simulations of communication systemsscholar.fju.edu.tw/課程大綱/upload/049469/content/961... ·...

1FJU-EE YUJL Simulations of Communication SystemsSimulations of Communication Systems Chapter0 - Page 1

Simulations of Simulations of Communication SystemsCommunication Systems

Download sitehttp://stu71.fju.edu.tw/guest/classsearch.jsp

FJU-EE YUJL Simulations of Communication SystemsSimulations of Communication Systems Chapter0 - Page 2

Simulations of Communication SystemsSimulations of Communication Systems

TextContemporary Communication Systems using MATLAB, John G. Proakis and M. SalehiSimulation and Software Radio for Mobile Communications, H. Harada and R. Prasad

Reference: PRINCIPLES OF COMMUNICATION SYSTEMS SIMULATION WITH WIRELESS APPLICATIONS, TRANTER 2004, PEARSON EDUCATION PTE. LTD.


OutlinesOutlinesSignals and Linear SystemsRandom ProcessesAnalog Modulation (Skip)Analog-to-Digital ConversionBaseband digital TransmissionDigital Transmission through Bandlimited ChannelsDigital Transmission via Carrier ModulationChannel Capacity and Coding (Skip)Spread Spectrum Communication SystemsRadio Communication Channel


GradeGrade

Homework (30%)Midterm Presentation (20%)Final Presentation and Report (50%)


Before startBefore start ourour coursecourse

What is a communication system ?What is difference between digital and analog communication system ?What kind of topics are in communication system ?


What is a communication system?What is a communication system?

What is electrical engineering (EE) ?Production and transmission of electrical energyTransmission of information

Communication systems areSystems designed to transmit information


Difference of comm. systemDifference of comm. systemIn electric energy system

Waveforms are usually knownIn communication systems

Waveform present at user is unknownOtherwise, no information would be transmitted, there would be no need for communication

The transmission of information implies the communication of message that are not known ahead of time (a priori)


Noise and comm. systemNoise and comm. system

If there were no noiseWe would communicate messages electrically to the destination using an infinitely small amount of power

Noise limits our ability to communicate


Goal of communication systemGoal of communication system

Transmit and Receive informationWith small power (energy)

SNR, CNRWith small error

Probability Error RateUnder noisy channel

AWGNFading Channel


Digital and Analog SourceDigital and Analog Source

Digital information sourceProduces a finite set of possible message

Ex: Typewriter

Analog information sourceProduces messages that are defined on a continuum

Ex: Microphone


Digital and Analog SystemDigital and Analog System

Digital communication systemTransfers information from a digital source to the sink

Analog communication systemTransfers information from an analog source to the sink


Digital communication systemDigital communication system

Digital waveformHave only a discrete set of values

Ex: Binary waveform has only two values

Analog waveformHave a continuous range of values

Electronic digital communication systemUsually have digital waveformsHowever, it can have analog waveforms

Ex: 1000Hz sine wave for binary 1 and 500Hz for binary 0Still called Digital communication system


Why Digital ?Why Digital ?Relatively inexpensive digital circuits may be usedPrivacy is preserved by using data encryptionData from voice, video and data sources may be merged and transmitted over a common digital transmission systemIn long distance system, noise does not accumulate from repeater to repeater


Why Digital ? (Cont.)Why Digital ? (Cont.)

Regenerative repeatersNoise are not accumulated


Why digital ? (Cont.)Why digital ? (Cont.)Errors in detected data may be small, even when a large amount of noise on the received signalError may often be corrected by the use of coding

Disadvantage of Digital communicationGenerally, more bandwidth is required than that for Analog systemsSynchronization is required

Digital System are becoming more and more popular !But we have to know analog system also !


Communication systemCommunication system

Three subsystemTransmitterChannelReceiver

Signal Processing

CarrierCircuits

TransmissionMedium

(Channel)

CarrierCircuits

Signal Processing

Transmitter

Receiver

Information input m(t)

s(t)

Noise n(t)

r(t)To userm(t)


Typical digital comm. systemTypical digital comm. system


Digital comm. transformationDigital comm. transformation

Transformation from one signal space to anotherSeven basic groups

Formatting and source codingModulation / demodulationChannel codingMultiplexing and multiple accessSpreadingEncryptionSynchronization


In this courseIn this course

Basic mathematical concepts of digital communication will be givenSome numerical examples using MATLAB is also givenBut, the implementation technique and application is weakly given

Please use references for these topics

MATLAB Tutorial

Referenced from Tolga Kurt [1]

23/09/2004 ELG 4174 2

Getting Started

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

analyzing datadeveloping algorithms and applications

23/09/2004 ELG 4174 3

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.

23/09/2004 ELG 4174 4

Defining vectors and matrices (2/2)

Definition with increments:

23/09/2004 ELG 4174 5

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)

23/09/2004 ELG 4174 6

Graphs

Using built in function

23/09/2004 ELG 4174 7

Writing a function (1/2)

23/09/2004 ELG 4174 8

Writing a function (2/2)

23/09/2004 ELG 4174 9

Calling a function

23/09/2004 ELG 4174 10

M-file example-1%%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)

23/09/2004 ELG 4174 11

M-file example-1% 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;

23/09/2004 ELG 4174 12

M-file example-1% 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

23/09/2004 ELG 4174 13

M-file example-2

f=(0:0.5*10^4:40*10^9);w=2*pi*f;h=exp(i*w.^2*(-2.7*10^-22));Hp=phase(h);hp=Hp*180/pi;plot(f, hp);xlabel('frequency');ylabel('Hp')

23/09/2004 ELG 4174 14

Useful Referenceshttp://www.genie.uottawa.ca/~ieee/pdffiles/matlab.ppthttp://webclass.ncu.edu.tw/~junwu/index.htmlhttp://www.cs.nthu.edu.tw/~jang/mlbook/http://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, Chapter2, H. Harada and R. Prasad

1

Signals and Linear SystemsSignals and Linear Systems

1.1 Linear Time Invariant (LTI) system1.2 Fourier Series1.3 Fourier Transform1.4 Power and Energy1.5 Lowpass Equivalent of Bandpass Signals

2

1.1 1.1 Linear Time InvariantLinear Time Invariant (LTI) system(LTI) system

Most communication channels, transmitter and receiverare well modeled as LTI systemLTI system means a time shift in the input signal causes the same time shift in the output signal

0 0then given ( ), output ( )x t t y t t

LTI System y(t)x(t)

3

Impulse ResponseImpulse Response

For arbitrary input

Impulse response

LTI System

LTI System ???

h(t)

4

Convolution IntegralConvolution Integral

For continuous LTI system

Input-output relationship of LTI systemh(t) is impulse response of the systemx(t) is input signaly(t) is output signal

( ) ( ) ( ) ( ) ( ) ( ) ( )y t x t h t h x t d x h t d

= = =

5

Convolution SumConvolution Sum

For discrete system

LTI Systemh(t)

At time t

( ) ( )t

ii

y t x h t i=

=

( ) ( )

( ) ( )

( )* ( )

n

ii

n

i

y n x h n i

x i h n i

x n h n

=

=

=

=

=

6

Time and Frequency domain ?Time and Frequency domain ?

Some signals are easier to visualize in the frequency domainSome signals are easier to visualize in the time domainEx: Sine wave

Frequency domain : 3 data (frequency, Amplitude, Phase)Time domain : lots of information to define accurately

Frequency domain analysis tools Fourier Series and Fourier Transform

7

1.2 1.2 FourierFourier SeriesSeriesFourier Series

An equation to calculate frequency, amplitude and phase of each sine wave needed to make up any periodic signals

FT (Fourier Transform)Extension of Fourier Series for Non-periodic signalsMathematical formula using Integrals

DFT (Discrete Fourier Transform)A discrete numerical equivalent using sums instead of integral

FFT (Fast Fourier Transform)Just a computational fast way to calculate DFT

8

Complex exponential input and LTIComplex exponential input and LTI

Input-output relation of LTI system

Let input be complex exponential :Then the output is

The output is complex exponential with the same frequency of input

( ) ( ) ( ) ( ) ( )y t x t h t h x t d

= =

02( ) j f tx t Ae =

0 0 02 ( ) 2 20( ) ( ) [ ( ) ] ( ) ( )

j f t j f j f ty t h Ae d h Ae d Ae H f x t

= = =

2( ) ( ) : Frequency response of ( )j fH f h Ae d h t

=

9

Exponential Fourier SeriesExponential Fourier Series

Any periodic signal x(t) with period T0 can be expressed by

where Fourier series coefficient xn is

is fundamental frequency is nth harmonics

is usually used

{ }0 02 2 /with basis : j nf t j nt Tn

e e

==02 /( ) j nt Tn

nx t x e

=

=

002 /

0

1 ( )T j nt T

nx x t e dtT

+ =

0 01f T=

0nf00

2Tor =

10

RealReal--valued periodic signalvalued periodic signal

x(t) should be periodic if Fourier series is appliedx(t) can be real or complex-valued signalIf x(t) is real-valued signal

Fourier series coefficient {xn} are complex{xn} are Hermitian symmetry

Real part(Magnitude) is evenImaginary part(Phase) is odd

0 00 02 / 2 / * *

0 0

1 1( ) [ ( ) ]

,

T Tj nt T j nt Tn n

n n n n

x x t e dt x t e dt xT T

x x x x

+ +

= = =

= =

11

Trigonometric Fourier seriesTrigonometric Fourier series

Can be applied to real, periodic signals

Using Hermitian symmetry

, 2 2

n n n nn n

a jb a jbx x +

= =

0 002 /

0 00 0

1 1( ) ( )[cos(2 / ) sin(2 / )]T Tj nt T

nx x t e dt x t nt T j nt T dtT T

+ += = 0 0

0 00 0

2 2( ) cos(2 / ) , ( ) sin(2 / )T T

n na x t nt T dt b x t nt T dtT T

+ += =

02 / 00 0

1( ) [ cos(2 / ) sin(2 / )]

