UNIT - 4 ANALYSIS OF DISCRETE TIME SIGNALS. Sampling Frequency Harry Nyquist, working at Bell Labs...

UNIT - 4

ANALYSIS OF DISCRETE TIME SIGNALS

Sampling Frequency

• Harry Nyquist, working at Bell Labs developed what has become known as the Nyquist Sampling Theorem:– In order to be ‘perfectly’ represented by its samples, a signal must be

sampled at a sampling rate (also called sampling frequency) equal to at least twice its highest frequency component

– Or: fs = 2f

– Note that fs here is frequency of sampling, not the frequency of the sample

How often do you sample? The sampling rate depends on the signal’s highest frequency (for baseband)

Sampling Rate Examples• Take Concert A: 440 Hz

– What would be the minimum sampling rate needed to accurately capture this signal?

– fs = 2 x 440 Hz = 880 Hz

• Take your telephone used for voice, mostly– Highest voice component is: 3000 Hz– Minimum sampling rate: fs = 2 x 3000 Hz = 6000 Hz– Real telephone digitization is done at 8000 Hz sampling rate (supporting a 4

kHz bandwidth). Why? Remember that Nyquist said “equal to at least twice…”

Undersampling and Oversampling• Undersampling

– Sampling at an inadequate frequency rate– Aliased into new form - Aliasing– Loses information in the original signal

• Oversampling – Sampling at a rate higher than minimum rate – More values to digitize and process– Increases the amount of storage and transmission– COST $$

Effects of Undersampling

Original waveform

Reconstructed waveform

DISCRETE TIME FOURIER TRANSFROM (DTFT)

• Definition of DTFT:– The Fourier transform (FT) of discrete – time signals is called Discrete Time

Fourier Transform (i.e DTFT).

Let x(n) = Discrete time signal

X(ej) = Fourier transform of x(n)

The Fourier transform of a finite energy discrete time signal, x(n) is defined as,

X(ej)= F{x(n)} =

FREQUENCY SPECTRUM• The FT X(ej) of a signal x(n) represents the frequency content of x(n).

By taking FT, the signal x(n) is decomposed into its frequency components. Hence X(ej) is also called frequency spectrum of discrete time signal or signal spectrum.

• Magnitude and Phase Spectrum• The X(ej is a complex value function of and so it can be expressed in

rectangular form as,

X(ej) = Xr(ej) + jXi(ej)

Where Xr(ej) = Real part of X(ej)

Xi(ej) = Imaginary part of X(ej)

• The polar form of X(ej) is,• X(ej) = |X(ej)|∟X(ej)

• Where, |X(ej)| = Magnitude spectrum

∟X(ej) = Phase spectrum

• The magnitude spectrum is defined as,• |X(ej)| = X(ej)X*(ej)

Derivation of the Discrete-time Fourier Transform

Recall DTFS pair

where

The limit of integration is over any interval of 2p in w

Periodic in w with period 2p

Thus,

DTFT Pair

Conditions for Convergence

Examples

0

,1

1

,1

1

a

a

IDTFT

6) Complex Exponentials

DTFT of Periodic Signals

Recall the following DTFT pair:

Represent periodic signal x[n] in terms of DTFS:

Example: A discrete-time Sine Function

Example: A discrete-time Periodic Impulse Train

The DTFS coefficients for this signal are:

ck

Properties of DTFT

Properties of DTFT

Convolution Property

Multiplication Property

PROPERTIES OF DTFT

The z-Transform

Content

• Introduction• z-Transform• Zeros and Poles• Region of Convergence• Important z-Transform Pairs• Inverse z-Transform• z-Transform Theorems and Properties• System Function

The z-Transform

Introduction

Why z-Transform?• A generalization of Fourier transform• Why generalize it?

– FT does not converge on all sequence– Notation good for analysis– Bring the power of complex variable theory deal

with the discrete-time signals and systems

The z-Transform

z-Transform

Definition

• The z-transform of sequence x(n) is defined by

n

nznxzX )()(

Let z = ej.

