part b chapter1 z-transform-modeemedia.ee.unsw.edu.au/contents/elec3104/lecturenotes/dsp3.pdf ·...

Chapter 1: Introduction to z-Transform

1.1 z-Transform1.2 Properties

1.2.1 Linearity1.2.2 Shifting1.2.3 Time Reversal1.2.4 Multiplication by Exponential Sequence1.2.5 Differentiation in the z-domain1.2.6 Discrete Convolution

1.3 Inverse z-transform1.3.1 Relationship between the z-transform and the

Laplace Transform1.4 Frequency Response Estimation1.5 Pole-Zero Description of Discrete-Time Systems1.6 A Second-Order Resonant System1.7 Problem Sheet B1Prof

r E. A

mbikair

UNSW, A

Chapter 1: Introduction to z-Transform [1,3]

fT πωθ 2==

Analogue Domain(Continuous-time Domain)

LaplaceTransform

ω-analoguefrequency

jωS-PlaneStable

Region

Discrete-time Domain

z-transform

z=ejθ

X(ejθ)

-π≤θ≤π

t = nT

Im(z) z-Plane

θ = πθ = -π Re(z)

stable region

|z|=1(unit circle)

-digital

frequency

θ -digital frequency

Profes

. Ambik

airaja

UNSW, A

The primary role of the Laplace transformin engineering is transient and stability analysis of causal LTI system described by differential equations.

The primary roles of the z-transform are the study of system characteristics and derivation of computational structures for implementing discrete-time systems on computers. The-transform is also used solve difference equations.

Profes

. Ambik

airaja

UNSW, A

1.1 z-TransformThe z-transform of a discrete-time signal

In causal systems, x[n] may be zero when n < 0

[ ]∑∞

−∞=

nznxzX )( z = rejθ, θ = digital frequencywhere z is a complex variable

two-sided transform

[ ]∑∞

nznxzX one-sided transform

Profes

. Ambik

airaja

UNSW, A

Clearly, the z-transform is a power series with an infinite number of terms and so may not converge for

all values of z.

The region where the z-transform converges is known as the region of convergence (ROC) and in this region the values of X(z) are finite

The step sequence: [ ]⎩⎨⎧

<∞≤≤

...11)( 21

0+++=⋅= −−

−∑ zzzzXn

This is a geometric series with a common ratio of z-1.The series converges if |z-1| < 1 or equivalently if |z| > 1Prof

r E. A

mbikair

UNSW, A

In this case, the z-transform is valid everywhere outside a circle of unit radius whose centre is at the origin (see below)

111...1)( 1

−=+++= −

−−

region of convergenceRe(z)

|z|=1 is a circle of unit radius referred to as the ‘unit circle’

Profes

. Ambik

airaja

UNSW, A

1...221)(2 121 =

−=+++=⇒= −

−−zXzLet

( ) ( ) ( ) KK +++=+++=⇒= −− 42115.0 1211

21zXzLet

⎭⎬⎫

⎩⎨⎧

|z| > 1 then X(z) converges and |z| < 1 then it diverges

So the Region Of Convergence (ROC) is seen to be bounded

by the circle |z| = 1, the radius of the pole of

Profes

. Ambik

airaja

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The delta sequence δ[n]:

[ ] [ ] 1}{0

== ∑∞

nznnZ δδ

==∑∞

zazazX

The geometric sequence: x[n] = an

or equivalently, |z| > |a|.

when a = 1, x[n] = 1 for n ≥ 0 ie. x[n] = u[n]

