Z-TRANSFORM

In today’s class…

Z-transform Unilateral Z-transform Bilateral Z-transform Region of Convergence

Inverse Z-transform Power Series method Partial Fraction method

Solution of difference equations

Need for transformation?

Why do we need to transform our signal from one domain to another?

Information available in one domain is not sufficient for complete analysis Looking at a sine wave in time-domain, we cannot really

know the frequency content

So we have to look into the frequency domain

An alternate domain may express the information more comprehensively A pole-zero map easily tells whether a systems is stable or

not

Z-transform

Digital counterpart for the Laplace transform used for analog signals

Mathematically defined as,

X (z) x[n] z n

n

This equation is in general a power series, where z is a complex variable.

Derivation…

The continuous-time Fourier transform of x(t) is given as,

Fxt xt e j 2ft dt

And the discrete-time Fourier transform of x[nT] is given as,

n

D xnT e F xnT j 2fnT

The Z-transform of x[n] is given as the Fourier transform of x[n] multiplied by rn

nT D D n x F r x nT

D x nT

D x nT F D r x nT n

r n x nT e j 2 fnT

n

Bi-lateral Z-transform

n

D

n

n

D

n D

x nT z n x nT

z re

x nT re x nT

x nT

j 2 fT

j 2 fT

x nT r n e j 2 fnT

Uni-lateral Z-transform

n

n D xnT z xnT

n0

X z xnT z n

n0

xnT xnT z n

where z-1 would show a delay by one sample time

Example 1: Find the z-transform of the following finite-length sequence

y nT

- 2 - 1 4 5 6 7

4 3 . 5

3 2 . 5

2 1 . 5

1 0 .5

0 0 1 2 3

0 0 0 2 0 4 3 2 0 0 ynT ynT 2 n 1T 4 n 3T 3 n 4T 2 n 5T

2 0 4 3 2

2 0 4 3 2 0 0

Y z 0

ynT 0

0 0z n

Y z ynT z n

n0

n0

Y z 0z 0 2z 1 0z 2 4z 3 3z 4 2z 5 0z 6 0z 7

Y z 2z 1 4z 3 3z 4 2z 5

So, Y(z) would exist on the entire z-plane except the point z=0

Z-transform as Rational Function

Q(z)

Often it is convenient to represent Z-transform X(z) as a rational function

X (z) P(z)

Where P(z) and Q(z) are two polynomials The values of z at which X(z) becomes zero (X(z) = 0) are called zeros The values of z at which X(z) becomes infinite (X(z) = ∞), are called poles

Significance of Poles & Zeros

Poles Roots of the denominator Q(z)

The point where H(z) becomes infinite

The point where H(ejw) shows a peak value

System may become unstable

Zeros Roots of the numerator P(z)

The point where H(z) becomes zero

The point where H(ejw) shows maximum attenuation

Convergence issues

A power series may not necessarily converge The infinite sum may not always be finite

The set of values of z for which the z-transform

converges is called Region of Convergence (RoC)

The convergence of X(z) depends only on z and it converges for

n | x [ n ] z | n

Replacing

n

n jw x [ n ] re

jw re z re j 2 fn

x [ n ] r n e jwn

| x [ n ] z n | n

X ( z )

1

n

x[n]r n n0

X (z) x[n]r n

n

This equation can be segmented into two parts, one for the right-sided (causal) signal and second for the left-sided (non-causal) signal

n0 n r

x[n] X (z) x[n]rn

n1

For X(z) to exist in a particular region (for certain values of z), both summations must be finite in that region

so that For the first summation, r should be small enough |x[-n]rn| converges when summed to infinite terms

For the second summation, r should be large enough so that |x[n]/rn| converges when summed to infinite terms

So, there are two circles with radius rL & rR for the sequence x[n]

If it is defined as a left-sided sequence (non-causal), then the second summation becomes zero (by definition), and the radius rL should be small enough to make the first summation converge