2j nt T

n n nn n

ax t x e a nt T b nt T

= =

= = + + AC componentDC component

*n nx x =

12

Polar form of Fourier seriesPolar form of Fourier series

Can be applied to real, periodic signalsUsing the relation:Define :

Then

2 2 , arctan nn n n nn

bc a ba

= + =

2 2cos sin cos( arctan )ba b a ba

+ = +

00 0

1

00

1

( ) [ cos(2 / ) sin(2 / )]2

cos(2 / )2

n nn

n nn

ax t a nt T b nt T

a c nt T

=

=

= + +

= + +

13

Discrete SpectrumDiscrete Spectrum

Relations between FS coefficientFor real-valued periodic signal

Discrete SpectrumMagnitude spectrum: Phase spectrum:

2Re[ ] , 2 Im[ ],

n n n n

n n n n

a x b xc x x

= =

= =

0.nx vs n or nf

0.nx vs n or nf

n

2n n

na jbx =

14

FS and realFS and real--even(odd) signaleven(odd) signal

Real and even signal:Then

All xn are realFourier series consists of all cosine functions

Real and odd signal:ThenAll xn are imaginaryFourier series consists of all sine functions

0

00

2 ( ) cos(2 / ) 0T

na x t nt T dtT

+= =

( ) ( )x t x t= 0

00

2 ( )sin(2 / ) 0T

nb x t nt T dtT

+= =

OddEven

( ) ( )x t x t =

15

IP1IP1--11: FS of a rectangular signal train: FS of a rectangular signal train

Periodic rectangular signal

Real and Even signal

0

00

,

( ) ( ) ,2 2

0,

A t tt Ax t A t tt

otherwise

-1 & x0 & x

23

IP1IP1--22: : magnitude and phase spectralmagnitude and phase spectral

25 20 15 10 5 0 5 10 15 20 250

0.02

0.04

0.06

0.08

0.1

0.12

0.14

Time

Am

plitu

deThe Discrete Magnitude Spectrum

25 20 15 10 5 0 5 10 15 20 254

2

0

2

4

Time

Am

plitu

de

The Discrete Phase Spectrum

24


Periodic Gaussian signal

t0l=0.1 tol=0.0001

2

21( )2

, for 6t

x tt e

=

20 15 10 5 0 5 10 15 200

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

Time

Amplit

ude

Gaussian signal waveform

25


2

21( )2

, for 6t

x tt e

=

25 20 15 10 5 0 5 10 15 20 250

0.02

0.04

0.06

0.08

0.1

Time

Am

plitu

deThe Discrete Magnitude Spectrum

25 20 15 10 5 0 5 10 15 20 254

2

0

2

4

Time

Am

plitu

de

The Discrete Phase Spectrum

26

1.2.1 1.2.1 Periodic Signals and LTI SystemsPeriodic Signals and LTI Systems

The output of LTI system with periodic inputFrom eq(1.2.3), if input is a complex exponential, the output ofLTI system is also a complex exponential with the same frequency as the input but the magnitude and phase are changedIf the input is a period signal, the output signal is also periodic

Then the relation between xn and yn ???

0 02 / 2 /( ) , ( )j nt T j nt Tn nn n

x t x e y t y e

= =

= = x(t) LTI System y(t)

27

Periodic Signals and LTI SystemsPeriodic Signals and LTI Systems

From convolution integral

Define transfer function of LTI

We haveIf we know xn (x(t)) and H(f) or h(t), we can get yn (y(t))

0

0 0 0

2 ( ) /

2 / 2 / 2 /

( ) ( ) ( ) ( )

{ ( ) }

j n t Tn

n

j n T j nt T j nt Tn n

n n

y t x t h d x e h d

x h e d e y e

=

= =

= =

= =

02 ( / )20

0

( ) ( ) ( ) ( ) ( ) j t n Tj ft nH f h t e dt H nf H h t e dtT

= = =

0

( )n nny x HT

=

28

IP1IP1--44: Filtering of Periodic Signals: Filtering of Periodic Signals

Triangle pulse and Filter

LTI SystemH(f)

H(f0)

H(f1)

H(f3)H(f5)

Sum of below

x0*H(f0)

x1*H(f1)

x3*H(f3)

x5*H(f5)

29

IP1IP1--44 : Filtering of Periodic Signals: Filtering of Periodic Signals

Fourier series of Lambda signal

Transfer function :using numerical approach (FFT)

00

0

/ 2 12 /

/ 2 10

2/ 2

1 1 1( ) ( ) ( )2 2

1 1[ ( )] sinc ( )2 2 2

T j nt T j nt j ntn T

f n

x x t e dt t e dt t e dtT

nF t

=

= = =

= =

Period =To=2 fundamental frequency=fo=0.5

{ }( ) ( ) f n F s nH n F H f T DFT h= = =

TS

T

FFS

DFT

h(t) H(f)

30


Selection of time parameter TAccording to property of DFT

Selection of time parameter TsDepends on the number of frequency bins we want to observeIf N is given, then

Usage of fftshift: generally, we want to display the frequency responses between

H=fft(h) returns H(0)~H(N-1) H1=fftshift(H) returns

1 1 and T TsF Fs

= =

( 2) ~ ( 2)H N H N

0 01 2 aWe want nd 1 2T f FF f = = = =

2 / 80 1/ 40 (in pp16)S SF N F T T N= = ==

( 2) ~ ( 2)H N H N

31

IP1IP1--44 : Filtering of Periodic Signals: Filtering of Periodic Signals1. % IP1-4 Chapter 1.2. echo on3. n=[-20:1:20];4. % FS coefficients of x(t) vector 5. x=.5*(sinc(n/2)).^2;6. % sampling interval7. ts=1/40;8. % time vector9. t=[-.5:ts:1.5];10. % impulse response

1. fs=1/ts;2. h=[zeros(1,20),t(21:61),zeros(1,

20)];3. % transfer function4. H=fft(h)/fs;5. % frequency resolution6. df=fs/80;7. f=[0:df:fs]-fs/2;8. % rearrange H9. H1=fftshift(H);10. y=x.*H1(21:61);11. % Plotting commands follow.

32


20 15 10 5 0 5 10 15 200

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Frequency

Am

plitu

de

Frequency Response of System

20 15 10 5 0 5 10 15 200

0.1

0.2

0.3

0.4

0.5

0.6

0.7

n

Am

plitu

de

Frequency Response of System

33


20 15 10 5 0 5 10 15 200

0.1

0.2

0.3

0.4

0.5

n

Am

plitu

deInput Signal

20 15 10 5 0 5 10 15 200

0.05

0.1

0.15

0.2

0.25

0.3

0.35

n

Am

plitu

de

Outptu Signal

34


Questions: What is the difference if we set ts=1/20 or ts=1/80. Can we still get the correct output signals?How to compute xn by using fft method? (xn is a periodic continuous signal)What is the difference if we select h(t), t=0~2 to perform the fftoperation ?

35

Homework #1Homework #1

Problems 1.1, 1.2, 1.3, 1.7, 1.8

36

1.3 1.3 Fourier TransformFourier Transform

FT is the extension of FS to non-periodic signalDefinition of Fourier transform

Fourier transform

Inverse Fourier transform

For a real signal x(t), X(f) is Hermitian Symmetry

Magnitude spectrum is even about the origin (f=0)Phase spectrum is odd about the origin

2( ) [ ( )] ( ) j ftX f F x t x t e dt

= =

1 2( ) [ ( )] ( ) j ftx t F X f X f e df

= =

*( ) ( )X f X f =

37

Properties of FTProperties of FTLinearity

DualityIf , Then

Time ShiftA shift in the time domain results in a phase shift in the frequency domain

Scaling: An expansion in the time domain results in a contraction in the frequency domain, and vice versa

1 2 1 2[ ( ) ( )] [ ( )] [ ( )]F x t x t F x t F x t + = +

( ) [ ( )]X f F x t= [ ( )] ( )F X t x f=

{ }0 02 20( ( )) ( ) : ( ) ( )j ft j ftF x t t e X f Note e X f X f = =

1[ ( )] ( ) , 0fF x at X aa a

=

38

Properties of FTProperties of FT

ModulationMultiplication by an exponential in the time domain corresponds to a frequency shift in the frequency domain

020

0 0 0

[ ( )] ( )1[ ( )cos(2 )] [ ( ) ( )]2

j f tF e x t X f f

F x t f t X f f X f f

=

= + +

x(t)

X(f)cos

f0f0-f0

-f0

39


DifferentiationDifferentiation in the time domain corresponds to multiplicationby j2f in the frequency domain

ConvolutionConvolution in the time domain is equivalent to multiplication in the frequency domain, and vice versa

[ ( )] 2 ( )

[ ( )] ( 2 ) ( )n

nn

dF x t j f X fdtdF x t j f X fdt

=

=

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

F x t y t X f Y fF x t y t X f Y f

==

40


Parsevals relation

Energy can be evaluated in the frequency domain instead of the time domain

Rayleighs relation

* *( ) ( ) ( ) ( )x t y t dt X f Y f df

=

2 2*( ) ( ) ( ) ( )x t x t dt x t dt X f df

= =

41

More on FT pairsMore on FT pairsSee Table 1.1 at page 20

Delta function FlatTime / Frequency shiftSin / cos input

impulses in the frequency domainsgn / unit step input

Rectangular sincLambda sinc2DifferentiationPulse train with period T0

Periodic signalimpulses in the frequency domain

42

FT of periodic signalsFT of periodic signalsFor a periodic signal with period T0x(t) can be expressed with FS coefficientTake FT

FT of periodic signal consists of impulses at multiples of the harmonics (fundamental frequency 1/T0) of the original signal

Ex:

02 /( ) j nt Tnn

x t x e

=

= 0

0

2 /

2 /

0

( ) [ ( )] [ ]