ROC: |z|>1Profes

. Ambik

airaja

UNSW, A

The complex exponential sequence: x[n] = ejnθ

1cos2)sincos(

))(()(

+−+−

++−−

=−−

−−

−∞

−∑

θθθ

θθθθθ

zeezezz

ezezezz

1cos2sin

1cos2)cos(}sin{cos

1cos2sin

1cos2)cos(}{

+−−

θθθθ

θθθ

zzzznjnZ

zzzzeZ nj

Profes

. Ambik

airaja

UNSW, A

Exploiting the linearity property

1cos2sin

1cos2)cos(}sin{cos

1cos2sin

1cos2)cos(}{

+−−

θθθθ

θθθ

zzzznjnZ

zzzzeZ nj

1cos2sin}{sin

1cos2)cos(}{cos

+−−

θθθ

zzzznZ

Profes

. Ambik

airaja

UNSW, A

1.2 Properties

1.2.1 Linearity [ ] [ ] )()( zbYzaXnbynax Z +⎯→←+

1.2.2 Shifting Property (Delay Theorem):

[ ] )(zXzknx kZ −⎯→←−

A very important property of the z-transform is the delay theorem. [ ]{ } ( )

[ ]{ } ( )zXznxZzXznxZ

Unit delay

x[n-1]

z-1X(z) Tx[n] x[n-1]

One sample delay

Profes

. Ambik

airaja

UNSW, A

1.2.3 Time reversal:

[ ] ( )11 −⎟⎠⎞

⎜⎝⎛⎯→←− zXor

zXnx z

Example: Find the z-transform of x[n] = -u[n].

[ ] [ ]zz

zznu−

⎯→←−−

⎯→← −

ROC: |z| < 1

Profes

. Ambik

airaja

UNSW, A

1.2.4 Multiplication by exponential sequence:

[ ] )( 1zaXnxa zn −⎯→←

[ ] )( zeXznxe jnj θθ −⎯→←

As a special case if x[n] is multiplied by ejnθ

1.2.5 Differentiation in the z-domain:

[ ] )(zXdzdzznnx −⎯→←

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. Ambik

airaja

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1.2.6 Discrete Convolution:If y[n] = x[n] * h[n] then Y(z) = X(z) . H(z)

[ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]⎭⎬⎫

⎩⎨⎧

−=⎭⎬⎫

⎩⎨⎧

−−=

⎭⎬⎫

⎩⎨⎧

−−⋅=

⎭⎬⎫

⎩⎨⎧

−−⋅=

∑∑

∑ ∑

−−∞

−∞=

−∞

−∞=

−∞

−∞=

zmhmukxku

knmLetzknhknukxku

zknhknukxku

knhknukxkuZnhnxZ

Profes

. Ambik

airaja

UNSW, A

[ ] [ ] [ ] [ ]

[ ] [ ]

)()()()(

zHzXzHzX

zmhzkx

zmhmuzkxku

⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

∑ ∑

∑∑∞

−−

−∞

−∞=

y[n] = x[n] * h[n]Y(z) = X(z) ⋅ H(z)

h[n]H(z)

Profes

. Ambik

airaja

UNSW, A

Example :

Concept of the transfer function

]2[]1[]2[]1[][][ 21210 −+−+−+−+= nybnybnxanxanxany

Take the z-transform of both sides:

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

( ) 22

−−

−−−−

−−++

++=−−

zbzbzazaa

zXzYzH

zazaazXzbzbzY

zzYbzzYbzzXazzXazXazY

Profes

. Ambik

airaja

UNSW, A

Example :Find the difference-equation of the following transfer function

First rewrite H(z) as a ratio of polynomials in z-1

2325)( 2 ++

( ) ( ) ( ) ( ) ( ) 2121

2523231

25)()()(

−−−−

−−

zzXzzXzzYzzYzYzz

zzzHzXzY

Take inverse z-transform

[ ] [ ] [ ] [ ] [ ]22132215 −−−−−+−=⇒ nynynxnxny

[ ] [ ] [ ] [ ] [ ]22152213 −+−=−+−+ nxnxnynyny

Profes

. Ambik

airaja

UNSW, A

Example : If x[n] = u[n] – u[n-10], find X(z)?

