ders notları toplu

LECTURE NOTES ON

DIGITAL SIGNAL

PROCESSING

Prof. Dr. A. Salim KAYHAN

Hacettepe University

Department of Electrical and Electronics Engineering

ANKARA-2014

ELE 407 Digital Signal Processing

Fall 2014

Place: E-? Time: Wednesday 13:00-15:50 Course Outline: W1-Sep.24 Review of Discrete-time signals, systems, Fourier, Z-tr. W2-Oct.01 Sampling, Decimation, Interpolation. W3-Oct.08 Frequency Response . W4-Oct.15 Relation Between Magnitude and Phase, Flow Graph

Realization. W5-Oct.22 Flow Graph Realization of FIR sys., Quantization. W6-Oct.29 No Class. (Holiday) W7-Nov.05 EXAM W8-Nov.12 Butterworth and Chebyshev Filter Design. W9-Nov.19 Butterworth and Chebyshev Filter, FIR Filter Design. W10-Nov26 Discrete-time Fourier Series, DFT, Properties of DFT. W11-Dec03 Convolution with DFT (Overlap-add/save). FFT. W12-Dec10 Fast Fourier Transform(FFT). W13-Dec17 EXAM. W14-Dec24 2-D Signal Processing. Textbook: Discrete-time Signal Processing, Oppenheim and Schafer, Prentice-Hall. Grading: 2Term Exams (50%), Final Exam (50%).

1Digital Signal Processing A.S.Kayhan

DIGITAL SIGNAL

PROCESSING

Textbook: Discrete-Time Signal Processing, Oppenheim and Schafer, Prentice-Hall.

Digital Signal Processing A.S.Kayhan

Course Outline

Review of Discrete-time signals, systems, Fourier tr.

Review of Z-transform

Sampling, Decimation, Interpolation

Time and frequency response of systems

Flow Graph Realization

Filter Design

Discrete-time Fourier Series, Discrete Fourier Transform

Fast Fourier Transform

2-D Signal Processing


SIGNALS:A signal is a function of time representing a physical variable, e.g. voltage, current, spring displacement, share market prices, number of students asleep in the class, cash in the bank account.

Continuous time signals : x(t), x(t1,t2,...)Speech signals, Image signals,Video signals, Seismic signals, Biomedical signals (ECG,EEG,...)

Discrete-time signals: x[n], x[n1,n2,...]Inherently discrete-time (Stock market)Discretized continuous-time(Most signals)


SIGNALS:Deterministic or Random


Speech data (8kHz):


Discrete-time signal example:Stock Market daily closing values


Annual max. temperature and precipitation for Ankara (data: MGM)


Graphical Representation of Signals

Continuous-timeSignal

Discrete-timeSignal


Reflection: x[-n]


Time-scaling: x(at)

Time-scaling in discrete-time is not straightforward.Discussed in Decimation and Interpolation.


Time-shift: x[n-no]


Basic Discrete-time Signals:Unit Step Function :

0,1

0,0][

n

nnu


Unit Impulse Function:

0,1

0,0][

n

nn

n

k

knu

nunun

][][

]1[][][

][]0[][][ nxnnx


Real Exponential: nn aCeCnx ][


Complex Exponential: nj oeAnx ][Sinusoidal Signal: )cos(][ nAnx o

}{][ )( nj oAenx Re


Periodicity Properties:

njnj oo ee )2(

Therefore, consider only 20 o

njNnj oo ee )(If periodic, then

N

m

e Nj o

2

1

0

m and N integer.


If periodic with fundamental period N, then fundamental frequency is N/2

If there are a few exponentials in the form:

nN

jk

k enx2

][

they are harmonically relatedThere are only N such distinct signals.


SYSTEM:Any process that transforms signals

10



Interconnection of Systems:Series(Cascade), Parallel, Series/Parallel.

11


Properties:Memoryless if output depends only on input at the same time.

Example: Discrete-time systems with memory:

n

k

kxny ][][

]1[][ nxny


Causality: Causal if output depends only on inputs at the present time and in the past.

12


M

Mk

knxM

ny ][12

1][

]1[][][ nxnxny

Example: Causal system:

Example: Noncausal systems:

][]1[][ nxnxny

t

dxC

ty )(1

)(


Stability: Stable if small inputs do not lead to diverging outputs.

13


]1[][][ nxnxnyExample: Stable system:

Example: Unstable systems:

].[][ if ],[)1(][][ nunxnunkxnyn

k


Time Invariance: Time-invariant if a time shift in input causes same time shift in output signal.

14


Example:Time-varying: ][][ nnxny

Consider two signals

][][],[ 121 onnxnxnx

][][ 11 nnxny

][][

][][

12

22

onnnxny

nnxny

shift ][1 ny

][][)(][ 211 nynnxnnnny ooo


Linearity: Linear if posses superposition property:Additivity and scaling (homogeneity):

1. Response to is 2. Response to is

][][ 21 nbxnax

][1 nax ][1 nay

Combining these two, we get:

][][ 21 nxnx

][][ 21 nbynay

][][ 21 nyny

Example: Linear: ][][ nnxny

Example: Nonlinear: 2])[(][ nxny

15


Linear Time-Invariant Systems:LTI systems can be analyzed in great detail.Many physical processess can be modeled by LTI systems.Unit impulse function will be used as building block and response of LTI systems to a unit impulse will be used to characterize such systems.


Input/output relationship for a LTI system is given as:

k

knhkxny ][][][

Convolution of x[n] and h[n]. h[n] is the impulse response.

][*][][ nhnxny

16


Example:


Properties:Commutative:

][*][][*][ nxnhnhnx

r

knr

k

rhrnxknhkx ][][][][

Use of this property:If it is easier; reflect and shift x[k] to obtain x[n-k] first, then multiply with h[k] to find convolution result at time n.

17


Associative:

][*])[*][(])[*][(*][ 2121 nhnhnxnhnhnx


Distributive:

][*][][*][])[][(*][ 2121 nhnxnhnxnhnhnx

18


Memory: System is memoryless if output depends only on input at same time.Then, system is memoryless if:

][][ nKnh then ].[][ nKxny

Similarly for continuous-time, memoryless if

).()( tKxty

)()( tKth hence


Causal: Consider convolution sum :

k

knxkhny ][][][

For a LTI system to be casual y[n] must not depend on input x[k] for k > n.If system is causal :

.0for ,0][ nnh

Then,

n

kk

knhkxknxkhny ][][][][][0

19


Stability: Stable if bounded inputs do not lead to diverging outputs. Assume

. allfor ,][ nBnx

Then,

k

knxkhny ][][][

k

knxkhny ][][][

with, Bknx ][

k

khBny ][][

System is stable iff,

k

kh ][


Example: A system with impulse response h[n]=[n-no] shifts the input

n n

onnnh 1][][

Example: Consider the fallowing accumulator

n

k

kxny ][][

Unstable][][0

n n

nunu

this system has unit step as the impulse response

20


Discrete-Time Fourier Transform (DTFT):For a discrete-time signal x[n], DTFT is defined as

n

njj enxeX ][)(

the inverse DTFT (synthesis equation) is defined as

2

)(2

1][ deeXnx njj

Differences between continuous-time FT and DTFT:X(ej) is periodic with 2. Integral in synthesis eq. is over 2 interval (0 to 2 or - to ).


