The document discusses various methods for finding the inverse z-transform, including inspection of z-transform pairs, partial fraction expansion, and power series expansion. It provides examples of using each method to find the inverse z-transform of given z-functions. It also discusses properties of the z-transform, such as time shifting and convolution, that can help in solving inverse problems. Sample problems demonstrate applying the techniques to compute inverse z-transforms and use properties to solve for sequences.

1. The inverse z transform
The inverse z-transform can be found by
one of the following ways
 Inspection method
 Partial fraction expansion
 Power series expansion
Each will be explained briefly next
1

2. Inverse z transform by
inspection method
The inspection method is based on the z
transform pair table.
In order to find the inverse z transform we
compare 𝑋(𝑧) to one of the standard
transform pairs listed in the z transform
pairs table
2

3. Inverse z transform by partial
fraction
If 𝑋(𝑧) is not in a form listed in the table of
z transform pairs we can use the partial
fraction method to simplify the function into
one of the standard forms listed in the z
transform pair table
3

4. Inverse z- transform example 1
Consider a sequence 𝑥[𝑛] with z transform
of 𝑋 𝑧 =
1
1−
1
4
𝑧−1 1−
1
2
𝑧−1
𝑧 >
1
2
Where the ROC is as shown
4

5. Inverse z- transform example 1
S𝑜𝑢𝑙𝑡𝑖𝑜𝑛
Note that 𝑋(𝑧) can be rewritten as X 𝑧 =
1
1−
1
4
𝑧−1 1−
1
2
𝑧−1
=
𝐴1
1−
1
4
𝑧−1
+
𝐴2
1−
1
2
𝑧−1
To find the constants 𝐴1 and 𝐴2 we use
the following 𝐴1 = 1 − 1
4
𝑧−1
𝑋 𝑧 | 𝑧−1=4 =
− 1
Similarly 𝐴2 = 1 − 1
2
𝑧−1
𝑋 𝑧 | 𝑧−1=2 = 2
5

6. Inverse z- transform example 1
Note that 𝑋(𝑧) Now can be rewritten as
X 𝑧 =
−1
1−
1
4
𝑧−1
+
2
1−
1
2
𝑧−1
The inverse z transform results in the 𝑥[𝑛]
shown below 𝑥 𝑛 = − 1
4
𝑛
𝑢 𝑛 + 2 1
2
𝑛
𝑢 𝑛
6

7. Inverse z transform by the
partial fraction with - 𝑀 ≥ 𝑁
The partial fraction method can be used to find the
inverse z-transform for rational functions with
numerator of order 𝑀 and denominator of order 𝑁
The partial fraction can be used only if the
numerator order 𝑀 is less than denominator order 𝑁
If the numerator order is greater than or equal the
denominator order then we use long division to
make the denominator order 𝑁 greater than the
numerator order 𝑀 before we can use the partial
fraction method
7

8. Partial fraction with - 𝑀 ≥ 𝑁
The long division converts the 𝑋(𝑧) function in
the following form
𝑋 𝑧 =
𝑟=0
𝑀−𝑁
𝐵𝑟 𝑧−𝑟 +
𝑘=1
𝑁
𝐴 𝑘
1 − 𝑑 𝑘 𝑧−1
Where 𝑀 is the numerator order, 𝑁 is the
denominator order, 𝐴 𝑘 are constants of the
partial fraction and 𝑑 𝑘 are the roots
8

9. Partial fraction example 2
Find the inverse z transform for the
sequence given by
If the ROC is as shown
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10. Partial fraction example 2
As it can be seen from 𝑋(𝑧) the order of
the numerator is equal to the order of the
denominator
Long division can be used to make the
order of the numerator less than the order
of the denominator as shown below
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11. Partial fraction example 2
Now the function 𝑋(𝑧) can be rewritten as
shown below
Or
Where 𝐵0 = 2
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12. Partial fraction example 2
The constants 𝐴1 and 𝐴2 can be found as
follows
𝑋(𝑧) can now be written as
12

13. Partial fraction example 2
Recall that from the z-transform pairs table
we have
Therefore 𝑥[𝑛] is given by
13

14. Partial fraction with multiple
poles and 𝑀 greater than 𝑁
If the 𝑋(𝑧) function contains multiple poles
and 𝑀 > 𝑁 as shown in this form
The coefficients 𝑐 𝑚 can be found by deriving
𝑋(𝑧) 𝑚 number of times as shown
14

