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FJU-EE – YUJL – Introduction DSP– Chapter1 - Page 1

CHAPTER Zero :

Outlines


PREFACE

This bookIntegrating Matlab with traditional topics on DSPUse Matlab to explore difficult topics and solve problems to gain insightMatlab provides a convenient tool so that many scenarios can be tried with easeEnhance the learning processAssume that the student is familiar with the fundamental of MATLAB


PREFACE

Organization of the BookChapter 1, IntroductionChapter 2, Discrete-time Signals and SystemsChapter 3, The Discrete-time Fourier AnalysisChapter 4, The Z-TransformChapter 5, The Discrete Fourier TransformChapter 6, Digital Filter StructuresChapter 7, FIR Filter DesignChapter 8, IIR Filter Design...... And more topics


CHAPTER One :Introduction

Introduce the discipline of Signal ProcessingDiscuss the advantages of DSP over analog signal processingIntroduce the MATLAB


INTRODUCTION

The field of DSP has grown to be important both theoretically and technologically.

Development and use of low-cost software and hardwareNew technologies and applications take advantage of DSP algorithmsMake DSP an integral part of any E.E. curriculum

What facilities do undergraduate students need in this DSP course ?

Advances in PCInteractive software, MATLAB


Why signals should be processed?

Signals are carriers of informationUseful and unwantedExtracting, enhancing, storing and transmitting the useful information

What signals are being processed?Analog Signal Processing vs. Digital Signal Processing


Why signals should be processed?

How are signal processed?ASP (Analog Signal Processing)Processed Using electrical networks containing active and passive circuit elements.Analog Signal - Continuously in time and amplitude

Ex) Radio & television receiver

Example: Analog Signal : xa(t) Analog signal processor ya(t) :

Analog Signal


DSP (Digital Signal Processing)

PrF ADC DSP DAC PoFAnalog Analog

Equivalent analog signal processor

PrF: antialiasing filtering to prevent aliasing

PoF: smoothing filter to smooth out the staircase waveform


DSP (Digital Signal Processing)

PrF : (Prefilter or antialiasing Filter) Conditions the analog signal to prevent aliasing

ADC :(Analog-to-Digital Converter) Produces a stream of binary number from analog Signal

DSP :(Digital Signal Processor)Heart of DSP

General-purpose computer, special-purpose processor digital hardware

DAC :(Digital-to-Analog Converter) Produce a staircase waveform from a sequence of binary numbers.

PoF : (Post filter)Smooth out staircase waveform into the desired analog Signal


Comparison of DSP over ASP

AdvantagesDeveloped using software running on a general-purpose computer with high testability and portability. DSP operations are base on additions and multiplications (so extremely stable processing, and stability independent of temperature)DSP operations can easily be modified in real time, by simple programming changes, or by reloading of registerDSP has lower cost due to VLSI technology. (memory, gates, microprocessor)

DisadvantagesLow speed in very high frequency

DSP ApplicationsConsumer electronics, Communications, Wireless telephones, Medical Image


Signal Analysis: measure signal properties and find what it is in the received signal

Spectrum (frequency/phase) analysisSpeech recognitionSpeaker verificationTarget detection

Signal Filtering: improve the signal quality from corrupted signals by signal-in-signal-out filter

Removal of unwanted background noise/interferenceSeparation of frequency bandsShaping of the signal spectrum

The two categories of DSP Tasks


What is DSP Used For?

……And much more!And much more!


Speech Processing

Speech coding/compressionSpeech synthesisSpeech recognition


Some Properties of Speech

The blue--- s---p--o---------t i-s--on--the-- k--ey a---g--ai----n------

“oo” in “blue”“o” in “spot”“ee” in “key”“e” in “again”“s” in “spot”“k” in “key”


Some Properties of Speech

“ee” in “key”“o” in “spot”“oo” in “blue” “e” in “again”

Vowels

“s” in “spot” “k” in “key”

Consonants

•Quasi-periodic

•Relatively high signal power

•Non-periodic (random)

•Relatively low signal power


Speech Coding

BTS(Base Transceiver)

BSC (Base Station Controller)

MSC(Mobile Switching Center) TRAU(Transcoder/Rate

Adapter Unit)

64 kbits/s

13 kbits/s 22.8 kbits/s


Try to predict the current sample value;Transmit the prediction error.

Speech Coding – Linear Prediction

ΣA(z)

s(n)+– d(n)

se(n)Σ…

d(n)

A(z)

++

se(n)

sr(n)


Standard music CD:Sampling Rate: 44.1 kHz16-bit samples2-channel stereoData transfer rate = 2×16×44,100 = 1.4 Mbits/s1 hour of music = 1.4×3,600 = 635 MB

Digital Audio


Key standards:MPEG: Layers I, II, and III (MP3); AAC.

used in DAB, DVDDolby AC3, Dolby Digital, Dolby Surround.

Typical bit rates for 2-channel stereo:64kbits/s to 384 kbits/s.

Subband- or transform-based, making use of perceptual masking properties.

Audio Coding (Cont’d)


Still Image Coding:JPEG (Joint Photographic Experts Group):

Discrete Cosine Transform (DCT) basedJPEG2000: Wavelet Transform based

Video Coding:MPEG (Moving Pictures Experts Group):

DCT-based,Interframe and intraframe prediction,Motion estimation.

Applications: Digital TV, DVD, etc.

Image/Video


JPEG Example

Original

JPEG (100:1)JPEG (4:1)


Self-learning: Filter coefficients adapt in response to training signal.

• Filter update: Least Mean Squares (LMS) algorithm

Adaptive Filtering

W(z) Σ+–x(n)

y(n)e(n)

d(n)

)()(2)()1( nnenn xww μ+=+


Echo cancellation (telephone lines)Used in modems (making Internet access possible!!)

Adaptive equalizationActive noise controlMedical signal processing

e.g. foetal heart beat monitoringImage analysis, e.g:

Face recognition,Optical Character Recognition (OCR);

Adaptive Filtering Applications


Restoration of old image, video, and audio signals;Analysis of RADAR data;Analysis of SONAR data;Data transmission (modems, radio, echo cancellation, channel equalization, etc.);Storage and archiving;Control of electric motors.

Adaptive Filtering Applications


Selecting a DSP – several choices:Fixed-point;Floating point;Application-specific devices(e.g. FFT processors, speech recognizers,etc.).

Main DSP Manufacturers:Texas Instruments (http://www.ti.com)Motorola (http://www.motorola.com)Analog Devices (http://www.analog.com)

DSP Devices & Architectures


FEW WORDS ABOUT MATLABMATLAB is an interactive, matrix –based system for scientific and engineering numeric computation and visualization

MATLAB Tutorial : Referenced from Tolga Kurt [1]

FEW WORDS ABOUT MATLAB


Getting Started

Matlab - Matrix laboratoryA high-level technical computing language and interactive environment Useful for :

analyzing datadeveloping algorithms and applications


Defining vectors and matrices –(1/2)

Defining a matrixRows are separated by semicolons.Semicolons will prevent the results display on the command window.To display type the name of the variable.


Defining vectors and matrices –(2/2)

Definition with increments:


Basic Operations

Element by element operations:A+B Element by element additionA-B Element by element subtractionA*B Matrix multiplicationA.*B Element by element multiplicationA/B A*inv(B)A./B Element by element divisionA^B Matrix powerA.' TransposeA' Complex conjugate transposeA^-1 Matrix inverse - inv(A)


Graphs

Using built in function


M-file example

%%Generate two random shifted BPSKclearclose allclclist=[.5 .6 .7 .8 .9 1]count=0;for k=list;

count=count+1R1=sqrt(k);R2=sqrt(1-k);mx2=0;itmax=100000;av=0;

for t=1:itmaxA=R1*(randint(64,1,2)*2-1);B=R2*(randint(64,1,2)*2-1);

Tran=A+B*j;

av=av+mean(abs(Tran).^2)/itmax;

endson(count)=10*log10(mx2/av);

endplot(list,son)


M-file example

% MATLAB script that generates the probability of error versus the signal to noise ratio.initial_snr=0;final_snr=12;snr_step=0.75; tolerance=eps; % Tolerance used for the integrationplus_inf=20; % This is practically infinitysnr_in_dB=initial_snr:snr_step:final_snr;for i=1:length(snr_in_dB),

snr=10^(snr_in_dB(i)/10);Pe_2(i)=1-quad8('bdt_int2',0,plus_inf,tolerance,[],snr,2);Pe_4(i)=1-quad8('bdt_int2',0,plus_inf,tolerance,[],snr,4);Pe_8(i)=1-quad8('bdt_int2',0,plus_inf,tolerance,[],snr,8);Pe_16(i)=1-quad8('bdt_int2',0,plus_inf,tolerance,[],snr,16);Pe_32(i)=1-quad8('bdt_int2',0,plus_inf,tolerance,[],snr,32);

end;


M-file example

% Plotting commands followt=snr_in_dB;semilogy(t,Pe_2,'-',t,Pe_4,'--',t,Pe_8,'-.',t,Pe_16,':',t,Pe_32,'-')legend('M=2','M=4','M=8','M=16','M=32',0);axis([initial_snr,final_snr,1e-16,1e0])

xlabel('Input SNR per Bit','fontsize',16,'fontname','Arial')ylabel('Symbol Error Rate (SER)','fontsize',16,'fontname','Arial')set(findobj('Type','line'),'LineWidth',2)set(findobj('fontname','Helvetica'),'fontname','Times new Roman','fontsize',16)title('Symbol error probability for biorthogonal signals')hold off


