1

7. Z-Transform

7.1 Why to learn z-Transform?

The Z-transform is the most general concept for the transformation of discrete-time

series, just like, the Laplace transform is the more general concept for the transformation

of continuous time processes.

For example, the Laplace transform allows you to transform a differential equation, and its corresponding initial and boundary value problems, into a space in which the equation

can be solved by ordinary algebra, the Z transform converts a difference equation into an

algebraic equation and hence is useful in solving difference equations.

The z-transform is an extension of the discrete timeFourier transform to a function that isdefined on regions of the complex plane.

Z-transform is useful in switching calculus problems into algebraic operations which will simplify the computations.

Special significance of z-transformis as a system designing tool, to facilitate designing of systems with the desired attributes that can be practically implemented.

It is used extensively today in the areas of applied mathematics, digital signal processing, control theory, population science, and economics. These discrete models are solved

with difference equations in a manner that is analogous to solving continuous models

with differential equations.

7.2 Z-Transform:

The z-transform of a discrete time signal, or a sequence x(n) is defined as

n

nznxzX )()(

-----7.1

LEARNING OBJECTIVES:

By the end of this chapter, the reader will be able to:

Describe use of z-Transform

Describe the difference between Fourier, Laplace, and z-transforms

List the properties of the z-transform

Identify the Fourier transform in the z-transform

Explain the relationship between z-transform pairs

Analyse ROC`s and stability of signal using Z Transform

2

Where Z is a complex variable. For any given sequence ),(nx the Z transform )(zX may or may

not converge. The set of values of z, or the set of points in the z-plane, for which )(zX converges

is said Region of Convergence (ROC) of ).(zX In general z-transform of a sequence has

associated with it a range of values of z for which z- transform converges.

The z transform is ‘unilateral’ or ‘one-sided’ and ‘bilateral’ or ‘two-sided’ hence,

0

)()(n

nznxzX

------7.2

7.3 Definition Based on the Laplace Transform: A linear system can be represented in the complex frequency domain (s-domain where s = +

j) using the Laplace Transform The Z-Transform is a special case of the Laplace transform and results from applying the

Laplace transform to a discrete-time signal.

Let x(t) be a causal function, i.e., x(t)=0 for t

3

And∠𝒛 = ɵ = 𝜔𝑇 = 𝜔2𝜋

𝜔𝑠= 2𝜋 (

𝜔

𝜔𝑠) so that when 𝜔 = 0, 𝜃 = 0; when 𝜔 = 𝜔𝑠 4⁄ , 𝜃 = 𝜋 2;⁄

when 𝜔 = 𝜔𝑠 2⁄ , 𝜃 = 𝜋; when 𝜔 = 3𝜔𝑠 4⁄ , 𝜃 = 3 𝜋 2;⁄ and when 𝜔 = 𝜔𝑠, 𝜃 = 2𝜋.

Figure 7.1s-plane and z-plane mapping

The jω axis of the s-plane is mapped into the circumference of the unit circle in the z-plane.

Moving a distance of 𝜔𝑠 parallel to the jω axis in the s-plane is equivalent to traversing the circumference of the unit circle in the z-plane

If 𝝈𝟏 < 0 then|𝒛| = 𝒆𝝈𝟏𝑻 < 1

and ∠𝒛 = ɵ = 𝜔𝑇 = 2𝜋 (𝜔

𝜔𝑠) Hence, points in the left half of the s-plane are mapped as points

inside the unit circle in the z-plane.

If 𝝈𝟏 > 0then|𝒛| = 𝒆𝝈𝟏𝑻 > 1and∠𝒛 = 2𝜋 (

𝜔

𝜔𝑠), Hence,points in the right half of the s-plane are

mapped as points exterior to the unit circle in the z-plane.

Recall, if a CT systemis stable; its poles lie in the left-half plane.Hence, a DT system isstable if

its poles areinside the unit circle.

The z-Transform behavesmuch like the Laplacetransform and can beapplied to difference

equationsto produce frequency and time domain responses.

7.4 Relationship to Fourier Transform:

Note that expressing the complex variable z in polar form reveals the relationship to the Fourier

transform:

n

nii

n

nini

n

n

ii

enxXeX

rifandernxreX

orrenxreX

)()()(

,1,)()(

,))(()(

Which is the Fourier transform of x(n).The z-transform of x(n) can be viewed as the Fourier

transform of x(n) multiplied by an exponential sequence r-n,

Z-transform reduces to Fourier Transform on the contour in the complex z-plane corresponding to a circle with a radius of unity.

Equivalent

im

Re

jω-axis

Re

im

s-plane z-plane

LHP RHP

|𝑍|=1

4

Some sequences that do not have Fourier transforms will have z-transforms, but thereverse is true as well.

For the Fourier transform to converge, the sequence must have finite energy, or:The Region of

convergence (ROC) is the set of points z in the complex plane, for which the summation is

bounded (converges):

n

nrnx )(

7.5 Z-Transform of basic discrete time signals:

1. Dirac delta function (or unit sample function):x[n] = [n]

1 )(0

0

n

n

n

n znznzX

Region of convergence: entire z-plane, since x(n)=0 for all n except n=0 for which it has a

value of 1

2. Shifted Unit sample sequencex[n] = [n-l]

We know that, 𝛿(𝑛 − 𝑙) = {1 𝑓𝑜𝑟 𝑛 = 𝑙0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

}

ln

n

n

n zzlnzlnzX

1

1

)(

Region of convergence: entire z-plane exceptz = 0

3. Unit step sequence x(n)=u(n)

We know that, 𝑢(𝑛) = {1 𝑓𝑜𝑟 𝑛 ≥ 00 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

}

Figure 7.2 Shaded region i.e; the exterior

of unit circle is ROC of U(z)

𝑋(𝑧) = ∑ 𝑥(𝑛)𝑧−𝑛∞

0

= ∑ 1𝑧−𝑛 =1

1 − 𝑧−1

∞

0

=𝑧

𝑧 − 1

OR

𝑋(𝑧) = ∑ 1 𝑧−𝑛∞

𝑛=0

1

1

3212111

21

1

1)(1)()(

)1()(

1

zzXzXzzX

zzzzzzzXz

zz

Provided|𝑧−1| < 1, i.e., R.O.C is |𝑧| > 1 The ROC in this case, is thus, the exterior of

the unit circle in the z-plane.

4. The ramp sequence x(n)= r(n)

z-plane

Re

im

1

5

𝑟(𝑛) = {𝑛 𝑓𝑜𝑟 𝑛 ≥ 00 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

}

𝑅(𝑧) = ∑ 𝑟(𝑛)𝑧−𝑛 = ∑ 𝑛𝑧−𝑛∞

𝑛=0

∞

𝑛=0

------7.3

We know that,

∑ 𝑧−𝑛∞

𝑛=0

=𝑧

𝑧 − 1; |𝑧| > 1

Differentiating the above both sides with respect to z we get,

− ∑ 𝑛𝑧−𝑛−1 = 𝑑

𝑑𝑧

∞

𝑛=0

[𝑧

𝑧 − 1] =

−1

(𝑧 − 1)2

−𝑧−1 ∑ 𝑛𝑧−𝑛 = −1

(𝑧 − 1)2

∞

𝑛=0

i.e., ∑ 𝑛𝑧−𝑛 =

𝑧

(𝑧 − 1)2

∞

𝑛=0

Substituting this in equation 7.3

𝑅(𝑧) = ∑ 𝑛𝑧−𝑛 =𝑧

(𝑧 − 1)2

∞

𝑛=0

; |𝑧| > 1

In this case too, ROC is the exterior of the unit circle in the z-plane.