[ ] ( )

j nt Tnn

j nt Tn nn n

X f F x t F x e

nx F e x fT

=

= =

= =

= =

00 0

1( ) ( ) ( ) ( )n n

nx t t nT X f fT T

= =

= =

43

FS FS versus FTversus FT

FS coefficient can be expressed using FTDefine truncated signal

FT of truncated signal:Expression of FS coefficient

0

0 0( ) ,( ) 2 2

0 ,T

T Tx t tx t

otherwise

< =

0 0( ) [ ( )]T TX f F x t=

0 00 0

0

0

0 0

2 / 2 /

0 00 0

2 /

0 0 0

1 1( ) ( )

1 1( ) ( )

T Tj nt T j nt Tn T

j nt TT T

x x t e dt x t e dtT T

nx t e dt XT T T

= =

= =

44

Spectrum of the signalSpectrum of the signal

Fourier transform of the signal is called the Spectrum of the signal

Generally complexMagnitude spectrumPhase spectrum

45

1.3.1 1.3.1 Sampling TheoremSampling Theorem

Basis for the relation between continuous-time signal and discrete-time signalsA bandlimited signal can be completely described in terms of its sample values taken at intervals Ts as long as Ts 1/(2W)

Ts

x(t)

fW-W

X(f)

1

46

Impulse samplingImpulse sampling

Sampled waveform

Take FT( ) ( ) ( ) ( ) ( )s s s

n nx t x t t nT x nT t nT

= =

= =

Ts

x(t)

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

1 1( ) ( ) ( )

s sn n

n ns s s s

X f F x t F x t t nT X f F t nT

n nX f f X fT T T T

= =

= =

= = =

= =

fW-W

X(f)

1/(2Ts)1/Ts

1/Ts

47

Reconstruction of signalReconstruction of signal

Low pass filterWith Bandwidth 1/(2Ts) and Gain of Ts

fW-W

X(f)

1/(2Ts)1/Ts

Low pass filter

( ) ( ),sX f T X f for f W= 1/(2W)Spectrum is overlappedWe can not reconstruct original signal with under-sampled valuesAnti-aliasing methods are needed

fW-W

X(f)

fW-W

X(f)

1/Ts

1/Ts

51

AliasingAliasing

We only sample the signal at intervalsWe dont know what happens between the samples

52

AliasingAliasing

We must sample fast enough to see the most rapid changes in the signal

This is Sampling theorem

If we do not sample fast enoughSome higher frequencies can be incorrectly interpreted as lower ones

53

AliasingAliasing

Called aliasing because one frequency looks like another

54

Discrete Discrete Time Time Fourier TransformFourier TransformZ-transform : Discrete Time Fourier Transform (DTFT) of discrete time sequence x[n]

From sampling theorem, we have FT versus DTFT relation

( ) [ ] nn

X z x n z =

=

2

( ) ( ) ( ) ( ) ( )

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

[ ] (DTFT)

1( ) ( ) ( ) (from Sampling Theorem)

s

s s sn n

s s s sn n

j fnTn

ns s

x t x t t nT x nT t nT

X f F x t F x nT t nT x nT F t nT

x n e

nX f X f X zT T

= =

= =

=

=

= =

= = =

=

=

22( ) ( ) [ ] sj fTs

j fnTz e n

X f X z x n e

= == =

2( ) ( ) [ ] ,sj fnTs s nX f T X f T x n e for f W

=

= =

55

Discrete Fourier TransformDiscrete Fourier TransformDFT of discrete time sequence x[n],n=0~N-1

Relation between DTFT and DFT

Relation between FT and DFT

Relation between FS and DFT

02 2 /0 1 1

0 0

( ) [ ] [ ] ( )

where

sN Nj kf nT j nk Nn n

s s

X f kf x n e x n e X k

f T T T N

= =

= = = =

= =

0

0

( ) ( ), , and ( ) ( )( ) ( )

s

s

X f T X f for f W X f kf X kX f kf T X k

= < = =

= =

( )

2 /1

2 /1

: ( ) [ ]

: [ ] 1 ( )

N j nk Nn

N j nk Nk

DFT X k x n e

IDFT x n N X k e

=

=

=

=

( )( ) ( )

00 0 0

0

1 ( ), and ( ) ( )

1 ( ) 1 ( )k T s

k s

x T X kf X f kf T X k

x T T X k N X k

= = =

= =

56

SummarySummary

2: ( ) ( ) j ftFT X f x t e dt

=

2 /1

: ( ) [ ]N j nk Nn

DFT X k x n e =

=

002 /

0

1: ( )T j kt T

kFS x x t e dtT

+ =

2: ( ) [ ] sj fnTn

DTFT X f x n e =

=

00

1 ( )kx X kfT=

0( ) ( )X f kf X k = =

( ) ( )sX f T X f=( )1 ( )kx N X k=

0( ) ( )sX f kf T X k= =

57

FFT in FFT in MMatlabatlabA sequence of length N=2m of samples of x(t) taken at Ts

Ts satisfies Nyquist conditionTs is called time resolution

FFT gives a sequence of length N of sampled Xd(f) in the frequency interval [0, fs=1/Ts]

The samples are apart byis called frequency resolutionFrequency resolution is improved by increasing N

/sf f N =f

58

DFT in DFT in MatlabMatlabFFT(Fast Fourier Transform) is an Efficient numerical method to compute DFT

See fft.m and fftseq.m, for more informationTime and frequency is not appeared explicitlyN is chosen to be N=2m

Zero padding is used if N is not power of 2

TS

T

FFS

DFT

h(t) H(f)Zero padding

59

Zero padding in Zero padding in FFTFFTPadding zeros at time resolution Ts

Maximum frequency is unchanged = FsNumber of sequence in changed from L to NFrequency resolution is increased from to

Let

The frequency response are

We conclude that

/sF F N =sF L

2 /1

12 / 2 ( / ) /1 1

( ) [ ]

( ) [ ] [ ] ( / )

L j nk LnN Lj nk N j n Lk N Ln n

X k x n e

X k x n e x n e X Lk N

=

= =

=

= = =

[ ], n=0,1, , -1[ ] [ ], n=0,1, , -1, and [ ] 0, n= , , -1, 2K

x n Lx n x n L x n L N N L= = =

(0) (0), (1) ( ), (2) (2 ), , ( 1) (( 1) )X X X X L N X X L N X N X N L N= = = =

60

IP1IP1--55: F: Fourier ourier TransfoermTransfoerm

Plot the frequency responses of x1(t) and x2(t)

Parameters setting:

Select t=[-5,5] as a time period for x1(t) and x2(t)

1/ 2, set 10 1/ 2 5 2 10 0.1default drequency resolution = 0.01

s sBW W F W tF Hz

= = = = =

61

IP1IP1--55: F: Fourier ourier TransfoermTransfoerm1. df=0.01;2. fs=10;3. ts=1/fs;4. t=[-5:ts:5];5. x1=(t+1).*(t>=-1 & t=0 & t=0 & t=1 & t

63

IP1IP1--66: F: Fourier Transformourier Transform

Plot the frequency response of x(t)

Parameters setting:

Select t=[-4,4] as a time period for x1(t) and x2(t)

1/ 4, set 10 1/ 4 2.5 2 5 0.2default drequency resolution = 0.01

s sBW W F W tF Hz

= = = = =

2 2 11 1 1

( )2 1 2

0

t tt

x tt t

+ = + < otherwise

2 2( ) 4sin (2 ) sin ( )X f c f c f=

64

IP1IP1--66: F: Fourier ourier TransfoermTransfoerm

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.50

0.5

1

1.5

2

2.5

3

Frequency

Magnitudepectrum of x(t) derived analytically

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.50

0.5

1

1.5

2

2.5

3

Frequency

Magnitudepectrum of x(t) derived numerically

65

Frequency domain analysis of LTI systemFrequency domain analysis of LTI system

The output of LTI system

Take FT (using convolution theorem)

Where the Transfer Function of the system

The relation between input-output spectra

( ) ( ) ( )y t x t h t=

( ) ( ) ( )Y f X f H f=

2( ) [ ( )] ( ) j ftH f F h t h t e dt

= =

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

Y f X f H fY f X f H f

=

= +

66

IP1IP1--77: : LTI system analysisLTI system analysis

Plot the frequency response of x(t)

Parameters setting:

Select t=[-5,5] as a time period for x(t)

5 0.2default drequency resolution = 0.01

s sF tF Hz

= =

2 2 01 0 1

( ) 2 2cos(0.5 ) 1 31 3 40

t tt

x t t tt

+ < = + < < otherwise

67


Ideal LPF

5 0 50

0.5

1

1.5

2

Time

Ampl

itude

Input Signal

4 2 0 2 40

2

4

6

8

Frequency

Ampl

itude

Input Signal

4 2 0 2 40

0.2

0.4

0.6

0.8

1

Frequency

Ampl

itude

LPF Frequency Response

5 0 50

0.5

1

1.5

2

2.5

Time

Ampl

itude

Output Signal

68


Selective frequency filter

5 0 50

0.5

1

1.5

2

Time

Ampl

itude

Input Signal

4 2 0 2 40

2

4

6

8

Frequency

Ampl

itude

Input Signal

5 0 50

0.2

0.4

0.6

0.8

1

Time

Ampl

itude

Impulse Response

10 5 0 5 100

0.5

1

1.5

2

2.5

3

Time

Ampl

itude

Output Signal

69


Remarks: how to compute output sequence precisely ?By FFT

By partially FFT (used in text)( ) [ ] ( ) [ ] ( ( ) ( ) and [ ] [ ])( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

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

o s

o

o o o

s s o o

x t x n X kf T X k x t X f x n X kh t H f H kfy t x t h t Y f X f H f Y kf X kf H kf

T Y k T X k H kf Y k X k H kf y n IDFT Y k

= = = =

= = =

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

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

o s

o s

o o o

s s s s

x t x n X kf T X k x t X f x n X kh t h n H kf T H ky t x t h t Y f X f H f Y kf X kf H kf

T Y k T X k T H k Y k T X k H k y n IDFT Y k

= == = =

= = =

70

IP1IP1--77: : LTI system analysisLTI system analysisRemarks: how to compute output sequence precisely ?