DSP example


More about Matlab

MATLAB Tutorial by張智星

[email protected]://neural.cs.nthu.edu.tw/jang/

Topics01-MATLAB小傳與外觀02-初探MATLAB03-二維平面繪圖09-矩陣的處理與運算14-MATLAB 的運算元15-M檔案


Useful Referenceshttp://www.genie.uottawa.ca/~ieee/pdffiles/matlab.ppthttp://neural.cs.nthu.edu.tw/jang/http://www.math.ohiou.edu/courses/matlab/Matlab Help http://webclass.ncu.edu.tw/~junwu/index.htmlhttp://caig.cs.nctu.edu.tw/course/NM/slides/MatlabTutorial.pdfhttp://users.ece.gatech.edu/~bonnie/book/TUTORIAL/tutorial.htmlContemporary communication systems using MATLAB / J. G. Proakis, M. Salehi.MATLAB programming for engineers / Stephen J. Chapman. Simulation and Software Radio for Mobile Communications, Chapter 2, H. Harada and R. Prasad …


http://www.math.ohiou.edu/courses/matlab/Math410, 1,2,3,4,5

Due date: 本章結束後當週星期五。

§ HomeWork


Chapter 2. Discrete-Time Signals and

Systems

Important types of signals and their operationsLinear and shift-invariant systemThe convolution and difference equation representationsRepresentations and implementation of signal and systems using MATLAB


§2.1 Discrete-Time Signals

Category of signals : Analog and discrete signalsAnalog signal xa(t)

t represents any physical quantity, time in sec.Discrete-Time Signals x(n) :

n is integer valued, represents discrete instances in time.A discrete-time signal, which is a number sequence, will be denoted by one of the following notations:

where the up-arrow indicates the sample at n=0.

( ) { ( )} { , ( 1), (0), (1), }x n x n x x x= = −↑


§2.1 Discrete-Time Signals

In Matlab, a finite-duration sequence representation requires two vectors, and each for x and n.

Example:

Then in Matlab :

Question: whether or not an arbitrary infinite-duration sequence can be represented in MATLAB?

( ) {2,1, 1, 0,1,4,3,7}x n = −↑

];7,3,4,1,0,1,1,2[];4,3,2,1,0,1,2,3[

−=−−−=

xn


Types of Elementary Sequences

Unit sample sequence δ(n)

Representation in MATLABn=[n1:n2]; x = zeros(1,n2n=[n1:n2]; x = zeros(1,n2--n1+1); x(n0n1+1); x(n0--n1+1)=1;n1+1)=1;n=-2:30; x=[zeros(1,2),1,zeros(1,30)];stem(n,x);n=[n1:n2]; x = [(n-n0)==0]; stem(n,x,’ro’);

{ }1, 0( ) ,0,0,1,0,0,0, 0n

nn

δ↑

=⎧= =⎨ ≠⎩

00

0

1,( )

0,n n

n nn n

δ=⎧

− = ⎨ ≠⎩


Types of Elementary Sequences

Unit sample sequence


Unit Step Sequence u(n)

Unit step sequence

Representation in MATLABn=0:30; x=[ones(1,31)]; stem(n,x);n=[n1:n2]; x=[(n-n0)>=0];

00

0

1, 0( ) ,0,0, 1,1,1,

0, 0

1,( )

0,

nu n

n

n nu n n

n n

≥ ⎧ ⎫⎧= =⎨ ⎨ ⎬


Unit Step Sequence u(n)

Unit step sequence


Real-valued exponential sequence


Rananx n ∈∀= ;,)(For Example: 100,)9.0()( ≤≤= nnx n

n=[0:10]; x=(0.9).^n; stem(n,x,’ro’)

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1



1. a=3/4;2. n=0:10;3. x=a.^n;4. subplot(2,1,1)5. stem(n,x);

1. a=4/3;2. n=0:10;3. x=a.^n;4. subplot(2,1,2)5. stem(n,x);

( ) , ;nx n a n a R= ∀ ∈


Complex-valued exponential sequence


nenx nj ∀= + ,)( )( 0ωσ

σ: Attenuation: 衰减因子Ω0: frequency in radians:

For Example: n=[0:10]; x=exp((2+3j)*n);


Sinusoidal Sequence


1. a=2;f0=4;phi=pi/6;2. w0=2*pi*f0;3. t=0:.001:1;4. x=a*sin(w0*t+phi);5. plot(t,x);6. ylable('x(t)');xlable('Time (s)');7. title('sinusoidal signal');

0( ) cos( ),x n A n nω θ= + ∀


Random Sequence

In Matlab two types of (pseudo-) random sequences are available.

The rand(1,N) generates a length N random sequence whose elements are uniformly distributed between[0,1].The randn(1,N) generates a length N Gaussian random sequence with mean 0 and variance 1.


Periodic sequence

A sequence x(n) is periodic if x(n)=x(n+N)The smallest integer N is called the fundamental periodFor example

A: xtilde=[x,x,x,x,x,x,x,x,x,x,x,x,x]B: xtilde=x’*ones(1,P); xtilde=xtilde(:); xtilde=xtilde’;


OPERATIONS ON SEQUENCES

Signal additionSignal multiplicationScalingShiftingFoldingSample summationSample productsSignal energySignal power


1. Signal addition

Signal addition is a sample-by-sample addition given by{x1(n)}+{x2(n)}={x1(n)+x2(n)}It is implemented in Matlab by the arithmetic operator ”+”. However, the lengths of x1(n) and x2(n) must be the same. The following function called the sigadd function, demonstrates these operations.1. function [y,n]=sigadd(x1,n1,x2,n2)2. n=min(min(n1),min(n2)): max(max(n1),max(n2));3. y1=zeros(1,length(n)); y2=y1;4. y1(find((n>=min(n1)) & (n=min(n2)) & (n


2. Signal multiplication

Signal multiplication is a sample-by-sample multiplication given by

{x1(n)}.{x2(n)}={x1(n)x2(n)}It is implemented in Matlab by the arithmetic operator ”.*”. Once again the similar restrictions apply for the .* operator . Therefore we have developed the sigmult function, which is similar to the sigadd function.1. function [y,n]=sigmult(x1,n1,x2,n2)2. n=min(min(n1),min(n2)) : max(max(n1),max(n2));3. y1=zeros(1,length(n)); y2=y1;4. y1(find((n>=min(n1)) & (n=min(n2)) & (n


3. Scaling

In this operation each sample is multiplied by a scalar α.α{x(n)}={αx(n)}An operator “*” is used to implement the scaling operation in Matlab.


4. Shifting

In this operation each sample of x(n) is shifted by an amount k to obtain x(n-k).

y(n)={x(n-k)}Or if we let m=n-k, y(m+k)={x(m)}

This is shown in the function sigshift. function [y,n] = sigshift(x,m,n0)% implements y(n) = x(n-n0)n = m+n0; y = x;


5. Folding

In this operation each sample of x(n) is flipped around n=0 to obtain a folded sequence x(-n).

y(n)={x(-n)}This is shown in the function sigfold.

function [y,n] = sigfold(x,n)% implements y(n) = x(-n)y = fliplr(x); n = -fliplr(n);%fliplr


6. Sample summation

It adds all sample values of x(n) between n1 and n2It is implemented by the sum(x(n1:n2)) function.

)(...)()( 212

1

nxnxnxnn

nn

++=∑=

=


7. Sample products

It multiplies all sample values of x(n) between n1 and n2.It is implemented by the prod(x(n1:n2)) function.

2

1 21

( ) ( ) ... ( )n

n

x n x n x n= × ×∏


8. Signal energy

The energy of a sequence x(n) is given by

The energy of a finite-duration sequence x(n) can be computed in MATLAB using

Ex=sum(x.*conj(x));% one approachEx=sum(abs(x).^2); % another approach

2*( ) ( ) ( )xn n

x n x n x nε∞ ∞

=−∞ =−∞

= =∑ ∑


9.Signal power

The average power of a periodic sequence withfundamental period N is given by

12

0

1 ( )N

xn

P x nN

−

=

= ∑


EXAMPLE 2.1

EXAMPLE 2.1 Generate and plot each of the following sequences over the indicated interval.

