Unit 5

Z- transformation

5.1 Z transform of Discrete time signal

The Z transform for discrete time system x(n ) is defined as

X(z) =    ------- (1) where z is a complex variable.

In polar form z can be expressed as

z  = r e jw  ---------------(2) where r is the radius of a circle.

For n≥ 0

X(z ) =       --------- (3)which is called one-sided z-transform.

By substituting z = r e jw

X(r e jw)  =    (r e jw) –n     ------- (4)

    e- jwn       ----- (5)

Equation(5) represents the Fourier transform of the signal x(n) r-n

Hence the inverse DTFT  X(r ejw) must be x(n) r-n.

x(n) r-n = 1/2π r e jw )  e jwn dw

On multiplying both sides by rn  we get

x(n) = 1/ 2 π   r ejw) (r ejw ) n dw                  [ z = r ejw

Let z= r e jw  and dw= dz/jz 

Dz = r ejw dw

Dw = dz/jre jw

Example

Find the z-transform for the sequence

x[n] = 2 + 3 + 5 + 2

Solution:

X(z) = 2 + 3 z-1 + 5 z -2 + 2 z -3   

Example

If X(z)= 4 – 5 z-2 + z-3 – 2z -4    then find x[n]

Solution:

x[n] = 4 - 5 + - 2

5.2 LTI system

For discrete time LTI system with Impulse response h[n]

For input z n

The output response y[n] = H(z) z n

Where

H(z) =   z -n

The Z-transform of general discrete time signal x[n] is defined as

X(z) =   z -n

5.3 Solution to difference equation

The z-transform can be used to convert a difference equation into an algebraic equation in the same manner that the Laplace converts a differential equation into an  algebraic equation

The difference equation DE contains the unknown function x(n) and shifted versions of it such as x(n−1) or x(n+3).  The solution of the equation is the determination of x(t). A linear DE has only simple linear combinations of x(n) and its shifts.

Z transform converts the difference into algebraic equation in z-domain.

Find the impulse response and step response for the following systems:

y(n) = -  ¾ y(n-1) + 1/8 y(n-2) = x(n)

y(n)  - ¾ y(n-1) + 1/8 y(n-2) = x(n)

Taking z-transform on both sides we get

Y(z) – ¾ [ z-1 Y(z) + y(-1) ] +1/8 [ z-2 Y(z) + z-1 y(-1)+y(-2)] = X(z)

Substituting y(-1)=y(-2)= 0

Y(z) -3/4 z-1 Y(z) + 1/8 z-2 Y(z) = X(z)

Y(z) =     1____________

1-      ¾ z-1 + 1/8 z-2

Impulse response

x(n) =        X(z) =1

Y(z) =    1____________           =      1____________ 

1-      ¾ z-1 + 1/8 z-2                 1-  ¾ z-1 + 1/8 z-2  

Y(z)  =  z__________       =        A___        +    B__

X(z) (z-1/2)(z-1/4)  (z-1/2)            (z-1/4)

By solving A=2 and B=-1.

Y(z) = 2 z          -       z

z-1/2       (z-1/4)

y(n) = 2 (1/2)n u(n) – (1/4) n u(n).

Step Response

x(n) = u(n)   X(z) = z/z-1

Y(z)  =         1_______

X(z)   1-3/4 z-1 + 1/8 z-2

Y(z) = z     + z2___________

z-1  z2 -3/4 z +1/8

Y(z)   =    z2___________

z     z2 -3/4 z +1/8

Y(z)   =    z2___________

z     (z-1)(z-1/2)(z-3/4)

=               A       +     B   +      C

z-1        z-1/2      z- 1/4

By solving A=8/3   B= -2  C= 1/3

Therefore

Y(z) = 8 . z      -2  z     + 1/3 z

3     z-1      z-1/2 z-1/4

y(n) = 8/3 u(n) – 2(1/2)n u(n) +1/3 (1/4) n u(n)

5.4 Application of Z transform to open loop system

Consider the following diagrams:

Y( s) = [Y(s)]*

Y(s) = [ X(s) G(s)]*

= [X G(s)]*

Taking z-transform on both sides we get

Y(z) = XG(z)

Y(s) = G1(s) G2(s) X*(s)

Y*(s) = [Y(s)]*

= X*(s) [G1(s)G2(s)]*

= X*(s) [G1G2(s)]*

Y(z) = X(z) G1G2(z)

5.5 Region of Convergence

Properties of Region of Convergence

1. The ROC is a ring or disk in z-plane centered at the origin.
2. The ROC cannot contain any poles.
3. If x(n) is finite duration casual sequence then the ROC is the entire z-plane except at z=0.
4. If x(n) is a finite duration anti-casual sequence then the ROC is the entire z-plane except at z=∞.
5. If x(n) is a finite duration two- sided sequence the ROC is entire z-plane except at z=0 and z=∞.
6. If x(n) is infinite duration two- sided sequence ROC will consist of a ring in z-plane bounded on the interior and exterior by a pole not containing any poles.
7. The ROC of an LTI stable system contains the unit circle
8. ROC must be connected region.

Find the z-transform and ROC of the signal

x(n) = an u(n)

Solution: 

X(z) =  

= an u(n) -----(1)    u(n) = 0 for n<0

1 for n≥0

= an ------- (2)

= n   ------- (3)

This is a geometric series of infinite length that is

a + ar + ar2 + ………….. ∞ = a /1-r if |r| <1

Then from equation (3) it converges when |az-1| < 1 or |z| >|a|

Therefore

X(z) = 1/ 1-az-1:  ROC    |z| > |a|

Find the z-transform of the signal x(n) =-b n u(-n-1). Find ROC

X(z) =  

X(z) =   bn u(-n-1)                                                                   u(-n-1) =0 for n ≥0

= 1 for n ≤ -1

= bn   = b-1  =

= b-1z/1- b-1z = z/ z-b = 1/ 1-bz-1                                                               |z| < |b|

Find the z-transform of x(n) = an u(n) – bn u(-n-1)

X(z) =  

=  n  + b-1

= z/z-a + z/z-b       ROC |a| < |z| < |b|

|b|< |a|

|b| >|a|

5.6 Z-domain analysis.

A block diagram consists of three elements, that is, the adder, the coefficient multiplier and the delayer. Figure shows these elements.

Example. Suppose that a causal linear time-invariant discrete-time system is characterized by

We obtain Y(z)=(1/4)z1Y(z)+(1/8)z2Y(z)+ X(z)(7/4)z1X(z)(1/2)z2X(z).

References:

      Signals and Systems by Simon Haykin

      Signals and Systems by Ganesh Rao

      Signals and Systems by P. Ramesh Babu

      Signals and Systems by Chitode