By discrete convolution (used by text, but the authors made a mistake

The above result is consistent to that by FFT. When the data and channel are transferred from continuous into discrete, the discrete output signal either by FFT multiplication or by directconvolution is equal to the continuous one up to a constant sampling period TS

( ) [ ], ( ) [ ]

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

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

s s sk

s s sn

x t x n h t h n

y t x t h t x h t d x kT h t kT T

y nT y n x k h n k T T x n h n

= =

= = =

71

Some remarks on short signalSome remarks on short signalFT works on signals of infinite durationBut, we only measure the signal for a short time

FFT works as if the data is periodic all the time

72

Frequency leakageFrequency leakageIf the period exactly fits the measurement time, the frequency spectrum is correctIf not, frequency spectrum is incorrect

It is broadened due to the window effect.

73

HomeworkHomework #2#2

Problems1.10, 1.12, 1.14, 1.15, 1.18

74

1.4 Power and Energy1.4 Power and EnergyDefinition of Power in communication systems

Instantaneous power

Average Power

Rms (Root mean square) value

Average Power for resistive load

Normalized PowerUsed by communication engineersR is assumed to be 1

Average Normalized Power

( ) ( ) ( )p t v t i t=/ 2 / 2

/ 2 / 2

1 1lim ( ) lim ( ) ( )T T

T TT TP p t dt v t i t dt

T T = =

/ 2 2

/ 2

1lim ( )T

rms TTX x t dt

T =

2/ 2 / 22 2 2

/ 2 / 2

11lim ( ) lim ( )T T rms

rms rms rmsT TT T

VP v t dt R R i t dt RI V IT T R

= = = = =

/ 2 2

/ 2

1lim ( )T

TTP w t dt

T =

75

DefinitionsDefinitions

Decibel gain of system

If dB is known

0 dB : 3 dB : -3dB :

( ) 10log( ) 10log( ) (unit: dB)outin

Paverage power outP dBaverage power in P

= =

( ) /1010P dBoutin

PP

=

out inP P=3/1010 2out in inP P P= =

3/10 1102out in in

P P P= =

76

DefinitionsDefinitions

Decibel SNR(signal-to-noise ratio)

Decibel power level with respect to 1mW

/ 2 2

/ 2

/ 2 2

/ 2

1lim ( )( / ) 10log( ) 10log( ) 20log( )

1lim ( )

T

TTsignal rms signaldB T

noise rms noiseTT

s t dtP VTS NP Vn t dt

T

= = =

3

( )10log( )10

30 10log( ( ))

actual power level WattdBm

actual power level Watt

=

= +

77

Power and EnergyPower and EnergyFor a real signal x(t), energy and power of x(t) are defined respectively by

Energy-type signal: A signal with finite energyEx: x(t) = (t)

Power-type signal: A signal with positive and finite powerEx: x(t) = cos(t)

All periodic signals are power-type signal.

2 ( )XE x t dt

=

/ 2 2

/ 2

1lim ( )T

X TTP x t dt

T =

78

Energy Spectral DensityEnergy Spectral Density

Energy spectral density of energy-type signal gives the distribution of energy at various frequency of signal For a real-valued signal x(t),

Autocorrelation function =ESD =Energy = 22 ( ) ( ) ( )X XE x t dt X f df g f df

= = =

2 *( ) ( ) [ ( ) ( )] [ ( ) ( ) ] [ ( )]X xg f X f F x t x t F x t x t d F R

= = = + =

( ) ( ) ( ) ( ) ( )xR x t x t d x t x t

= + =

79

PowerPower Spectral DensitySpectral Density

For a real-valued power-type signal x(t), Autocorrelation function =PSD =Power =

PSD is always a real nonnegative function of frequencyFind average normalized power

( )X XP S f df

=

( ) [ ( )]X xS f F R =

/ 2

/ 2

1( ) lim ( ) ( )T

x TTR x t x t d

T

= +

/ 2 2

/ 2

1( ) (0) lim ( )T

x x x TTP S f df R x t dt

T

= = =

80

More on PSDMore on PSD

For a periodic x(t) with period T0 and FS coefficient xn

PSD is Discrete spectrum !All powers are concentrated at the harmonics

Power at the nth harmonics at n/T0 is

20( ) { ( )} ( )x x kkS f F R x f kf = =

2nx

0 0

0

/ 2 / 2 2 / 2 ( ) /

/ 2 / 2

2 2

1 1( ) lim ( ) ( )T T j nt T j k t T

x n kT TT n k

j kfkk

R x t x t d x e x e dtT T

x e

+

= =

= + =

=

81

PSD and LTI systemPSD and LTI system

Output ESD

Output PSD

H(f)x(t) y(t)Sx(f) Sy(f)

2( ) ( ) ( )y xg f H f g f=

2( ) ( ) ( )y xS f H f S f=

82

PSD for discrete time signalsPSD for discrete time signals

For an aperiodic sampled signal { , x[-1], x[0], x[1], }

For a periodic sampled signal:

2

2

[ ]

1lim [ ]2 1

X sn

N

X N n N

E T x n

P x nN

=

=

=

=+

12

01

2

0

[ ]

1 [ ]

N

X snN

Xn

E T x n

P x nN

=

=

=

=

83

IP1IP1--88: : Power SpectrumPower Spectrum

Plot the PSD of x(t)

Parameters setting:

Spectrum.m : compute Sx(f) using Welch PSD method with nonoverlapped window nonparametric PSD estimatorpsd=spectrum(x,1024) : returns two vectors, one is mean vector and another is variance vector.specplot(psd,fs) : plot spectrum of data in psd with three curves, psd(mean), psd(mean+variance) and psd(mean-variance)

0.001, [0 : :10]st t ts= =

cos(2 47 ) cos(2 219 ) 0 10( )

0t t t

x t +

= otherwise

84

IP1IP1--88: : Power SpectrumPower Spectrum

0 100 200 300 400 50010

20

1010

100

1010

Pxx X Power Spectral Density

Frequency

No noise

0 100 200 300 400 50010

5

100

105


Frequency

SNR=10dB

0 100 200 300 400 50010

4

102

100

102

104


Frequency

SNR=0dB

0 100 200 300 400 50010

2

100

102

104


Frequency

SNR=10dB

85

1.51.5 Lowpass Equivalent of Lowpass Equivalent of BandpassBandpass SignalsSignals

Bandpass signalAll frequency components are located in the neighborhood of a central frequency f0

Lowpass signalAll frequency components are located around the zero frequency

f0-f0

2WX(f)

-W 0 W

X(f)BP signal LP signal

86

Why?Why?

To increase the efficiency of communication system, bandpass signal is usedBut, the analysis and design is usually done with lowpass signal

87

Analytic signalAnalytic signal

Define analytic signal as (in time domain)

where is Hilbert transform of x(t)Hilbert transform

whereH(f) = F[h(t)] corresponds to a 90 phase shift

( ) ( ) ( )z t x t jx t= +

( )x t

( ) ( ) ( )x t x t h t= ( ) 1h t t=

, 0( ) sgn( )

, 0j f

H f j fj f >

= =

89

Lowpass EquivalentLowpass Equivalent

Move central frequency of analytic signal to zero by using modulation

f0-f0

Z(f)

f0-f0

Xl(f)

0

0 1 0 0( ) ( ) 2 ( ) ( )lX f Z f f u f f X f f= + = + +

Analytic signalLP equivalent signal

90

Lowpass EquivalentLowpass Equivalent

In time domain

From analytic signal, we have

In general, xl(t) is a complex signal

02( ) ( ) j f tlx t z t e=

02( ) ( ) ( ) ( ) j f tlz t x t jx t x t e= + =

0

0

2

2

( ) Re[ ( ) ]

( ) Im[ ( ) ]

j f tl

j f tl

x t x t e

x t x t e

=

=

( )( ) ( ) ( ) ( ) j tl c sx t x t jx t V t e= + =

Quadrature (sine) component of x(t)In-phase(cosine) component of x(t)

( )( ) arctan( )

s

c

x ttx t

= 2 2( ) ( ) ( )c sV t x t x t= +

91

Cartesian Coordinate SystemCartesian Coordinate System

From

We have

Or

020 0( ) ( ) ( ) ( ) [ ( ) ( )][cos(2 ) sin(2 )]

j f tl c sz t x t jx t x t e x t jx t f t j f t

= + = = + +

0 0

0 0

( ) ( ) cos(2 ) ( )sin(2 )( ) ( )sin(2 ) ( ) cos(2 )

c s

c s

x t x t f t x t f tx t x t f t x t f t

= = +

0 0

0 0

( ) ( ) cos(2 ) ( )sin(2 )( ) ( ) cos(2 ) ( )sin(2 )

c

s

x t x t f t x t f tx t x t f t x t f t

= +=

92

PPolarolar Coordinate SystemCoordinate System

From , we have

Then

Envelope is independent of f0Phase depends on f0

0 02 2 ( )( ) ( ) ( ) ( ) ( )j f t j f t j tlz t x t jx t x t e V t e e = + = =

0

0

2 ( )0

2 ( )0

( ) Re[ ( ) ] ( ) cos(2 ( ))

( ) Im[ ( ) ] ( )sin(2 ( ))

j f t j t

j f t j t

x t V t e e V t f t tx t V t e e V t f t t

= = +

= = +2 2( ) ( ) ( )

( )( ) arctan( 2 )( ) o

V t x t x tx tt f tx t

= +

=

( )( ) ( ) j tlx t V t e=

93

IP1IP1--99: : BP to LP transformationBP to LP transformation

Given BP signal x(t)( ) sinc(100 )cos(2 200 )x t t t=

94


2 1.5 1 0.5 0 0.5 1 1.5 21

0.5

0

0.5

1BP Signal

time

inpu

t sig

nal x

(t)