% a) x(n) = 2*delta(n+2) - delta(n-4), -5


EXAMPLE 2.1

% c) x(n) = cos(0.04*pi*n) + 0.2*w(n); 0


EXAMPLE 2.1

−5 0 5−2

−1

0

1

2

3Sequence in Example 2.1a

n

x(n)

0 5 10 15 20

0

2

4

6

8

10

Sequence in Example 2.1b

n

x(n)

0 10 20 30 40 50

−1

−0.5

0

0.5

1

Sequence in Example 2.1c

n

x(n)

−10 −5 0 5−1

0

1

2

3

4

5

6Sequence in Example 2.1d

n

xtild

e(n)


EXAMPLE 2.2

Let x(n)={1,2,3,4,5,6,7,6,5,4,3,2,1}. Determine and plot the following sequences.

figure(1); clfn = -2:10; x = [1:7,6:-1:1];% a) x1(n) = 2*x(n-5) - 3*x(n+4)1. [x11,n11] = sigshift(x,n,5); [x12,n12] = sigshift(x,n,-4);2. [x1,n1] = sigadd(2*x11,n11,-3*x12,n12);3. subplot(2,1,1); stem(n1,x1); title('Sequence in Example 2.2a')4. xlabel('n'); ylabel('x1(n)'); axis([min(n1)-1,max(n1)+1,min(x1)-

1,max(x1)+1])5. set(gca,'XTickMode','manual','XTick',[min(n1),0,max(n1)])


EXAMPLE 2.2

% b) x2(n) = x(3-n) + x(n)*x(n-2)1. [x21,n21] = sigfold(x,n); [x21,n21] = sigshift(x21,n21,3);2. [x22,n22] = sigshift(x,n,2); [x22,n22] = sigmult(x,n,x22,n22);3. [x2,n2] = sigadd(x21,n21,x22,n22);4. subplot(2,1,2); stem(n2,x2); title('Sequence in Example 2.2b')5. xlabel('n'); ylabel('x2(n)'); axis([min(n2)-1,max(n2)+1,0,40])6. set(gca,'XTickMode','manual','XTick',[min(n2),0,max(n2)])


EXAMPLE 2.2

−6 0 15

−20

−15

−10

−5

0

5

10

15Sequence in Example 2.2a

n

x1(n

)

−7 0 120

10

20

30

40Sequence in Example 2.2b

n

x2(n

)


EXAMPLE 2.3

Generate the complex-valued signalx(n) = exp((-0.1+j0.3)n), for -10≦n ≦ 10

And plot its magnitude, phase , the real part, and the imaginary part in four separate subplots.Solution : figure(1); clf

1. n = [-10:1:10]; alpha = -0.1+0.3j;2. x = exp(alpha*n);3. subplot(2,2,1); stem(n,real(x));title('real part');xlabel('n')4. subplot(2,2,2); stem(n,imag(x));title('imaginary part');xlabel('n')5. subplot(2,2,3); stem(n,abs(x));title('magnitude part');xlabel('n')6. subplot(2,2,4); stem(n,(180/pi)*angle(x));title('phase part');xlabel('n')


EXAMPLE 2.3

−10 −5 0 5 10−3

−2

−1

0

1

2real part

n−10 −5 0 5 10−2

−1.5

−1

−0.5

0

0.5

1imaginary part

n

−10 −5 0 5 100

0.5

1

1.5

2

2.5

3magnitude part

n−10 −5 0 5 10

−200

−100

0

100

200phase part

n


Some useful results

Unit sample synthesisAny arbitrary sequence x(n) can be synthesized as a weighted sum of delayed and scaled unit sample sequences, such as

Even and odd synthesisA real-valued sequence x0(n) is called even (symmetric) if

xe(n)=xe(-n)A real-valued sequence x0(n) is called odd (antisymmetric) if

x0(n)=x0(-n)

∑∞

−∞=

−=k

knkxnx )()()( δ


Some useful results

Then the arbitrary real-valued sequence x(n) can be decomposed into its even and odd components.

x(n)=xe(n)+x0(n)where the even and odd parts are given by

xe(n)=1/2[x(n)+x(-n)] and x0 (n)=1/2[x(n)-x(-n)] We will use this decomposition in studying properties of the Fourier transform. Using MATLAB operations discussed so far, we can obtain the following evenoddfunction.


Evenodd functionfunction [xe, xo, m] = evenodd(x,n)

1. % Real signal decomposition into even and odd parts 2. % [xe, xo, m] = evenodd(x,n)3. if any(imag(x) ~= 0)4. error('x is not a real sequence')5. end6. m = -fliplr(n);7. m1 = min([m,n]); m2 = max([m,n]); m = m1:m2;8. nm = n(1)-m(1); n1 = 1:length(n);9. x1 = zeros(1,length(m));10. x1(n1+nm) = x; x = x1;11. xe = 0.5*(x + fliplr(x));12. xo = 0.5*(x - fliplr(x));


EXAMPLE 2.4

Let x(n)=u(n)-u(n-10). Decompose x(n) into even and odd components.Solution :

1. n = [0:10];2. x = stepseq(0,0,10)-stepseq(10,0,10);3. [xe,xo,m] = evenodd(x,n);4. subplot(1,1,1)5. subplot(2,2,1); stem(n,x); title('Rectangular pulse')6. xlabel('n'); ylabel('x(n)'); axis([-10,10,0,1.2])7. subplot(2,2,2); stem(m,xe); title('Even Part')8. xlabel('n'); ylabel('xe(n)'); axis([-10,10,0,1.2])9. subplot(2,2,4); stem(m,xo); title('Odd Part')10. xlabel('n'); ylabel('xe(n)'); axis([-10,10,-0.6,0.6])


EXAMPLE 2.4

−10 −5 0 5 100

0.2

0.4

0.6

0.8

1

1.2Rectangular pulse

n

x(n)

−10 −5 0 5 100

0.2

0.4

0.6

0.8

1

1.2Even Part

n

xe(n

)

−10 −5 0 5 10

−0.4

−0.2

0

0.2

0.4

0.6Odd Part

n

xe(n

)


The geometric series

A one-side exponential sequence of the form {an, n>=0}, where a is an arbitrary constant, is called a geometric series. The series converges for |a|


Correlations of sequences

It is a measure of the degree to which two sequences are similar. Given two real-valued sequences x(n) and y(n) of finite energy,

the cross correlation of x(n) and y(n) is a sequence rxy(l) defined as

When y(n)=x(n) is called autocorrelation and is defined by

∑∞

−∞=

−=n

yx lnynxlr )()()(,

∑∞

−∞=

−=n

xx lnxnxlr )()()(,

The index l is called the shift or lag parameter.


§2.2 DISCRETE SYSTEMS

Mathematically, discrete system is an operation T[.]y(n) = T [ x(n)]x(n): excitation, input signaly(n): response, output signal

Discrete systems are broadly classified into linear and nonlinear systems. We will deal mostly with linear systems.A discrete system T[.] is a linear operator L[.] if and only if L[.] satisfies the principle of superposition, namely,

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

2121

22112211

nxnxaanxLanxLanxanxaL

∀+=+


Linear System

The output y(n) of a linear system to an arbitrary input x(n) is given by

The response can be interpreted as the response of a linear system at time n due to a unit sample at time k. It is called an impulse response and is denoted by h(n,k). The output then is given by the superposition summation

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

y n L x n L x k n k x k L n kδ δ+∞ +∞

=−∞ =−∞

⎡ ⎤= = − = −⎢ ⎥⎣ ⎦∑ ∑

)]([ knL −δ

(( ) ( ) ( , ) [, ( ], ))n

y n x k h n k L nn k kh δ∞

=−∞

== −∑


Linear time-invariant (LTI) sysem

Linear time-invariant (LTI) system A linear system in which an input-output pair, x(n) and y(n), is invariant to a shift k in time.y(n) = L[x(n)] y(n-k) = L[x(n-k)]For an LTI system the L[.] and the shifting operators are reversible as shown below.

h(n) ≣h(n,0)=L[δ(n)] h(n-k) = L[δ(n-k)]

L[.]x(n) y(n) Shift by k y(n-k)

Shift by k x(n-k) L[.] y(n-k)x(n)


Linear time-invariant (LTI) sysem

The output of an LTI system to input x(n) is given by

The impulse response of an LTI system is given by h(n). The mathematical operation shown above is called a linear convolution sum and is denoted by y(n)=x(n)*h(n). Hence an LTI system is completely characterized in the

time domain by the impulse response h(n) as shown below.

( ) ( ) ( , ) ( ) ( ) ( ) ( )n n

y n x k h n k x k h nn hk x n∞ ∞

=−∞ =−∞

≡= ∗= −∑ ∑

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


Stability

A system is said to be bounded-input bounded-output (BIBO) stable if every bounded input produces a bounded output.An LTI system is BIBO stable if and only if its impulse response is absolutely summable

BIBO Stability ( ) ( ) , ,

( )

x n y n x y

h n∞

−∞

←→ < ∞⇒ < ∞ ∀

←→ < ∞∑


Causality

This important concept is necessary to make sure that systems can be built.

A system is said to be causal if the output at index n0 depends only on the input up to and including the index n0The output does not depend on the future values of the input.An LTI system is causal if and only if the impulse responsesatisfies h(nh(n)=0 for n


§2.3 CONVOLUTION

Convolution for an LTI system can be evaluated in many different ways

If the sequences are mathematical functions, then we can analytically evaluate x(n)*h(n) for all n to obtain a functional formof y(n)

Graphical interpretation, folded-and-shifted versionMatlab implementation1. function [y,ny]=conv_m(x,nx,h,nh)2. nyb = nx(1)+nh(1); nye = nx(length(x))+nh(length(h));3. ny = [nyb:nye];4. n = conv(x,h)

( ) ( ) * ( ) ( ) ( )k

y n x n h n x k h n k∞

=−∞

= −∑


EXAMPLE 2.5

Let the rectangular pulse x(n)=u(n)-u(n-10) of example 2.4 be an input to an LTI system with impulse response

Determine the output y(n)Solution : The input x(n) and impulse response h(n) are shown in figure 2.5. From (2.11)

It is almost a geometric series sum except that the term u(n-k) takes different values depending on n and k. There are three differentconditions under which u(n-k) can be evaluated.