Example:Consider 1],[][ anuanx n

n

njnj nueaeX ][)(

jn

njj

aeaeeX

1

1)()(

0

)cos(21

1)(

2

2

aaeX j

)cos(1

)sin(tan)( 1

a

aeX j

)sin())cos(1(

1)(

jaaeX j

21



Common DTFT Pairs:

)2(2)( allfor ,1][ leXnnx j

1)(][][ jeXnnx

)cos(][ nnx o

)]2()2([)( lleX ol

oj

nj oenx ][

l

oj leX )2(2)(

22


Convergence: Analysis equation will converge if x[n] is absolutely summable or it has finite energy

n

nx ][

n

nx2

][

Example:Consider

Nn

Nnnx

,0

,1][

)2/sin(

))2/1(sin(

1

1)(

)12(

N

e

eeeeX

j

NjNj

N

Nn

njj


23


Properties of DTFT:Linearity: DTFT is linear operator:

)()(][][

)(][

)(][

jj

j

j

eYbeXanbynax

eYny

eXnx

Time and frequency shifting:

)(][

)(][

)(][

)( oo

o

jnj

jnjo

j

eXnxe

eXennx

eXnx


Symmetry: If x[n] is real, then

)()( * jj eXeX

Real part is even, imaginary part is odd function of .Similarly, magnitude is even, phase is odd funtion.

Time and frequency scaling: Time scaling is not straightforward for discrete-time signals. Consider special case; x[an], a = -1

)(][ jeXnx

)()(

k of multiplenot isn if 0,

k of multiple isn if],/[][

1

1

jkj eXeX

knxnx

24


Parsevals relation: Energy is equal on both time and frequency domains:

2

22

)(2

1][ deXnx j

n

Since the right hand term is the total energy in frequency domain, is called energy spectral density function (or spectrum).

2)( jeX

Convolution: Convolution in time corresponds to multiplication in frequency domain.

)()()(][*][][ jjj eHeXeYnhnxny


nj

n

j enheH

][)(

where, is the frequency response defined as)( jeH

Example:Consider a LTI sytem withThen the frequency response is

][][ onnnh

onjj eeH )(

For any input x[n] with Fourier transform Fourier transform of output

)( jeX

)()( jnjj eXeeY o

The output is equal to the input with a time shift

].[][ onnxny

25


Example:Consider a LTI sytem with |a|,|b|

26


Z- Transform :For a discrete-time signal x[n], z-transform is defined as

n

nznxzX ][)(

z is a general complex variable shown in polar form as jerz

We get Fourier transform as a special case when r=1


Convergence: Z-transform exists if

n

nrnx |][|

Poles-Zeros: consider X(z) as a rational function

)(

)()(

zQ

zPzX

where P(z) and Q(z) are polynomials in z. Values of z making X(z)=0 are called zeros of X(z) (roots of P(z)).Values of z making X(z)= are called poles of X(z) (roots of Q(z)).

27



28


||||,1

1)( 1 azaz

z

azzX

n

n

az || 1For convergence

0

1 )(][)(n

n

n

nn azznuazX

][][ nuanx nConsiderExample:


||||,1

11)( 1 azaz

z

zazX

n

n

za || 1For convergence

0

1 )(1)(n

nzazX

]1[][ nuanx nConsiderExample:

29


||||||,)( bzabz

z

az

zzX

Then

abnubnuanx nn ],1[][][

ConsiderExample:

||||,][ azaz

znua n

||||,]1[ bzbz

znub n


30


Properties of convergence:ROC is a ring or disk centered at origin.DTFourier transform exists if ROC includes Unit CircleROC can not contain any polesIf x[n] has finite duration ROC is entire plane.If x[n] is right-sided, ROC is outward from outermost poleIf x[n] is left-sided, ROC is inward from innermost poleIf x[n] is two-sided, ROC is a ring not containing any pole.


Example:

],1[)2(][)4.0(][ nununx nn

|2||||4.0|,24.0

)(

zz

z

z

zzX

31


Transfer Function: Z-transform of impulse response h[n]

n

nznhzH ][)(

If impulse response h[n] is causal then ROC will be outward.If the system is causal and stable, all the poles will be inside the unit circle and ROC will include the unit circle.


Inverse Z-Transform:Power Series Expansion: Consider

12

2

11

2

1)( zzzzX

remember

1012 ]1[]0[]1[]2[

][)(

zxzxzxzx

znxzXn

n

therefore]1[

2

1][]1[

2

1]2[][ nnnnnx

32


Example (Long Division): |z| > |a|

33221

11

1

1)( zazaaz

azzX

therefore ][][ nuanx n

Example (Long Division): |z| < |a|

33221)( zazazaza

zzX

therefore ]1[][ nuanx n


Partial fraction expansion: Consider

kdzkk

N

k k

k

k

N

k

k

M

k

o

oN

k

kk

M

k

kk

zXzdAzd

A

zd

zc

a

b

za

zbzX

|)()1(,1

)1(

)1()(

1

11

1

1

1

1

0

0

then

N

k

nk nudAnx k

1

][][

33


Example : Consider

.2

1||,

)21

1)(41

1(

1)(

11

zzz

zX

)21

1()41

1()(

1

2

1

1

z

A

z

AzX

2|)()2

11(

1|)()4

11(

2/11

2

4/11

1

z

z

zXzA

zXzA

][4

1][

2

12][ nununx

nn


Properties of Z-Transform:Linearity:

)()(][][

)(][

)(][

zYbzXanbynax

zYny

zXnx

Time shift:

x

n

RROC

zXzzYnnxny

zXnxo

)()(][][

)(][

0

yx RRROC

34


Multiplication by Exponential:

xo

ono

RzROC

zzXnxz

zXnx

||

)/(][

)(][

Conjugation:

xRROC

zXnx

)(][ ***


Time reversal:

xRROC

zXnx

/1

)/1(][ ***

xRROC

zXnx

/1

)/1(][

Convolution:

yx RRROC

zYzXnxny

)()(][*][

35


Example: ][][ nuanx n ][][ nunh

||||,)( azaz

zzX

|1|||,

1)(

z

z

zzH

][*][][ nhnxny

11

2

11

1

1

1)(

|1|||,)1)((

)()()(

az

a

zazY

zzaz

zzHzXzY

y[n] is obtained as:

])[][(1

1][ 1 nuanu

any n


Sampling: Consider a signal x[n] which is obtained by taking samples of a continuous time signal xc(t):

)(][ nTxnx cwhere T is the sampling period, its reciprocal, is the sampling frequency.

Tf s1

36



37


where has the quantized samples and xB[n] has the coded samples (such as 2s complement).

])[(][ nxQnx


38



Frequency Domain Representation of Sampling: Consider the sampled signal xs(t) obtained from xc(t) by multiplying with periodic impulse train:

n

nTtts )()(

then

n

ccs nTttxtstxtx )()()()()(

The discrete time signal is x[n]= xc(nT). The Fourier transform of s(t) is

TkTS s

ks

2,)(

2)(

39


Then, Fourier transform of xs(t) is

k

scs kXTSXX )(

1)(*)(

2

1)(

Ns 2


Ns 2

We observe that when the sampling frequency is not chosen high enough, copies of spectrum overlap; this is called aliasing (distortion). The minimum sampling frequency (rate) to avoid aliasing is 2N (Nyquist rate) :

Ns T 2

2

40


If sampling rate is greater than Nyquist rate, then the original signal xc(t) may be obtained without distortion using a low-pass filter.


Fourier Tr. of xs(t) may be written as

n

njj enxeX ][)(

n

Tnjcs enTxX )()(

On the other hand DTFT of x[n] is

Comparing these equations we observe that )(|)()( TjT

js eXeXX

k

cj

T

k

TX

TeX )

2(

1)(

n

cs nTttxtx )()()(

k

scTj kX

TeX )(

1)(

41


Example:Consider

ttx oc cos)(


)cos()( ttx or

))cos(()( ttx osr

42


Example: Consider

4000,cos)( ooc ttxwith sampling period T=1/6000, we obtain

nnTnx3

2cos4000cos][

Nyquist sampling condition is satisfied , there is no aliasing.

80002120002

os T


Example: Consider

16000,cos)( ooc ttxwith sampling period T=1/6000, we obtain

nnn

nnTnx

3

2cos6000/40002cos

6000/16000cos16000cos][

160002120002

os T

Nyquist sampling condition is not satisfied , there is aliasing.