500 400 300 200 100 0 100 200 300 400 5000

1

2

3

4

5

6x 10

3

frequency

mag

nitu

de s

pect

rum

95


2 1 0 1 20.5

0

0.5

1Inphase component,f0=200

time

inpu

t sig

nal r

eal

2 1 0 1 20

0.2

0.4

0.6

0.8

1

1.2Envelope component,f0=200

time

inpu

t sig

nal i

mag

500 400 300 200 100 0 100 200 300 400 5000

0.002

0.004

0.006

0.008

0.01

0.012LP signal

frequency

mag

nitu

de s

pect

rum

96


2 1 0 1 21

0.5

0

0.5

1Inphase component,f0=200

time

inpu

t sig

nal r

eal

2 1 0 1 20

0.2

0.4

0.6

0.8

1

1.2Envelope component,f0=200

time

inpu

t sig

nal i

mag

500 400 300 200 100 0 100 200 300 400 5000

0.002

0.004

0.006

0.008

0.01

0.012LP signal

frequency

mag

nitu

de s

pect

rum

1

Baseband Digital TransmissionBaseband Digital Transmission

5.2 Binary Signal Transmission5.3 Multiamplitude Signal Transmission5.4 Multidimensional Signals

2

5.2 Binary Signal Transmission5.2 Binary Signal Transmission

Data Rate: R [bps] Channel noiseAWGN n(t)

Binary Input01101001 Mapping

0

1

0 ( ), 01 ( ), 0

b

b

s t t Ts t t T

( ) ( ) ( )0,1 , 0

i

b

r t s t n ti t T

= +=

Bit time intervalTb = 1/R

Bit time intervalTb = 1/R

S0 S1 S1 S0 S1Receiver will determinewhether 0 or 1 is sent by observing r(t)

Optimum receiver: Minimize the

probability of error

3

Optimum receiver for AWGN channelOptimum receiver for AWGN channel

Consists of 2 Blocks

Get informationusing received signal(Only two signalss0(t) and s1(t) are expected)

CorrelatorOr

Matched FilterDetector

BinaryDigitalSignal

+AWGN

OutputData

Determine whichsignal is receivedusing the output ofCorrelator orMatched filter

4

Signal CorrelatorSignal Correlator

Cross-correlates the received signal r(t) with the two possible transmitted signal s0(t) and s1(t)

0( )

td

0( )

td

Detectorr(t) 01

( )( )

s ts t

OutputData

0 00( ) ( ) ( )

tr t r s d =

1 10( ) ( ) ( )

tr t r s d = Samples at t=Tb

5

Illustrative Problem 5.1Illustrative Problem 5.1: : orthogonal binary signalorthogonal binary signal

Two possible transmitted signal s0(t) and s1(t), orthogonal binary signal

Correlator output (at t = T0)

A

Tb

A Tb

S0(t) S1(t)

-A

0( )b

Td

0( )b

Td

r(t)=S0(t)+n(t) 0

1

( )( )

s ts t

20 0 00 0

20 0

( ) ( ) ( )b bT T

b

r s d s n d

A T n n

= +

= + = +

1 0 1 10 0

1

( ) ( ) ( ) ( )b bT T

r s s d s n d

n

= +

=

orthogonalSignal energy

Noise component

6


Output of correlator (Noise free case)

0( )

td

0( )

td

r(t)=S0(t)

0

1

( )( )

s ts t Tb

Tb

/2

0( )

td

0( )

td

r(t)=S1(t)

0

1

( )( )

s ts t

Tb

Tb

/2

7


Noise effects AWGN : n0, n1 are Gaussian process

With zero mean

And variances

0[ ( )] 0, [ ( ) ( )] ( )2

NE n t E n n t t = =

0 0 00 0

1 1 10 0

[ ] [ ( ) ( ) ] ( ) [ ( )] 0

[ ] [ ( ) ( ) ] ( ) [ ( )] 0

b b

b b

T T

T T

E n E s n d s E n d

E n E s n d s E n d

= = =

= = =

( )

2 2

0 0

00 0 0 0

2 20 0 00 0 0

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

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

( ) ( ) ( ) , 1,22 2 2

b b

b b b b

b b b

T T

i i i i

T T T T

i i i i

T T T

i i

E n E s t s n t n dtd

Ns t s E n t n dtd s t s t dtd

N N Ns t t d dt s t dt i

= =

= =

= = = =

8


pdf of r0 and r1 2 200 0

( ) / 2

( | ( ))1

2r

p r s t

e

=

0( )b

Td

0( )b

Td

r(t)=S0(t)+n(t)

0

1

( )( )

s ts t

r

2 21

1 0

/ 2

( | ( ))1

2r

p r s t

e

=

0( )b

Td

0( )b

Td

r(t)=S1(t)+n(t)

0

1

( )( )

s ts t

r

0 1( | ( ))p r s t1 1( | ( ))p r s t

r0

r1

r0

r1 0

0

9

Matched FilterMatched Filter

Alternative to signal correlator

Identical output to correlator

Matched Filterh(t)=s(Tb-t)

s(t)+

n(t)

y(t)

0

0 0

( ) ( ) ( )

( ) ( ) ( ) ( )

b

b

b b

b

T

bt T

T T

bt T

y T s h t d

s s T t d s s d

=

=

=

= + = =

; convolution

10

Illustrative Problem 5.2Illustrative Problem 5.2

Output of matched filter

Tb 2Tb

r(t)=S0(t) Tb

TbTb Tb 2Tb

/2

r(t)=S1(t) Tb

TbTb

Tb 2Tb

/2

Tb 2Tb

hi(t)=si(Tb-t)

t

t

t

t

11

The detectorThe detector

Decides the transmitted signal is s0(t) or s1(t) by observing the output of correlator (r0 or r1)

Optimum DetectorDetector that has minimum probability of error

12

Illustrative Problem 5.3Illustrative Problem 5.3 ((Bdt_f57Bdt_f57.m).m)

Optimal Detector : if binary signals are equally probable and have equal energies. 0 ( s0(t) is sent ) if r0 > r1 1 ( s1(t) is sent ) if r0 < r1

Probability of error If s0(t) is transmitted, but it detects s1(t)

r0 > r1r0 < r1

r0r1

0

1

r

0 0( | ( ))p r s t1 0( | ( ))p r s t

Probability of error, Pe(0)r1

13


Probability of error If s0(t) is transmitted

BER:

is called SNR0N

0 0

1 1

r nr n=+=

2

0

1 0 0 1 0 1 0

/ 2

/0 0

(0) ( | sent) ( ) ( )

1 12 22

e

x

N

P P r r s P n n P n n

e dx Q erfcN N

= > = > + = >

= =

14


Same result can be obtained when s1(t) is transmitted Pe(1) Pe=Pe(0)P(0)+Pe(1)P(1)=Pe(0)[P(0)+P(1)]=Pe(0) See Figure 5.7, page 180 (Bdt_f57.m)

Probability of error for orthogonal signals Pe decreases exponentially as the SNR increasesQ(.) and erfc(.)

2

2

21( ) , probability of (0,1) in [ , )2

2 1( ) , probability of (0, ) in ( , ]&[ , )2

1( ) ( )2 2

( ) 2 ( 2 )

t

x

t

x

Q x e dt N x

erfc x e dt N x x

xQ x erfc

erfc x Q x

=

=

=

=

15


16


1. initial_snr=0;2. final_snr=15;3. snr_step=0.25; 4. snr_in_dB=initial_snr:snr_step:final_snr;5. for i=1:length(snr_in_dB),6. snr=10^(snr_in_dB(i)/10); 7. Pe(i)=Qfunct(sqrt(snr)); 8. echo off;9. end;10. echo on;11. semilogy(snr_in_dB,Pe);12. xlabel('Input SNR per Bit')13. ylabel('Bit Error Rate (BER)')14. set(findobj('Type','line'),'LineWidth',2)

17

Monte Carlo SimulationMonte Carlo Simulation

Sometimes the analysis of detector performance is difficult to performThe probability of errors can be estimated by using software to model the received sequences, make decisions by the receiver and count the number of error bits.There are many ways to perform the Monte-Carlo simulations, depending on the parts in communication systems you want to evaluate.The number of samples (N) required to estimate an error probability P(m) based on Monte-Carlo method satisfies N>>1/P(m). A rule of thumb is N>10/P(m). (Sec2.7)

18


Simulation Model: see Fig. 5.8Uniform random

number generator

Binary data source

0 / E1 / E

r0r1

detectorOutput

data

Compare

Error counter

Gaussian randomnumber generator

n0

Gaussian random number generator

n10 0

01 1

if (t) is sentr n

sr n= +

=

0 01

1 1

if (t) is sentr n

sr n=

= +

19


Simulation Model: see Fig. 5.8 Define SNR :

Signal generator using uniform random number with a range of [0,1]

Noise generator:

Simulation result : smldPe54.mTheoretical result :

2 0 ( ) 22 2 2N SNR

SNR SNR = = = =

100

ln10 ( )( )10 10

, ( ) 10log

10SNR dBSNR dB

SNR SNR dB SNRN

SNR e

= =

= =

( )0

12 2e

SNRP Q Q SNR erfcN

= = =

Used in textbook

Suitable in multi -user scenario

20


12. theo_err_prb(i)=Qfunct(sqrt(SNR)); 13. echo off ;14. end;15. echo on;16. % Plotting commands follow17. semilogy(SNRindB1,smld_err_prb,'*r');18. hold19. semilogy(SNRindB2,theo_err_prb,'-b');20. xlabel('Input SNR per Bit')21. ylabel('Bit Error Rate (BER)')22. set(findobj('Type','line'),'LineWidth',2)

1. SNRindB1=0:1:12;2. SNRindB2=0:0.1:12;3. for i=1:length(SNRindB1),4. % simulated error rate5. smld_err_prb(i)=smldPe54(SNRindB1(i)); 6. echo off ;7. end;8. echo on ;9. for i=1:length(SNRindB2),10. SNR=exp(SNRindB2(i)*log(10)/10); 11. % theoretical error rate