9 9( )0 0

( ) ( ) ( ) (2.15)

(1)(0.9) ( ) (0.9) (0.9) ( )k

n k n kk k

y n x k h n k

u n k u n k

∞

=−∞

− −= =

= −

= − = −

∑∑ ∑

)()9.0()( nunh n=


EXAMPLE 2.5

Case 1 n


EXAMPLE 2.5

−5 0 5 10 15 20 25 30 35 40 45 500

1

2Input Sequence

n

x(n

)

−5 0 5 10 15 20 25 30 35 40 45 500

1

2Impulse Response

n

h(n

)

−5 0 5 10 15 20 25 30 35 40 45 500

5

Output Sequence

n

y(n

)


EXAMPLE 2.6

The above example can also be done by using graphical convolution. In this method h(n-k) is interpreted as a folded-and-shift version of h(k). The output y(n) is obtained as a sample sum under the overlap of x(k) and h(n-k). Given a sequence x(n), what are the following sequences like?

x(n-3)x(n+2)x(-n)x(-n-3)x(-n+2)

shift

folded


EXAMPLE 2.6

Given the two sequences x(n)=[3,11,7,0,-1,4,2 ], -3≦n≦3 andh(n)=[2,3,0,-5,2,1], -1≦n≦4, determine the convolution y(n)=x(n)*h(n).Solution : we plot x(k) and h(k), the original sequences first. Thensequences h(-k), h(-1-k), … etc are plotted.

Note that the beginning point (first nonzero sample) of y(n) is given by n=-3+(-1)=-4, while the end point (the last nonzero sample) is given by n=3+4=7. the complete output is given

{ }2,8,3,22,18,41,5,51,6,47,31,6)( −−−−=ny

( ) ( ) ( )

( 1) ( ) ( 1 ) 3 ( 5) 11 0 7 3 0 2 6

(2) ( ) (2 ) 41

k

k

k

y n x k h n k

y x k h k

y x k h k

∞

=−∞

∞

=−∞

∞

=−∞

= −

− = − − = × − + × + × + × =

= − =

∑∑∑


EXAMPLE 2.6

−5 0 5

−5

0

5

10

x(k) and h(k)

k

solid: x dashed: h

−5 0 5

−5

0

5

10

x(k) and h(−k)

k

n=0

solid: x dashed: h

−5 0 5

−5

0

5

10

x(k) and h(−1−k)

k

n=−1

solid: x dashed: h

−5 0 5

−5

0

5

10

x(k) and h(2−k)

k

n=2

solid: x dashed: h


EXAMPLE 2.6

Given the two sequences x(n)=[3,11,7,0,-1,4,2 ], -3≦n≦3 andh(n)=[2,3,0,-5,2,1], -1≦n≦4, determine the convolution y(n)=x(n)*h(n).

Solution :

{ }2,8,3,22,18,41,5,51,6,47,31,6)( −−−−=ny

3

3( ) ( ) ( )

ky n x k h n k

=−= −∑

( ) ( ) ( 3) 3 ( 2) 11 ( 1) 7 (0) 0 (1) 1 (2) 4 (3) 2( 1) 2 6 22 14 0 2 8 4(0) 3 9 33 21 0 3 12 6(1) 0 0 0 0 0 0 0 0(2) 5 15 55 35 0 5 20 10(3) 2 6 22 14 0 2 8 4(4) 1 2 11 7 0 1 4 2

h n x n x x x x x x xhhhhhh

− = − = − = = = − = =− = −= −== − − − − − −= −= −


MATLAB IMPLEMENTATION

Matlab provides a built-in function called conv.m that computes the convolution between two finite duration sequence. The conv function assumes that the two sequences begin at n=0 and is invoked by

y=conv(x,h);For example, to do the convolution in Ex. 2.5, we could use

x=[3,11,7,0,-1,4,2];h=[2,3,0,-5,2,1];y=conv(x,h)

We obtain the correct convolution results but lose the time index.


MATLAB IMPLEMENTATION

Let {x(n); nxb≦n≦ nxe} and {h(n); nhb≦n≦ nhe} be two finite-duration sequences. Referring to Example 2.6, we observe that the beginning and end points of y(n) are

nyb=nxb+nhb and nye=nxe+nheA simple extension of the conv function, called conv_m, which performs the convolution of arbitrary support sequences can now be designed.

1. function [y,ny] = conv_m(x,nx,h,nh)2. nyb = nx(1)+nh(1); nye = nx(length(x)) + nh(length(h));3. ny = [nyb:nye];4. y = conv(x,h);


EXAMPLE 2.7

Perform the convolution in Example 2.6 using the conv_mfunction.

x = [3, 11, 7, 0, -1, 4, 2]; nx = [-3: 3];h = [2, 3, 0, -5, 2, 1]; nh = [-1: 4];[y,ny] = conv_m(x,nx,h,nh)Hence y(n)={6,31,47,6,-51,-5,41,18,-22,-3,8,2}

An alternate method in Matlab can be used to perform the convolution. This method uses a matrix-vector multiplication approach, which is explored in Problem 2.17.


SEQUENCE CORRELATIONS REVISITED

The correlation can be computed using the conv function if sequences are of finite duration.

Example 2.8: we will demonstrate one application of the crosscorrelation sequence. Let x(n)=[3,11,7,0,-1,4,2] be a prototype sequence, and let y(n) be its noise-corrupted-and-shifted versionwhere ω(n) is Gaussian sequence with mean 0 and

variance 1. Compute the crosscorrelation between y(n) and x(n).

( ) ( ) ( ) ( ) ( ( )) ( )* ( )

( ) ( ) ( ) ( ) ( ( )) ( )* ( )

yxk k

xxk k

r n y k x k n y k x n k y n x n

r n x k x k n x k x n k x n x n

+∞ +∞

=−∞ =−∞

+∞ +∞

=−∞ =−∞

= − = − − = −

= − = − − = −

∑ ∑

∑ ∑

)()2()( nnxny ω+−=


Example 2.8

Solution Form the construction of y(n) it follows that y(n) is “similar” to x(n-2) and hence their crosscorrelation would show the strongest similarity at l=2. To test this out using Matlab, let us compute the crosscorrelation using two different noise sequences.

x = [3, 11, 7, 0, -1, 4, 2]; nx=[-3:3]; % given signal x(n)[y,ny] = sigshift(x,nx,2); % obtain x(n-2)w = randn(1,length(y)); nw = ny; % generate w(n)[y,ny] = sigadd(y,ny,w,nw); % obtain y(n) = x(n-2) + w(n)[x,nx] = sigfold(x,nx); % obtain x(-n)[rxy,nrxy] = conv_m(y,ny,x,nx); % cross-corrlationsubplot(1,1,1);subplot(2,1,1);stem(nrxy,rxy)axis([-4,8,-50,250]);xlabel('lag variable l')ylabel('rxy');title('Crosscorrelation: noise sequence 1')gtext('Maximum')


Example 2.8

−4 −2 0 2 4 6 8−50

0

50

100

150

200

250

lag variable l

rxy

Crosscorrelation: noise sequence 1

Maximum

−4 −2 0 2 4 6 8−50

0

50

100

150

200

250Maximum

lag variable l

rxy

Crosscorrelation: noise sequence 2


Example 2.8

From Figure 2.8 we observe that the crosscorrelationindeed peaks at l=2, which implies that y(n) is similar to x(n) shifted by 2. This approach can be used in applications like radar signal processing in identifying and localizing targets.It should be noted that the signal-processing toolbox in Matlab also provides a function called xcorr for sequence correlation computations. In its simplest form, we use

>>xcorr(x,y) to compute the crosscorrelation between vectors x and y, while>>xcorr(x) to compute the autocorrelation of vector x.


§2.4 DIFFERENCE EQUATIONS

An LTI discrete system can also be described by a linear constant coefficient difference equation (LCCDE) of the form

if aN≠0, then the different equation is of order N. This equation describes a recursive approach for computing the current output, given the input values and previously computed output values. In practice this equation is computed forward in time, from n=-∞to ∞. Therefore another form of this equation is

nmnxbknyaN

k

M

mmk ∀−=−∑ ∑

= =

),()(0 0

∑∑==

−−−=N

kk

M

mm knyamnxbny

10)()()(


DIFFERENCE EQUATIONS

A solution to this equation can be obtained in the form

Where yH(n) is homogeneous part of the solution, is given by

zk, k=1,….N are N roots (also called natural frequencies) of the characteristic equation

If the root zk satisfy the condition |zk|


Matlab IMPLEMENTATION

A routine called filter is available to solve difference equations numerically, given the input and the difference equation coefficients.

y=filter(b,a,x)where

b=[b0,b1,…,bM] and a=[a0,a1,…,aN] are the coefficient arrays from the equation given in (2.19), and x is the input sequence array. The output y has the same length as input x. One must ensure that the coefficient a0 not be zero.

To compute impulse response, MATLAB provides the function impz(b,a,n), where n is a vector for time range.


EXAMPLE 2.9

Given the following difference equation y(n)-y(n-1)+0.9y(n-2)=x(n);

a. Calculate and plot the impulse response h(n) at n=-20, …, 100.b. Calculate and plot the unit step response s(n) at n=-20,…, 100.c. Is the system specified by h(n) stable?

Solutiona=[1,-1,0.9];b=1; n=[-20:120];x=impseq(0,-20,120);h=filter(b,a,x);subplot(2,1,1);stem(n,h)axis([-20,120,-1.1,1.1])title('Impulse Response');xlabel('n');ylabel('h(n)')

h=impz(b,a,n)


EXAMPLE 2.9

% Part b)x=stepseq(0,-20,120); s=filter(b,a,x);subplot(2,1,2);stem(n,s);axis([-20,120,-.5,2.5])title('Step Response');xlabel('n');ylabel('s(n)')The plot of the unit step response is shown in Figure 2.9

% Part c) Observe that h(n) is practically zero for n>120. Hence the system stability can be determined by sum(abs(h)).An alternate approach is using MATLAB’s root function.

z=roots(a); magz=abs(z); subplot


EXAMPLE 2.9

−20 0 20 40 60 80 100 120

−1

−0.5

0

0.5

1

Impulse Response

n

h(n)

−20 0 20 40 60 80 100 120−0.5

0

0.5

1

1.5

2

2.5Step Response

n

s(n)


EXAMPLE 2.10

Let us consider the convolution given in Example 2.5. The input sequence is of finite duration x(n)=u(n)-u(n-10) while the impulse response is of infinite duration h(n)=(0.9)nu(n). Determine y(n)=x(n)*h(n)Solution:

If one of the sequences is of infinite length, it is more suitable to use filter instead of conv function for numerical evaluation of the convolution. Given the impulse response h(n), one can compute a difference equation and then y(n) can be obtained from the filter function.