43


Reconstruction from Samples:Use ideal LPF to recover original signal


With

n

r nTthnxtx )(][)(

Ttth sinc)(

n

r T

nTtnxtx sinc][)(

n

ncs

nTtnx

nTtnTxtx

)(][

)()()(

44



Discrete-time Processing of Cont.-time Signals:

)()(

)(][

THeH

nThTnh

cj

c

with h(t) is impulse response of continuous-time system

][ nh

45


Changing the Sampling Rate:Downsampling and Decimation:Sampling rate can be reduced by further sampling in discrete-time:

)(][][ nMTxnMxnx cd

Let MTT '

Then )'(][ nTxnx cd


If NcX for ,0)( NMTT 2

2

'

2 and

orNM

22

(N is BW of x[n])

Then xc(t) can be recovered without distortion.

Remember that

k

cj

T

k

TX

TeX )

2(

1)(

Similarly

k

cj

d T

k

TX

TeX )

'

2

'(

'

1)(

k

cj

d MT

k

MTX

MTeX )

2(

1)(

46


Let

otherwise,0

,2,,0],[][1

MMnnxnx

neM

nxnxM

l

Mnlj ,1

][][1

0

/21

( [ ]=Discrete Fourier Series rep. of impulse train with period M )

][][][ 1 MnxMnxnx d

nj

n

nj

nd

jd eMnxenxeX

][][)( 1

MknMnk /

Mkj

k

jd ekxeX

/1 ][)(


1

0

//2

/1

0

/2

][1

1][)(

M

l k

MkjMlkj

Mkj

k

M

l

Mlkjjd

eekxM

eeM

nxeX

1

0

)2

()(

1)(

M

l

M

l

Mjj

d eXMeX

Combining the exponential terms

First obtain then shift by increments to get)( Mj

eX

2

)( )/2'( MljeX

47



48


IfNM

22


A general system for downsampling called decimator is

49


Upsampling and Interpolation:Sampling rate can be increased by obtaining intermediate values:

2,/'),'(][ LLTTnTxnx cigiven ).(][ nTxnx c

,2,,0),/(]/[][ LLnLnTxLnxnx ci Consider the following system


The block with inserts (L-1) zeros between samples:L

otherwise,0

2,,0,/][

LLnLnxnx e

or

.][ kLnkxnxk

e

In frequency domain:

n

nj

k

je ekLnkxeX

)(

)()( Ljk

kLjje eXekxeX

50


L=2


k

i LkLn

LkLnkxnx

/)(

/)(sin][][

Ln

Lnnh

/

/sin

51


Changing the sampling rate by noninteger factor:Following system may be used


52


End of Part 1


DIGITAL SIGNAL

PROCESSING

Part 2


Frequency Analysis of LTI Systems:the Input/Output (I/O) relation of a LTI system is given by convolution:

k

knxkhnhnxny ][][][*][][

In z-domain)()()( zHzXzY

jez

jjj eHeXeY

)()()(

On the unit circle

where is the frequency response of the system.

)( jeH


)()()( jjj eHeXeY

Magnitude is

where is the magnitude response of the system.

)( jeH

Phase is )()()( jjj eHeXeY

where is the phase response of the system.

)( jeH


Discrete-time ideal filters: LPF, HPF, BPF:

These are not realizable. Why?


Phase distortion and delay:Assume that ][][ did nnnh

thendnjj

id eeH )(

ord

jid

jid neHeH

)(,1)(

Linear phase distortion causes simple delay.

Group delay: It measures linearity of the phase response.Consider nnsnx ocos][][ where is a narrowband (slowly varying) signal.][ ns


Assume around o

dojj neHeH )(,1)(

Then doood nnnnsny cos][][

Thus, group delay of a system is

)(arg)]([

jj eHd

deHgrd

where is the continuous phase. )(arg jeH



Example: Consider the following filter


Input is the following signal


Pulses are at 85.0,5.0,25.0


Systems with Difference Equation Models:Assume that the Input/Output (I/O) relation of a system is given by a constant coefficient difference equation:

M

kk

N

kk knxbknya

00

][][

Applying z-transform, using linearity and shifting properties, we obtain

M

k

kk

N

k

kk zXzbzYza

00

)()(

)()(00

zXzbzYzaM

k

kk

N

k

kk


Then, the transfer function (or system function) is

N

k

kk

M

k

kk

za

zb

zX

zYzH

0

0

)(

)()(

We can write the transfer function in terms of its poles, dk, and zeros, ck, as

N

kk

M

kk

o

o

zd

zc

a

bzH

1

1

1

1

)1(

)1()(


Example: Consider system function

)4

31)(

2

11(

)1()(

11

21

zz

zzH

)(

)(

83

41

1

21)(

21

21

zX

zY

zz

zzzH

then

)()8

3

4

11()()21( 2121 zYzzzXzz

]2[8

3]1[

4

1][

]2[]1[2][

nynyny

nxnxnx


Stability and Causality: A system is causal if ROC for the transfer function H(z) is outward.A sytem is stable if ROC for the transfer function H(z) includes the unit circle.


Inverse Systems :Let be the inverse system of , then )( zH)( zH i

1)()()( zHzHzG i

then

)(

1)(,

)(

1)(

j

jii eH

eHzH

zH

Not all systems have an inverse. Ideal LPF does not have an inverse, we can not recover high frequency components.Now, consider

N

kk

M

kk

o

o

zd

zc

a

bzH

1

1

1

1

)1(

)1()(

nnhnhng i *Eq.(1)


Then the inverse system has

M

kk

N

kk

o

oi

zc

zd

b

azH

1

1

1

1

)1(

)1()(

Poles become zeros of the inverse system, zeros become poles. For Eq.(1) to hold, ROC of and must overlap. If is causal, ROC is

)( zH i )( zH)( zH

kdz max

Some part of ROC of must overlap with this.)( zH i


Example: Suppose

9.0,9.01

5.0)(

1

1

zz

zzH

The inverse system is

1

1

1

1

21

8.12

5.0

9.01)(

z

z

z

zzH i

Two possible ROC:

2z nununh nni 11 28.1121which is stable but noncausal.

2z 128.12 112

nununh nni

unstable but causal.


Example: Suppose

9.0,9.01

5.01)(

1

1

zz

zzH

1

1

5.01

9.01)(

z

zzH i

The inverse system is

The only choice for ROC is overlaps with is 9.0z

5.0z

then

15.09.05.0 1 nununh nni

which is both stable and causal.

10


Impulse Response for Rational System:Assume that a stable LTI system has a rational transfer function.

N

kk

M

kk

o

o

zd

zc

a

bzH

1

1

1

1

)1(

)1()(

N

k k

kNM

NMifr

rr zd

AzBzH

11

00 1

)(

then

N

k

nkk

NM

rr nudArnBnh

10

1

Impulse response is of infinite length, called Infinite Impulse Response (IIR) system.


If the system has no poles, then

M

k

kk zbzH

0

)(

M

kk knbnh

1

Impulse response is of finite length, called Finite Impulse Response (FIR) system.