0 11)1()( −

−−

== ∑ zzzzX

Example:

∑∑

−∞=

⎟⎠⎞

⎜⎝⎛−=

⎟⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛−=

−−−=

If y[n] = -αnu[-n-1], find Y(z)?

Profes

. Ambik

airaja

UNSW, A

u[n+1]

u[-(n+1)]

11)( 1

−= −

The sum converges provided ie. |z| < |α|

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. Ambik

airaja

UNSW, A

Depict the ROC and pole and zero locations in the z-plane

Re(z)α

Profes

. Ambik

airaja

UNSW, A

1.2.6 Discrete ConvolutionCompute the convolution y[n] of the digital signals given byx1[n] = [1, -2, 1];x2[n] = 1 for 0 ≤ n ≤ 5, x2[n] = 0 elsewherey[n] = x1[n] * x2[n] ⇒ Y(z) = X1(z) ⋅X2(z)X1(z) = 1 –2z-1 + z-2

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

Y(z) = X1(z) ⋅ X2(z)= 1-z-1 – z-6 + z-7

Inverse z-transform

y[n] = [1, -1, 0,0,0,0,-1,1]Profes

. Ambik

airaja

UNSW, A

Example :

Determine the system function H(z) of the system shown below:

x[n]y[n]

aay[n-1]

y[n] = x[n] + ay[n-1]

X(z) Y(z)

aaz-1Y(z)

Y(z) = X(z) + az-1Y(z)

)()()( −−==

azzXzYzH Prof

r E. A

mbikair

UNSW, A

Profes

. Ambik

airaja

UNSW, A

Profes

. Ambik

airaja

UNSW, A

Basic z-Transforms

|z|>rrnsin(nθ) u[n]

|z| > rrncos(nθ) u[n]

|z| > 1sin(nθ) u[n]

|z|>1cos(nθ) u[n]

|z| > |α|nαn u[n]

|z| > |α|αn u[n]

|z| > 1u[n]

all z1δ[n]

ROCTransformSignal

−− z

−− zα

)1( −

− zzαα

cos21cos1

−−

+−−

cos21sin

−−

+− zzz

cos21cos1

−−

+−−

zrrzrzθ

cos21sin1

−−

+−−

zrrzrzθ

θProfes

. Ambik

airaja

UNSW, A

z-Transform Properties

X(z)⋅Y(z)x[n]*y[n]

anx[n]

z-kX(z)X[n-k]

aX(z) + bY(z)ax[n] + by[n]

X(z)X[n]

Transformsignal

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

)(zXdzdz−Prof

r E. A

mbikair

UNSW, A

If N →∞ , |z-1| < 1

1)1(21

11...1 −

−−−−−

−−

=++++=zzzzzS

−∞ −=

Profes

. Ambik

airaja

UNSW, A

1.3 Inverse z-Transform

The inverse z-transform-Partial fraction

The inverse z-transform allows us to recover the discrete-time sequence.

x[n] = Z-1[X(z)]

where X(z) is the z-transform of x[n].

Profes

. Ambik

airaja

UNSW, A

Example :Find x[n] for the following:

⎥⎥⎥⎥

⎢⎢⎢⎢

⎟⎠⎞

⎜⎝⎛

−−

⎟⎠⎞

⎜⎝⎛

⎥⎥⎥⎥

⎢⎢⎢⎢

⎟⎠⎞

⎜⎝⎛−

⎟⎠⎞

⎜⎝⎛

⎥⎦⎤

⎢⎣⎡

−−=

−−= −−

75.054

5.075.0)5.0)(75.0(375.025.0

375.025.01)(

[ ] 0,)5.0()75.0(54

75.054][ 11

>−−=

⎥⎦⎤

⎢⎣⎡

+−⎥⎦

⎤⎢⎣⎡

−= −−

nnProfes

. Ambik

airaja

UNSW, A

Power series Method

Residue Method - evaluating the contour integral

1.3 Inverse z-Transform

Profes

. Ambik

airaja

UNSW, A

1.3.1 Relationship between the z-transform and Laplace transform.

If we let z = esT, then z = e(σ +jω)T (T is the sampling period)

z = eσ T⋅ ejωT = eσ T⋅ ejθ -π ≤ θ ≤ π

Thus |z| = eσ T and

θπω ===∠sf

fTz 2 (θ = digital frequency)

Profes

. Ambik

airaja

UNSW, A

As ω varies from -∞ to ∞ the s-plane is mapped to the z-plane as shown in Figure 2.10.