( ) ( )j j n

n

X e x n e

Fourier Transform

z-Plane

Re

Im

z = ej

n

nznxzX )()(

( ) ( )j j n

n

X e x n e

Fourier Transform is to evaluate z-transform on a unit circle.

Fourier Transform is to evaluate z-transform on a unit circle.

z-Plane

Re

Im

X(z)

Re

Im

z = ej

Periodic Property of FT

Re

Im

X(z)

X(ej)

Can you say why Fourier Transform is a periodic function with period 2?Can you say why Fourier Transform is a periodic function with period 2?

The z-TransformZeros and Poles

Definition

• Give a sequence, the set of values of z for which the z-transform converges, i.e., |X(z)|<, is called the region of convergence.

n

n

n

n znxznxzX |||)(|)(|)(|

ROC is centered on origin and consists of a set of rings.

ROC is centered on origin and consists of a set of rings.

Example: Region of Convergence

Re

Im

n

n

n

n znxznxzX |||)(|)(|)(|

ROC is an annual ring centered on the origin.

ROC is an annual ring centered on the origin.

xx RzR ||r

}|{ xx

j RrRrezROC

Stable Systems

• A stable system requires that its Fourier transform is uniformly convergent.

Re

Im

1

Fact: Fourier transform is to evaluate z-transform on a unit circle.

A stable system requires the ROC of z-transform to include the unit circle.

Example: A right sided Sequence

)()( nuanx n

1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7-8

n

x(n)

. . .

Example: A right sided Sequence

)()( nuanx n

n

n

n znuazX

)()(

0n

nn za

0

1)(n

naz

For convergence of X(z), we require that

0

1 ||n

az 1|| 1 az

|||| az

az

z

azazzX

n

n

10

1

1

1)()(

|||| az

aa

Example: A right sided Sequence ROC for x(n)=anu(n)

|||| ,)( azaz

zzX

Re

Im

1 aaRe

Im

1

Which one is stable?Which one is stable?

Example: A left sided Sequence

)1()( nuanx n

1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7-8n

x(n)

. . .

Example: A left sided Sequence

)1()( nuanx n

n

n

n znuazX

)1()(

For convergence of X(z), we require that

0

1 ||n

za 1|| 1 za

|||| az

az

z

zazazX

n

n

10

1

1

11)(1)(

|||| az

n

n

n za

1

n

n

n za

1

n

n

n za

0

1

aa

Example: A left sided Sequence ROC for x(n)=anu( n1)

|||| ,)( azaz

zzX

Re

Im

1 aaRe

Im

1

Which one is stable?Which one is stable?

The z-TransformRegion of

Convergence

Represent z-transform as a Rational Function

)(

)()(

zQ

zPzX where P(z) and Q(z) are

polynomials in z.

Zeros: The values of z’s such that X(z) = 0

Poles: The values of z’s such that X(z) =

Example: A right sided Sequence

)()( nuanx n |||| ,)( azaz

zzX

Re

Im

a

ROC is bounded by the pole and is the exterior of a circle.

Example: A left sided Sequence

)1()( nuanx n|||| ,)( az

az

zzX

Re

Im

a

ROC is bounded by the pole and is the interior of a circle.

Example: Sum of Two Right Sided Sequences

)()()()()( 31

21 nununx nn

31

21

)(

z

z

z

zzX

Re

Im

1/2

))((

)(2

31

21

121

zz

zz

1/31/12

ROC is bounded by poles and is the exterior of a circle.

ROC does not include any pole.

Example: A Two Sided Sequence

)1()()()()( 21

31 nununx nn

21

31

)(

z

z

z

zzX

Re

Im

1/2

))((

)(2

21

31

121

zz

zz

1/31/12

ROC is bounded by poles and is a ring.

ROC does not include any pole.

