z - transform

Z-TRANSFORM

Introduction to Z - Transform

A generalize form of Discrete Time Fourier Transform (DTFT)

Why z-Transform?

In many cases, DTFT of a sequence does not exist.

Permits simple algebraic manipulations

Provide access to transient behavior that monitors all states stability of a system

Definition

For a sequence z-transform G(z) is :

Where z = Re(z) + jIm(z) is a continuous complex variable.

Definition (cont’d)

Then we express in polar form , then, earlier equation reduces to :

G(

Compare to DTFT:

G(

Tells us that the equation for the z-transform is actually a DTFT of the modified sequence


Location of point z in the complex z-plane gives a geometrical interpretation of the z-transform

Location of in the complex z-plane is at the tip of the vector of length originating at = 0.

Evaluation of Fourier Transform is made by traversing the unit circle (clockwise or counter clockwise).

Complex z-plane

𝑧=𝑟 𝑒 𝑗 𝜔


Condition of convergence of the infinite series:

For a sequence we will be having G(z) and its corresponding ROC

ROC:

Example 1

Determine of the causal sequence and its ROC.

=

= , for

In term of z, ROC :

ROC in the complex z-plane

𝛼

Example 2

Determine of the anti-causal sequence

and its ROC.

=

= , for

In term of z, ROC :

ROC in the complex z-plane

𝛼

Rational z-Transforms

All pertinent z-transforms are rational function to , or ratio of polynomials and in :

Questions:

1. For the sequence , is given by:

a) d)

b)

c)

2. For the sequence , which of the following ROC of is true?

a)

b)

c)

d)

Region of Convergence of a

z-Transform

𝑋 (𝑧 )𝑛= ∑𝑛=−∞

∞

𝑥 (𝑛) 𝑧−𝑛

Possibilities Region of Convergence

•Right-Sided Signal

• Left-Sided Signal

• Two-Sided Signal

Right-Sided Signal

• A right-sided sequence values only for n > 0

Right-Sided Signal

• ROC at |z| > x(n)

Right-Sided Signal

• Example

X[n] u[n]

u[n]𝑧

Right-Sided Signal

• The z-transform of u[n] is with an ROC at |z| > 0.5

Left-Sided Signal

• A left-sided sequence values only for n < 0.

Left-Sided Signal

• ROC at |z| < x(n)

Left-Sided Signal• Example

X[n] u[-n - 1]

-

-

1 -

u[-n - 1]𝑧

Left-Sided Signal

• The z-transform of u[-n - 1] is with an ROC at |z| < 0.5

Two-sided Signal

• +

Two-sided Signal• Example

X[n] u[-n - 1]

-

-

u[-n - 1]𝑧

Two-sided Signal

• The ROC at 0.5 < |z| < 0.75

Question

1. Determine the possibilities of ROC.

A. Right-Sided Signal

B. Left-Sided Signal

C. Two-Sided Signal

D. All the above

2.

A left-sided sequence x[n] values only for?

A. n<0.

B. n>0

C. n=0

D. n≠0

INVERSE Z-TRANSFORM

For a give sequence x[n] we can define Z(x[n])

Where z is

Hence general expression for the inverse z-transform

Where C is a closed-contour that lies inside the unit circle on the z-plane, and encircles the point z = {0, 0}.

The process by which a Z-transform of a time namely X(z), is returned to the time domain

INVERSE TRANSFOR BY LOOK-UP TABLE

Example: Determination of the inverse transform h[n] using look up table

H(Z)=

Hence H(z) can be written as =

From the table we get h[n]=nu[n].

INVERSE TRANSFORM BY PARTIAL-FRACTION EXPRESSION

A rational G(z) can be expressed as: G(z) = where P(z) and D(z) are the polynomials in 1 . If the degree M of the numerator polynomial P(z) is greater than or equal to the degree N of the denominator polynomial D(z), we can divide P(z) by D(z) and re-express G(z) as:

Where the degree of the polynomial P (z) 1 is less that that of D(z). and G(z) =

is called a proper fraction.

INVERSE TRANSFORM BY PARTIAL-FRACTION EXPRESSION

Let z –transform of a causal sequence h[n] be given by

H(Z)= =

PARTIAL-FRACTION EXAMPLE

H(z) is then of the form +

Obtain A= H(z)|z=.2 = |z=0.2 =2.75

B== |z=-0.6 =-1.75

Substituting the value H(z) = +

h[n]=2.75(u[n] - 2.75(u[n]

INVERSE Z-TRASNFORM USING LONG DIVISION

This method can be used to determine finite number of the coefficient of the inverse z-transform {g[n]} beginning with sample n=0.