Example: Consider a causal system with][]1[][ nxnayny

If then stable and the impulse response is

1a

.nuanh n

11


Example: Consider .

otherwise,0

0,

Mna

nhn

Then

1

11

0 1

1)(

az

zazazH

MMM

n

nn

Zeros at

.,,1,0,12

Mkaez Mk

j

k

Difference equation is

However, we can also write

1][]1[][ 1 Mnxanxnayny M

M

k

k knxany1


Frequency Response for Rational System Functions:Assume that a stable LTI system has a rational transfer function. Then frequency response is obtained by evaluating it on the unit circle:

N

k

kjk

M

k

kjk

j

ea

eb

eH

0

0)(

N

k

jk

M

k

jk

o

oj

ed

ec

a

beH

1

1

)1(

)1()(

12


The magnitude response of the system is

N

k

jk

M

k

jk

o

oj

ed

ec

a

beH

1

1

1

1)(

The magnitude-squared response of the system is

)()()( *2 jjj eHeHeH

N

k

jk

jk

M

k

jk

jk

o

oj

eded

ecec

a

beH

1

*

1

*2

2

11

11)(


Log magnitude or Gain in decibels(dB) :

N

k

jk

M

k

jk

o

oj

ed

eca

beH

110

1101010

1log20

1log20log20)(log20

Attenuation in dB = - Gain in dB

Note that :

)(log20

)(log20)(log20

10

1010

j

jj

eX

eHeY

13


Phase response for rational system function:

N

k

jk

M

k

jk

o

oj

ed

eca

beH

1

1

1

1)(

Group delay for rational system function:

N

k

jk

M

k

jk

j

edd

d

ecd

deH

1

1

]1arg[

]1arg[)(grd


Frequency Response of A single Zero:Consider transfer function of a system as

11)( azzH

then with jrea jjeHjjj ereeeHeH

j 1)()( )(

then magnitude is

and magnitude in dB is

)cos(21)( 22

rreH j

)]cos(21[Log20)(Log20 21010 rreH j

14


then phase is

)cos(1

)sin(tan)( 1

r

reH j

group delay is derivative of the (unwrapped) phase function

d

eHdeHgrd

jj ))(()]([

Example: Consider two cases : r=0.9, =0 and r=0.9, =:


15


magnitudes of vectors give the magnitude response

31

3)( vv

v

e

reeeH

j

jjj

jezz

azzH

)(


phases of vectors give the phase response

31313

)(

vv

e

reeeH

j

jjj

16


Example: Consider a second order system with

11

1

11

1)(

zrezre

zezH

jj

j

21

3)(vv

veH j


jez

j zHeH

)()(

:

17


Example: Consider a third order system with

211

211

7957.04461.11683.01

0166.11105634.0)(

zzz

zzzzH


Relation Between Magnitude and Phase:In general, knowledge about magnitude does not provide information about phase, and vice versa.But, for rational system functions, with some additional information such as number of poles and zeros, magnitude and phase responses provide information about each other.Consider

jez

jjj

zHzH

eHeHeH

|)/1()(

)()()(

**

*2

with

N

kk

M

kk

o

o

zd

zc

a

bzH

1

1

1

1

)1(

)1()(

18


N

kk

M

kk

o

o

zd

zc

a

bzH

1

*

1

*

**

)1(

)1()/1(

Then

N

kkk

M

kkk

o

o

zdzd

zczc

a

bzHzHzC

1

*1

1

*12

**

)1)(1(

)1)(1()/1()()(

Notice that poles and zeros of C(z) occur in conjugate reciprocal pairs. If one pole/zero is inside the unit circle there is another outside.


If H(z) is causal and stable, then all poles must be inside the unit circle, with this we can identify the poles. But zeros of H(z) can not be uniquely identified from zeros of C(z) with this constraint alone.

Example: Consider two stable systems with

14/14/

11

1 8.018.01

5.0112)(

zeze

zzzH

jj

and

14/14/

11

2 8.018.01

211)(

zeze

zzzH

jj

19


Pole/zero plots are


zezezeze

zzzz

zHzHzC

jjjj 4/4/14/14/

11

**111

8.018.018.018.01

5.01125.0112

)/1()()(

zezezeze

zzzz

zHzHzC

jjjj 4/4/14/14/

11

**222

8.018.018.018.01

211211

)/1()()(

Observe that (with )

zzzz 21215.015.014 11

then

)()( 21 zCzC

1224 zz

20


Example: Consider pole/zero plot of C(z) for a system, determine H(z).

For a causal and stable system, poles of H(z) are: p1, p2,p3.For real ak, bk, zeros/poles occur in complex conjugate pairs.


All-Pass Systems:Consider following stable system function

1

*1

1)(

az

azzH ap

j

jj

j

jj

ap ae

aee

ae

aeeH

1

)1(

1)(

**

then constant )(but ,1)( jap

jap eHeH

This is called an all-pass system .A general all-pass system has the following form

cr M

k kk

kkM

k k

kap zeze

ezez

zd

dzAzH

11*1

1*1

11

1

)1)(1(

))((

1)(

21


Example: Consider pole/zero plot of a typical all-pass system


Example: First order all-pass system with a real pole: z=0.9 (z=-0.9)

22


Example: Second order all-pass system with poles:4/9.0 jez


Notice that group delay for causal all-pass systems are positive (unwrapped/continuous phase is nonpositive).All-pass systems may be used as phase compensators.They are also useful in transforming lowpass filters into other frequency-selective forms.

23


Block-Diagram Representation:LTI systems with difference equation represetation (rational system function) may be imlemented by converting to an algorithm or structure that can be realized in desired technology. These structures consists of basic operations of addition, multiplication by a constant and delay.In block diagram representation:


Example: Consider the second order system :][]2[]1[][ 21 nxbnyanyany o

with

22

111

)(

zaza

bzH o

Block diagram representation is

24


For a system with higher order difference equation

M

kk

N

kk knxbknyany

01

][][][

with system function

N

k

kk

M

k

kk

za

zbzH

1

0

1)(

Rewriting the equation as

][][

][][][

1

01

nvknya

knxbknyany

N

kk

M

kk

N

kk

where

M

kk knxbnv

0

][][


N+MDelayelement

Direct Form I:

25


Previous diagram is implementation of

M

k

kkN

k

kk

zbza

zHzHzH0

1

21

1

1)()()(

or

)()()()(0

1 zXzbzXzHzVM

k

kk

)(1

1)()()(

1

2 zVza

zVzHzY N

k

kk


We can rearrange the system function

)(1

1)()()(

1

2 zXza

zXzHzW N

k

kk

)()()()(0

1 zWzbzWzHzYM

k

kk

In the time domain

M

kk knwbny

0

][][

][][][1

nxknwanwN

kk

26


M = N


Direct Form II:

Max(N,M)Delay elements

27


Example:Consider the system with

21

1

9.05.11

21)(

zz

zzH


Flow Graph Representation:Similar to the block diagram representation:

28



)()()( 41 zXzWzW )()( 12 zWzW )()()( 23 zXzWzW

)()( 31

4 zWzzW

)()()( 42 zWzWzY


1

1

1)(

z

zzH

)(1

)(1

1

zXz

zzY

29


Structures for IIR Systems:Some considerations:Computaional complexity(no. of multiplication, delay )Finite precision, Ease of implementation, etc.

Direct Forms:We have already seen direct forms.

Cascade Form:We factor numerator and denominator polynomials of H(z)

21

21

1

1*1

1

1

1

1*1

1

1

)1)(1()1(

)1)(1()1()( N

kkk

N

kk

M

kkk

M

kk

zdzdzc

zgzgzfAzH


We have cascade of first or second order subsystems

s

ss

N

k kk

kk

o

N

k kk

kkokN

kk

zaza

zbzbb

zaza

zbzbbzHzH

12

21

1

22

~1

1

~

12

21

1

22

11

1

1

1

1)()(

30



11

11

21

21

25.015.01

11

125.075.01

21)(

zz

zz

zz

zzzH


Parallel Form:H(z) may be written as sum of subsystems

crf N

k kk

kkN

kk

k

kN

kk zdzd

zeB

zc

AzCzH

01*1

1

00

1

)1)(1(

)1(

1)(

N

k kk

kokN

kk zaza

zeezCzH

f

02

21

1

11

0

1

1)(

31




112121

25.01

25

5.01

188

125.075.01

21)(

zzzz

zzzH

32


Transposed Forms :Reverse the directions of all branches while keeping the values as they were and reverse roles of input and output. Transposed forms may be useful in finite precision implementation.

Example:Consider the second order system with


Structures for FIR Systems:Consider an FIR system with following input-output relation

][][0

knxbnyM

kk

Observe that the impulse response of this system is

otherwise,0

0,][

Mnbnh n

Direct form (or tapped delay line or transversal filter) :

33


Transposed direct form structure:


ss N

kkkok

N

kk zbzbbzHzH

1

22

11

1

)()()(

Cascade form:

34


Linear-Phase FIR Systems:Finite impulse response (FIR) systems with linear-phase have symmetry properties such as

MnnMhnh 0],[][

MnnMhnh 0],[][or

For M even and odd. Thus there are four types of linear phase FIR filters.