15. Inverse z transform by using
power series expansion
From the definition of the z-transform we
can write the z-transform as
𝑋 𝑧 =
𝑛=−∞
∞
𝑥[𝑛]𝑧−𝑛
= ⋯ + 𝑥 −2 𝑧2 + 𝑥 −1 𝑧1 + 𝑥 0 + 𝑥 2 𝑧−1 + 𝑥 2 𝑧−2 + ⋯
This is known as Laurent series
From this series we can find the sequence
𝑥[𝑛] as illustrated by the next example
15

16. Inverse z transform by using power
series example 3
Find the inverse z-transform for the
sequence defined by
𝑋 𝑧 = 𝑧2
1 − 1
2
𝑧−1
1 + 𝑧−1
1 − 𝑧−1
Solution
Note the sequence 𝑋[𝑧] can be expanded
as 𝑋 𝑧 = 𝑧2
− 1
2
𝑧 − 1 + 1
2
𝑧−1
If we compare 𝑋 𝑧 with the Laurent series
we can extract 𝑥[𝑛] as follows
16

17. Inverse z transform by using
power series example 3
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18. Inverse z transform by using
power series example 4
Consider the z transform defined by
𝑋 𝑧 =
1
1−𝑎𝑧−1 𝑓𝑜𝑟 𝑧 > 𝑎
Find 𝑥[𝑛] by using long division
18

19. Inverse z transform by using
power series example 4
Solution
This series reduces to 𝑥 𝑛 = 𝑎 𝑛
𝑢[𝑛]
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20. Inverse z transform by using
power series example 5
Find the inverse z transform of the
sequence defined by
𝑋 𝑧 =
1
1 − 𝑎𝑧−1
𝑓𝑜𝑟 𝑧 < 𝑎
20

21. Inverse z transform by using
power series example 5
Solution
Because the region of convergence, the
sequence is a left-sided
The solution can be obtained by long division as
indicated
21

22. Z-transform properties
The z-transform has many useful properties
similar to Fourier transform properties
These properties can be used to find the
inverse z-transform for certain complex z
functions as it will be demonstrated in the
examples
These properties are
 Linearity
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23. z transform properties
 Time shifting
 Multiplication by an exponential
 Differentiation of 𝑋(𝑧)
 Conjugation of a complex sequence
 Time reversal
 Convolution of a sequence
 Initial value theorem
These properties are summarized in the table
shown in the next two slides
23

24. Z-transform properties table
24

25. Z-transform properties table
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26. Example 6
Determine the z-transform and the ROC for
the sequence 𝑥 𝑛 = 3(2) 𝑛
𝑢 𝑛 − 4(3) 𝑛
𝑢 𝑛
Solution
We can divide 𝑥[𝑛] into two different
functions 𝑥1 𝑛 = (2) 𝑛
𝑢 𝑛 𝑎𝑛𝑑 𝑥2 𝑛 =
(3) 𝑛
𝑢 𝑛
Now 𝑥[𝑛] can be rewritten as 𝑥 𝑛 = 3𝑥1 𝑛 −
4𝑥2 𝑛
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27. Example 6
From the z-transform table we have
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28. Example 7
Determine the z-transform of 𝑥[𝑛] = 𝑢[−𝑛]
Solution
By using the time reversal property we
have
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29. Example 8
Compute the convolution of the following two
sequences using the z transform
𝑥1 𝑛 = 1, −2, 1
𝑥2 𝑛 =
1, 0 ≤ 𝑛 ≤ 5
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
Solution
Note that the z transform of each of the previous
sequences is given by 𝑋1 𝑧 = 1 − 2𝑧−1 + 𝑧−2
and 𝑋2 𝑧 = 1 + 𝑧−1
+ 𝑧−2
+ 𝑧−3
+ 𝑧−4
+ 𝑧−5
29

30. Example 8
If we multiply 𝑋1(𝑧)𝑋2(𝑧) we get the
following answer
The inverse z-transform which is the
convolution of 𝑥 𝑛 = 𝑥1 𝑛 ∗ 𝑥2 𝑛 is given
by
30

31. Example 9
Find the inverse z transform for the
function defined by
𝑋 𝑧 = log 1 + 𝑎𝑧−1
𝑧 > 𝑎
solution
31

32. Example 9
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