EXAMPLE 2.10

From the h(n) expression,

h(n) is the output of an LTI system when the input is δ(n). Hence substituting x(n) for δ(n) and y(n) for h(n), the difference equation is y(n)-0.9y(n-1)=x(n).Now the filter(.) function can be used to compute the convolution indirectly.

[ ]

1

( ) (0.9) ( )(0.9) ( 1) (0.9)(0.9) ( 1) (0.9) ( 1)

( ) (0.9) ( 1) (0.9) ( ) (0.9) ( 1)(0.9) ( ) ( 1) (0.9) ( ) ( )

n

n n

n n

n n

h n u nh n u n u n

h n h n u n u nu n u n n nδ δ

−

=

− = − = −

− − = − −

= − − = =


EXAMPLE 2.10

Program :b = [1]; a = [1,-0.9];n = -5:50; x = stepseq(0,-5,50) - stepseq(10,-5,50);y = filter(b,a,x);subplot(1,1,1);subplot(2,1,2); stem(n,y); title('Output sequence')xlabel('n'); ylabel('y(n)'); axis([-5,50,-0.5,8])

−5 0 5 10 15 20 25 30 35 40 45 50

0

1

2

3

4

5

6

7

8Output sequence

n

y(n)


Zero-Input and Zero-State Responses

In DSP the difference equation is generally solved forward in time from n=0. Therefore initial conditions on x(n) and y(n) are necessary to determine the output for n>=0.

Subject to the initial conditions:

The solution can be expressed in alternative form

0,)()()(0 1

≥−−−= ∑ ∑= =

nknyamnxbnyM

m

N

kkm

}1);({}1);({ −≤≤−−≤≤− nMnxandnNny

)()()( nynyny ZSZI +=

If


Zero-Input and Zero-State Responses

yZI(n): zero-input solutionA solution due to the initial conditions alone

yZS(n): zero-state solutionA solution due to input x(n) alone

Chapter 4 will discuss this topic.


Digital Filters

Discrete-time LTI systems are also called digital filter.Classification

FIR filter & IIR filterFIR filter

Finite-duration impulse response filterCausal FIR filter :

h(0)=b0,…,h(M)=bMNonrecursive or moving average (MA) filterDifference equation coefficients, {bm} and {a0=1}Implementation in Matlab: conv(x,h); filter(b,1,x) (difference?)

∑=

−=M

mm mnxbny

0)()(


Infinite Impulse Response (IIR) Filters

Infinite-duration impulse response filterDifference equation

Recursive filter, in which the output y(n) is recursively computed from its previously computed valuesAutoregressive (AR) filter

∑=

=−N

kk nxknya

0)()(


ARMA filter

Generalized IIR filter

It has two parts: MA part and AR partAutoregressive moving average filter, ARMASolution

filter(b,a,x); % given {bm}, {ak}

0 1

( ) ( ) ( ), 0M N

m km k

y n b x n m a y n k n= =

= − − − ≥∑ ∑


Homework

Textbook: pp7 to pp39Exercises:

P2.5, P2.10, P2.15, P2.16, P2.17, P2.19


Chapter 3. The Discrete-Time Fourier

Analysis

Discuss discrete-time signal and system representation in the frequency domain.Study the sampling and reconstruction of analog signals.


§3.0 Introduction

A linear and time-invariant system can be represented using its response to the unit sample sequence.

h(n) is called as the unit impulse responsey(n)=x(n)*h(n): system response

The convolution representation is based on the fact that any signal can be represented by a linear combination of scaled and delayed unit samples.We can also represent any arbitrary discrete signal as a linear combination of basis signals introduced in Chapter 2.


Each basis signal set provides a new signal representation.Each representation has some advantages and disadvantages depending upon the type of system under consideration.When the system is linear and time-invariant, only one representation stands out as the most useful. It is based on the complex exponential signal set {ejw} and is called the discrete-time Fourier Transform.

§3.0 Introduction


If x(n) is absolutely summable, i.e., , the DTFT is defined by

F[.] transforms a discrete signal x(n) into a complex-valued continuous function X of real variable w, called a digital frequency, which is measured in radians.

Time domain Frequency domainDiscrete ContinuousReal valued Complex-valuedSummation IntegralThe range of w: -∞~∞. The integral range of w: -π~ π

1

: ( ) [ ( )] ( )

1: ( ) [ ( )] ( )2

jw jwn

n

jw jw jwn

DTFT X e F x n x n e

IDTFT x n F X e X e e dwπ

ππ

+∞−

=−∞

+−

−

= =

= =

∑

∫

∑+∞∞− ∞


Determine the discrete-time Fourier transform of

Result visualization with MatlabComplex function: magnitude and angle; real/imaginary part with respect to wThe range of w:showing interval: [0,pi]

)()5.0()( nunx n=

~−∞ +∞

Example 3.1

( )0.5

jwjw

jw

eDTFT X ee

⇒ = =−


Example 3.1

0 0.2 0.4 0.6 0.8 1

0.8

1

1.2

1.4

1.6

1.8

2

frequency in pi units

Magnitude Part

Mag

nitu

de

0 0.2 0.4 0.6 0.8 1−0.7

−0.6

−0.5

−0.4

−0.3

−0.2

−0.1

0


Angle Part

Rad

ians

0 0.2 0.4 0.6 0.8 1

0.8

1

1.2

1.4

1.6

1.8

2


Real Part

Rea

l

0 0.2 0.4 0.6 0.8 1−0.7

−0.6

−0.5

−0.4

−0.3

−0.2

−0.1

0


Imaginary Part

Imag

inar

y


Periodicity:The DTFT is periodic in w with period 2πImplication: we need only one period for analysis and not the whole domain

Symmetry:For real-valued x(n), X is conjugate symmetric.Implication: to plot X, we now need to consider only a half of X: [0, π]

)()( ]2[ π+= wjjw eXeX

)()( * jwjw eXeX =−

Two important properties

Re[ ( )] Re[ ( )]Im[ ( )] Im[ ( )]| ( ) | | ( ) |

( )] ( )

jw jw

jw jw

jw jw

jw jw

X e X e even symmetryX e X e odd symmetry

X e X e even symmetryX e X e odd symmetry

−

−

−

−

=

= −

=

∠ = −∠


Matlab Implementation

If x(n) is of infinite duration, then Matlab can not be used directly to compute X from x(n).We can use DTFT to evaluate the expression X over [0, π] frequencies and then plot its magnitude and angle (or real and imaginary parts) as in Example 3.1.For a finite duration, the DTFT can be implemented as a matrix-vector multiplication operation if X is evaluated at equispaced frequency between [0, π].


Matlab Implementation

If x(n) has N samples over n1≦n≦nN and we want to evaluate X at wk=π/(Mk), k=0, 1, … M.

Stack X(ejwk) into vector form

1

( / )

1

( / )( / )

1

( ) ( ) ( ,:)

( ,:) [ ][ ( ) ( )]

pk

N

N j M knjwpp

j M knj M kn

TN

X e x n e k

k e ex n x n

π

ππ

−

=

−−

= =

=

=

∑ W xWx

0

( / )

[ ( ) ( )]

{ , 1 , 0,1, , }

exp , , column vector

M

p

jw jw T

j M kn

T T

X e X e

e p N k M

jM

π

π

−

= =

= ≤ ≤ =

⎡ ⎤⎛ ⎞= − ∈⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

X Wx

W

x kn k n


Example 3.4

If x(n) is of infinite duration, x(n)={1,2,3,4,5}, -1≦n≦3, determine DTFT.Solution: the vectors are in row vector form1. n=-1:3; x=1:5;2. K=0:500; w=(pi/500)*k;3. X=x*(exp(-j*pi/500).^(n’*k);4. magX=abs(X); angX=angle(X);5. realX = real(X); imagX = imag(X);6. subplot(2,2,1); plot(k/500,magX); grid;7. xlabel(‘frequency in pi unit’); title(‘Magnitude Part’);


Example 3.4

0 0.2 0.4 0.6 0.8 12

4

6

8

10

12

14

16


Magnitude Part

Mag

nitu

de

0 0.2 0.4 0.6 0.8 1−4

−2

0

2

4


Angle Part

Rad

ians

0 0.2 0.4 0.6 0.8 1−5

0

5

10

15


Real Part

Rea

l

0 0.2 0.4 0.6 0.8 1−10

−8

−6

−4

−2

0

2

4


Imaginary Part

Imag

inar

y


Example 3.5

Let x(n)= [0.9exp(jπ/3)n], 0≦n≦10. Determine X(ejwk) and study its periodicity.The DTFT of complex-valued signal is periodic in w but it is not conjugate-symmetric.

