5.4_Inverse Z Transform.pptx

U18ECT3101
SIGNALS AND SYSTEMS
Ms. A. Kalaiselvi, AP II
Department of ECE

The Z Transform
Part I
Inverse Z - Transform
17-Jan-23 U18ECT3101 - Signals and Systems The Inverse Z Transform 2

17-Jan-23 U18ECT3101 - Signals and Systems The Inverse Z-Transform 3
Outline
• Z transform
• Forward Transform
• Pole-zero plot
• Region of Convergence (RoC)
• Properties of Z transform
• Inverse Transform

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U18ECT3101 - Signals and Systems The Z Transform Properties
4
U18ECT3101 - Signals and Systems The Inverse Z Transform
The Inverse Z transform
• Given X(z), find the sequence x[n] that has X(z) as its z-transform.
• By Z- Transform pair x[n] is given by
• We need to specify both algebraic expression and ROC to make the inverse Z-transform unique.
• Techniques for finding the inverse z-transform:
 Inspection Method
 Power series Method
 Residue Method,
 Long Division Method
 Partial Fraction methods
𝒙 𝒏 =
𝟏
𝟐𝝅𝒋 𝑪
𝒛𝒏−𝟏𝑿 𝒛 𝒅𝒛

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5
5.4.1 Inspection Method
In order to find the inverse z transform we compare 𝑋(𝑧) to one of the standard transform pairs
Eg. Find the inverse z-transform of
From the transform pair
x[n] = 0.5nu[n].
  1
5
0
1
1



z
z
X
.

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6
5.4.2 Power Series Expansion
𝑋 𝑧 =
𝑛=−∞
∞
𝑥 𝑛 𝑧−𝑛
By Definitions The z-Transform of a signal x[n] is given by
𝑋(𝑧) =. . . +𝑥 −2 𝑧2
+ 𝑥 −1 𝑧 + 𝑥 0 + 𝑥 1 𝑧−1
+ 𝑥 2 𝑧−2
+. . .
Expanding the above equation
Any particular value of the sequence is determined by finding the coefficient of the appropriate power of 𝑧−1
This method is used .
• For Finite length sequence
• X(z) expressed as polynomial in positive and negative powersof z.

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Example: 1
 Find the inverse z-transform of
Soln:
By direct expansion of X(z), we have
Thus,
𝑋 𝑧 = 𝑧2
1 − 0.5𝑧−1
1 + 𝑧−1
1 − 𝑧−1
𝑋 𝑧 = 𝑧2
− 0.5𝑧 − 1 + 0.5𝑧−1
𝑥 𝑛 = 𝛿 𝑛 + 2 − 0.5𝛿 𝑛 + 1 − 𝛿 𝑛 + 0.5𝛿 𝑛 − 1
𝑥 𝑛 =
1 𝑛 = −2
−0.5
−1
0.5
𝑛 = −1
𝑛 = 0
𝑛 = 1
Or

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8
Example: 2
Find the inverse z-transform of
Using the power series expansion for log(1+x) with |x|<1, we obtain
Thus
    a
z
az
z
X 

 
1 1
log
   







1
1
1
n
n
n
n
n
z
a
z
X
   










0
0
1
1
1
n
n
n
a
n
x
n
n
/

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5.4.3 Inverse Z-Transform of Rational Functions
If X(z) is expressed as a ratio of two polynomials, then X(z) is a rational function of z , where
𝑋 𝑧 =
𝑁 𝑧
𝐷 𝑧
In such case, the inverse z-transform is obtained by
 Long Division Method (Power series)
 Partial Fraction Expansion
 Residue Method
Note:
• For the above method , the ROC must be specified.
• Roc decides whether the given signal is causal, non causal or two sided

5.4.3.2 Inverse Z-Transform by Long Division
• Consider:
Solution:
4
2
1
)
( 3
2




z
z
z
z
X
1
1
1
2
2
3
4
3
4
2
1
4
2










z
z
z
z
z
z
z
3
2
1
3
2
1
3
2
1
1
2
2
3
3
0
12
6
4
12
6
3
4
3
4
2
1
4
2

























z
z
z
z
z
z
z
z
z
z
z
z
z
z
4
3
2
1
4
3
2
4
3
1
3
2
1
3
2
1
1
2
2
3
4
3
0
16
20
6
16
8
4
12
6
4
12
6
3
4
3
4
2
1
4
2






































z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
...
]
4
[
4
]
3
[
3
]
1
[
1
]
[
0
]
[
...
4
3
0
)
( 4
3
2
1