Example: A Finite Sequence

10 ,)( Nnanx n

nN

n

nN

n

n zazazX )()( 11

0

1

0

Re

ImROC: 0 < z < ROC does not include any pole.

1

1

1

)(1

az

az N

az

az

z

NN

N

1

1

N-1 poles

N-1 zeros

Always StableAlways Stable

Properties of ROC• A ring or disk in the z-plane centered at the origin.• The Fourier Transform of x(n) is converge absolutely iff the ROC includes

the unit circle.• The ROC cannot include any poles• Finite Duration Sequences: The ROC is the entire z-plane except possibly

z=0 or z=. • Right sided sequences: The ROC extends outward from the outermost

finite pole in X(z) to z=. • Left sided sequences: The ROC extends inward from the innermost

nonzero pole in X(z) to z=0.

More on Rational z-Transform

Re

Im

a b c

Consider the rational z-transform with the pole pattern:

Find the possible ROC’s

Find the possible ROC’s

More on Rational z-Transform

Re

Im

a b c

Consider the rational z-transform with the pole pattern:

Case 1: A right sided Sequence.

More on Rational z-Transform

Re

Im

a b c

Consider the rational z-transform with the pole pattern:

Case 2: A left sided Sequence.

More on Rational z-Transform

Re

Im

a b c

Consider the rational z-transform with the pole pattern:

Case 3: A two sided Sequence.

More on Rational z-Transform

Re

Im

a b c

Consider the rational z-transform with the pole pattern:

Case 4: Another two sided Sequence.

The z-TransformImportant

z-Transform Pairs

Z-Transform Pairs

Sequence z-Transform ROC)(n 1 All z

)( mn mz All z except 0 (if m>0)or (if m<0)

)(nu 11

1 z

1|| z

)1( nu 11

1 z

1|| z

)(nuan 11

1 az

|||| az

)1( nuan 11

1 az

|||| az

Z-Transform Pairs

Sequence z-Transform ROC)(][cos 0 nun 21

0

10

]cos2[1

][cos1

zz

z1|| z

)(][sin 0 nun 210

10

]cos2[1

][sin

zz

z1|| z

)(]cos[ 0 nunr n 2210

10

]cos2[1

]cos[1

zrzr

zrrz ||

)(]sin[ 0 nunr n 2210

10

]cos2[1

]sin[

zrzr

zrrz ||

otherwise0

10 Nnan

11

1

az

za NN

0|| z

The z-TransformInverse z-Transform

The Inverse Z-Transform• Formal inverse z-transform is based on a Cauchy

integral• Less formal ways sufficient most of the time

– Inspection method– Partial fraction expansion– Power series expansion

• Inspection Method– Make use of known z-transform pairs such as

• Example: The inverse z-transform of

az az11

nua 1Zn

nu21

nx 21

z z

21

1

1zX

n

1

Inverse Z-Transform by Partial Fraction Expansion• Assume that a given z-transform can be expressed as

• Apply partial fractional expansion

• 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

N

0k

kk

M

0k

kk

za

zbzX

s

1mm1

i

mN

ik,1k1

k

kNM

0r

rr

zd1

Czd1

AzBzX

Partial Fractional Expression

• Coefficients are given as

s

1mm1

i

mN

ik,1k1

k

kNM

0r

rr

zd1

Czd1

AzBzX

kdz

1kk zXzd1A

1idw

1sims

ms

msi

m wXwd1dwd

d!ms

1C

Example: 2nd Order Z-Transform

– Order of nominator is smaller than denominator (in terms of z-1)

