find the inverse z-transform of
1
1
1
X ( z)
,
1
1
1
1 az 1 bz 1 cz
we perform a partial fraction expansion:
A
B
C
X ( z)
,
1
1
1
1 az
1 bz
1 cz
where
1
A (1 az ) X ( z )
z a
1
B (1 bz ) X ( z )
z b
1
C (1 cz ) X ( z )
etc.
z c
,
,
,
The resultant inverse z-transform is
x[n] Aan Bbn Cc n u[n].
Example: Find the inverse z-transform of
z 1
X ( z) e .
using the complex inversion integral.
Solution: We take the MacLaurin series for ez-1
e
z 1
1
2
3
1 z z z .
1
2
1
3!
The inverse z-transform follows rather naturally:
1
x[n] .
n!
We could obtain this result through the complex
inversion integral, or we could recognize that
x[n]=1/n! corresponds to the coefficients of the ztransform.