5. The exponential sequence: x(n) = an

𝑥(𝑛) = {𝑎𝑛 ; 𝑛 ≥ 00 ; 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

}

Figure7.3 ROC of Z-transform of an

𝑋(𝑧) = ∑ 𝑥(𝑛)𝑧−𝑛∞

𝑛=0

= ∑ 𝑎𝑛𝑧−𝑛∞

𝑛=0

𝑋(𝑧) = ∑(𝑎𝑧−1)𝑛∞

𝑛=0

=1

1 − 𝑎𝑧−1

Provided,|𝑎𝑧−1| < 1, 𝑖. 𝑒. , |𝑧| > |𝑎|

𝑋(𝑧) =𝑧

𝑧 − 𝑎; |𝑧| > |𝑎|

Case1: if a is purely real, say , a=e-α where is purely real, then 𝑋(𝑧) =𝑧

𝑧−𝑒−𝛼; |𝑧| > 𝑒−𝛼

Case 2: if a is purely imaginary, say, a= e-jω then 𝑋(𝑧) =𝑧

𝑧−𝑒−𝑗𝜔; |𝑧| > 1

Case 3: if a is purely complex number, say, a= e-(σ+jω) then 𝑋(𝑧) =𝑧

𝑧−𝑒−(𝜎+𝑗𝜔); |𝑧| > 𝑒−𝜎

6. Discrete sinusoidal and Cosinusodial:x(n)= sinωn and x(n)= cos ωn

z-plane

Re

im

a

6

𝑥(𝑛) = {𝑐𝑜𝑠𝜔𝑛 ; 𝑛 ≥ 0

0 ; 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒}

2sin ;

2cos

sincos ;sincos

j

eet

eet

tjtetjte

tjtjtjtj

tjtj

21

1

21

1

211

11

11

11

11

cos21

cos1

)cos21(2

cos22

)(1

)(2

2

1

)1)(1(

11

2

1

1

1

1

1

2

1)(

])[][(2

1

2][cos)(

zTz

Tz

zz

z

zzeze

zeze

zeze

zeze

zezezX

eZeZee

ZnZzX

jj

jj

jj

jj

jj

njnjnjnj

= 𝑧2 − 𝑧𝑐𝑜𝑠𝜔

𝑧2 − 2𝑧𝑐𝑜𝑠𝜔 + 1 ; |𝑧| > 1

Similarly, it can be shown that 𝑍[𝑠𝑖𝑛 𝜔𝑛] = 𝑧𝑠𝑖𝑛𝜔

𝑧2−2𝑧𝑐𝑜𝑠𝜔+1 ; |𝑧| > 1

7.6 ROC and its significance:

The ROC is an annular ring in the z-plane centered about the origin (which is equivalent to a

vertical strip in the s-plane).

When we deal with sequences in general, it is necessary to take into account the ROC of an X(z)

while determining its IZT. This is because, different sequences may have the same expression

for X(Z) but with different ROC`s. the following example illustrate this point.

Example 7.1 Determine X(z) if x(n)=u(n)

𝑋(𝑧) = ∑ 𝑥(𝑛)𝑧−𝑛∞

0

= ∑ 1𝑧−𝑛 =1

1 − 𝑧−1

∞

0

=𝑧

𝑧 − 1

Figure 7.4 ROC of X(z) when x(n)=u(n)

Provided |𝑧−1| < 1, i.e., R.O.C is |𝑧| > 1 As shown in figure 7.4 the ROC in this case, is

thus, the exterior of the unit circle in the z-plane.

Example 7.2 Determine X(z) if x(n) = -u-(-n-1)

z-plane

Re

im

1

pole X

7

𝑋(𝑧) = ∑ 𝑥(𝑛)𝑧−𝑛∞

0

= − ∑ 1𝑧−𝑛−1

−∞

,

Figure 7.5 ROC and pole location of X(z)

when x(n)=-u(-n-1)

Putting k=-n and

𝑋(𝑧) = − ∑ 1𝑧𝑘∞

𝑘=0

= 1 − ∑ 𝑧𝑘∞

𝑘=0

= 1 −1

1 − 𝑧=

𝑧

𝑧 − 1,

Provided |𝑧| < 1 As shown in figure 7.5 the ROC in this case, is thus,

interior of the unit circle in the z-plane.

Thus the two sequences x(n)=u(n) and x(n)= -u(-n-1) have the same expression 𝑧

𝑧−1 for their

X(z). However, their ROCs are different while the example 7.1 is right sided sequence, and

example 7.2 is left sided sequence.

It is clear from these examples that one must consider the ROC to determine the correct inverse

Z Transform in this context it is essential to note the following properties of ROC`s.

7.7 Properties of the ROC: The ROC does not contain any poles (similar tothe Laplace transform).

If x[n] is of finite duration, then the ROC is the entirez-plane except possiblyz = 0 and/or z =:

)Re()()()(][]1[][

)Re()()()(0][]1[][

1)()()(1][][][

][

1

sesXTttxzROCzzXnnx

sesXTttxzROCzzXnnx

sXttxzallROCzXnnx

znxX(z)

sT

sT

n

n

:CT:DT

If x[n]is a right-sided sequence, and if |z| = r0is in the ROC, then all finite values of zfor which |z|

>r0are also in the ROC.

If x[n]is a left-sided sequence, and if |z| = r0is in the ROC, then all finite values of zfor which |z|

8

right-sided left-sided two-sided Figure 7.6 ROC for right-sided, left-sided and two-sided sequence.

7.8 Stability Criteria:

We know that necessary and sufficient condition for a causal LTI system to be stable is

∑ |ℎ(𝑛)|∞0 < ∞ The system function of the causal LTI system is written as 𝐻(𝑧) = ∑ ℎ(𝑛)𝑧−𝑛∞𝑛=0 . Taking magnitude on both sides,

|𝐻(𝑧)| = |∑ ℎ(𝑛)𝑧−𝑛∞

𝑛=0

| ≤ ∑|ℎ(𝑛)𝑧−𝑛|

∞

𝑛=0

∑|ℎ(𝑛)𝑧−𝑛|

∞

𝑛=0

= ∑|ℎ(𝑛)||𝑧−𝑛|

∞

𝑛=0

Evaluating above equation on unit circle, we get,

|𝐻(𝑧)| ≤ ∑|ℎ(𝑛)|

∞

𝑛=0

< ∞

Therefore, for a stable system, ROC of system function includes unit circle.

For a causal system, ROC is exterior to the circle of radius r, consequently, a causal and stable

system must have transfer function |𝑧| > 𝑟 < 1. Since ROC can not contain any pole of H(z), a causal LTI system is BIBO stable if and only if all poles are inside unit circle.

Let us see some examples to understand the stability and the ROC.

Example 7.3 Determine X(z) if 𝑥(𝑛) = 𝑎𝑛𝑢(𝑛) and comment for 𝑎 > 0 and 𝑎 < 0 .

For a>0 :

𝑋(𝑧) =

1

1 − 𝑎𝑧−1 ,

For |𝑧| > |𝑎|

(a) If the ROC is outside the unit circle, the signal is unstable.

(b) 1for1

1];[][

1

z

zX(z)nunx

(c) If the ROC includes the unit circle, the signal is stable.

9

(a) (b) (c) Figure 7.7 (a) for a=1.1 (b) for a=1 (c) for a=0.9

For a |𝑎|

(a) If the ROC is outside the unit circle, the signal is unstable.

(b) 1for1

1];[][

1

z

zX(z)nunx

(c) If the ROC includes the unit circle, the signal is stable.

(a) (b) (c)

Figure 7.8 (a) for a= -1.1 (b) for a= -1 (c) for a= -0.9

Example 7.4 ]1[][ nuanx n (left sided sequence)

?1

)(11

]1[)(

1

1

11

za

za

zaznuazXn

n

n

nn

Figure 7.9 ROC for left--sided sequence

If : azza or,,11

az

z

az

za

za

zaza

za

zaX(z)

1

1

1

11

1

1

1

1

1

1

1

1

1

1

11

z-plane

Re

im

1 X

10

i.e., Here the z-Transform is the same for example 7.3 and 7.4, but the region of convergence is

different.