Consider an example

PARTIAL-FRACTION EXPANSION USING MATLAB

The M-file residuez can be used to develop the partial fraction expansion of the z-transform and to convert a z-transform expressed in partial-fraction form to its original form

Syntax : [r,p,k]=residuez(num,den)

example: using MATLAB we determine the partial fraction of z-transform G(z)

G(z) =

>>num=[18];

>>den=[18 3 -4 -1]; [r,p,k]=residuez(num,den)

INVERSE TRANSFORM VIA POWER SERIES EXPANSION

If the z-transform is given as a power series in the form:

Example of finite length sequence

𝑋 (𝑧 )=𝑧 2(1−12𝑧− 1)(1+𝑧−1)(1− 𝑧− 1)

Multiply the above z.-transform 𝑥 (𝑧 )=𝑧 2−12𝑧−1+

12𝑧−1

By inspection the corresponding sequence is

Inverse z-transform using MATLAB

The inverse of rational z-transform sequence can also be calculated using matlab. To end this both M-file impz and M-file filter , impz is used to determine the first 11 coefficients of the inverse z-transform

Example:[h,t] = impz(b,a) returns the impulse response of the filter with numerator coefficients, b, and denominator coefficients, a. impz chooses the number of samples and returns the response in the column vector, h, and the sample times in the column vector, t. t = [0:n-1]' and n = length(t) is computed automatically

H(Z) =

s + 2 ------------------ s^2 + 0.4 s - 0.12

>>den=[1 0.4 -0.12];>>num=[1 2];

Colum 1 through 7 1.0000 1.6000 -0.5200 0.4000 -0.2224 0.1370 -0.0815Colum 8 through 11 0.0490 -0.0294 0.0176 -0.0106

Questions:

1. The general formulae for inverse z-transform can be expressed as:

a) x[n] =

b) x[n]=

c) x[n]=

Questions:

2.The inverse z-transform of 1 is :

a)δ[n]

b)u[n]

c)1

Z-TRANSFORM THEOREMLet G[z] denotes the z-transform of g[n] and ROC of g[z] is Rg.

For properties involve two sequence and associated z-transform, the transform pairs will be denoted as:

Example

Continue….

Computation of the convolution sum of finite-length sequence

Questions: q1

Questions:q2

The region of convergence of conjugated g[n] is

a) Rg

b) 1/Rg

c) Rg, except possibly the point z=0 or

d) ||Rg

TRANSFER FUNCTION

The transfer function of an LTI system is simply the z-transform of its impulse response:

TRANSFER FUNCTION = H(z)

= Z{h[n]}

= Z {IMPULSE RESPONSE}

= h[0] + h[1] + h[2] + . .

That’s it?! What is the use of that??....

System Transfer Functions

o Why should you care about z-transforms and transfer functions? Because of problems like this:

Given that the response of an LTI system to input u[n] is y[n] = d [n] − d[n − 1]. Compute the response of the system to input u[n], That is:

If x[n] = *U[n] → LTI → y[n] = d [n] − d[n − 1]

Then x[n] = u[n]→ LTI → y[n] =?

By the end of this presentation, we will solve this problem, and quite easily. How? Listen carefully…...

Z-Transform Relations

1. Convolution:

Assuming x[n] and y[n] are causal

Where:

X(z) = Z{x[n]} and Y (z) = Z{y[n]}

Then:

Z{x[n] ∗ y[n]} = X(z)Y(z)

Z-Transform Relations

2. Time shift:

If x[n]→ X[z]

what is the Z-transform of x[n-1]?

Let yn = x[n-1]

Z(yn) = X(z)

WAYS TO SOLVE PROBLEMS USING

Z-T-FUNCTION

Input- Output Pairs

Block Diagrams

Input-Output Pairs

Suppose that the response of a given LTI system to input x[n] is output y[n]:

x[n] → LTI → y[n] = h[n] ∗ x[n]

→ Y (z) = H(z)X(z)

H(z) =

EXAMPLES…..• Given that the response of an LTI system to input

x[n] = u[n] is y[n] = d [n] − d[n − 1].

• Compute the step response, which is the system’s response to a step input u[n]

• H(z) =Y (z)/X(z)

= [1 − ] / [ = 2 − 3+ → h[n] = {2,−3, 1}

• Now we could compute the step response using convolution

Y (z) = H(z)X(z) = [2 − 3+ ]*[1/(1 - )]

= 2 − → y[n] = {2,−1}

BLOCK DIAGRAMSExample: Find the T-Function of the system.

Cont…….. yn = 0.48y[n-2] - 0.2y[n-1] + x[n] + 0.5x[n-1]

Y(z) = Y(z)[0.48- 0.2] + X(z)[1 + 0.5]

H(z)=Y[z]/X(z)

= [1 + 0.5] / [1- 0.48- 0.2]

Questions:

1.The transfer function of LTI system is simply the z-transform of its impulse response.

a. True b. False

2.One among the ways used to compute the step response of the system in Z- Transform is:

a. Input-output pairs

b. Linear equation

c. Laplace transform

z - transform

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