The entire jω axis in the s-plane is mapped onto the unit circle. The left-hand s-plane is mapped to the inside of the unit circle and the right-hand s-plane maps to the outside of the unit circle.

Profes

. Ambik

airaja

UNSW, A

s-plane

z-plane

θ = 0

If θ = πUnit circle|z|=1

θ = πθ = -π

(Half thesampling frequency)

Figure 2.10. Mapping of the s-plane to the z-plane.

Profes

. Ambik

airaja

UNSW, A

In terms of frequency response, the jω axis is the most important in the s-plane. In this

case σ = 0 and the frequency points in the s-plane are related to points on the z-plane unit circle by z = eσT.ejωT = 1 ejθ

θjez =

Profes

. Ambik

airaja

UNSW, A

1.4 Frequency Response Estimation

There are many instances when it is necessary to evaluate the frequency response of discrete-time systems. The frequency response of a system can be readily obtained from its z-transforms.

For example, if we set z = ejθ, that is evaluate the z-transform around the unit circle, we obtain the Fourier Transform of the system.

πθπθ θ <<==

- , |)()( jezzHH

Fourier transform of the discrete-time systemProfes

. Ambik

airaja

UNSW, A

Example :

Find H(θ). {H(θ)-the frequency response}

Fourier transform of the discrete-time system.

6.0say 01 , 1

1)( 1 =<<−

= − aaaz

( ) πθπθ θ ≤≤−==

|)( jezzHH

Profes

. Ambik

airaja

UNSW, A

( ) ( ) ( ) θθθ θθ sincos1

1|jaaae

zHH jez j+−

== −=

( ) ( ) 222 cos211

sincos1

1)(aaaa

=θθθ

θ = ωT; θ = 2π f/fs:; θ = π ⇒ f = fs/2fs= sampling frequency

|H(θ)|

-π πθ

-fs/2 fs/20 f (analogue frequency)

digital frequency

a−11

Profes

. Ambik

airaja

UNSW, A

1.5 Pole-Zero Description of Discrete-Time Systems

The zeros of a z-transform H(z) are the values of zfor which H(z)=0. The poles of a z-transform are

the values of z for which H(z)=∞ . If H(z) is a rational function , then

))......()(()).......()((

......1...

)()()(

pzpzpzzzzzzza

zbzbzazaa

zXzYzH

−−−−−−

++++++

== −−

−−

Profes

. Ambik

airaja

UNSW, A

The complex quantities (or may be real) z1, z2, z3 ….

are called zeros of H(z) and the complex quantities

(or may be real) p1, p2, p3 … are called the poles of

H(z). We thus see that H(z) is completely

determined , except for the constant a0, by the

values of poles and zeros

The information contained in the z-transform can be conveniently displayed as a poles-zero diagram (see figure in the next slide)

Profes

. Ambik

airaja

UNSW, A

|z|=1 0.5

Re(z)-1

In the diagram, ‘X’ marks the position of a pole and ‘O’denotes the position of a zero.

The poles are located at z = 0.5 ± 0.5j and z = 0.75, a single zero is at z = -1.Prof

r E. A

mbikair

UNSW, A

An important feature of the pole-zero diagram is the unit circle |z|=1. The pole-zero diagram provides an insight into the properties of a given discrete-time system.

From the locations of the poles d zeros we can infer the frequency response of the system as well as its degree of stability.