Analyzing the above example for different values of ‘a’ we get,

(a) (b) (c)

Figure 7.10 (a) for a=1.1 (b) for a=1 (c) for a=0.9

(a) If the ROC is includes unit circle, the signal is stable.

(b) 1for1

1];[][

1

z

zX(z)nunx

(c) If the ROC includes the unit circle, the signal is unstable.

7.9 Properties of z-transform:

(i) Linearity If 𝑥1(𝑛) ↔ 𝑋1(𝑧) with ROC =R1 and 𝑥2(𝑛) ↔ 𝑋2(𝑧) with ROC =R2 then

𝑎1𝑥1(𝑛) + 𝑎2𝑥2(𝑛) ↔ 𝑎1𝑋1(𝑧) + 𝑎2𝑋2(𝑧) with ROC containing R1 ∩ R2

Example 7.5: Determine the Z-Transform of the signal x[n] = [3(2n) – 4(3n)]u[n]:

Solution:

3:;31

14

21

13]34)2(3[

1

1]][[

11

1

zROCzz

z

aznuaz

nn

n

Example 7.6: Determine the Z-Transform of the signal cos (w0n) u[n]

Solution:

11

1:;cos21

cos1

1

1

2

1

1

1

2

1][cos

2

1

2

1][cos

2

0

1

0

1

110

0

00

00

zROCzwz

wz

zezenunwz

eenunw

jwjw

njwnjw

(ii) Time Shifting Property:

If 𝑥(𝑛) ↔ 𝑋(𝑧) with ROC =R then 𝑥(𝑛 − 𝑘) ↔ 𝑧−𝑘𝑋(𝑧) with ROC=R except for possible addition or deletion of origin or infinity.

Proof:

since

n

nzknxknxz ][]][[

then the change of variable m = n-k produces

)(][

][]][[ )(

zXzzmxz

zmxknxz

k

m

mk

m

km

Example 7.7: Find the Z-Transform of a unit step function. Use time shifting property to find z-

transform of u[n] – u[n-N].

Solution:

The z-transform of u[n] can be found as

1

21

0

1

1.......1

][]][[

zzz

zznunuzn

n

n

n

Now the z-transform of u[n]-u[n-N] may be found as follows:

01:;1

1

1

1

1

1]][][[

1

11

NifzROCz

z

zz

zNnunuz

N

N

(iii) Scaling in the z-domain: If 𝑥(𝑛) ↔ 𝑋(𝑧) with ROC = R

then 𝑎𝑛𝑥(𝑛) ↔ 𝑋(𝑎−1𝑧) for any constant a, real or complex, with ROC : |𝑎|𝑅 Proof:

zaXzanxznxanxazn

nn

n

nn 11][][][

Example 7.8: Determine the Z-Transform of the signal 𝑎𝑛𝑐𝑜𝑠𝜔0𝑛 𝑢(𝑛) Solution: since

12

2

0

1

0

1

0cos21

cos1][)[cos(

zwz

wznunwz

azROCzawaz

waznunwaz n

:;cos21

cos1]][cos[

22

0

1

0

1

0

(iv) Time reversal: If 𝑥(𝑛) ↔ 𝑋(𝑧)with ROC = R

then 𝑥(−𝑛) ↔ 𝑋(𝑧−1) with ROC=1/R

Proof:

m m

mm

n

n zXzmxzmxznxnxz )(][][][]][[ 11

Example 7.9: Determine the Z-Transform of u[-n].

Solution:Since 𝑍{𝑢(𝑛)} =1

1−𝑧−1 ROC: |𝑧| > 1

Therefore,

𝑍{𝑢(−𝑛)} =1

1−𝑧 with ROC: |𝑧| ≤ 1

(v) Differentiation in the z – Domain: If 𝑥(𝑛) ↔ 𝑋(𝑧)with ROC = R

then 𝑛𝑥(𝑛) ↔ −𝑧𝑑

𝑑𝑧𝑋(𝑧)with ROC=R

Example 7.10: Determine the Z-Transform of thesignal 𝑥[𝑛] = 𝑛𝑎𝑛𝑢[𝑛] Solution:

azROC

az

az

azdz

dznunaz

azROCaz

nuaz

n

n

:;11

1]][[

:;1

1]][[

21

1

1

1

(vi) Convolution and Correlation: To study the LTI systems, convolution plays important role. Shifting, multiplications and

summation are operations in computation of convolution. Correlation which is very much similar

to convolution provides information about the similarity between the two sequences. It is used in

Radars, digital communication and mobile communication etc. The main application of

correlation is that the incoming/received signal is correlated with standard signals and signal of

this set which has maximum correlation with the incoming/received signal is detected.

(a) Convolution of two sequences:

Let 𝑥1(𝑛) and 𝑥2(𝑛) be two sequences with 𝑋1(𝑧) and 𝑋2(𝑧) as their respective Z- Transforms. i.e., 𝑥1(𝑛) ↔ 𝑋1(𝑧) with ROC =R1and 𝑥2(𝑛) ↔ 𝑋2(𝑧) with ROC =R2

Then 𝑥1(𝑛)| ∗ 𝑥2(𝑛) = 𝑋1(𝑧)𝑋2(𝑧)with ROC at least R1∩ R2. Proof:

The convolution of x1[n] and x2[n] is defined as

13

k

knxkxnxnxnx ][][][*][][ 2121

The z-transform of x[n] is

n

n

kn

n zknxkxznxzX 21 ][][)(

Upon interchanging the order of the summation and applying the time shifting property, we

obtain

zXzXzkxzXzknxkxzXk

kn

nk

121221 ][)(

Example 7.11: Compute the convolution of the signals x1[n] = [1, -2, 1] and

elsewhere

nnx

,0

50,1][2

Solution:

𝑋1(𝑧) = 1 − 2𝑧−1 + 𝑧−2

𝑋2(𝑧) = 1 + 𝑧−1 + 𝑧−2 + 𝑧−3 + 𝑧−4 + 𝑧−5

Now 𝑋(𝑧) = 𝑋1(𝑧)𝑋2(𝑧) = 1 − 𝑧−1 + 𝑧−6 + 𝑧−7

Hence 𝑥(𝑛) = [1, −1, 0, 0, 0, 0, −1, 1]

Note: You should verify this result from the definition of the convolution sum.

(b) Correlation of two sequences:

If 𝑥1(𝑛) and 𝑥2(𝑛) be two sequences with 𝑋1(𝑧) and 𝑋2(𝑧) as their respective Z- Transforms. i.e., 𝑥1(𝑛) ↔ 𝑋1(𝑧) with ROC =R1 and 𝑥2(𝑛) ↔ 𝑋2(𝑧) with ROC =R2 then,

)(z(z)XXm)(n(n)xxmr z

n

xx

1

2121)(21

Proof:the following is the correlation of two sequences 𝑥1(𝑛)𝑥2(𝑛)

𝑟𝑥1𝑥2(𝑚) = ∑ 𝑥1(𝑛)𝑥2[(𝑛 − 𝑚)]

∞

𝑛=−∞

… . … 7.4

Arranging term 𝑥2(𝑛 − 𝑚) as 𝑥2[−(𝑚 − 𝑛)] in above equation we get

𝑟𝑥1𝑥2(𝑚) = ∑ 𝑥1(𝑛)𝑥2[−(𝑚 − 𝑛)]

∞

𝑛=−∞

… … .7.5

Therefore the RHS of equation 7.5represents the convolution 𝑥1(𝑛) and 𝑥2(−𝑚) can be written as

𝑟𝑥1𝑥2(𝑚) = 𝑥1(𝑚) ⊗ 𝑥2(−𝑚)

𝑍[𝑟𝑥1𝑥2(𝑚)] = 𝑍[𝑥1(𝑚) ⊗ 𝑥2(−𝑚)]