−2 −1.5 −1 −0.5 0 0.5 1 1.5 20

2

4

6

8

frequency in units of pi

|X|

Magnitude Part

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1

−0.5

0

0.5

1


radian

s/pi

Angle Part


Example 3.6

Let x(n)= 2n, -10≦n≦10. Determine X(ejwk) and study its periodicity.

The DTFT of real-valued signal is periodic and conjugate-symmetric. Therefore for real sequence we will plot their Fourier transformmagnitude and angle responses from 0 to pi.

−2 −1.5 −1 −0.5 0 0.5 1 1.5 20

5

10

15


|X|

Magnitude Part

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1

−0.5

0

0.5

1


radian

s/pi

Angle Part


§3.2 The properties of the DTFT

1. Linearity:The DTFT is a linear transformation.

2. Time shifting:A shift in the time domain corresponds to the phase shifting.

3. Frequency shifting:Multiplication by a complex exponential corresponds to a shift in the frequency domain.

1 2 1 2[ ( ) ( )] [ ( )] [ ( )]F x n x n F x n F x nα β α β+ = +

[ ( )] ( )jw jwkF x n k X e e−− =

0 0( )[ ( ) ] ( )jw n j w wF x n e X e −=


The properties of the DTFT

4. Conjugation:Conjugation in the time domain corresponds to the folding and conjugation in the frequency domain.

5. Folding:Folding in the time domain corresponds to the folding in the frequency domain

6. Symmetries in real sequence:Implication: If the sequence x(n) is real and even, then X is also real and even.

* *[ ( )] ( )jwF x n X e−=

[ ( )] ( )jwF x n X e−− =

( ) ( ) ( )

[ ( )] Re[ ( )]

[ ( )] Im[ ( )]

e ojw

ejw

o

x n x n x n

F x n X e

F x n j X e

= +

=

=


The properties of the DTFT

7. Convolution:

8. MultiplicationPeriodic convolution (discussed in Chapter 5)

9. Energy

1 2 1 2 1 2[ ( ) * ( )] [ ( )] [ ( )] ( ) ( )jw jwF x n x n F x n F x n X e X e= =

1 2 1 2

( )1 2

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

(1/ 2 ) ( ) ( )j j wF x n x n F x n F x n

X e X e dθ θπ θ−⋅ = ⊗

= ∫

2 2

2

1| ( ) | | ( ) | : Parseval's Theorem2

| ( ) |( ) : energy density spectrum

jwx

jw

x

x n X e dw

X ew

π

πε

π

π

+∞ +

−−∞

= =

Φ =

∑ ∫


Example 3.9

Let x(n)= cos(nπ/2), 0≦n≦100 and y(n)=ejnπ/4 x(n). Verify the frequency shift property by using Matlab.

−1 −0.5 0 0.5 10

10

20

30

40

50

60


|X|

Magnitude of X

−1 −0.5 0 0.5 1−1

−0.5

0

0.5

1


radi

ands

/pi

Angle of X

−1 −0.5 0 0.5 10

10

20

30

40

50

60


|Y|

Magnitude of Y

−1 −0.5 0 0.5 1−1

−0.5

0

0.5

1


radi

ans/

pi

Angle of Y


Example 3.12

Let x(n)= sin(nπ/2), -5≦n≦10. Verify the symmetry property by using Matlab.

−1 −0.5 0 0.5 1−2

−1

0

1

2


Re(X

)

Real part of X

−1 −0.5 0 0.5 1−10

−5

0

5

10


Im(X

)

Imaginary part of X

−1 −0.5 0 0.5 1−2

−1

0

1

2


XE

Transform of even part

−1 −0.5 0 0.5 1−10

−5

0

5

10


XO

Transform of odd part


§3.3 The frequency domain representation of LTI system

The Fourier transform representation is the most useful signal representation for LTI systems.It is due to the following result:

Response to a complex exponential ejw0nResponse to sinusoidal sequencesResponse to arbitrary sequences

Those responses are called the stead-state responses of systems.


Response to a complex exponential


Frequency response: the DTFT of an impulse response is called the frequency response (or transfer function) of an LTI system and is denoted by H().

The output sequence is the input exponential sequence modified by the response of the system at frequency w0

)(nhnjwe 0njw

wwnjw enhFenh 0

0

0 ]|)]([[*)( ==

( ) [ ( )] ( )jw jnwn

H e F h n h n e∞=−∞

= = ∑

)( jweHnjwe 0 njwjw eeH 00 )( ×



Response to multiple complex exponential sequences

In general, the frequency response H is a complex function of w.

The magnitude |H| is called the magnitude (gain) response function, and the the angle is called the phase response function.

)(),( nheH jw∑k

njwk

keA ∑k

njwjwk

kk eeHA )(


Response to sinusoidal sequences

Response to x(n)= Acos(w0n+θ0),

)(nh0 0cos( )w n θ+( ) ?y n =

0 0

0 0

0 0

0 0

( ) cos( )

( ) | ( ) | cos( ( ))

( ) cos( )

( ) | ( ) | cos( ( ))

jw jw

k k kk

jw jwk k k

k

x n A w n

y n A H e w n H e

x n A w n

y n A H e w n H e

θ

θ

θ

θ

= +

= + +∠

= +

= + +∠

∑

∑


Response to arbitrary sequences

Let X=F[x(n)] and Y=F[y(n)]. Then using convolution property, we have

)(nh)(nx )(*)()( nxnhny =

)(),( nheH jw)( jweX )()()( jwjwjw eXeHeY =


Example 3.13

Plot the magnitude and the phase responses of a system characterized by h(n)=(0.9)nu(n).

1( )1 0.9

jwjwX e e−

=−

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

2

4

6

8

10


|H|

Magnitude Response

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.4

−0.3

−0.2

−0.1

0


Phas

e in p

i Rad

ians

Phase Response


Frequency response function from difference equations

To evaluate its frequency response , we would need the impulse response h(n).How do we do when an LTI system is represented by the difference equation?

We know that when , then y(n) must be

∑ ∑= = −=−+Nl

Mm ml mnxblnyany 1 0 )()()(

jwnenx =)(jwnjw eeH )(

∑∑

=−

=−

+= N

ljwl

l

Mm

jwmmjw

eaeb

eH0

0

1)(Therefore,


Example 3.15

An LTI system is specified by y(n)=0.8y(n-1)+x(n). (a.) Determine X(ejw) and (b.) Plot the steady-state response to x(n)=cos(0.05nπ)u(n)

0 10 20 30 40 50 60 70 80 90 100−1

−0.5

0

0.5

1

n

x(n)

Input sequence

0 10 20 30 40 50 60 70 80 90 100−5

0

5

n

y(n)

Output sequence


Example 3.16

We can generalize the frequency response by using a simple matrix-vector multiplication.

If X is evaluated at equispaced frequency between [0, π] with K points

0 1 0 1

( ) exp( ). / exp( )[ , , ]; [ , , ][0,1,2, , ]; [0,1,2, , ][0,1,2, , ]; * /

jw T T

M N

H e j jb b b a a a

M NK pi K

= − −= == == =

b m w a n wb am nk w k

0

0

( )1

M jwmmjw m

N jwnln

b eH e

a e

−=

−=

=+

∑∑


Example 3.16

A 3rd-order lowpass filter with difference equation

1. b = [0.0181, 0.0543, 0.0543, 0.0181];2. a = [1.0000, -1.7600, 1.1829, -0.2781];3. m = 0:length(b)-1; l = 0:length(a)-1;4. K = 500; k = 1:1:K;5. w = pi*k/K; % [0, pi] axis divided into 501 points.6. num = b * exp(-j*m'*w); % Numerator calculations7. den = a * exp(-j*l'*w); % Denominator calculations8. H = num ./ den;9. magH = abs(H); angH = angle(H);

( ) 0.0181 ( ) 0.0543 ( 1) 0.0543 ( 2) 0.0181 ( 3)1.76 ( 1) 1.1829 ( 2) 0.2781 ( 3)

y n x n x n x n x ny n y n y n

= + − + − + −+ − − − + −


Example 3.16

A 3rd-order lowpass filter with difference equation

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1


|H|

Magnitude Response

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1

−0.5

0

0.5

1


Phas

e in

pi R

adia

ns

Phase Response


§3.4 Sampling and reconstruction of analog signals

Analogy signals can be converted into discrete signals using sampling and quantization operations: analogy-to-digital conversion, or ADCDigital signals can be converted into analog signals using a reconstruction operation: digital-to-analogy conversion, or DACUsing Fourier analysis, we can describe the sampling operation from the frequency-domain view-point, analyze its effects and then address the reconstruction operation.We will also assume that the number of quantization levels is sufficiently large that the effect of quantization on discrete signals is negligible.


Sampling

Let xa(t) be analog signal. We have FT/IFT pair

Assume xa(t) is Absolutely integrableΩ(or F) is an analogy frequency in radians/secω (or f) is a digital frequency radians/secAnalogy and digital frequencies are related through sampling rate Fs=1/Ts, ω=Ω/Fs, f=F/Fs.