If x[n] is defined as a right-sided sequence (causal), then the first summation becomes zero (by definition), and the radius rR

should be large enough to make the second summation converge

rrRR

rr L

Example 2: Find the z-transform of the following finite-length sequence

0.2

0.4

0.6

0.8

1

x[n] 2 [n 2] [n 1] 2 [n] [n 1] 2 [n 2] The z-transform of this sequence is given as,

it is clear to see that the sequence does not have any poles

-1 -0.5 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0 4

0 Real Part

Imag

inar

y P

art

(denominator is 1), it has 4 zeros

It can be observed that X(z) becomes undetermined at z = 0 and z = ∞, so the RoC is entire z-plane except at z = 0 and z = ∞

X ( z) 2 z 2 z 2 z 1 2 z 2

Example 3: Find the z-transform of the following right-sided sequence

x[n] anu[n]

X (z) anu(n)z n

n

an z n

n0

(az 1 )n

n0

For convergence we require X (z)

z 1

n0 1 az z a

1 (az 1 )n X (z)

Now, X(z) will not exist for z=a & RoC is entire z-plane except z=aHowever, since the z-plane is a circle so we have to use the following condition (the sequence is right-sided)

| z || a |

Example 4: Find the z-transform of the following right-sided sequence

x[n] anu[n 1]

X (z) anu(n 1)z n

n

1

an z n

n

an zn

n1

1 an zn

n0

z

1 n0 1 a z z a

1 (a1 z)n 1 X (z) 1

For convergence we require X (z)

Now, X(z) will not exist for z=a & RoC is entire z-plane except z=a However, since the z-plane is a circle so we have to use the following condition (the sequence is left-sided)

| z || a |

Concepts

From the two examples we observe that the closed form equations for the z-transform of causal & non- causal signals come out to be same

This creates an ambiguity about the existence of

their z-transform

Therefore, we require complimentary information

apart from the closed form equations, i.e. the RoC

Properties of RoC

Property 1: The RoC is a ring or disk in the z-plane centred at the origin; i.e., 0 rR z rL

Property 2: The RoC cannot contain any poles

Property 3: If x[n] is a finite-duration sequence i.e.

a sequence that is zero except in a finite interval

N1 n N2 , then the RoC is the entire z-plane except possibly z=0 and z=∞

Property 4: If x[n] is a right-sided sequence i.e. a sequence that is zero for, , the RoC extends outward from the outermost (i.e. largest magnitude) finite pole in X(z) to z=∞

Property 5: If x[n] is a left-sided sequence i.e. a

n N1

sequence that is zero for, , the RoC extends outward from the outermost (i.e. largest magnitude) finite pole in X(z) to z=0

n N2

( n ) 1

( n m ) z m All z except 0 (if m>0) or (if m<0)

Z-transform pairs

Sequence z-transform RoC

All z

a n u ( n ) 1

1 az 1 | z | | a |

anu(n 1) 1

1 az 1 | z || a |

nanu(n) (1 az 1 )2

az 1

| z || a |

1

[sin 0 n]u(n) 0

0

1 [2 cos ]z 1 z 2

[sin ]z 1 | z | 1

[cos 0 n]u(n) 0

1[2 cos ]z 1 z 2

0

1[cos ]z 1 | z | 1

Sequence z-transform RoC

n [r sin 0n]u(n) 0 1 2 2

1[2r cos 0 ]z r z [r sin ]z 1

| z | r

n [r cos 0 n]u(n) 0

0

1[2r cos ]z 1 r 2 z 2

1[r cos ]z | z | r

Example 5: Find the RoC of x[n] (0.5)n u[n] (0.4)n u[n]

Using the properties of z-transform we get

1 1

1 0.4z 1 1 0.5z 1 X (z)

z z z(z 0.4) z(z 0.5)

z 0.5 z 0.4 (z 0.5)(z 0.4)