21

Illustrative Problem smldPe54Illustrative Problem smldPe54

21. % matched filter outputs22. if (dsource(i)==0),23. r0=E+gngauss(sgma);24. r1=gngauss(sgma); % if the source output is "0"25. else26. r0=gngauss(sgma);27. r1=E+gngauss(sgma); % if the source output is "1"28. end;29. % detector follows30. if (r0>r1),31. decis=0; % decision is "0" 32. else33. decis=1; % decision is "1" 34. end;35. if (decis~=dsource(i)), % if it is an error, increase the

error counter36. numoferr=numoferr+1;37. end;38. end;39. p=numoferr/N; % probability of error e

1. function [p]=smldPe54(snr_in_dB)2. % [p]=smldPe54(snr_in_dB)3. % SMLDPE54 finds the probability of error for the given4. % snr_in_dB, signal to noise ratio in dB.5. E=1;6. SNR=exp(snr_in_dB*log(10)/10); % signal to noise ratio7. sgma=E/sqrt(2*SNR); % sigma, standard deviation of noise8. N=10000;9. % generation of the binary data source10. for i=1:N,11. temp=rand; % a uniform random variable over (0,1)12. if (temp

23

Antipodal Signals Antipodal Signals for binary signal transmissionfor binary signal transmission

Antipodal If one signal is negative of other

Choose s0(t) = s(t), s1(t) = -s(t) to transmit binary signal, where s(t) is arbitrary with energy . Then received signal is

Are there any advantages using antipodal signal instead of orthogonal signal ?

Tb

TbTb

Tb

( ) ( ) ( ), 0 br t s t n t t T= +

24

Optimum Receiver with Antipodal signalOptimum Receiver with Antipodal signal

Matched Filter or Correlator

0( )

td

( )bs T t

r(t)=s(t)+n(t)

( )s t detectorOutput decision

Sample at t = Tb

correlator

Matched Filter

Tb

Tb

-

Correlator outputs(t) -s(t)

25

Noise component with Antipodal signalsNoise component with Antipodal signals

Let be transmittedOutput of correlator (or Matched filter)

Statistics of n and rn is Gaussian, with zero mean E[n]=0With Variance

Therefore 0 0(0, ) and ( , )2 2

N Nn N r N

( ) ( ) ( )r t s t n t= +

2

0( ) ( )b

T

br A T s t n t dt n= + = + Noise componentSignal energy

2 2

0 0

20 0 0

0 0 0

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

( ) ( ) ( ) ( )2 2 2

b b

b b b

T T

T T T

E n s t s E n t n dtd

N N Ns t s t dtd s t dt

= =

= = =

26

Optimum detectorOptimum detector

Optimal detector and pdf

0( )

tdr(t)=

s(t)+n(t)( )s t

r>0r

27

BER & Benefit of antipodal signalBER & Benefit of antipodal signal

Probability of error : if binary signals are equally probable and have equal energies

Orthogonal signal case

Antipodal signal is 3dB more efficient than orthogonal signal Antipodal signal gives better performance (3dB) with same signal energy Same performance with Antipodal signal energy /2 as orthogonal signal with

energy .

2 2

2

0 ( ) / 20 1

/ / 2

0

1(0) (1) (0) ( 0 | ( ) sent)2

1 2( ) ( )2

re e e e

r

P P P PP P P r s t e dr

e dr Q QN

= + = = < =

= = =

0

( )eP Q N

=

28


Binary Antipodal Simulation: see Fig. 5.13

Uniform random number generator

Binary data source

Compare

Error counter

detector E rn


Output data

r n= +

29


Binary Antipodal Simulation: (ip_05_05a.m) Define SNR :


Noise generator:

Simulation result : (smldPe55a.m, content has been modified)Theoretical result :

2 0 ( ) 22 2 2N SNR

SNR SNR = = = =

100

ln10 ( )( )10 10

, ( ) 10log

10SNR dBSNR dB

SNR SNR dB SNRN

SNR e

= =

= =

( ) ( )0

2 122e

P Q Q SNR erfc SNRN

= = =

30

Illustrative Problem smldPe5Illustrative Problem smldPe55 (modified)5 (modified)

1. function [p]=smldPe55(snr_in_dB)2. E=1;3. SNR=exp(snr_in_dB*log(10)/10); % signal-to-noise ratio4. sgma=E/sqrt(2*SNR); % sigma, standard deviation of noise5. N=1000000;6. temp1=rand(N,1);7. temp2=find(temp1>0.5);8. dsource=zeros(N,1);9. dsource(temp2)=1;10. % The detection, and probability of error calculation follows.11. temp3=sgma*randn(N,1); 12. r=ones(N,1)*(-E);13. r(temp2)=E;14. r=r+temp3;15. decis=zeros(N,1);16. decis(find(r>0))=1;17. numoferr=sum(abs(dsource-decis));18. p=numoferr/N % probability of error

31


0 2 4 6 8 1010

6

105

104

103

102

101

Input SNR (E/No)(dB)

Bit E

rror R

ate

(BER

)

Simulation ResultTheoretical Result

32

OnOn--Off signals Off signals for binary signal transmissionfor binary signal transmission

On-Off signalsNo signal for 0Arbitrary signal s(t) for 1

Received waveform

Are there any advantages using On-Off signal ?

( ), if 0 is trasnmitted( )

( ) ( ), if 1 is trasnmittedn t

r ts t n t

= +

TbTb

0 1

33

Optimum detectorOptimum detector

Detector and pdf

0( )

tdr(t)=

0,s(t)( )s t

r>r

35

Illustrative Problem 5.Illustrative Problem 5.66

On-Off signaling simulation

Uniform random number generator

Binary data source

Compare

Error counter

detector0 01 E

rn


Output data

36


On-Off signaling simulation (ip_05_06a.m) Define SNR :


Noise generator:

Simulation result : (smldPe56.m)Theoretical result :

2 0 ( ) 22 2 2N SNR

SNR SNR = = = =

100

ln10 ( )( )10 10

, ( ) 10log

10SNR dBSNR dB

SNR SNR dB SNRN

SNR e

= =

= =

02 2e

SNRP Q QN

= =

37


0 5 10 1510

5

104

103

102

101

100

Input SNR (E/No)(dB)

Bit E

rror R

ate

(BER

)


38

Signal Constellation Diagram for Binary SignalSignal Constellation Diagram for Binary Signal

Performance comparison

(0,)s1(t)

(,0)s(t)

(-,0)-s(t)

(,0)s0(t)Orthogonal

(0,0)00On-Off

(,0)s(t)Antipodal

Avg PowerPeCoordinatesEnergySignalsType

0

QN

02Q

N

0

2QN

2

(,0)(-,0) (,0)(0,0) (,0)

(0,)

Antipodal On-Off Orthogonal

s(t)s0(t)

s(t)

s1(t)

39

Illustrative Problem 5.Illustrative Problem 5.7 (Orthogonal signaling)7 (Orthogonal signaling)

Noise effects and Constellation : (ip_05_07a.m) If s0(t) is transmitted

If s1(t) is transmitted

Can you explain why performance of Antipodal > Orthogonal > On-Off using this figure ?

0 0

1 1

r nr n= +=

0 0

1 1

r nr n== +

40


Noise effects and Constellation : See Figure 5.18

1 0.5 0 0.5 1 1.5 21

0.5

0

0.5

1

1.5

2Signal constellation of orthogonal signals, 2=0.1

Coefficient in axis X

Coef

ficie

nt in

axis

Y

Signal 1Signal 2

41



1 0.5 0 0.5 1 1.5 21

0.5

0

0.5

1

1.5



Coef

ficie

nt in

axis

Y

Signal 1Signal 2

42



1 0.5 0 0.5 1 1.5 21

0.5

0

0.5

1

1.5



Coef

ficie

nt in

axis

Y

Signal 1Signal 2

43


1. sigma=[0.1 0.3 0.5];2. for ii=1:length(sigma)3. n0=sigma(ii)*randn(100,1);4. n1=sigma(ii)*randn(100,1);5. n2=sigma(ii)*randn(100,1);6. n3=sigma(ii)*randn(100,1);7. x1=1.+n0;8. y1=n1;9. x2=n2;10. y2=1.+n3;11. tt=num2str(sigma(ii));12. figure13. plot(x1,y1,'o',x2,y2,'*')14. axis('square')

15. grid on16. axis([-1 2 -1 2])17. title(['Signal constellation of orthogonal

signals, {\sigma}^{2}=',num2str(sigma(ii))],'fontsize',15)

18. xlabel('Coefficient in axis X','fontsize',15)19. ylabel('Coefficient in axis Y','fontsize',15)20. set(findobj('Type','line'),'LineWidth',2.0)21. set(findobj('fontname','Helvetica'),'fontna

me','Arial','fontsize',15)22. legend('Signal 1','Signal

2','fontsize',15,'fontname','Arial')23. hold off24. aa=['IP_05_07' num2str(ii)];25. print('-depsc','-tiff',aa)26. end

44


5.3 Multiamplitude Signal Transmission

45

Binary Vs. Binary Vs. MultiamplitudeMultiamplitude signalsignal

Binary signal1 bit of informationTwo level to represent 0 and 1

If we use multiple amplitude levelsWe can transmit multiple bits per signal waveform

Questions :Probability of Error Vs. Bandwidth ???