Sampling: Sample xa(t) at sampling interval Ts sec apart to obtain the discrete-time signal x(n)

1( ) ( ) ( ) ( )2

IFTj t j ta a a aFT

X j x t e dt x t X j e dπ

+∞ +∞− Ω Ω

−∞ −∞⎯⎯→Ω = = Ω Ω←⎯⎯∫ ∫

( ) ( )1 2( ) ( )

a s

s ans s s

x n x t nT

X X nT T T

ω πω

= =

⇒ = −∑


Sampling


Sampling

Fig3.10


Sampling

X is a countable sum of amplitude-scaled, frequency-scaled, and translated version of xa(t) The above relation is known as the aliasing formulaSuppose signal band is limited to ±Ώ0,If Ts is small, Ώ0Tsπ Ώ0>π/Ts, orF0= Ώ0/2π > 1/2Ts = Fs/2 Fs


Sampling

Sampling:

1( ) ( ) ( ) ( )

1( ) ( ) ( )

1 2( )

s a a sns

s s a s sns s

ans s s

X X P X nT

X X X F nF T

X nT T T

ω ω

ω π

Ω = Ω ∗ Ω = Ω − Ω

Ω⇒ ≡ = − Ω

= −

∑

∑

∑

2 21( ) ( ) , 1/

1 1( ) ( ) or ( ) ( ), 2

( ) ( ) ( ) ( ) ( ) ( )

s sj nF t j nF ts n s s

n n ns

s s s sn ns s

s a s a

p t t nT c e e F TT

P n P F F nF FT T

x t x t p t X X P

π πδ

δ δ π

= − = = =

Ω = Ω − Ω = − Ω =

= ⇒ Ω = Ω ∗ Ω

∑ ∑ ∑

∑ ∑


Ban-limited signal

A signal is band-limited if there exists a finite radians frequency Ώ0 such that Xa(Ώ) is zero for |Ώ|> Ώ0.

The frequency F0= Ώ0 /2π is called the signal bandwidth in HzReferring to Fig.3.10, if π> Ώ0Ts , or Fs>2 F0 then

ssssa

s

jw

TTw

TTwjX

TeX ππ ≤


Sampling Principle

A band-limited signal xa(t) with bandwidth F0 can be reconstructed from its sample values x(n)=xa(nTs) if the sampling frequency Fs=1/Ts is greater than twice the bandwidth F0 of xa(t) , Fs >2 F0.

Otherwise aliasing would result in x(n). The sampling rate of 2F0 for an analog band-limited signal is called the Nyquist rate.


Example 3.17

Let xa(t)=exp(-1000|t|). Determine and plot its Fourier transform.

Ideal :

Using DTFT :

Selection of Δt : Xa(Ω)≒0 for Ω≧(2π)2000 or F≧2000. Therefore, Δt should be ≦1/4000. We set Δt=0.00005Selection of time period of xa(t) : -5m≦t≦5m since exp(-5)≒0.

( ) ( ) ( )j t j m ta a amX j x t e dt x m t e t+∞ − Ω − Ω Δ

−∞Ω = ≈ Δ Δ∑∫

2

0.002( )1 ( /1000)a

X Ω =+ Ω


Example 3.17

1. % Analog Signal2. Dt = 0.00005; t = -0.005:Dt:0.005; xa = exp(-1000*abs(t));3. %% Continuous-time Fourier Transform4. Wmax = 2*pi*2000; K = 500; k = 0:1:K; W = k*Wmax/K; 5. Xa = xa * exp(-j*t'*W) * Dt; Xa = real(Xa);6. W = [-fliplr(W), W(2:501)]; % Omega from -Wmax to

Wmax7. Xa = [fliplr(Xa), Xa(2:501)];8. subplot(2,1,1);plot(t*1000,xa);9. xlabel('t in msec.'); ylabel('xa(t)'); title('Analog Signal')10. subplot(2,1,2);plot(W/(2*pi*1000),Xa*1000);11. xlabel('Frequency in KHz'); ylabel('Xa(jW)*1000')12. title('Continuous-time Fourier Transform')


Example 3.17

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

t in msec.

xa(t)

Analog Signal

−2 −1.5 −1 −0.5 0 0.5 1 1.5 20

0.5

1

1.5

2

Frequency in KHz

Xa(

jW)*

1000

Continuous−time Fourier Transform


Example 3.18

Let xa(t)=exp(-1000|t|). Determine and plot its Fourier transform.

Ideal :

Using DTFT :

Selection of Δt : Xa(Ω)≒0 for Ω≧(2π)2000 or F≧2000. Therefore, Δt should be ≦1/4000. We set Δt=0.00005Selection of time period of xa(t) : -5m≦t≦5m since exp(-5)≒0.

( ) ( ) ( )j t j m ta a amX j x t e dt x m t e t+∞ − Ω − Ω Δ

−∞Ω = ≈ Δ Δ∑∫

2

0.002( )1 ( /1000)a

X Ω =+ Ω


Example 3.18a% Analog SignalDt = 0.00005; t = -0.005:Dt:0.005; xa = exp(-1000*abs(t));% Discrete-time SignalTs = 0.0002; n = -25:1:25; x = exp(-1000*abs(n*Ts));% Discrete-time Fourier transformK = 500; k = 0:1:K; w = pi*k/K;X = x * exp(-j*n'*w); X = real(X);w = [-fliplr(w), w(2:K+1)]; X = [fliplr(X), X(2:K+1)];subplot(2,1,1);plot(t*1000,xa);xlabel('t in msec.'); ylabel('x1(n)');title('Discrete Signal'); hold onstem(n*Ts*1000,x); gtext('Ts=0.2 msec'); hold offsubplot(2,1,2);plot(w/pi,X);xlabel('Frequency in pi units'); ylabel('X1(w)')title('Discrete-time Fourier Transform')


Example 3.18a

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

t in msec.

x1(n

)Discrete Signal

Ts=0.2 msec

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

Frequency in pi units

X1(

w)

Discrete−time Fourier Transform


Example 3.18b% Analog SignalDt = 0.00005; t = -0.005:Dt:0.005; xa = exp(-1000*abs(t));% Discrete-time SignalTs = 0.001; n = -5:1:5; x = exp(-1000*abs(n*Ts));% Discrete-time Fourier transformK = 500; k = 0:1:K; w = pi*k/K;X = x * exp(-j*n'*w); X = real(X);w = [-fliplr(w), w(2:K+1)]; X = [fliplr(X), X(2:K+1)];subplot(2,1,1);plot(t*1000,xa);xlabel('t in msec.'); ylabel('x1(n)');title('Discrete Signal'); hold onstem(n*Ts*1000,x); gtext('Ts=0.2 msec'); hold offsubplot(2,1,2);plot(w/pi,X);xlabel('Frequency in pi units'); ylabel('X1(w)')title('Discrete-time Fourier Transform')


Example 3.18b

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

t in msec.

x2(n

)

Discrete Signal

Ts=1 msec

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

Frequency in pi units

X2(

w)

Discrete−time Fourier Transform


Reconstruction

From the sampling theorem, the analog signal can be reconstructed from its sampled version through the process at the bottom if there is no aliasing effect.

Lowpass filter band-limited to the [-Fs/2,Fs/2] bandThe ideal interpolation is not practically feasible because the entire system is noncausal and hence not realizable.

Impulse train conversion

Ideal lowpass filter)(nx

)(txa

+−+++−+=−∑+∞

−∞=)()1()()0()()1()()( ss

ns TtxtxTtxnTtnx δδδδ

∑+∞

−∞=−=

nssa nTtFnxtx )]([sinc)()(


Reconstruction

Ideal lowpass filter is not practically realizable since the system is noncausalFig 3.14


Reconstruction

Derivation :

/ 2

/ 2

/ 2

/ 2

( ) { ( )} (1/ 2 ) ( )exp( )

From sampling theorem, we know ( ) ( ), / 2 .

( ) { ( )} (1/ 2 ) ( )exp( )

( ) ( / 2 ) ( ) exp( ( ))

s

s

s

s

s s a s sn

a s s s

a a a

a s a s sn

a

X FT x t x nT jn T

X T X

Then x t IFT X X j t d

x t T x nT j t nT d

x

π

π

π

Ω

−Ω

Ω

−Ω

Ω = = − Ω

Ω = Ω Ω ≤ Ω

= Ω = Ω Ω Ω

⇒ = Ω − Ω

⇒

∑

∫∑ ∫

/ 2exp( ( ))( ) ( / 2 ) ( )

( )2sin(( / 2)( ))( ) ( / 2 ) ( )

( )sin( ( ))( ) ( ) ( )sinc( ( ))

( )

s

ss a s

n s

s sa s a s

n s

s sa a s a s s s

n ns s

j t nTt T x nTj t nT

t nTx t T x nTt nT

F t nTx t x nT x nT F t nTF t nT

π

π

ππ

±Ω

Ω −=

−Ω −

⇒ =−

−⇒ = = −

−

∑

∑

∑ ∑

( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

sn

s a a s a s sn n

p t t nT

x t x t p t x t t nT x nT t nT

δ

δ δ

= −

= = − = −

∑

∑ ∑


Practical D/A converters

Zero-order-hold (ZOH) interpolation:In this interpolation a given sample value is held for the sample interval until the next sample is received.It can be obtained by filtering the impulse train through an interpolating filter of the formThe resulting signal is a piecewise-constant (staircase) waveform which requires an appropriately designed analog post-filter for accurate waveform reconstruction.

0

ˆ ( ) ( ), ( 1)1 0

( )0 otherwise

a s s

s

x t x n nT n n Tt T

h t

= ≤ ≤ +

≤ ≤⎧= ⎨⎩



Zero-order-hold (ZOH) interpolation:

ZOH Post-Filter)(nx )(txa)(ˆ txa

/ 2ˆ ( ) ( ) ( )

/ 2( )

a an

an

t Tx t x nT t nT rectT

t nT Tx nT rectT

δ −⎛ ⎞= − ∗ ⎜ ⎟⎝ ⎠

− −⎛ ⎞= ⎜ ⎟⎝ ⎠

∑

∑



First-order-hold (FOH) interpolation: In this case the adjacent samples are joined by straight lines.