)2/sin(

)2/5sin()( 2

jj eeH

otherwise,0

40,1][

nnh

Example: Consider

35


Example: Consider

]2[]1[2][][ nnnnh

)].cos(1[2

2

21)(

1

111

21

j

jjj

jjj

e

eee

eeeH


Assume M is even

M

Mk

M

k

M

k

knxkh

MnxMhknxkh

knxkhny

12/

12/

0

0

][][

]2/[]2/[][][

][][][

With, in the last term, k=M-l, l=0M/2-1

12/

0

12/

0

][][

]2/[]2/[][][][

M

l

M

k

lMnxlMh

MnxMhknxkhny

36


][][ nMhnh If

]2/[]2/[

])[][]([][12/

0

MnxMh

kMnxknxkhnyM

k

If ][][ nMhnh

])[][]([][12/

0

kMnxknxkhnyM

k


Finite-Precision Numerical Effects:Input Quantization:We have seen earlier that continuous-time signals are sampled, quantized and coded first

37


Example:


In twos complement binary system leftmost bit is the sign bit. If we have (B+1)-bit twos complement fraction of the following form

Baaaa 210 Value is B

Baaaa 2222 22

11

00

Example: Binary Code Numeric value

110 4/3

101 2/1

011 4/1

38


Quantizer step size is12

2

B

mX

With][][

^^

nxXnx Bmwhere

)complement s(two'1][1^

nx B

Analysis of Quantization Errors:We observe that the quantized samples are in general different from the true values. The difference is the quantization error

].[][][^

nxnxne

and2/][2/ ne


A simplified model of quantizer is

Assumptions about e[n]:e[n] is stationarye[n] is uncorrelated with x[n]e[n] is white noisee[n] is uniformly distributed

39


Example:

3bit quantizer

Error for 3bit

Error for 8bit

Sinusoidal signal


Mean value of e[n] is zero

012/

2/

deee

Variance (or power) of e[n] is

12

1 22/

2/

22

deee

For (B+1)-bit quantizer with full-scale value Xm

12

2 222 mXB

e

40


Signal-to-Noise Ratio (SNR) is defined as the ratio of signal variance (power) to noise variance. Expressed in dB

x

m

m

xB

e

x

XB

X

10

2

22

102

2

10

log208.1002.6

212log10log10SNR

If

6dB-SNRSNR,2/

dB 6SNRSNR,1

xx

BB


Coefficient Quantization in IIR Systems:Consider

N

k

kk

M

k

kk

za

zbzH

1

0

1)(

If the coefficients are quantized, we get

N

k

kk

M

k

kk

za

zbzH

1

^

0

^

^

1)(

wherekkkkkk aaabbb

^^

,

41


Each pole or zero will be affected by all the errors in the coefficient quantization. If the poles (or zeros) are close to each other (clustered), then quantization of coefficients may cause large shifts of poles (or zeros).Direct form structures are more sensitive to coefficient quantization than the other forms (parallel, cascade,...)


Example: Consider a bandpass IIR elliptic filter of order 12 implemented in cascade form of 2nd order subsystems and direct form.

42


Passband cascade unquantized

Passband cascade 16-bit


Passband parallel 16-bit

Direct form 16-bit

43


Poles of Quantized 2nd order subsystems :Because of robustness, parallel and cascade forms are used more than direct forms.We can further improve the robustness, by improving implementation of the 2nd order subsystems. Consider the following implementation in direct form:


When coefficients are quantized, a finite number of pole possitions possible.

4bits 7bits

Circles correspond to r2, vertical lines to 2rcos

44


Consider the following coupled form


4bits 7bits

45


Coefficient Quantization in FIR Systems:Consider FIR system with transfer function

M

n

nznhzH0

][)(

If][][][

^

nhnhnh

)()(][)(0

^

zHzHznhzHM

n

n

then

where

M

n

nznhzH0

][)(and

)()()(^

jjj eHeHeH


Example: Consider FIR filter of order 27

46



Effects of Round-off Noise:Analysis of Direct Form IIR Structures:Consider Nth-order difference equation for Direct Form I

M

kk

N

kk knxbknyany

01

][][][

Assume that all signal values and coefficients are represented by (B+1)-bit fixed point binary numbers. Therefore, each multiplication is followed by a quantizer, then

M

kk

N

kk knxbQknyaQny

01

^^

][][][

47



An alternative representation is as following

48


Rounding or truncation of a product bx[n] is represented by a noise source of the form

][][][ nbxnbxQne

Assumptions about e[n]s:Each e[n] is stationaryEach e[n] is uncorrelated with x[n] and other e[n]sEach e[n] is uniformly distributed

For (B+1)-bit rounding

2/2][2/2 BB ne


For (B+1)-bit truncation

0][2 neB

Mean and variance for rounding are

0e 122 22

B

e

Mean and variance for truncation are

2

2 Be

12

2 22B

e

49


Now, lets try to determine effect of quantization noise on the output of the system. We can redraw the system as

][][][][][][ 43210 nenenenenene


Since all the noise sources are

12

255

22222222

043210

B

eeeeeee

For the general Direct Form I case

12

21

22

B

e NM

Now, we observe that

][][][1

neknfanfN

kk

For rounding mean of the output noise is zero.The variance for rounding or truncation is

22

2 ][12

21

n

ef

B

f nhNM

is impulse response for ][ nh ef )(/1)( zAzH ef

50


Example: Consider the following system

.1,1

)(1

aazb

zH

][][ nuanh nef

Then the noise variance(power) at the output is

2222

2

1

1

12

22

12

22

aa

Bn

n

B

f


Analysis of Direct Form FIR systems:Consider

M

k

knxkhny0

][][][

12

21,][][][

22

0

B

f

M

kk Mnenenf

51


End of Part 2


DIGITAL SIGNAL

PROCESSING

Part 3


IIR Filters:Infinite impulse response (IIR) filters have rational transfer functions as

N

k

kk

M

k

kk

za

zbzH

0

0)(

Some IIR filter types are: Butterworth, Chebyshev, Elliptical. IIR filters have to be stable. IIR filters may not have linear-phase.


Filter Specifications (Low-pass):

)(log20 10

Design analog filter transform to Digital filter


Analog Butterworth Filters:Approximate ideal LPF with magnitude-squared response

c

cKH,0

,)(

2

by the following rational function:

0,11

)(22

1

221

212

nn

n

nn b

bb

aaKH

|H()|2 must be even function of , and denominator degree must be higher than degree of numerator (lowpass).


.1,,2,1, niba ii

.1)( 22

nnb

KH

Then, we have

For maximal flatness at the origin, =0, The first 2n-1 derivatives of |H()|2 must be zero. This requires

Maximal flatness also at , =, requires

0,11

)(22

1

221

212

nn

n

nn b

bb

bbKH

.1,,2,1,0 niba ii


.

1

1)( 2

2

n

o

H

Then, we have the Butterworth response

Let |H(=0)|2 = 1. Then

At the half-power frequency

n

onn

on

bb

H

2

2

2 1

1

1

2

1)(

.11

)(2

2

nnb

H


Consider the filter specifications:

.dB)(log20 10 H

.dB1log10

2

10

n

o

n

o

2

10/ 110 no 2/110/ 110

Using max and pn

o

p

2

10/ 110 max

Using min and sn

o

s

2

10/ 110 min

To find n:


Dividing these equations

n

p

s

2

10/

10/

110

110max

min

Taking logarithm, we get the filter order as:

p

s

nlog2

110

110log 10/

10/

max

min


then, set the denominator to zero

We can normalize the frequency so that o = 1. Then

To find the filter poles, we let =s/j,

nnn sjssHsH

22 )1(1

1

)/(1

1)()(

.11

)(2

2

nH

1)1(0)1(1 22 nnnn ss

If n is even, then

.12,...,1,0,

.12,...,1,0,1

)2

21(

)2(2

nkes

nkes

n

kj

k

kjn


If n is odd, then

.12,...,1,0,

.12,...,1,0,1 22

nkes

nkes

n

kj

k

kjn

Note that all the poles lie on a circle with radius 1, because the frequency is normalized. (If we want to design analog filter we need to correct this). Also, choose the poles in the left-half plane to get a stable filter.