𝑍[𝑟𝑥1𝑥2(𝑚)] = 𝑍[𝑥1(𝑚)] ⊗ 𝑍[𝑥2(−𝑚)]

14

𝑍[𝑥1(𝑚)] = 𝑋1(𝑧) and 𝑍[𝑥2(−𝑚)] = 𝑋2(𝑧−1)

𝑍[𝑟𝑥1𝑥2(𝑚)] = 𝑋1(𝑧)𝑋2(𝑧−1)

Example 7.12: Solve the following

If𝑋+(𝑧) = 𝑍{𝑥(𝑛)}then

(𝑖)𝑍{𝑥(𝑛 − 𝑚)} = 𝑧−𝑚 {𝑋+(𝑧) + ∑ 𝑥(−𝑘)𝑧𝑘

𝑚

𝑘=1

}

(𝑖𝑖)𝑍{𝑥(𝑛 + 𝑚)} = 𝑧𝑚 {𝑋+(𝑧) − ∑ 𝑥(𝑘)𝑧−𝑘

𝑚−1

𝑘=0

}

Proof:

(𝑖)𝑍{𝑥(𝑛 − 𝑚)} = 𝑧−𝑚 {𝑋+(𝑧) + ∑ 𝑥(−𝑘)𝑧𝑘

𝑚

𝑘=1

}

𝑍{𝑥(𝑛 − 𝑚)} = ∑ 𝑥(𝑛 − 𝑚)

∞

𝑛=0

𝑧−𝑛 = 𝑧−𝑚 ∑ 𝑥(𝑛 − 𝑚)

∞

𝑛=0

𝑧−(𝑛−𝑚)

𝑃𝑢𝑡 𝑛−m=l

Therefore , 𝑍{𝑥(𝑛 − 𝑚)} = 𝑧−𝑚 ∑ 𝑥(𝑙)

∞

𝑙=−𝑚

𝑧−𝑙

= 𝑧−𝑚 {∑ 𝑥(𝑙)𝑧−(𝑙)∞

𝑙=0

+ ∑ 𝑥(𝑙)𝑧−(𝑙)−1

𝑙=−𝑚

}

Replacing l by –k,

𝑍{𝑥(𝑛 − 𝑚)} = 𝑧−𝑚{𝑋+(𝑧) + ∑ 𝑥(−𝑘)𝑧𝑘

𝑚

𝑘=1

}

(ii)

𝑍{𝑥(𝑛 + 𝑚)} = ∑ 𝑥(𝑛 + 𝑚)

∞

𝑛=0

𝑧−𝑛 = 𝑧𝑚 ∑ 𝑥(𝑛 + 𝑚)

∞

𝑛=0

𝑧−(𝑛+𝑚)

𝑃𝑢𝑡 𝑛+m=l

Therefore , 𝑍{𝑥(𝑛 + 𝑚)} = 𝑧𝑚 ∑ 𝑥(𝑙)

∞

𝑙=𝑚

𝑧−𝑙

= 𝑧𝑚 {∑ 𝑥(𝑙)𝑧−(𝑙)∞

𝑙=0

− ∑ 𝑥(𝑙)𝑧−(𝑙)𝑚−1

𝑙=1

}

𝑍{𝑥(𝑛 + 𝑚)} = 𝑧𝑚{𝑋+(𝑧) − ∑ 𝑥(𝑘)𝑧−𝑘

𝑚−1

𝑘=1

}

(vii) The Initial Value Theorem:

If x[n] is causal then )(lim]0[ zXxz

15

Proof:

....]2[]1[]0[][)( 21

0

zxzxxznxzXn

n

Obviously, as z , z-n 0 since n >0, this proves the theorem.

(viii) Final Value Theorem if x(n) is causal and 𝑥(𝑛) ↔ 𝑋(𝑧) and if all the poles of 𝑋(𝑧) lie strictly inside the unit circle except possibly for first order pole at z=1, then

𝑥(∞) = lim𝑛→∞

𝑥(𝑛) = lim𝑧→1

(1 − 𝑧−1)𝑋(𝑧)

Proof:consider the transform of {𝑥(𝑛 + 1) − 𝑥(𝑛)}

𝑍{𝑥(𝑛 + 1)} − 𝑍{𝑥(𝑛)} = ∑[𝑥(𝑛 + 1) − 𝑥(𝑛)]𝑧−𝑛∞

𝑛=0

= [𝑧. 𝑋(𝑧) − 𝑧. 𝑥(0)] − 𝑋(𝑧)

(𝑧 − 1)𝑋(𝑧) − 𝑧. 𝑥(0) = ∑[𝑥(𝑛 + 1) − 𝑥(𝑛)]𝑧−𝑛∞

𝑛=0

Now, dividing throughout by z,

[(𝑧 − 1)

𝑧] 𝑋(𝑧) − 𝑥(0) = ∑[𝑥(𝑛 + 1) − 𝑥(𝑛)]𝑧−(𝑛+1)

∞

𝑛=0

[(𝑧 − 1)

𝑧] 𝑋(𝑧) = 𝑥(0) + lim

𝑘→∞∑[𝑥(𝑛 + 1) − 𝑥(𝑛)]𝑧−(𝑛+1)

𝑘

𝑛=0

Taking the limit on both sides as 𝑧 → 1,

lim𝑧→1

[(𝑧 − 1)

𝑧] 𝑋(𝑧) = lim

𝑧→1[𝑥(0) + lim

𝑘→∞∑[𝑥(𝑛 + 1) − 𝑥(𝑛)]𝑧−(𝑛+1)

𝑘

𝑛=0

]

= 𝑥(0) + [𝑥(1) − 𝑥(0)] + [𝑥(2) − 𝑥(1)] + ⋯

= lim𝑘→∞

𝑥(𝑘)

= lim

𝑧→1[𝑧 − 1

𝑧] 𝑋(𝑧) = 𝑥(∞)

This theorem enables us to find steady state value of x(n) without solving the sequence.

Remark: (i) If the poles of X(z) are outsidethe unit circle in the z-plane, the 𝑥(∞) term will be infinitely large; hence the requirement that they lie strictly inside the unit circle.

(ii) Even if X(z) has a simple pole at z=1, it will cancle out with the (z-1) in the numerator of

(𝑧−1

𝑧) 𝑋(𝑧) and so as long as this pole at z=1 is restricted to be a simple pole,lim

𝑧→1[

𝑧−1

𝑧] 𝑋(𝑧)will

be finite.

Example7.13: Find 𝑥(∞) when X(z) is given by 𝑋(𝑧) =𝑧+2

(𝑧−0.8)2

16

Solution:

Answer: Given that,𝑋(𝑧) =𝑧+2

(𝑧−0.8)2

By inspection we can find that, ROC of X(z) is |𝑧| > 0.8 and (z-1)X(z) has no pole on or outside unit circle. Therefore, by final value theorem of z transforms,

𝑥(∞) = lim𝑧→1

(𝑧 − 1)𝑧 + 2

(𝑧 − 0.8)2= 0

Example7.14: Find 𝑥(∞) when X(z) is given by X(z)=𝑧+1

3(𝑧−1)(𝑧+0.9)

Solution:

Answer: Given that,𝑋(𝑧) =𝑧+1

3(𝑧−1)(𝑧+0.9)

By inspection we can find that, ROC of X(z) is |𝑧| > 1 and (z-1)X(z) has no pole on or outside unit circle. Therefore, by final value theorem of z transforms,

𝑥(∞) = lim𝑧→1

(𝑧 − 1)𝑧 + 1

3(𝑧 − 1)(𝑧 + 0.9)= lim

𝑧→1

𝑧 + 1

3(𝑧 + 0.9)= 0.3508

Example7.15: Find 𝑥(∞) when X(z) is given by X(z)=3𝑧+4

(𝑧−1)(𝑧+1)

Solution: By inspection we can find that, (z-1)X(z) has no pole on unit circle.