For a stable system, all the poles must lie inside the unit circle. Zeros may lie inside, on, or outside the unit circle.Prof

r E. A

mbikair

UNSW, A

Example :Determine the transfer function H(z) of a discrete-time system with the pole-zero diagram shown below:

0.5|z| =1 0.5

Re(z)-1

5.01)1(

)5.05.0()5.05.0()1()1()(

−−

−−+

+−−−+−

jzjzjzjzKzH

Profes

. Ambik

airaja

UNSW, A

Example :

Determine the pole-zero plot:

azzzH−

Profes

. Ambik

airaja

UNSW, A

Example :

Determine the pole-zero plot :

)().........)(()(

)( −−

−−−=

−−

zMzzzzz

azzazzH

The zero z = a cancels the pole at z = a. Thus, H(z)has M-1 zeros and M-1 poles as shown in the diagram below for M = 8.

8 polesProfes

. Ambik

airaja

UNSW, A

Consider a system, H(z) with two complex conjugate poles in the z-plane :

Re(z)θθ

rep−=

1 Poles

)(01 zeroz =

Profes

. Ambik

airaja

UNSW, A

A typical transfer function might be:

( ) ( )

⎥⎥⎥⎥

⎢⎢⎢⎢

−−

⎥⎦⎤

⎢⎣⎡

−−=

−−

θθθθ

θθjj

rezjrz

rezrezzzH

⎥⎦⎤

⎢⎣⎡

−−

−= −−− 11 1

1)sin2(

1)(zrezrerj

zH jj θθθProfes

. Ambik

airaja

UNSW, A

( ) ( )[ ][ ]njnj

rererj

sin21)(

θ = frequency of oscillation

[ ]θθ

nrnh n sinsin

1)( 1−=

This is the impulse response of the 2nd order system with complex poles

Profes

. Ambik

airaja

UNSW, A

We note that the impulse response will decay away to zero provided r is less than one. [ r < 1]

Recall that r is also the distance from the origin in

the z-plane to the poles p1 or p2, so that system will be stable if the poles in the z-plane lie inside the unit circle.

Profes

. Ambik

airaja

UNSW, A

Example :Poles inside unit circle

Exponential decay sinewave (r<1)Stable system

[ ]θθ

nrnh n sinsin

1)( 1−=

Poles on theunit circle

Profes

. Ambik

airaja

UNSW, A

Poles outside unit circleθ

Exponential increasing (r>1)

[ ]θθ

nrnh n sinsin

1)( 1−=

one real Pole inside the unit circle rn, θ = 0

Profes

. Ambik

airaja

UNSW, A

Note :

A system that is both stable and casual must have all its poles inside that unit circle within the z-plane.

We cannot have a pole outside the unit circle, since the inverse transform of a pole located outside the circle will contribute either a right sided increasing exponential term, which is not stable, or a left-sided decaying exponential term that is not causal.

Profes

. Ambik

airaja

UNSW, A

1.6 A second order Resonant System (Complex Poles)

+y[n]x[n]

sincos

sincos0

θθθ

jrrrep

jrrrepj

+==−

11 bzbz

= −−Profes

. Ambik

airaja

UNSW, A

All pole system has poles only (without counting the zeros at the origin)

(B) cos2)(

))(())(()(

rzeerzzzH

rezrezz

bzbzzzH

++−=

−−=

−−

θθθ

01 cos2 θrb −= 22 rb =

10 2 b

bCos −=∴ θ

02πθ =

Comparing (A) and (B), we obtain

θ0 = resonant frequencyProfes

. Ambik

airaja

UNSW, A

We can derive H(θ) and the magnitude from

1)( −− ++=

zbzbzH

0.99-1.41

0.9-1.34

0.7-1.16

0.5-0.94

b2 = r2b1

⎥⎥⎦

⎢⎢⎣

⎡ −=

)cos(2

Profes

. Ambik

airaja

UNSW, A

40πθ =4

π− ππ−

Profes

. Ambik

airaja

UNSW, A

-3 -2 -1 0 1 2 3-20

Magnitude response

I II IIIIV

Profes

. Ambik

airaja

UNSW, A

Example :sketch the magnitude response for the system having the transfer function

The system has a zero at z = -1 & poles at

)9.01()9.01(

1)(1414

−−−

−−

zzHjj ππ

49.0πj

∴ Magnitude response will be zero at θ = π and large at θ0= ± π /4 because the poles are close to the unit circle.