It is clear that the RoC is given by | z || 0.4 | and | z || 0.5 |

So we can conclude that the RoC is | z || 0.5 |

Example 6: Find the RoC of x[n] (0.5)n u[n] (0.9)n u[n 1]

Using the properties of z-transform we get

1 1

1 0.9z 1 1 0.5z 1 X (z)

z z z(z 0.9) z(z 0.5)

z 0.5 z 0.9 (z 0.5)(z 0.9)

The RoC due to the first part is | z || 0.5 | since it is a right-sided sequence

however, the second part is a left-hand sequence, therefore its RoC is

| z || 0.9 |

So we can conclude that the RoC for X(z) is | 0 .5 | | z | | 0 .9 |

Inverse Z-transform Power Series method

Simple Tedious for large n Not accurate

Partial Fraction method

Complicated More accurate

IZT: Power Series method

In this method we divide the numerator of a rational Z-transform by its denominator

The basic idea is

“Given a Z-transform X(z) with its corresponding RoC, we can expand X(z) into a power series of the form

n

which converges in the given RoC”

n X (z) cn z

Since RoC is the exterior of the circle, so we expect a right-sided sequence, so we seek an expansion in the negative powers of z By dividing the numerator of X(z) by its denominator, we obtain the power series

Example 7: Find the Inverse Z-transform of X(z) 1

11.5z10.5z2 X(z) RoC |z|>1

1 1 3 z 1 7 z 2 15 z 3 31 z 4 ...

2 4 8 16 1 3 z 1 1 z 2

2 2

x[n] = [1, 3/2, 7/4, 15/8, 31/16,…. ]

Since RoC is the interior of the circle, so we expect a left-sided sequence, so we seek an expansion in the positive powers of z By dividing the numerator of X(z) by its denominator, we obtain the power series

Example 8: Find the Inverse Z-transform of X(z) 1

11.5z10.5z2 X(z) RoC |z|<1

1 2 z 2 6 z 3 14 z 4 30 z 5 ... 1 3 z 1 1 z 2

2 2

x[n] = […., 30, 14, 6, 2, 0, 0]

IZT: Partial Fraction method

Steps to follow Eliminate the negative powers of z for the z-transform

function X(z)

Determine the rational function X(z)/z (assuming it is

proper), and apply the partial fraction expansion to the determined rational function X(z)/z using formulae in table (next slide)

Partial fraction(s) and formulas for constant(s)

Partial fraction with the first-order real pole:

A z p z z p A (z p) X (z) |

Partial fraction with the first-order complex poles:

Az A* z

z p z p * z z p A (z p) X (z) |

Partial fraction with mth-order real poles:

Ak A1

(z p)k

z p (z p)2

Ak 1 k z z p

A

X (z) | (z p)k 1 d

(k 1)! dzk 1

k 1

, A A*

An example for Simple Real Poles

An example for Multiple Real Poles

𝑓 𝑛 =[9(0.3)n­8(0.2)n+2n(0.2)n]u(n) 

Pulse Transfer Function

Pulse transfer function H(z) is defined as the ratio of the Z-transform of the input x[n] to the Z-transform of the output y[n]

Y z X z H z

Derivation

l k

y n bi xn i ai y n i i 0 i 1

Applying Z-transform and moving the terms of ‘y’ to one side

i i i b X zz Y za z Y z

l

i i0

k

i1

l i i

l i i

k i i

b z a z X z

b z a z X z z Y z Y

i i0

k

i i1

i0 i1

Y z1

i

l i bi z

X z Y z

H z k

i1

i0

1 ai z

l ibi z

Y z H z i0 k

a2 0.02 a1 0.1

b0 2 b1 1 1

Example 8: Find the Pulse Transfer function of the difference equation

yn 0.1yn 1 0.02 yn 2 2xn xn 1

i i X z 1 a z

i1

1

i 2

i i1

i

a z

bi z H z i0

1 0 1

1 0.1z 1 0.02z 2

2 z 1 0.1z 1 0.02z 2

2z 1z H z