Example7.16: Find the final value of21

1

8.08.11

2)(

zz

zzX

Solution:

21

111

8.08.11

21)(1

zz

zzzXz

1

1

11

11

5.01

2

5.011

21

z

z

zz

zz

The final value theorem yields

102.0

2

8.01

2lim][

1

1

1

z

zy

z

Example7.17: Determine the Z-Transform of the signal y[n] = [1, 2, 5, 7, 0, 1]

Solution:Y(z) = z2 + 2z + 5 + 7z-1 + z-3

ROC: entire z-plane except z = 0 and z =±.

Example7.18: Determine the Z-Transform of the signalx[n] = [1, 2, 5, 7, 0, 1]

Solution:X(z) = 1 + 2z-1+ 5z-2 + 7z-3 + z-5,

ROC: entire z plane except z = 0

Example7.19: Determine the Z-Transform of the signal z[n] = [0, 0, 1, 2, 5, 7, 0, 1]

Solution: z-2 + 2z-3 + 5z-4 + 7z-5 + z-7, ROC: all z except z=0

Example7.20: Determine the Z-Transform of the signal p[n] = [n]

Solution:P(z) = 1, ROC: entire z-plane.

17

Example7.21: Determine the Z-Transform of the signal q[n] = [n – k], k> 0 Solution:Q(z) = z-k, entire z-plane except z=0.

Example7.22: Determine the Z-Transform of the signal r[n] = [n+k], k > 0 Solution:R(z) = zk,

ROC: entire z-plane except z = ±.

Example7.23: Determine the Z-Transform of the signal x[n] = (1/2)nu[n]

Solution:

1

2

1

0

1

0

2

11

1

.......2

1

2

11

2

1

2

1

][)(

z

zz

zz

znxzX

n

n

n

n

n

n

n

ROC: ||1

2𝑧−1| < 1, or equivalently 𝑧 >

1

2

Example7.24: Determine the Z-Transform of the signalx[n] = anu[n]

Solution:

|||:|

1

1

.......1

)(

1

211

0

1

0

azROC

az

azaz

azzazX

n

n

n

n

n

Example 7.25: Determine the Z-Transform of two-sided exponential sequence

1-n-u2

1-

3

1n

nunx

n

Solution:

11

1

0

1

0

1

3

11

1

3

11

3

1

3

1

3

1

zz

zz

zn

n

;……….

z

zROC

3

1

;1 3

1: 1

18

11

0

11

11

2

11

1

2

11

2

1

2

1

2

1

zz

zz

zn

n

; …………

z

zROC

2

1

;1 2

1: 1

2

1

3

1

12

12

2

11

1

3

11

1

11zz

zz

zz

zX

Figure 7.21 ROC two sided sequence

Example 7.27: Express the Z Transform of 𝑦(𝑛) = ∑ 𝑥(𝑘)𝑛𝑘=−∞

Solution: we can write the given expression as,

𝑦(𝑛) − 𝑦(𝑛 − 1) = ∑ 𝑥(𝑘)

𝑛

𝑘=−∞

− ∑ 𝑥(𝑘)

𝑛−1

𝑘=−∞

= 𝑥(𝑛)

By taking z transform on both the sides, (1 − 𝑧−1)𝑌(𝑧) = 𝑋(𝑧)

𝑌(𝑧) =𝑋(𝑧)

(1 − 𝑧−1)

Example 7.28: Find the final value of 21

1

8.08.11

2)(

zz

zzX

Solution:

21

111

8.08.11

21)(1

zz

zzzXz

1

1

11

11

5.01

2

5.011

21

z

z

zz

zz

The final value theorem yields

102.0

2

8.01

2lim][

1

1

1

z

zy

z 7.10 Inverse z-transform:

z-plane

Re

im

pole

X X oo

𝟏

𝟐 −

𝟏

𝟑

𝟏

𝟏𝟐

19

In the previous sections we defined z transform, its relation to Laplace and Fourier Transform

and found out the Z Transform of some signals and, then studied some Z Transform properties

and theorems. These properties and theorems simplifies the procedure to find Z Transform of

more complicated signals. Now to recover the time domain signal from Z domain signal,

i.e.,Inverse Z-Transform may be found by using any of the following methods:

(i) Power series method, or long division method (ii) Partial fraction expansion method (iii) Complex inversion integral method, or residue method

(a) Power Series Method:The long division method is quite simple but doesn’t give a closed form expression for the time signal. it is useful only if x(n) is a sequence which is either

purely right sided or left sided.The z Transform is expanded as

2112 2 1 0 1 2 zxzxxzxzxzX Z-transforms of this form can generally be inversed easily,especially useful for finite-length

series.the method is best illustrated by example 7.29 and 7.30

(b) Partial fraction expansion method: Assume that a given z-transform can be expressed as

N

k

k

k

M

k

k

k

za

zb

zX

0

0

Apply partial fractional expansion

s

mm

i

mN

ikk k

kNM

r

r

r

zd

C

zd

AzBzX

11

,11

0 11 First term exist only if M>N, Br is obtained by long division

Second term represents all first order poles

Third term represents an order s pole, there will be a similar term for every high-order pole

Each term can be inverse transformed by inspection

Coefficients are given as

kdz

kk zXzdA 11

1

11!

1

idw

s

ims

ms

ms

i

m wXwddw

d

dmsC

This method is Easier to understand with examples and is best illustrated by example 7.32 and

7.33

Remark: before doing partial-fraction expansion, make sure the z-transform is in proper

rational function of 𝒛−𝟏

(c) Complex inversion integral method: The inversion integral method requires a knowledge of the theory of complex complex variables,

but is quite powerful and useful.

20

Properties of z transform and z transform of some basic functions is listed in the following table,

this can be used as ready reference to solve inverse z transform problems.

Table 7.1

Sequence z-Transform Region of Convergence )(n 1 all z

)(nun 11

1 z

z

)1( nun 11

1 z

z

)(nun n 21

1

)1(

z

z

z

)1( nun n 21

1

)1(

z

z

z

)()cos( 0 nun 210

1

0

)(cos21

)(cos1

zz

z

1z

)()sin( 0 nun 210

1

0

)(cos21

)(cos1

zz

z

1z

Example 7.29: Consider the following algebraic expression for Z-Transform X(z) of a signal

x[n].

𝑋(𝑧) =1 + 𝑧−1

1 + (1

3)𝑧−1

Assuming ROC to be|𝑧| > 1/3, use long division method to determine x(0), x(1) and x(2).

Solution:

1 +1

3𝑧−1 )1 + 𝑧−1(1 +

2

3𝑧−1 −

2

9𝑧−2

1 +1

3𝑧−1

_________ 2

3𝑧−1

2

3𝑧−1+

2

9𝑧−2

-____-_______

-2

9𝑧−2

-2

9𝑧−2 −

2

27𝑧−3

_____+_____+__

2

27𝑧−3

By observing quotient we can write x(0)=1, x(1)=2/3 and x(2)=-2/9

21

Example 7.30: Consider the following algebraic expression for Z-Transform X(z) of a signal

x[n].

𝑋(𝑧) =1 + 𝑧−1

1 + (1

3)𝑧−1

Assuming ROC to be |𝑧| < 1/3, use long division method to determine x(0), x(-1) and x(-2).