46

Signal waveform Signal waveform with 4 amplitude levelswith 4 amplitude levels

Normalized pulse whose energy is 1

Consider 4 equally spaced Amplitude

4 waveform called PAM(Pulse Amplitude Modulated) signal

See figure 5.19

1 ,0( )

0 ,

t TTg totherwise

=

(2 3) , 0,1,2,3or { 3 , , ,3 }

m

m

A m d mA d d d d= =

=

( ) ( ), 0m ms t A g t t T=

47


Signal Constellation DiagramOne dimensional signal : g(t) is basis

-3d -d d 3d

S0(t) S1(t) S2(t) S3(t)00 01 11 10

Euclidean distance = 2d

t0T

s t0 ( )

3dT

0 t

T

s t1( )

dT

0 tT

s t2 ( )dT

t0 T

s t3( )3dT

g(t)

48


Transmit 2 bits of information per waveformAssign information to waveform

00 s0(t)01 s1(t)11 s2(t)10 s3(t)

Definition of SymbolEach pair of information bits {00, 01, 10, 11} is called a symbolTime duration T is called Symbol Duration

If Bit rate R =1/Tb, then Symbol interval T = 2Tb and baud rate = 1/T=R/2

49

More on Symbol and Symbol intervalMore on Symbol and Symbol interval

Bit rate R=1/Tb

InformationSource

0100110 ModulatorM=2k

Symbol

00011110

s0(t)s1(t)s2(t)s3(t)

Tb 2Tb . T=2Tb

Ex: 9600bps = 4800sps (baud rate)

50

Received SignalReceived Signal

AWGN channel assumption

Receiver determines which of 4 signal waveform was transmitted by observing the received signal r(t)Optimum receiver is deigned by minimizing the probability of error

( ) ( ) ( ), 0,1,2,3, 0ir t s t n t m t T= + =

White Gaussian processWith power spectrum N0/2

51

Optimum receiver Optimum receiver for AWGN channelfor AWGN channel

Implementation

Correlator

Correlator(Matched filter)

AmplitudeDetector

r(t) DecisionOutput

0( )

td

r(t)

g(t)

Samples at t=T

0

0

2

0 0

( ) ( )

{ ( ) ( )} ( )

( ) ( ) ( )

T

T

i

T T

i

i

r r t g t dt

A g t n t g t dt

A g t dt g t n t dt

A n

=

= +

= +

= +

52

Optimum receiver Optimum receiver for AWGN channelfor AWGN channel

Statistics of correlator output Noise component : Gaussian with Zero mean and Variance of

Received signal

0 0

2 2 20 0

0

[ ] [ ( ) ( ) ] ( ) [ ( )] 0

[ ] ( )2 2

T T

T

E n E g t n t dt g t E n t dt

N NE n g t dt

= = =

= = =

-3d -d d 3d

2 2( ) / 21( | ( )) , 3 , , ,32

ir Ai iP r s t e A d d d d

= =

53

Optimum DetectorOptimum Detector

If all symbols are equally probable , the optimum detector is the smallest distance criterion, where the thresholds are set at (-2d,0,2d) min , 0,1,2,3i iiA D r A i= = =

Average Probability of Symbol Error

22 2

/ 24 2/

0

3 3 1 3 3 2( )4 2 2 22

xm d

d dP P r A d e dx Q QN

= > = = =

-3d -d d 3d-2d 0 2d

54

Probability of ErrorProbability of Error

Using Average transmitted signal energy / symbol (bit):

SER, see figure 5.21

2

40

3 22

dP QN

=

42 2 2

01

2

1 ( ) 54 5

225

T avav i i k

i k

avbav avb

P s t dt d d

d

=

= = = =

= =

40

43 3 42 5 2 5

avbP Q Q SNRN

= =

55

Probability of ErrorProbability of Error

0 2 4 6 8 10 1210

4

103

102

101

100

Input SNR (Eavb/No)(dB)

Bit E

rror R

ate

(BER

)

Simulation Result

56

Illustrative Problem 5.Illustrative Problem 5.8 8 44--ary PAM simulation ary PAM simulation : see Fig. 5.: see Fig. 5.2222

Figure 5.22: Block diagram of four level PAM for Monte Carlo Simulation

UniformRNG

Gaussian randomNumber Generator

compare

Error counter

detectorMapping to Amplitude levels +

X Am r

2(0, )N

mAmr A n= +

57



Define SNR :

Noise generator:

Simulation result : ip_05_08a.m, smldPe58.m

2 22 25 4(set 1) or (set 1)

4 5d SNRd d

SNR = = = =

ln10 ( )

2

( )10 10

0

2

2 02

10

25

2

45

SNR dBSNR dBavb

avb

SNR eN

d d SNRN

= = =

=

=

=

0 0.25 0.5 0.75 1

S0(t) S1(t) S2(t) S3(t)-3d d -d 3d

58


0 2 4 6 8 10 1210

4

103

102

101

100


Bit E

rror R

ate

(BER

)


59


1. % MATLAB script for Illustrated Problem 8, Chapter 5.2. clear 3. echo on4. SNRindB1=0:1:12; 5. SNRindB2=0:0.1:12; 6. for i=1:length(SNRindB1),7. % simulated error rate 8. smld_err_prb(i)=smldPe58(SNRindB1(i)); 9. echo off;10. end;11. echo on;12. for i=1:length(SNRindB2),13. % signal to noise ratio14. SNR_per_bit=exp(SNRindB2(i)*log(10)/10); 15. % theoretical error rate16. theo_err_prb(i)=(3/2)*Qfunct(sqrt((4/5)*SNR_per_bit)); 17. echo off;18. end;19. echo on;

20. figure21. semilogy(SNRindB2,theo_err_prb,'r-','LineWidth',2.0);22. xlabel('Input SNR

(Eavb/No)(dB)','fontsize',15,'fontname','Arial')23. ylabel('Bit Error Rate (BER)','fontsize',15,'fontname','Arial')24. set(findobj('Type','line'),'LineWidth',2.0)25. set(findobj('fontname','Helvetica'),'fontname','Arial','fontsiz

e',15)26. legend('Simulation Result','Theoretical

Result','fontsize',15,'fontname','Arial')27. hold off28. print -depsc -tiff IP_05_08a29. figure30. semilogy(SNRindB1,smld_err_prb,'b*','LineWidth',2.0);31. hold32. semilogy(SNRindB2,theo_err_prb,'r-','LineWidth',2.0);33. xlabel('Input SNR

(Eavb/No)(dB)','fontsize',15,'fontname','Arial')34. ylabel('Bit Error Rate (BER)','fontsize',15,'fontname','Arial')35. set(findobj('Type','line'),'LineWidth',2.0)36. set(findobj('fontname','Helvetica'),'fontname','Arial','fontsiz

e',15)37. legend('Simulation Result','Theoretical

Result','fontsize',15,'fontname','Arial')38. hold off39. print -depsc -tiff IP_05_08b

60

Illustrative Problem 5.Illustrative Problem 5.8 8 smldPe58smldPe58.m.m

1. function [p]=smldPe58(snr_in_dB)2. % [p]=smldPe58(snr_in_dB)3. % SMLDPE58 simulates the probability of error for the

given4. % snr_in_dB, signal to noise ratio in dB.5. d=1;6. SNR=exp(snr_in_dB*log(10)/10); % signal to noise ratio per

bit7. sgma=sqrt((5*d^2)/(4*SNR)); % sigma, standard deviation of

noise8. N=10000; % number of symbols being simulated9. % Generation of the quaternary data source follows.10. for i=1:N,11. temp=rand; % a uniform random variable over

(0,1)12. if (temp

61

MM--aryary PAM PAM Signal waveforms with multiple amplitude levelsSignal waveforms with multiple amplitude levels

General case : M=2k M-ary signal waveforms

where g(t) is a baseband pulse waveform with unit energy

Each signal waveform conveys k=log2M bits of information If bit rate Rb=1/Tb, then symbol (baud) rate is Rs=1/T=1/kTb=R/k

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

m m

m

s t A g t t T m MA m M d m M

= = = + =

(-M+1)d -3d -d d 3d 5d (M-1)d

Euclidean distance = 2d

62


Optimum receiverA single correlator (or Matched filter)An amplitude detectorCompute Euclidean distance for m=0,1,2,,M-1

( ) ( ) ( ), 0,1, , 1, 0ir t s t n t i M t T= + = White Gaussian process with power spectrum density= N0/2

0 0

22 0

22

, (0, ), ( , )2 2

( )1( | ( )) exp , 222

i i

ii

N Nr A n n N r N A

r A Np r s t

= +

= =

63


Probability of Error in M-level PAM system Using the BER analysis of binary signaling scheme, we have

Furthermore1 1

2 2 2 2 2

0 02 2

2 22

22

2 22 2 2

00

1 1 1(2 1) ( 1)3

3 log( 1)3log ( 1)

6 log 6log , ( 1) ( 1)

M M

av ii i

av avbavb

avb avb

E A i M d M dM ME E MM dE dK M ME M EMd SNR SNR

NM N M

= =

= = + =

= = =

= = =

1

0

(0) ( 1), ( ) 2 , 1,2, , 2

1assumi(2 2 n)( ) , g

e e e

M

e iM ii

d dP Q P M P i Q i M

M dP P PM

P i QM

=

= = = =

= =

=

64

Illustrative Problem 5.Illustrative Problem 5.99 MM--aryary PAM simulationPAM simulation

Probability of Error in M-level PAM system Finally

figure5_24.m, theoretical result in figure 5.24 for M=2,4,8,16 IP_05_09.m(+smldPe59.m) Monte-Carlo simulation in fig 5.25 for M=16 Note the settings for signal and noise generator

22

6(log )2( 1)( 1)M

MMP Q SNRM M

=

2 22

0 22

2

( 1)given , then 6 log

85If 16, then 8

avbE d MSNRN SNR M

dMSNR

= =

= = A bad programming setting

65

Theoretical BER simulation (figure_05_24.m)Theoretical BER simulation (figure_05_24.m)

0 5 10 15 20 2510

8

106

104

102

100


Sym

bol E

rror R

ate

(SER

)

M=2M=4M=8M=16

M,SER

66

Illustrative Problem 5.Illustrative Problem 5.9 9 : Monte: Monte--Carlo trialsCarlo trials

5 10 15 20 2510

8

106

104

102

100


Bit E

rror R

ate

(BER

)

M=16 PAM


67


1. function [p]=smldPe59(snr_in_dB)2. % [p]=smldPe59(snr_in_dB)3. % SMLDPE59 simulates the error probability for the given4. % snr_in_dB, signal to noise ratio in dB.5. M=16; % 16-ary PAM6. d=1;7. SNR=exp(snr_in_dB*log(10)/10); % signal to noise ratio per bit8. sgma=sqrt((85*d^2)/(8*SNR)); % sigma, standard deviation of noise9. N=10000; % number of symbols being simulated10. % generation of the quarternary data source11. for i=1:N,12. temp=rand; % a uniform random variable over (0,1)13. index=floor(M*temp); % the index is an integer from 0 to M-1,

where14. % all the possible values are equally likely15. dsource(i)=index;16. end;17. % detection, and probability of error calculation18. numoferr=0;19. for i=1:N,20. % matched filter outputs21. % (2*dsource(i)-M+1)*d is the mapping to the 16-ary constellation22. r=(2*dsource(i)-M+1)*d+gngauss(sgma); 23. % the detector 24. if (r>(M-2)*d),25. decis=15;26. elseif (r>(M-4)*d),27. decis=14;28. elseif (r>(M-6)*d),29. decis=13;30.