 



n
n
n
n
n
x
z
z
z
z
z
X





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5.4.3.3 Inverse Z-Transform by Partial Fraction Expansion
Partial Fraction method
• used to find inverse z-transform for rational functions
• With corresponding ROC
• 𝑀 – order of numerator polynomial
• 𝑁 - Order of denominator polynomial
 If 𝑀 ≤ 𝑁 and denominator is factorizable and has ‘k’ distinct real roots then
 The coefficients A, B, C, D are given by
𝑋 𝑧 =
𝑁(𝑧)
𝐷(𝑧)
=
𝑎0 + 𝑎1𝑧−1 + 𝑎2𝑧−2 + ⋯ + 𝑎𝑀𝑧−𝑀
𝑏0 + 𝑏1𝑧−1 + 𝑏2𝑧−2 + ⋯ + 𝑏𝑀𝑧−𝑁
𝑋 𝑧
𝑧
=
𝐴1
𝑧 − 𝑧1
+
𝐴2
𝑧 − 𝑧2
+ ⋯ +
𝐴𝑘
𝑧 − 𝑧𝑘
𝐴1 = 𝑧 − 𝑧1
𝑋 𝑧
𝑧 𝑧→𝑧1
, 𝐴2 = 𝑧 − 𝑧2
𝑋 𝑧
𝑧 𝑧→𝑧2
⋯ 𝐴𝑘 = 𝑧 − 𝑧𝑘
𝑋 𝑧
𝑧 𝑧→𝑧𝑘

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 If denominator polynomial has repeated roots where
 The coefficeints are determined by
 𝑀 ≥ 𝑁 use long division to make the denominator order 𝑁 greater than the numerator order 𝑀
and apply partial fraction method.
𝑋 𝑧
𝑧
=
𝐴1
𝑧 − 𝑧1
+
𝐴2
𝑧 − 𝑧1
2
+ ⋯ +
𝐴𝑘
𝑧 − 𝑧1
𝑚
𝐴𝑘 =
1
𝑚 − 𝑘 !
𝑑𝑚−𝑘
𝑑𝑧𝑚−𝑘
𝑧 − 𝑧1
𝑚
𝑋 𝑧
𝑧 𝑧→𝑧1
𝑋 𝑧 =
𝑟=0
𝑀−𝑁
𝐵𝑧−𝑟
+
𝑘=1
𝑁
𝐴𝑘
1 − 𝑑𝑘𝑧−1

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Example: Find the inverse z- Transform of X(z)

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Example :2
Solution:
Find the inverse z-transform of
First eliminate the negative power of z.
Dividing both sides by z,
Finding the constants:
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑏𝑦 𝑧

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Example 3 :
Solution:
Find 𝑦 𝑛 if
Dividing Y(z) by z,
We first find B:
Applying the partial fraction expansion,
Next find A:

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Using the polar form,
Now we have:
Therefore, the inverse z-transform is:

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Example 4:
Solution:
Find 𝑥 𝑛 if
Dividing both sides by z:
Where,
Using the formulas for mth-order,
𝑋 𝑧
𝑧
=
𝐴1
𝑧 − 𝑧1
+
𝐴2
𝑧 − 𝑧1
2
+ ⋯ +
𝐴𝑘
𝑧 − 𝑧1
𝑚
, 𝐴𝑘 =
1
𝑚 − 𝑘 !
𝑑𝑚−𝑘
𝑑𝑧𝑚−𝑘
𝑧 − 𝑧1
𝑚
𝑋 𝑧
𝑧 𝑧→𝑧1
m=3,k=2,z1=0.5

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Then,
From Table,
Finally we get,

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5.4.3.4 Inverse Z-Transform by Residue Method
Residue Theorem
Inverse Z-Transform obtained by evaluating the contour integral:
• Where C is the path of integration enclosing all the poles of X(z).
Cauchy’s residue theorem:
• Sum of the residues of z n-1X(z) at all the poles inside C
• Every residue Ck, is associated with a pole at pk
• m is the order of the pole at z=pk
• For a first-order pole:
𝒙 𝒏 =
𝟏
𝟐𝝅𝒋 𝑪
𝒛𝒏−𝟏
𝑿 𝒛 𝒅𝒛

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Example:
Find the inverse z transform :
Single pole @ z=0.5, second-order pole @ z=1

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Example:
Find the inverse z transform :
Single pole @ z=0.5, second-order pole @ z=1
x(n)=2[(n-1)+(0.5)n]

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U18ECT3101 - Signals and Systems The Z Transform
Properties
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Properties
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Properties
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Properties
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Properties
26

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The Inverse Z transform
Comparison of the inverse z-transform Methods
• Power series:
Does not lead to a closed form solution, it is simple, easy computer implementation
• Partial fraction, residue:
 Closed form solution,
 Need to factorize polynomial (find poles of X(z))
 May involve high order differentiation (multiple poles)
• Partial fraction : Useful in generating the coefficients of parallel structures for digital filters.
• Residue method : widely used in the analysis of quantization errors in discrete-time systems.

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