Solution: 1

3𝑧−1 + 1 )1 + 𝑧−1(3𝑧 +3+9𝑧2

−1 − 3𝑧 _________ 𝑧−1 − 3𝑧 −𝑧−1 -3 ______ -3z-3 3z-9𝑧2 _____+_____+__ -3-9𝑧2

By observing quotient we can write x(0)=3, x(-1)=3 and x(-2)= 9

Example 7.31: Determine the inverse Z-Transform of

112

11 1112

zzzzzX

Solution:

12

1112

2

11

2

1z

112

11

zz

zzzzzX

12

11

2

12 nnnnnx

Hence, the time domain discrete signal is given as,

20

12

101

12

121

n

n

n

n

n

nx

Example 7.32: Determine the inverse Z-Transform of

22

2

1z :ROC

2

11

4

11

1

11

zz

zX

Solution: Order of numerator is smaller than denominator (in terms of z-1) hence, No higher

order pole

1

2

1

1

2

11

4

11 z

A

z

AzX

Figure 7.22 ROC Shaded region extends to infinity

1

4

1

2

11

1

4

11

1

4

1

1

1

z

zXzA

2

2

1

4

11

1

2

11

1

2

1

1

2

z

zXzA

2

1z

2

11

2

4

11

1

11

zz

zX

nu4

1-

2

12

n

nunx

n

ROC extends to infinity which Indicates right sided sequence

Example 7.33: Determine the inverse Z-Transform of

1z

12

11

1

2

1

2

31

21

11

21

21

21

zz

z

zz

zzzX

Solution: To solve it by partial fraction method first numerator is simplifiedbyLong division to

obtain Bo

15

23

2

1212

3

2

1

1

12

1212

z

zz

zzzz

,

12

11

512

11

1

zz

zzX

z-plane

Re

im

1

2

1

4

X X

23

1

2

1

1

1

2

11

2

z

A

z

AzX

92

11

2

1

1

1

z

zXzA

811

1

2

zzXzA

1z 1

8

2

11

92

11

zz

zX

n8u-2

192 nunnx

n

Figure 7.23 ROC Shaded region extends to

infinityIndicates right-sides sequence

Example 7.34: Determine the inverseZ-Transform of

21 5.05.11

1)(

zzzX

Solution: By dividing the numerator of X(z) by its denominator, we obtain the power series

...11

1 416313

8152

471

23

2

211

23

zzzz

zz

𝑥(𝑛) = [1, 3 2⁄ , 7 2,⁄ 15 8,⁄ 31 16,⁄ ]

Example 7.35:Determine theinverse Z-Transform of

21

1

22

4)(

zz

zzX

Solution: By dividing the numerator of X(z) by itsdenominator, we obtain the power series

𝑥(𝑛) = [2, 1.5,0.5, 0.25, … . . ]

Example 7.36: Find the signal corresponding to the Z-Transform

21

3

32)(

zz

zzX

Solution:

5.015.0

5.05.1

5.0

32)(

2321

3

zzzzzzzz

zzX

5.04

1

113

5.01

5.0)(22

zzzzzzzz

zX

5.0)4(

1

13)(

z

z

z

z

zzX

Or 11

1

5.01

14

1

13)(

zzzzX

z-plane

Re

im

1

2

1 X X

24

][5.04][]1[][3][ nununnnx n

Example 7.37: Find Inverse Z Transform of 𝐹(𝑍) =2𝑍

(𝑍−2)(𝑍−1)2

Solution: By Partial Fraction Expansion

2

321

211212

2

z

zk

z

zk

z

zk

zz

zzF

To find k1 multiply both sides of the equation by (z-2), divide by z, and let z2

232

121

2

1

2

1

2

z

zzk

z

zzkzk

z

z

232

121

2

1

2

1

2

z

zk

z

zkk

z

2

2

3

2

21

2

21

2

1

2

1

2

zzz

z

zk

z

zkk

z or

k1 = 2

Similarly to find k3 multiply both sides by (z-1)2, divide by z, and let z1

zkzkz

zk

z32

2

1 12

1

2

2

…….7.6 And

k3 = -2

Finding k2 requires going back to Equation 7.6 above and taking the derivative of both sides

zkzkz

zk

z32

2

1 12

1

2

2

222

122

12

2

12

2

2k

z

z

z

zk

z

Now again let z1

k2 = -2

Hence f(n)=𝑓(𝑛) = 2(2𝑛)𝑢(𝑛) − 2(1𝑛)𝑢(𝑛) − 2

Example 7.38: Find the signal corresponding to the Z-Transform

211 2.012.011

)(

zz

zY

Solution:

23

2.02.0)(

zz

zzY

21

2

1

2

2

2

z

z

z

z

z

zzF

25

222

2.0

1.0

2.0

75.0

2.0

25.0

2.02.0

)(

zzzzz

z

z

zY

22.01.0

2.0

75.0

1

25.0)(

z

z

z

z

z

zzY

211

2.01.0

112.01

2.0

2.01

175.0

2.01

125.0

z

z

zz

][2.05.0][2.075.0][2.025.0][ nunnununy nnn

Example 7.39: Find inverse Z-Transform of 𝑋(𝑧) = (1

1024) [

1024−𝑧−10

1−0.5𝑧−1] , |𝑧| > 0

Solution:

𝑋(𝑧) = (1

1024) [

1024−𝑧−10

1−0.5𝑧−1]=

1−(0.5𝑧)−10

1−0.5𝑧= ∑ (0.5𝑧−1)𝑘9𝑘=0

Therefore, [𝑛] = (0.5)𝑛 ; 0 ≤ n ≤ 9

Example 7.40: Determine the sequence 𝑥[𝑛] associated with Z-Transform using residue method.

1) 𝑋(𝑧) =10𝑧

(𝑧−1)(𝑧−2)

2)𝑋(𝑧) =(1 − 𝑒−𝑎)𝑧

(𝑧 − 1)(𝑧 − 𝑒−𝑎)

Solution: 1) The discrete sequence can be obtained from its z-transform X(z) by using inverse integral

𝑥[𝑛] =1

2𝜋𝑗∮ 𝑋(𝑧)𝑧𝑛−1𝑑𝑧

Let 𝑋(𝑧)𝑧𝑛−1 = 𝐺(𝑧)

𝑥[𝑛] =1

2𝜋𝑗∮ 𝐺(𝑧)𝑑𝑧 = 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑠𝑖𝑑𝑢𝑒𝑠 𝑜𝑓 𝐺(𝑧)𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 𝑝𝑜𝑙𝑒𝑠 𝑜𝑓 𝐺(𝑧)

Here, 𝐺(𝑧) =10𝑧𝑛

(𝑧−1)(𝑧−2)

G(z) has two poles at z=1 and z=2

X[n]=residue of G(z) at z=1+Residue of G(z) at z=2

Residue of G(z) at (z=1)=𝑅𝑧=1 = (𝑧 − 1)𝐺(𝑧)|𝑧 = 1

=10𝑧𝑛

(𝑧−2)|𝑧 = 1

= -10

Residue of G(z) at (z=2)=𝑅𝑧=2 = (𝑧 − 2)𝐺(𝑧)|𝑧 = 2 =10 2𝑛

Therefore, 𝑥[𝑛] = −10 +10 2𝑛=10(2𝑛 − 1); n≥0

2) The discrete sequence can be obtained from its z-transform X(z) by using inverse integral

26

𝑥[𝑛] =1

2𝜋𝑗∮ 𝑋(𝑧)𝑧𝑛−1𝑑𝑧

Let 𝑋(𝑧)𝑧𝑛−1 = 𝐺(𝑧)

𝑥[𝑛] =1

2𝜋𝑗∮ 𝐺(𝑧)𝑑𝑧 = 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑠𝑖𝑑𝑢𝑒𝑠 𝑜𝑓 𝐺(𝑧)𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 𝑝𝑜𝑙𝑒𝑠 𝑜𝑓 𝐺(𝑧)

Here, 𝐺(𝑧) =(1−𝑒−𝑎)𝑧𝑛

(𝑧−1)(𝑧−𝑒−𝑎)