Example: Design specs. for a Butterworth LP filter:At f = 2000 Hz, max= 3dBAt f = 3000 Hz, min= 10dBThen, the filter order n isn 2.7154 n = 3.Poles are at

.5,...,1,0,3 kesk

j

k

.

))23

21

())(23

21

()(1(

1

)1)(1(

1)(

2

jsjss

ssssH


Analog Chebyshev Filters:We can generalize Butterworth response as:

.)(11

)(1

1)(

22

2

nn F

H

Fn() is a function of . Now, consider:

))(coscos()cos(

)(cos)cos(1

1

xnny

xx

n

This is the Chebyshev polynomial of order n. When

))(coshcosh()(,1 1 xnxCx n

When x 1 , Cn(x) 0.


Cn(x) has zeros in -1< x < 1 :

-1 0 1-1

0

1

x

C1(x)

-1 0 1-1

0

1

x

C2(x)

-1 0 1-1

0

1

x

C3(x)

-1 0 1-1

0

1

x

C4(x)

-1 0 1-1

0

1

x

C5(x)

.)(,1)(

),()(2)(

1

11

xxCxC

xCxCxxC

o

nnn


Use Cn(.) in :

)(1

1)(

22

2

nCH

))(coscos()(,1

))(coshcosh()(,11

1

nC

nC

n

n

2 1 , is known as the ripple factor.


Behaviour at =0:

even isn if,)1(

1)0(,1)(

odd isn if,1)0(,0)(

2

22

22

HC

HC

n

n

Behaviour at =1:

)1(

1)1(

n allfor ,1)(

2

2

2

H

C n


110)(1log10 10max/22 nCThe attenuation is

When .dB01.31)(22 nC

This defines the half-power frequency (3dB cut-off) hp.Then,

).1()).1

(cosh1

cosh(

)1

(cosh1

)(cosh

))(coshcosh(1

)(

1

11

1

hphp

hp

hphpn

n

n

nC


The specifications for a Chebyshev filter are max, min, s(p=1). Bandpass is between 0 1 rad/s.To find n:

)(110

)(1log102210/

22min

minsn

sn

C

C

With 110 10max/2

110

110))(coshcosh(

10/

10/1

max

min

sn

Finally,

)(cosh

110110cosh

1

10/

10/1

max

min

s

n


To find the filter poles, we let =s/j,

)/(1

1)()(

22 jsCsHsH

n

then

1

)/(0)/(1 22 jjsCjsC nn

Letjsjvuw /)cos()cos(

wjs )/(cos 1

With

)sinh()sin()cosh()cos()/(

)cos())/(coscos()/( 1

nvnujnvnujsC

nwjsnjsC

n

n

)cosh(2

)cos(,2

)cos( xee

jxee

xxxjxjx

10


With

1

)/( jjsC n then

1

)sinh()sin(

0)cosh()cos(

nvnu

nvnu

cosh(nv) can never be zero, therefore cos(nu) must be zero. It is possible for

.12,,1,0),12(2

,2

5,

2

3,

2

nkkn

u

nnnu

k

k


For these values of u, sin(nu) = 1, then

.)1

(sinh1 1 an

v k

Remember that

jvuwjs )/(cos 1then

jakn

jwjs kk 122cos)cos(

The poles are :12,,1,0, nkjs kkk

)cosh()2

12cos(

)sinh()2

12sin(

an

k

an

k

k

k

Choose left half poles.

11


Discrete Time Filter Design(IIR):There are 3 approches:1-Sampling (Impulse invariance)2-Bilinear transformation3-Optimal procedures

Impulse invariance method:We can sample the impulse response hc(t) of an analog filter with desired specifications as (Td is sampling interval):

.][ dcd nThTnh then

).2

()( kTT

HHdk d

c


If dc TH /,0)( then

.),()(

d

c THH

Analog and digital frequencies have a linear relation: . dT

Consider the transfer function of a system expressed as

,)(1

N

k k

kc ss

AsH

Then, the impulse response is

.0,)(1

teAthN

k

tskc

k

12


The impulse response of the discrete time filter is

.)(

)(

1

1

nueATnh

nueATnThTnh

N

k

nTskd

N

k

nTskddcd

dk

dk

Transfer (or system) function of the discrete time filter is

.1

)(1

1

N

kTs

kd

ze

ATzH

dk

Poles are .dk Tsk ezss

If Hc(s) is stable (k < 0), then H(z) is also stable (|z| < 1).


Example: Transfer function of analog filter is

.

)23

21

(

12

121

)23

21

(

12

121

1

1

)1)(1(

1)(

2

js

j

js

j

s

ssssH c

then

.

1

)12

121

(

1

)12

121

(

1)(

)2

3

2

1(1)2

3

2

1(1

1

jT

d

jT

d

Td

dd

d

ez

jT

ez

jT

ez

TzH

13


Bilinear Transformation:Transformation needed to convert an analog filter to a discrete time filter must have following properties:1- j axis of the s-plane must be mapped onto the unit circle of the z-plane,2- stable analog filters must be tranformed into stable discrete time filters (left hand plane of the s-plane must be mapped into inside the unit circle of the z-plane).Following BT satisfies these conditions:

./1

/1

1

11

1

Ks

Ksz

z

zKs


Let jrezjs ,

then

.)/()/1(

)/()/1(22

222

KK

KKr

Therefore

)u.c. theinside(1)planeLH(0 r

)u.c. the(1)axis (0 rj

)u.c. theoutside(1)planeRH(0 r

14



Now, let jezjs ,

then.

)(

)(

1

12/2/2/

2/2/2/

jjj

jjj

j

j

eee

eeeK

e

eKj

).2/tan()2/cos(

)2/sin(

jKKjj

or).2/tan( K

15


Observations:1- According to Taylor series expansion of tan(), for small

.2/24/2/ 3 KK 2- For high frequencies, the relation is nonlinear causing a distortion called warping effect. Therefore, BT is usually used for the design of LPF to avoid this.


Discrete Butterworth Filter Design:

We convert a frequency normalized analog Butterworth filterto a discrete filter using the BT.

Normalized half power frequecy HP=1 is mapped into the discrete half power frequency HP. To do that, we set K to

).2/cot()2/tan(/)1( HPHPHPbK

16


Then, we use ).5.0tan(/)5.0tan()5.0tan( HPbK

in nnH 2

2

1

1)(

And get the discrete Butterworth Low Pass Filter as:

n

HP

nH 22

)5.0tan()5.0tan(

1

1)(

Using the design specifications, we find the filter order as

)5.0tan()5.0tan(log2

110110log 10/

10/

max

min

p

s

n


The half-power freq. is

npHP 2/110/

1

110

)5.0tan(tan2

max

).5.0cot( HPbK

Example: Consider the second order analog filter

)12(

1)(

22

sssH

Applying the BT, we get (Kb = 1)

2929.0)1716.0(

)1()(

2

2

2

z

zzH

17


Remarks:1- Given the specs.:

a) find the filter order nb) obtain the analog filter H(s)(using LHP poles)c) calculate HP and Kbd) use biliear transformation to find H(z).

2- H(z) is BIBO stable, because H(s) is stable.3- Applying the BT to high order filters may be tedious. Therefore, first express H(s) as product or some of first and second order functions, then apply the BT.