No higher order pole

1

2

1

1

z21

1

A

z41

1

AzX

1

41

21

1

1zXz

41

1A1

41

z

11

2

21

41

1

1zXz

21

1A1

21

z

12

21

z :ROC z

21

1z41

1

1zX

11

Example Continued

• ROC extends to infinity – Indicates right sided sequence

21

z z

21

1

2

z41

1

1zX

11

nu41

-nu21

2nxnn

Example #2

• Long division to obtain Bo

1z z1z

21

1

z1

z21

z23

1

zz21zX

11

21

21

21

1z5

2z3z

21z2z1z

23

z21

1

12

1212

11

1

z1z21

1

z512zX

1

2

1

1

z1A

z21

1

A2zX

9zXz21

1A

21

z

11

8zXz1A1z

12

Example #2 Continued

• ROC extends to infinity– Indicates right-sides sequence

1z z1

8

z21

1

92zX 1

1

n8u-nu21

9n2nxn

Inverse Z-Transform by Power Series Expansion

• The z-transform is power series

• In expanded form

• Z-transforms of this form can generally be inversed easily

• Especially useful for finite-length series

n

nz nxzX

2112 z 2xz 1x 0xz 1xz 2xzX

12

1112

z21

1z21

z

z1z1z21

1z zX

• Example,

12

1112

z21

1z21

z

z1z1z21

1z zX

1n21

n1n21

2nnx

2n0

1n21

0n1

1n21

2n1

nx

Z-Transform Properties: Linearity• Notation

• Linearity

– Note that the ROC of combined sequence may be larger than either ROC– This would happen if some pole/zero cancellation occurs– Example:

• Both sequences are right-sided• Both sequences have a pole z=a• Both have a ROC defined as |z|>|a|• In the combined sequence the pole at z=a cancels with a zero at z=a• The combined ROC is the entire z plane except z=0

xZ RROC zXnx

21 xx21

Z21 RRROC zbXzaXnbxnax

N-nua-nuanx nn

Z-Transform Properties: Time Shifting

• Here no is an integer– If positive the sequence is shifted right– If negative the sequence is shifted left

• The ROC can change the new term may– Add or remove poles at z=0 or z=

• Example

xnZ

o RROCzXznnx o

41

z z

41

1

1z zX

1

1

1-nu41

nx1-n

Z-Transform Properties: Multiplication by Exponential• ROC is scaled by |zo|

• All pole/zero locations are scaled

• If zo is a positive real number: z-plane shrinks or expands

• If zo is a complex number with unit magnitude it rotates

• Example: We know the z-transform pair

• Let’s find the z-transform of

xooZn

o RzROC z/zXnxz

1z:ROC z-1

1nu 1-

Z

nure21

nure21

nuncosrnxnjnj

on oo

rz zre1

2/1

zre1

2/1zX

1j1j oo

Z-Transform Properties: Differentiation

• Example: We want the inverse z-transform of

• Let’s differentiate to obtain rational expression

• Making use of z-transform properties and ROCCopyright (C) 2005 Güner Arslan

351M Digital Signal Processing 86

x

Z RROC dz

zdXznnx

az az1logzX 1

1

11

2

az11

azdz

zdXz

az1az

dzzdX

1nuaannx 1n

1nuna

1nxn

1n

Z-Transform Properties: Conjugation

• Example

Copyright (C) 2005 Güner Arslan

351M Digital Signal Processing 87

x**Z* RROC zXnx

nxZz nxz nxzX

z nxz nxzX

z nxzX

n

n

n

n

n

n

n

n

n

n

Z-Transform Properties: Time Reversal

• ROC is inverted• Example:

• Time reversed version of

Copyright (C) 2005 Güner Arslan

351M Digital Signal Processing 88

x

Z

R1

ROC z/1Xnx

nuanx n

nuan

111-

1-1

az za-1

za-az1

1zX

Z-Transform Properties: Convolution• Convolution in time domain is multiplication in z-domain• Example:Let’s calculate the convolution of

• Multiplications of z-transforms is

• ROC: if |a|<1 ROC is |z|>1 if |a|>1 ROC is |z|>|a|• Partial fractional expansion of Y(z)