Profes

. Ambik

airaja

UNSW, A

-π π-π/4 π/4

|H(θ)|

Magnitude Response π/40.9

θ=0θ = -π

θ = π

Profes

. Ambik

airaja

UNSW, A

Example :Sketch the approximate magnitude response from the pole-zero map given below:

|z|= 1

- 0.8Re(z)

Im(z)dB

-π -π/2 0 π/2 π

-fs/2fs/2

Profes

. Ambik

airaja

UNSW, A

Example :

Sketch an approximate magnitude response from the pole-zero map given below:

- 0.5 Re(z)1

-π -π/2 0 π/2 π θ

|H(θ)| in dB

Profes

. Ambik

airaja

UNSW, A

Summary of Part B Chapter 1At the end of this chapter, it is expected that you should know:

The properties of z transforms and their application.

Discrete convolution in the time and z domains.

How to find the inverse z transform, given a transfer function.

The difference between a z transform and a Laplace transform and when to use each.

Estimation of frequency response from a transfer function

Hand-calculate magnitude and phase responses for simple transfer functions and plot them.Prof

r E. A

mbikair

UNSW, A

The pole-zero description of a discrete time system.

Given a pole-zero diagram, transfer function or difference equation, how to find any of the other representations and discuss the system’s stability with reference to the pole-zero diagram.

How to derive the resonance frequency equation for a second-order resonance system (a complex pole pair).

Profes

. Ambik

airaja

UNSW, A

Examples

Profes

. Ambik

airaja

UNSW, A

A digital filter structure is shown below. Determine the Transfer function H(z).

X(z) a0

)()()(

−−

==zbzb

zaazHzXzY

Profes

. Ambik

airaja

UNSW, A

Re(z)1

-π -π/2 0 π/2 π θ

|H(θ)|

Profes

. Ambik

airaja

UNSW, A

Question 5(b)Determine the magnitude response of the following filter and show that it has an all-pass

characteristic.

[4 marks]1)1(

<−−

azazazH

filter Pass -All

=−−+−−+

−−

=→−−

−−

1)()().(

1)()1(

θθθ

aeaeaaeaeaHHH

aeeaHzzH

θ-π π

Profes

. Ambik

airaja

UNSW, A

A first-order digital filter is described by the system function :

1111)( −−

−++−

Draw a canonic realisation of the transfer function H(z).

+ Y(z)

1-a/1-b

Profes

. Ambik

airaja

UNSW, A

Sketch roughly the magnitude response corresponding to the pole-zero pattern given below:

Question 3

Profes

. Ambik

airaja

UNSW, A

Re(z)1

Profes

. Ambik

airaja

UNSW, A

The difference equation of a digtal filter is given by y(n) = x(n)-x(n-8)- y(n-2 )

Find the transfer function for the above filter.

11).1(

)()()()(

−−

−−=

zzYzzXzXzY

Profes

. Ambik

airaja

UNSW, A

A first-order digital filter has a transfer function given by

Determine the impulse response of the above digital filter H(z) and show that the filter is stable if a < 1.

11)( −−

zazkzH

⎥⎦

⎤⎢⎣

⎡−

1 111)(

[ ])1()()( 1 −+= − nuanuaknh nn

if a< 1, then h(n) decays to zero.Therefore, stable system.

Profes

. Ambik

airaja

UNSW, A

part b chapter1 z-transform-modeemedia.ee.unsw.edu.au/contents/elec3104/lecturenotes/dsp3.pdf ·...

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