Example: Design specs. for a LP digital Butterworth filter:At f = 0 Hz, 1= 18dBAt f = 2250 Hz, 2= 21dBAt f = 2500 Hz, 3= 27dBSampling freq. fsmp= 9000 Hz(ej0)= 18dBmax = 2 1=3dBmin = 3 1=9dB smp

ss

smp

pp f

f

f

f 2,2

Filter order n 5.52,n = 6, Kb = 1 and a gain G for = 18dB

)02.0.0)(17.0)(59.0(

)1(0037.0)(

222

6

6

zzz

zzH

18


Discrete Chebyshev Filter Design:

We find the constant K by transforming (normalized passband freq.) P=1 into the discrete passband frequency P. We get

)5.0tan(/)5.0tan()5.0tan(/1 PPcK

And get the discrete Chebyshev Low Pass Filter as:

))5.0tan(/)5.0(tan(1

1)(

22

2

pnCH


Using C1(1)=1, we let max, p

110

1log10

10/

2max

min

Also using min, sWe find the filter order

)]5.0tan(/)5.0[tan(cosh

110110cosh

1

10/

10/1

max

min

ps

n

19


Example: Design specs. for a LP digital Chebyshev filter:At f = 0 Hz, 1= 0dBAt f = 2250 Hz, max= 3dBAt f = 2500 Hz, min= 10dBSampling freq. fsmp= 9000 Hz

Filter order n = 3, Kc = 1

)54.0)(72.015.0(

)1(09.0)(

2

3

3

zzz

zzH

smp

ss

smp

pp f

f

f

f 2,2


Example: Design specs. for a LP digital Chebyshev filter:At = 2/5 rad, max= 1dBAt = /2 rad, min= 9dB

Filter order n = 3

)472.0)(619.0505.0(

)1(736.0)(

2

3

3

zzz

zzH

Example: Design specs. for a LP digital Butterworth filter:At = /2 rad, max= 3dB = 0 rad, = 0dBAt = 5/9 rad, min= 10dB

Filter order n = 7, Kb = 1

)052.0)(232.0)(636.0(

)1(01656.0)(

222

7

7

zzzz

zzH

20


Frequency Transformations:

The objective is to obtain transfer functions of other types of dicrete filters from already available prototype Low Pass Filters using tranformation functions g(z) as

))(()( zgHzH LP

Example: Design specs. for a LP digital Chebyshev filter:At f = 0 Hz, = 0dBAt f = 2250 Hz, 2= 21dBAt f = 2500 Hz, 3= 27dBSampling freq. fsmp= 9000 Hz


39.0802.0691.0

)1(09.0)(

23

3

zzz

zzH LPF

21


Now, we want a HPF with cutt-off at 3.6kHz, we use following transformation

)5095.01(

)5095.0(1

11

z

zz

6884.0102.2361.2

)133(0066.0)(

23

23

zzz

zzzzH HPF


Design of FIR Filters by Windowing:

Ideal frequency response and corresponding impulse response functions are .),( nheH djd

Generally hd[n] is infinitely long. To obtain a causal practical FIR filter, we can truncate it as

otherwise.,0

0, Mnnhnh d

In a general form we can write it as nwnhnh d

Where w[n] is window function.

22


In frequency domain, we have

.)()(2

1)(

deWeHeH jjdj


Some commonly used window functions are Rectangular, Bartlett, Hamming, Hanning, Blackman, Kaiser.

Some important factors in choosing windows are: Main lobe width and Peak side lobe level. For rectangular:

23


Main lobe width and transition region; Peak side lobe level and oscillations of filter are related.


24



25


Discrete Time Fourier Series (DTFS):

Consider a periodic sequence with period N:

.][11

0

2~~

N

k

nkN

jekX

Nnx

Which can be represented by a Fourier series as:

.][21

0

~~ nkN

jN

n

enxkX

The DTFS coefficients are obtained as:

.~~

Nnxnx

nx~


Let NjN eW

2

The DTFS analysis and synthesis equations are:

.][1

0

~~nk

N

N

n

WnxkX

.][11

0

~~

N

k

nkNWkXN

nx

Example: Consider the periodic impulse train:

r

rNnnx ][~

The DTFS

. allfor 1][1

0

~

kWnkX nkNN

n

26


Similarly, ][~~

lkXnxW nlN

Properties: 1- Linearity: It is linear.2- Shift of a sequence: If ][

~~

kXnx

then ][~~

kXWmnx kmN

3- Periodic convolution: Consider two periodic sequences with period N

][~

1

~

1 kXnx ][~

2

~

2 kXnx

Then,

][][][~

2

~

1

~

3 kXkXkX .][][1

0

~

2

~

13

~

N

m

mxmnxnx


Example: Consider periodic convolution of two sequences:

27


Sampling the Fourier Transform:

Consider signal x[n] with DTFT X() and assume by sampling X() , we get

kN

XXkXk

N

2|)(][ 2

~

][~

kX could be the sequence of DTFS coefficients of a periodic signal which may be obtained as

.][11

0

~~

N

k

nkNWkXN

nx


Substituting,

k

Nm

mj XkXemxX 2

~

|)(,

.*~

rr

rNnxrNnnxnx then,

28


If N is too small, then aliasing occurs in the time domain.

If there is no aliasing, we can recover x[n].


Discrete Fourier Tranform (DFT):

DFT is obtained by taking samples of the DTFT. Remember DTFT is defined as:

n

njj enxeX ][)(

We take samples of X(ej) at uniform intervals as:

.1,1,0,)(][ 2

NkeXkXk

N

j

We define: N

j

N eW2

29


Then the DFT is defined as (analysis equation):

N

n

knNWnxkX

0

][][

for k=0,1,...,N-1. The inverse DFT is defined as (synthesis equation):

N

n

knNWkXN

nx0

][1

][

for n=0,1,...,N-1.


Example:N=5

30


N=10


Properties:1- Linearity: DFT is a linear operation.2- Circular shift:

].[10,2

kXeNnmnxkm

Nj

N

31


3- Circular convolution:

kXkXkX

mxmnxnxnxnxN

mN

213

1

021213 ][][

Example:


Example: N=L

32


N=2L


4- Multiplication(Modulation):

.213213 kXkXkXnxnxnx

Linear Convolution Using the DFT:

Since there are efficient algorithms to take DFT, like FFT, we can use it instead of direct convolution as:

1. Compute N point DFTs, 2. Multiply them to get 3. Compute the inverse DFT, to get:

. and 21 kXkX .213 kXkXkX

nxnxnx 213

33


Linear Convolution of Finite Length Signals:Consider two sequences x1[n] of length L and x2[n] of

length P, and x3[n] = x1[n]* x2[n].Observe that x3[n] will be of length (L+P-1).

Circular Convolution as Linear Convolution with Aliasing :

Consider again x1[n] of length L and x2[n] of length P, and x3[n] = x1[n]* x2[n].

)()()( 213 jjj eXeXeX


Taking N samples .213 kXkXkX

Now, taking the inverse DFT, we have

otherwise.,0

10,33

NnrNnxnx

rp

nxnxnx p 213

This circular convolution is identical to the linear convolution corresponding to X1(ej) X2(ej), if N (L+P-1).

34


Example:


Example:

35



LTI Systems Using the DFT :Consider x[n] of length L and h[n] of length P, and y[n] =

x[n]* h[n] will be of length (L+P-1).To obtain the result of linear convolution using the DFT,

we must use N ( (L+P-1))-point DFTs. For that x[n] and h[n] must be augmented with zeros (zero-padding).

Overlap-Add Method :Consider h[n] of length P, and length of x[n] is much

grater than P.We can write x[n] as combination of length L segments:

,otherwise,0

10,

LnrLnx

nx r

r

r rLnxnx

36


Since the system is LTI then

nhnxnyrLnyny rrr

r *,

Each output segment yr[n] can be obtained using (L+P-1) point DFT.

Example:


37


Overlap-Save Method :Consider again h[n] of length P, and length of x[n] is much

grater than P.In this method L-point circular convolution (or DFT) is

used. Since the first (P-1) points will be incorrect, input segments must overlap.