G(z) has two poles at z=1 and z=𝑒−𝑎

x[n]=residue of G(z) at z=1+Residue of G(z) at z=𝑒−𝑎

Residue of G(z) at (z=1)=𝑅𝑧=1 = (𝑧 − 1)𝐺(𝑧)|𝑧 = 1

=(1−𝑒−𝑎)𝑧𝑛

(𝑧−𝑒−𝑎)|𝑧 = 1

= 1

Residue of G(z) at (z=𝑒−𝑎)=𝑅𝑧=𝑒−𝑎 = (𝑧 − 𝑒−𝑎)𝐺(𝑧)|𝑧 = 𝑒−𝑎

=(1−𝑒−𝑎)𝑧𝑛

(𝑧−1)|𝑧 = 𝑒−𝑎

= -𝑒−𝑎𝑛

Therefore, 𝑥[𝑛] = 1 − 𝑒−𝑎𝑛; n≥0

Example 7.41: Find inverse Z Transform of

2

1

2

3

12)(

2

2

zz

zzzX

This isRatio of polynomial z-domain functions, Divide through by the highest power of z then

we get,

21

21

2

1

2

31

21)(

zz

zzzX

Factor denominator into first-order factors

11

21

12

11

21)(

zz

zzzX

Use partial fraction decomposition to get first-order terms

1

2

1

10

1

2

11

)(

z

A

z

ABzX

Find B0 by polynomial division

27

15

23

2

1212

3

2

1

1

12

1212

z

zz

zzzz

Expressing in terms of B0we get,

11

1

12

11

512)(

zz

zzX

Solving for A1 and A2we get,

8

2

1

121

2

11

21

921

441

1

21

1

1

21

2

2

1

21

1

1

1

z

z

z

zzA

z

zzA

Then express X(z) in terms ofB0, A1, and A2

11 1

8

2

11

92)(

z

z

zX i.e,

nununnxn

82

1 9 2

Example 7.42: Find Z-Transform of the autocorrelation of 𝑥[𝑛] = 𝑎𝑛𝑢[𝑛], |𝑎| < 1 Let 𝑟𝑥𝑥(𝑙) be the autocorrelation of the given sequence x[n].

Z{𝑟𝑥𝑥(𝑙)} = X(z).X(𝑧−1) = (

1

1−𝑎𝑧−1) (

1

1−𝑎𝑧) =

1

1−𝑎(𝑧+𝑧−1)+𝑎2 ROC:|𝑎| < |𝑧| < 1/|𝑎|

Example 7.43: Find the Z-Transform of 𝑥[𝑛] = (𝑛 + 1)𝑢[𝑛] Answer: x[n] can be written as, 𝑥[𝑛] = ∑ 𝑢[𝑘]𝑛𝑘=0

Z-transform of which is X(z)=1

1−𝑧−1𝑈(𝑧)=(

1

1−𝑧−1) (

1

1−𝑧−1) ROC: |𝑧| < 1

Example 7.44: Find Z-Transform of 𝑥[𝑛] = 𝑎𝑛 ; 0 ≤ 𝑛 ≤ 𝑀 − 1

Solution:𝑋(𝑧) = ∑ 𝑎𝑛𝑀−10 𝑧−𝑛 = ∑ (𝑎𝑧−1)𝑛𝑀−10 =

1−(𝑎𝑧−1)𝑀

1−𝑎𝑧−1=

(𝑧𝑀−𝑎𝑀)

𝑧𝑀−1(𝑧−𝑎)

Note- Here, X(z) has M zeros which are located at 𝑧𝑘 = 𝑎𝑒𝑗2𝜋𝑘/𝑀 where k is varying from o to

M-1.

As far as pole is concerned, there are M poles out of which M-1 poles are located at zero and

remaining pole is at z=a.

28

Example 7.45: Find 𝑧−1 [𝑧2

(𝑧−𝑎)(𝑧−𝑏)]

Solution:

𝑧−1 [𝑧2

(𝑧−𝑎)(𝑧−𝑏)] = 𝑧−1 [

𝑧

𝑧−𝑎∙

𝑧

𝑧−𝑏]

= 𝑧−1 [𝑧

𝑧−𝑎] ∙ 𝑧−1 [

𝑧

𝑧−𝑏]

= 𝑎𝑛 ∗ 𝑏𝑛

= ∑ 𝑎𝑚𝑛𝑚=0 𝑏𝑛−𝑚

= ∑ 𝑎𝑚𝑛𝑚=0 𝑏𝑛𝑏−𝑚

= 𝑏𝑛 ∑ 𝑎𝑚𝑛𝑚=01

𝑏𝑚

= 𝑏𝑛 ∑ (𝑎

𝑏)

𝑚𝑛𝑚=0

= 𝑏𝑛 [1 + (𝑎

𝑏) + (

𝑎

𝑏)

2+ ⋯ + (

𝑎

𝑏)

𝑛]

= 𝑏𝑛 [(

𝑎

𝑏)

𝑛+1−1

(𝑎

𝑏−1)

] [∵ 1 + 𝑎 + 𝑎2 + ⋯ + 𝑎𝑛−1 =𝑎𝑛−1

𝑎−1]

= 𝑏𝑛 [(

𝑎𝑛+1

𝑏𝑛+1−1)

(𝑎−𝑏

𝑏)

]

= 𝑏𝑛 [(

𝑎𝑛+1−𝑏𝑛+1

𝑏𝑛+1)

(𝑎−𝑏

𝑏)

]

= 𝑏𝑛 [(

𝑎𝑛+1−𝑏𝑛+1

𝑏𝑛)

(𝑎−𝑏)]

= 𝑎𝑛+1−𝑏𝑛+1

(𝑎−𝑏)

𝑖𝑒. , 𝑧−1 [𝑧2

(𝑧−𝑎)(𝑧−𝑏)] =

𝑎𝑛+1−𝑏𝑛+1

(𝑎−𝑏)

Example 7.46: Find the inverse Z-transform of 𝑧2

(𝑧−𝑎)2 using convolution theorem.

Solution:

𝑍−1 [𝑧2

(𝑧−𝑎)2] = 𝑍−1 [(

𝑧

𝑧−𝑎) ∙ (

𝑧

𝑧−𝑎)]

= 𝑎𝑛 ∗ 𝑎𝑛 𝑤ℎ𝑒𝑟𝑒 ∗ 𝑑𝑒𝑛𝑜𝑡𝑒𝑠 𝑐𝑜𝑛𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

= ∑ 𝑎𝑘𝑛𝑘=0 𝑎𝑛−𝑘

= ∑ 𝑎𝑘𝑛𝑘=0 𝑎𝑛𝑎−𝑘

= ∑ 𝑎𝑛𝑛𝑘=0

29

= 𝑎𝑛 ∑ (1)𝑛𝑘=0

= 𝑎𝑛[1 + 1 + 1 + ⋯ (𝑛 + 1)𝑡𝑖𝑚𝑒𝑠]

= 𝑎𝑛 + 𝑎𝑛 + 𝑎𝑛 + ⋯ [(𝑛 + 1)𝑡𝑖𝑚𝑒𝑠]

= (𝑛 + 1)𝑎𝑛

𝑖𝑒. , 𝑍−1 [𝑧2

(𝑧−𝑎)2] = (𝑛 + 1)𝑎𝑛

𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟕. 𝟒𝟕: Find the inverse Z − transform of 𝑍−1 [𝑧

𝑧2+7𝑧+10].