Copyright (C) 2005 Güner Arslan

351M Digital Signal Processing 89

2x1x21

Z21 RR:ROC zXzXnxnx

nunx and nuanx 2n

1

az:ROC az11

zX 11

1z:ROC z1

1zX 12

1121 z1az11

zXzXzY

1z :ROC asume az11

z11

a11

zY 11

nuanua1

1ny 1n

The z-Transformz-Transform Theorems and Properties

Linearity

xRzzXnx ),()]([Z

yRzzYny ),()]([Z

yx RRzzbYzaXnbynax ),()()]()([Z

Overlay of the above two

ROC’s

Shift

xRzzXnx ),()]([Z

xn RzzXznnx )()]([ 0

0Z

Multiplication by an Exponential Sequence

xx- RzRzXnx || ),()]([Z

xn RazzaXnxa || )()]([ 1Z

Differentiation of X(z)

xRzzXnx ),()]([Z

xRzdz

zdXznnx

)()]([Z

Conjugation

xRzzXnx ),()]([Z

xRzzXnx *)(*)](*[Z

Reversal

xRzzXnx ),()]([Z

xRzzXnx /1 )()]([ 1 Z

Real and Imaginary Parts

xRzzXnx ),()]([Z

xRzzXzXnxe *)](*)([)]([ 21R

xj RzzXzXnx *)](*)([)]([ 21Im

Initial Value Theorem

0for ,0)( nnx

)(lim)0( zXxz

Convolution of Sequences

xRzzXnx ),()]([Z

yRzzYny ),()]([Z

yx RRzzYzXnynx )()()](*)([Z

Convolution of Sequences

k

knykxnynx )()()(*)(

n

n

k

zknykxnynx )()()](*)([Z

k

n

n

zknykx )()(

k

n

n

k znyzkx )()(

)()( zYzX

The z-TransformSystem Function

Shift-Invariant System

h(n)h(n)x(n) y(n)=x(n)*h(n)

X(z) Y(z)=X(z)H(z)H(z)

Shift-Invariant System

H(z)H(z)X(z) Y(z)

)(

)()(

zX

zYzH

Nth-Order Difference Equation

M

rr

N

kk rnxbknya

00

)()(

M

r

rr

N

k

kk zbzXzazY

00

)()(

N

k

kk

M

r

rr zazbzH

00)(

Representation in Factored Form

N

kr

M

rr

zd

zcAzH

1

1

1

1

)1(

)1()(

Contributes poles at 0 and zeros at cr

Contributes zeros at 0 and poles at dr

Stable and Causal Systems

N

kr

M

rr

zd

zcAzH

1

1

1

1

)1(

)1()( Re

ImCausal Systems : ROC extends outward from the outermost pole.

Stable and Causal Systems

N

kr

M

rr

zd

zcAzH

1

1

1

1

)1(

)1()( Re

ImStable Systems : ROC includes the unit circle.

1

Example

Consider the causal system characterized by

)()1()( nxnayny

11

1)(

azzH

Re

Im

1

a

)()( nuanh n

Determination of Frequency Response from pole-zero pattern

• A LTI system is completely characterized by its pole-zero pattern.

))(()(

21

1

pzpz

zzzH

Example:

))(()(

21

1

00

0

0

pepe

zeeH jj

jj

0je

Re

Im

z1

p1

p2

Determination of Frequency Response from pole-zero pattern

• A LTI system is completely characterized by its pole-zero pattern.

))(()(

21

1

pzpz

zzzH

Example:

))(()(

21

1

00

0

0

pepe

zeeH jj

jj

0je

Re

Im

z1

p1

p2

|H(ej)|=?|H(ej)|=? H(ej)=?H(ej)=?

Determination of Frequency Response from pole-zero pattern

• A LTI system is completely characterized by its pole-zero pattern.

Example:0je

Re

Im

z1

p1

p2

|H(ej)|=?|H(ej)|=? H(ej)=?H(ej)=?

|H(ej)| =| |

| | | | 1

2

3

H(ej) = 1(2+ 3 )

Example

11

1)(

azzH

Re

Im

a

0 2 4 6 8-10

0

10

20

0 2 4 6 8-2

-1

0

1

2

dB