We can write each L sample segment of x[n] as :

10,)1()1( LnPPLrnxnx r

11,

,)1()1(

LnPnyny

PPLrnyny

rpr

rr


Example:

38


Efficient Computation of the DFT :Remember the definition of the DFT and inverse DFT

1

0

][][N

n

knNWkXnx

1

0

][][N

n

knNWnxkX

N2 comlex multiplications and N(N-1) additions or 4N2 real mult. and 4N(N-2) real additions necessary:

N

n

knN

knN

knN

knN

Wnx

Wnxj

Wnx

Wnx

kX0

}Re{}][Im{

}Im{}][Re{

}Im{}][Im{

}Re{}][Re{

][


To improve the efficiency, we can use some properties as:

nNkN

knN

nNkN

knN

knN

nNkN

WWW

WWW)()(

*)(

)2

)()1

}.Re{}][Re{}][Re{}Re{}][Re{}Re{}][Re{ )(

knN

nNkN

knN

WnNxnx

WnNxWnx

Example:

39


Fast Fourier Transform(FFT):Fast Fourier Transform algorithms are used to implement DFT efficiently to reduce the number of multiplication and addition operations.Two main algorithms are:Decimation in time and Decimation in frequency. Both these algorithms require that N=2m.The number of operations using either of these will require (N/2)log2N complex multiplications and Nlog2Nadditions.Direct implementation of DFT requires N2 complex multiplications and additions. For N=1024= 210

N2=1048576 but Nlog2N=10240.


Decimation in time FFT Algorithm:Assume N=2m

1,,1,0,][][1

0

NkWnxkXN

n

knN

Which can be written as

oddn

knN

evenn

knN WnxWnxkX ][][][

With n = 2r (even) and n = 2r+1 (odd)

.

]12[)(]2[

)(]12[)(]2[][

12/

02/

12/

02/

12/

0

212/

0

2

kHWkG

WrxWWrx

WrxWWrxkX

kN

N

rN

kN

N

rN

N

r

rkN

kN

N

r

rkN

rkrk

40


.][ kHWkGkX kN

Requires N+2(N/2)2 multiplications


Since G[k] and H[k] requires 2(m-1)-point DFTs, each one can be similarly decomposed until we reach 2-point DFTs.

41



Butterfly:(in place comp.)

Bit reversed order

42


Decimation in frequency FFT Algorithm:Again assume N=2m

1,,1,0,][][1

0

NkWnxkXN

n

knN

then)12/(,,1,0,][]2[

1

0

2

NrWnxrXN

n

rnN

1

2/

212/

0

2 ][][]2[N

Nn

rnN

N

n

rnN WnxWnxrX

12/

0

)2/(212/

0

2 ]2/[][]2[N

n

NnrN

N

n

rnN WNnxWnxrX

12/

0

12/

02/2/ ][])2/[][(]2[

N

n

N

n

rnN

rnN WngWNnxnxrX


Similarly for

.][

])2/[][(]12[

12/

02/

12/

02/

N

n

rnN

nN

N

n

rnN

nN

WWnh

WWNnxnxrX

)12/(0 Nr

43


Procedure is repeated until 2-point DFTs


Computation of Inverse DFT:

1,,1,0,][1

][0

NnWkXN

nxN

n

knN

1,,1,0,][][0

NkWnxkXN

n

knN

][DFT1][1][ *0

** kXN

WkXN

nxN

n

knN

** ][DFT1][ kXN

nx

44


End of Part 3


DIGITAL SIGNAL

PROCESSING

Part 4

J.S. Lim, Two Dimensional Signal Processing, in Advanced Topics In Signal Processing, Prentice-Hall.


Signals:Impulses: The 2-D impulse is,

otherwise,0

0,1, 2121

nnnn


Any sequence may be represented as a linear combination of shifted impulses:

].,[],[, 221121211 2

knknkkxnnxk k

21, nnx


Line impulses are:

otherwise,0

0,1, 1121

nnnnx T


or

otherwise,0

0,1, 2221

nnnnx T


or

otherwise,0

,1, 212121

nnnnnnx T


Step sequence:

otherwise,0

0,,1, 2121

nnnnu

].,[,1

1

2

2

2121

n

k

n

k

kknnu

or

1,11,,1,, 2121212121 nnunnunnunnunn


Separable sequences:A separable sequence can be written as

221121, nfnfnnx The impulse is separable

2121, nnnn Periodic sequences:

is periodic with period N1xN2 if 21,~ nnx 22121121 ,~,~,~ NnnxnNnxnnx

Example: Periodic with 2x4

2121 )2/(cos,~ nnnnx


Linearity: A system is linear if

Systems:The relation between the input and the response of the system is given by

.,, 2121 nnxFnny

21 , nnx

.,,,, 212211212211 nnybnnyannxbnnxaF where .2,1,,, 2121 innxFnny ii

Time Invariance: A system is time-invariant if

.,, 22112211 mnmnymnmnxF


Convolution:For a LSI system with impulse response, , the input/output relation is given by

Where * denotes 2-D convolution.

Convolution with a delayed impulse :

21 , nnh

].,[],[

,*,,

221121

212121

1 2

knknhkkx

nnhnnxnny

k k

].,[,*, 2211221121 mnmnxmnmnnnx


Example:


].,[],[, 221121211 2

knknhkkxnnyk k


If the impulse response is separable

221121, nhnhnnh then

1

1 2

1 2

21111

22221111

2221112121

,][

][],[][

][][],[,

k

k k

k k

nkfknh

knhkkxknh

knhknhkkxnny


Stability: A system is BIBO stable iff a bounded input leads to a bounded output. If

then the system is BIBO stable.

1 2

],[ 21n n

nnh


Discrete Space Fourier Transform (2D F.T.):

.),()2(

1,

,),(

2121221

2121

2211

1 2

2211

1 2

ddeeXnnx

eennxX

njnj

njnj

n n

is periodic with 2 in both variables.),( 21 X

).2,(),2(),( 212121 XXX

2D Fourier Transform and its inverse are defined as


Example:


Frequency Response:

If the input to Linear Shift Invariant (LSI) system is

2211],[ 21njnj eennx

then 2211),(],[ 2121

njnj eeHnny

where

22111 2

2121 ,),(kjkj

k k

eekkhH

is the frequency response function of the system.


Example:Given LPF

otherwise,0

and,1),( 2121

baH

This function can be written as

).()(),( 221121 HHH

10


.)sin()sin(

][][],[2

2

1

1221121

n

bn

n

annhnhnnh

then


Example:Given impulse response and magnitude response of a 2D LPF as

11


Original LP-Filtered HP-Filtered


Example:Given frequency response of a 2D LPF as

2122

21

22

21

21,and,0

,1),(

c

cH

12


where J1(x) is the Bessel function.

2221122

21

212

],[ nnJnn

nnh cc

then the impulse response sequence is


2D Z-Transform:

211 2

212121 ,),(nn

n n

zznnxzzX

2D Z-Transform is defined as

where z1 and z2 are complex variables. The space represented by (z1, z2) is four-dimesional (4-D).

13


Example:Given 2D sequence 2121 ,],[ 21 nnubannx nn

then

bzazbzazz

b

z

a

zznnubazzX

n n

nn

nn

n n

nn

2112

110 0 21

212121

and,1

1

1

1

,),(

1 2

21

21

1 2

21


Inverse Z-Transform:2-D polynomials, in general, can not be factored as a product of lower order polynomials, therefore the partial fraction expansion is not a general procedure for 2-D signals.

Linear, Constant Coefficient Difference Eq.:LCCDE is given in the following form

221121),(

221121),(

,,,,2121

knknxkkbknknykkakk Rkk R BA

where a and b are known, with boundary conditions.

14


Example:Given an IIR filter with a first quadrant impulse response as

12

11

21 5.01

1),(

zzzzH

then

12

1121

2121 5.01

1

),(

),(),(

zzzzX

zzYzzH

),(),(5.0),( 211

21

12121 zzXzzzzYzzY

],[]1,1[5.0],[ 212121 nnxnnynny


End of Part 4

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