Solution:

𝐿𝑒𝑡 𝑋(𝑧) = 𝑧

(𝑧+5)(𝑧+2)

Solving by partial fraction,

𝑋(𝑧)

𝑧=

1

(𝑧+5)(𝑧+2)=

𝐴

(𝑧+2)+

𝐵

(𝑧+5)

⟹ 1 = 𝐴(𝑧 + 5) + 𝐵(𝑧 + 2)

𝑃𝑢𝑡 𝑧 = −2 , 𝑤𝑒 𝑔𝑒𝑡

1 = 𝐴(−2 + 5) + 𝐵(0)

1 = 3𝐴

𝐴 =1

3

𝑃𝑢𝑡 𝑧 = −5 , 𝑤𝑒 𝑔𝑒𝑡

1 = 𝐴(0) + 𝐵(−5 + 2)

1 = −3𝐵

𝐵 =−1

3

∴ 𝑋(𝑧)

𝑧=

1

3(𝑧 + 2)−

1

3(𝑧 + 5)

𝑋(𝑧) = 𝑧

3(𝑧 + 2)−

𝑧

3(𝑧 + 5)

⟹ 𝑍{𝑥(𝑛)} = 𝑧

3(𝑧 + 2)−

𝑧

3(𝑧 + 5)

⟹ 𝑥(𝑛) = 1

3𝑧−1 [

𝑧

(𝑧 + 2)] −

1

3𝑧−1 [

𝑧

(𝑧 + 5)]

= 1

3(−2)𝑛 −

1

3(−5)𝑛

30

= 1

3[(−1)𝑛(2𝑛 − 5𝑛)]

= (−1)𝑛

3[2𝑛 − 5𝑛]

𝑖𝑒. , 𝑍−1 [𝑧

𝑧2 + 7𝑧 + 10] =

(−1)𝑛

3[2𝑛 − 5𝑛]

Exercise:

Review Question:

1) Define Z- transform.

2) What is the need for Z-transform?

3) Explain applications of Z –transform.

4) Define Differentiation property.

5) Define time reversal property.

6) Give expression for z transform.

7) Give expression for inverse z transform.

8) Define region of convergence.

9) How z-transform is related to Fourier transform.

10) Explain about the roc of causal and anti-causal infinite sequences?

11) What are the properties of ROC?

Summary The basic mathematical tool discussed in this chapter was the z‐transform. A useful table of z‐transforms was developed, and techniques for inverting the z‐transform were presented. A number of useful z‐transform theorems and concepts were then presented. These included the definition of linearity, the initial value and final value theorems, and a

discussion of the very important ROC and stability.

31

12) Explain the linearity property of the z transform

13) State the time shifting property of the z transform.

14) State the scaling property of the z transform

15) State the time reversal property of the z transform

16) Explain convolution property of the z transform

17) What are the different methods of evaluating inverse z-transform?

18) Give the Z-transform of unit sample sequence 𝛿(𝑛).

19) Define zeros.

20) Define poles.

21. Explain all the properties of z transform

22. Explain how the causalty of the system can be studied from region of convergence.

23. Explain power series method to find inverse Z-Transform

24. Explain partial fraction expansion method to find inverse Z-Transform

25. Explain residue method to find inverse Z-Transform.

26. Explain residue theorem.

27. How stability of a system can be evaluated using Z Transform.

28. Derive condition for stability using Z transform.

29. Explain unilateral Z transform.

30. How difference equations can be solved using Z Transform.

31. Derive the time advanced & delay property for unilateral Z-Tranform

Problems:

32

1 Determine the cross correlation sequence lrXY of the sequences

x (n)={…..0,0,2,-1,3,7,1,2,-3,0,0,…..} and y (n)={…..0,0,1,-1,2,-2,4,1,-2,5,0,0…..}

2 Determine the z-transform of the signal

0,0

0,

n

nnunx

n

n .

3 Determine the transform of the signal

1,

0,01

nnunx

n

n

.

4 Determine the z-transform of the signal 1 nubnunx nn .

5 Determine the z-transform and the ROC of the signal nunx nn 3423 6 Determine the z-transform of the signals nunnx 0cos

7 Determine the z-transform of the signals nunnx 0sin

8 Determine the transform of the signal

elsewhere

Nnnx

,0

10,1.

9 Determine the autocorrelation sequence of the signal 11, anuanx n .

10 Evaluate the inverse z transform of ||||,

1

11

azaz

zX

using the

complex inversion integral.

11 Determine the inverse z- transform of

21 5.05.01

1

zz

zX ,

When (a) ROC: |z|>1

(b) ROC: |z|

33

e. 𝑥(𝑛) = 4 𝑒−3𝑛 sin(0.1𝜋𝑛) 𝑢(𝑛)

16 Using the properties of the z-transform, find the z-transform for each of the

following sequences

a. 𝑥(𝑛) = 𝑢(𝑛) + (0.5)𝑛 𝑢(𝑛)

b. 𝑥(𝑛) = 𝑒−3(𝑛−4) cos(0.1𝜋(𝑛 − 4)) 𝑢(𝑛 − 4),

where 𝑢(𝑛 − 4) = 1 for 𝑛 ≥ 4 and 𝑢(𝑛 − 4) = 0 for 𝑛 < 4.

17 Given two sequences

𝑥1(𝑛) = 5 𝛿(𝑛) − 2 𝛿(𝑛 − 2)

𝑥2(𝑛) = 3 𝛿(𝑛 − 3)

a. determine the z-transform of convolution of the two sequences using the

convolution property of the z-transform, 𝑋(𝑧) = 𝑋1(𝑧)𝑋2(𝑧)

b. determine convolution by the inverse z-transform from the result in part (a),

𝑥(𝑛) = 𝑍−1(𝑋1(𝑧)𝑋2(𝑧))

18 Using the Table 7.1 and z-transform properties, find the inverse z-transform for each

of the following functions,

a. 𝑋(𝑧) = 4 −10𝑧

𝑧−1−

𝑧

𝑧+0.5

b. 𝑋(𝑧) =−5𝑧

(𝑧−1)+

10𝑧

(𝑧−1)2+

2𝑧

(𝑧−0.8)2

c. 𝑋(𝑧) =𝑧

𝑧2+1.2𝑧+1

d. 𝑋(𝑧) =4𝑧−4

𝑧−1+

𝑧−1

(𝑧−1)2+ 𝑧−8 +

𝑧−5

𝑧−0.5

19 Using the partial fraction expansion method and Table 5.1, find the inverse z-

transform for each of the following functions,

a. 𝑋(𝑧) =1

𝑧2−0.3𝑧−0.04

b. 𝑋(𝑧) =2

(𝑧−0.2)(𝑧+0.4)

c. 𝑋(𝑧) =𝑧

(𝑧+0.2)(𝑧2−𝑧+0.5)

d. 𝑋(𝑧) =𝑧(𝑧+0.5)

(𝑧−0.1)2(𝑧−0.6)

20 A system is described by the difference equation

𝑦(𝑛) + 0.5𝑦(𝑛 − 1) = 2 (0.8)𝑛𝑢(𝑛)

determine the solution when the initial condition is 𝑦(−1) = 2.

21 A system is described by the difference equation

𝑦(𝑛) − 0.5𝑦(𝑛 − 1) + 0.06 𝑦(𝑛 − 2) = (0.4)𝑛−1𝑢(𝑛 − 1)

34

determine the solution when the initial condition are 𝑦(−1) = 1 and 𝑦(−2) = 2.

22 Given the following difference equation with the input-output relationship of a

certain initially relaxed system (all initial conditions are zero),

𝑦(𝑛) − 0.7𝑦(𝑛 − 1) + 0.1 𝑦(𝑛 − 2) = 𝑥(𝑛) + 𝑥(𝑛 − 1)

a. Find the impulse response sequence 𝑦(𝑛) due to the impulse sequence 𝛿(𝑛).

b. find the output response of the system when the unit step function 𝑢(𝑛) is

applied

23 Given the following difference equation with the input-output relationship of a

certain initially relaxed DSP system (all initial conditions are zero),

𝑦(𝑛) − 0.4𝑦(𝑛 − 1) + 0.29 𝑦(𝑛 − 2) = 𝑥(𝑛) + 0.5𝑥(𝑛 − 1)

a. Find the impulse response sequence 𝑦(𝑛) due to the impulse sequence 𝛿(𝑛).

b. find the output response of the system when the unit step function 𝑢(𝑛) is

applied