ˆ ( ) ( ) ( ) ( / )

( ) (( ) / )

a a s s sn

a s s sn

x t x nT t nT tri t T

x nT tri t nT T

δ= − ∗

= −

∑

∑

noncausal

equivalent to

ˆ ( ) ( ) [ ( 1) ( )], ( 1)at nTx t x n x n x n nT t n T

T−

= + + − ≤ ≤ +




ˆ ( ) ( ) ( ) (( ) / )

( ) (( ) / )

a a s s s sn

a s s s sn

x t x nT t nT tri t T T

x nT tri t nT T T

δ= − ∗ −

= − −

∑

∑

causal




1

1

ˆ ( ) ( ) ( ) ( )

( ) ( )

a a s sn

a s sn

x t x nT t nT h t

x nT h t nT

δ= − ∗

= −

∑

∑

causal and predictive FOH

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

≤≤−

≤≤+

=

otherwise0

21

01

)(1 sss

ss

TtTTt

TtTt

th



Cubic-order-hold (COH) interpolationThis approach uses spline interpolation for a smoother, but not necessarily more accurate, estimate of the analog signal between samples.Hence this interpolation does not require an analog post-filterThe smoother reconstruction is obtained by using a set of piecewise continuous third-order polynomials called cubic splines

2 30 1 2 3( ) ( ) ( )( ) ( )( ) ( )( ) ,

for ( 1)where ( ), 0 ~ 3 are coefficients depended on sample values

a s s s

s s

i

x t n n t nT n t nT n t nTnT t n Tn i

α α α α

α

= + − + − + −≤ ≤ +=


Example 3.19

Reconstruction x1(t) in Example 3.18 by sinc()Ts = 0.0002; Fs = 1/Ts; n = -25:1:25; nTs = n*Ts;x = exp(-1000*abs(nTs));Dt = 0.00005;t = -0.005:Dt:0.005;xa = x * sinc(Fs*(ones(length(nTs),1)*t-nTs'*ones(1,length(t))));error = max(abs(xa - exp(-1000*abs(t))))

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t in msec.

xa(t)

Reconstructed Signal from x1(n) using sinc function


Example 3.20

Reconstruction x2(t) in Example 3.18 by sinc()Ts = 0.0002; Fs = 1/Ts; n = -25:1:25; nTs = n*Ts;x = exp(-1000*abs(nTs));Dt = 0.00005;t = -0.005:Dt:0.005;xa = x * sinc(Fs*(ones(length(nTs),1)*t-nTs'*ones(1,length(t))));error = max(abs(xa - exp(-1000*abs(t))))

−5 −4 −3 −2 −1 0 1 2 3 4 5−0.2

0

0.2

0.4

0.6

0.8

1

1.2

t in msec.

xa(t)

Reconstructed Signal from x2(n) using sinc function


Example 3.21

Reconstruction x1(t) in Example 3.18 by ZOH and FOHZOH: stairs(nTs*1000,x); FOH: plot(nTs*1000,x);

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

t in msec.

xa(t)

Reconstructed Signal from x1(n) using zero−order−hold

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

t in msec.

xa(t)

Reconstructed Signal from x1(n) using first−order−hold


Example 3.22

Reconstruction x1(t) and x2(t) by cubic splineCubic spline : spline(nTs,x,t)

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

t in msec.

xa(t)

Reconstructed Signal from x1(n) using cubic spline function

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

t in msec.

xa(t)

Reconstructed Signal from x2(n) using cubic spline function


HomeWork

Textbook: pp40 to pp79Exercises:

P3.2; P3.4; P3.14; P3.20; P3.22


Chapter 4

The bilateral z-TransformProperty of z-TransformInversion of z-TransformSystem representation in the z-domainSolution of difference equations


§4.0 Review

The discrete-time Fourier transform approach for representing discrete signals is using complex exponential sequence.Advantages for LTI system

It describes systems in the frequency domain using the frequencyresponse function H.The computation of the sinusoidal steady-state response is great facilitated by the use of H.Response to any arbitrary absolutely summable sequence x(n) can easily be computed in the frequency domain by multiplying the transform X and the frequency response H.


Review

Shortcomings to the FTThere are many useful signals in practice, such as u(n), nu(n), for which the DTFT does not exist.The transient response of a system due to initial conditions or due to changing inputs cannot be computed using the DTFT approach.

z-transform: To address the above two problems, z-transform is proposed which is an extension of the DTFT:

Bilateral (two-sided) version provides another domain in which a large class of sequence and systems can be analyzed.Unilateral (one-sided) version can be used to obtain system response with initial conditions or changing inputs.


§4.1 The Bilateral z-Transform

Definition of z-transform (ZT) and inversion of ZT (IZT):

where z is a complex variable and C is counterclockwise contour encircling the origin and lying in the ROC.

Region of convergence (ROC) : The set of z values for which X(z) exists is called the ROC and is given by

for some positive numbers Rx- and Rx+.

∫

∑−−

∞

−∞=

−

=Ζ=

=Ζ=

c

n

n

n

dzzzXj

zXnx

znxnxzX

11 )(21)]([)(

)()]([)(

π

+−


Comments

The complex variable z is called the complex frequency given by , where |z| is the attenuation and w is the real frequency;Since the ROC is defined in terms of the magnitude |z|, the shape of the ROC is an open ring. Note that Rx- may be equal to 0 and/or Rx+ could possibly be infinity;If Rx+


Example 4.1

Let .(This sequence is called a positive-time sequence). Then

Note: if X(z)=B(z)/A(z), then the roots of B(z) are called zeros and the roots of A(z) are called poles. A pole-zero diagram of X(z) is plotted in z-plane in which zeros are denoted by ‘o’ and poles by ‘×’.

1( ) ( ),0nx n a u n a= < < ∞

1 10 0

1

1( )1

;if 1 or :

nn n

RxRx

a zX z a zz az z a

a z a ROC a zz

∞ ∞−

−

+−

⎛ ⎞= = = =⎜ ⎟ − −⎝ ⎠

< > ⇒ < < ∞

∑ ∑


Example 4.2

Let .(This sequence is called a negative-time sequence). Then

2 ( ) ( 1),0nx n b u n b= − − − < < ∞

1 1

21

0

2

( )

1 1 11

: 0

n nn n

n

Rx Rx

b zX z b zz b

z zzb z bb

ROC z b

− − ∞−

−∞ −∞

∞

− +

⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= − = − =⎜ ⎟ −⎝ ⎠ −

⇒ ≤ <

∑ ∑ ∑

∑


Example 4.3

Let (This sequence is called a two-side sequence). Then using the above two examples.

3 1 2( ) ( ) ( ) ( ) ( 1)n nx n x n x n a u n b u n= + = − − −

1

30

( ) , 1: , 2 :

; 3 : 1 2

n n n n z zX z a z b z ROC z a ROC z bz a z b

z z ROC ROC ROCz a z b

∞ −− −

−∞

⎧ ⎫ ⎧ ⎫= − = > +


Properties of ROC

The ROC is always bounded by a circle since the convergence condition is on the magnitude |z|;The ROC for right-sided sequences (n=0, x(n) is a causal sequence)The ROC for left-sided sequences (n>n0, x(n)=0) is always inside of a circle of radius Rx+. (if n0


Properties of ROC

The ROC for finite-duration sequences(nn2, x(n)=0) is the entire z-plane. If n10, thenz=0 is not in the ROC;

Ex. ROC: entire Z-plane except z=∞

Ex.ROC: entire Z-plane except z=0, z=∞

Ex.ROC: entire Z-plane except z=0

The ROC cannot include a pole since X(z) diverges there;There is at least one pole on the boundary of a ROC of a rational X(z); The ROC is one contiguous region, the ROC does not come in pieces.

x(n) is finite

5 4 3 2( ) 3 5 3X z z z z z z= + + + +

2 1 1 2( ) 3 5 3X z z z z z− −= + + + +

1 2 3 4 5( ) 3 5 3X z z z z z z− − − − −= + + + +


§4.2 Important properties of the z-transform

Linearity:

Time Shifting:

Frequency Shifting:

Folding n ↔ inverting z

21:);()()]()([ 22112211 xx ROCROCROCzXazXanxanxaZ ∩+=+

xn ROCROCzXznnxZ :);()]([ 00

−=−

||:;)]([ abyscaledROCROCazXnxaZ x

n ⎟⎠⎞

⎜⎝⎛=

xROCInvertedROCzXnxX :);/1()]([ =−


Important properties of the z-transform

Complex conjugation

Differentiation in the z-domain

Multiplication

Convolution

xROCROCzXnxZ :);()]([*** =

xROCROCdzzdXznnxZ :;)()]([ −=

21:)/()(

21)]()([ 12121 xxC ROCROCROCdvvvzXvXj

nxnxZ ∩= ∫ −π

21:)()()](*)([ 2121 xx ROCROCROCzXzXnxnxZ ∩=


Example 4.4

Let and DetermineSolution:

From the definition of the z-transform we observe thatx1(n)={2,3,4} and x2(n)={3,4,5,6}

Then the convolution of the above two sequences will give the coefficients of the required polynomial product.x1=[2 3 4]; x2=[3 4 5 6];x3=conv(x1,x2)x3 =

6 17 34 43 38 24

211 432)(

−− ++= zzzX 3212 6543)(−−− +++= zzzzX

);()()( 213 zXzXzX =


Example 4.5

Let and DetermineS

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