31. elseif (r>(M-8)*d),32. decis=12;33. elseif (r>(M-10)*d),34. decis=11;35. elseif (r>(M-12)*d),36. decis=10;37. elseif (r>(M-14)*d),38. decis=9;39. elseif (r>(M-16)*d),40. decis=8;41. elseif (r>(M-18)*d),42. decis=7;43. elseif (r>(M-20)*d),44. decis=6;45. elseif (r>(M-22)*d),46. decis=5;47. elseif (r>(M-24)*d),48. decis=4;49. elseif (r>(M-26)*d),50. decis=3;51. elseif (r>(M-28)*d),52. decis=2;53. elseif (r>(M-30)*d),54. decis=1;55. else56. decis=0; 57. end;58. if (decis~=dsource(i)), % if it is an error, increase the error counter59. numoferr=numoferr+1;60. end;61. end;62. p=numoferr/N; % probability of error estim

68


5.4 Multidimensional Signals

69

Multidimensional Vs. Multidimensional Vs. MultiamplitudeMultiamplitude

Multiamplitude signal is One-dimensional

With One basis signal: g(t)

Multidimensional signalCan be defined using multidimensional orthogonal signals

-3d -d d 3d

00 01 10 11

70

Multidimensional Orthogonal SignalsMultidimensional Orthogonal Signals

Many ways to constructIn this text, Construction of a set of M=2k waveforms si(t), i=1,2,,M-1Mutual OrthogonalityEqual EnergyUsing Mathematics

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

1,where , is Kronecker delta

0,

T

i k ik

ik

s t s t dt i k M

i ki k

= =

==

71

Example of M=2Example of M=222

Multiamplitude Vs. Multidimensional

All signals have identical energy

See figure 5.27, page 208, for signal constellation M=2, 3

s3s2s1s0

t

3dd

-d-3d

22

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

T

iA Ts t dt i MM

= = =

s0 s1 s2 s3

t or f

A

72

ExampleExampless

4 orthogonal equal energy signal waveforms

Signal constellation M=2, 3

AT

s t3( )

34T t

AT

s t2 ( )

34TT

2t

T4

T

s t0 ( )

At

AT

s t1( )

T4

T2

t

M=2

E s0s1

E

M=3

E s0s1

E

E

2s

73

ExampleExampless

M-orthogonal equal energy signal waveforms can be represented by a set of M-dimensional orthogonal vectors. The waveform basis contains the unit-energy signal waveforms and the coordinates are shown in the following as M-dimensional orthogonal vectors

0

1

2

1

( ,0,0,0, ,0)

(0, ,0,0, ,0)

(0,0, ,0, ,0)

(0, ,0,0,0, )M

s E

s E

s E

s E

=

=

=

=

74

Receiver for AWGN ChannelReceiver for AWGN Channel

Received signal from AWGN channel

Receiver decides which of M signal waveform was transmitted by observing r(t)Optimum receiver minimizes Probability of Error

( ) ( ) ( ), 0,1,..., 1, 0ir t s t n t m M t T= + =

White Gaussian processWith power spectrum N0/2

75

Optimum Receiver Optimum Receiver for AWGN Channelfor AWGN Channel

Needs M Correlators (or Matched Filters)

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

T

i ir r t s t dt i M= =

0( )

td

r(t)s0(t)

Samples at t=T

0( )

td

sM-1(t)Detector

OutputDecision

r0

rM-1

76

Signal CorrelatorsSignal Correlators

Let s0(t) be transmitted

Statistics of noise component (Gaussian Process)Zero mean and Variance

0 0 0 00 0

20 0 00 0

( ) ( ) { ( ) ( )} ( )

( ) ( ) ( )

T T

T T

r r t s t dt s t n t s t dt

s t dt s t n t dt n

= = +

= + = +

0 00 0

00 0

( ) ( ) { ( ) ( )} ( )

( ) ( ) ( ) ( ) , 0

T T

i i

T T

i i i

r r t s t dt s t n t s t dt

s t s t dt s t n t dt n i

= = +

= + =

Orthogonal

2 2 20 0

0

2 20

( ) ( )2 2

( , ), (0, )

T

i i

i

N NE n s t dt

r N r N

= = =

77

pdfpdf of correlator outputof correlator output

Let s0(t) is transmitted

Mean of correlator outputE[r0] = , E[ri] = 0

0( )

td

s0(t)+n(t)

s0(t) Samples at t=T

0( )

td

sM-1(t)

r0

rM-1

2 20

0 0

( ) / 2

( | ( ))1

2r

P r s t

e

=

2 2

0

/ 2

( | ( ))1

2i

i

r

P r s t

e

=

0

78


For equiprobable symbol, the optimum detector is the smallest distance criterion or the largest correlator output.Let s0(t) be transmittedThe probability of correct decision Is probability that r0 > ri

Symbol error probability

are linearly independent

0 1 0 2 0 1( , ,..., )c MP P r r r r r r = > > >

0 1 2 1, , ,..., Mr r r r 1

0 1 0 2 0 1 1 0 01

1 0

( , ,..., ) ( | ( ))

( )

Mc M i i

Mi i

P P r r r r r r P r r s tP r r

=

=

= > > > = >

= >

0 1 0 2 0 11 1 ( , ,..., )M c MP P P r r r r r r = = > > >

79


For a fixed r00

0 0 0 0( | ) 1 ( | ) 1i irP r r r P r r r Q < = > =

[ ]1

1 1 00 1 0 0 1 0 0

10

0 0

( | ) ( | ) 1 ( | ) 1

( | ) 1 ( | ) 1 1

MM Mi i i i

M

rP c r P r r r P r r r Q

rP e r P c r Q

= =

= > = < =

= =

r

20 (0, )r N 2(0, )ir N

0 r00 0( | )iP r r r>

80


Using Bayes theorem

Then

2 20

0 0 0

( ) 20

0

/

0( / ) ( ) ( / ) ( )

where 1( )2

r

MP P e r P r P e r P r dr

P r e

= =

=

{ }

2 20

20

1( ) / 20

0

( 2 / ) / 21

11 12

1 1 [1 ( )]2

Mr

M

y NM

rP Q e dr

Q y e dy

=

=

0

20

0

where , 2

Note : ( ), , for binary caseb bP Q

ryN

N

=

= =

In text, use bdt_int.m to describe the integration elements

81

Symbol Symbol errorerror rate & Bit error raterate & Bit error rate

Converting probability of symbol error to probability of a binary digit error.For equiprobable orthogonal signals, all symbol errors are equiprobable.The number of possible error symbols is M-1. If the symbol is in error, the bit error probability is

Average bit error probability12

2 1

k

b MkP P

=

Symbol error probabilityBit error probability

1 12 21 2 1

K K

KM

=

82

Theoretical BER simulationTheoretical BER simulation

Bdt_f527.m SNR settings : SNR per bit

Use bdt_int.m to describe the integration elements

Figure 5.29 : We can reduce SNR per bit to get a given probability of error by increasing the number M of waveforms.

{ } 20

1

( 2 / ) / 21

22 1

1 1 [1 ( )]2

k

b Mk

y NMM

P P

P Q y e dy

=

=

ln10 ( )( )10 10

20

0

2

Given 10 , th

2log 2 lo

e

nd g

n

a

SNR dBSNb

b

R dB

bEE E

ESNR

K E M SNR MN

eN

= = =

= = =i

Bdt_int.m

83

Theoretical BER simulationTheoretical BER simulation

M,BER, WHY???

84

Illustrative problemIllustrative problem 5.10 : Monte5.10 : Monte--Carlo trialsCarlo trials

M=4 Orthogonal signaling simulation: see figure 5.30

En0 r0

n1 r1

n20

0

0 n3

Output decision

Uniform RNG

Error counter

Comparesi with ^si

Mapping to signal points

Gaussian RNG

detector

Gaussian RNG

Gaussian RNG

Gaussian RNG

2r

3r

is

si

85


M=4 Orthogonal signaling simulation: IP_05_10.mSignal model IUsing si(t) as matched filter

SNR settings : SNR per bit

2 2 20 0

0

2 2

or

( ) ( )2 2

( , ) or (0, )

i i i i

T

i i

i i

r n r nN NE n s t dt

r N r N

= + =

= = =

ln10 ( )( )10 10

02

22

02

Given 10 , then

log and2 2

2 log

SNR dBSNR dBb

bb

ESNR eN

EN EEE E MSN NRR

ES M

===

= = =

=

i

86


Signal model IIUsing as matched filter

SNR settings : SNR per bit

2 2 20 0

0

2 2

or

( ) ( )2 2

( , ) or (0, )

i i i i

T

i i

i i

r E n r nN NE n t dt

r N E r N

= + =

= = =

ln10 ( )( )10 10

0

02

22

Given 10 , then

2 l

log ao

nd2 2 g

SNR dBSNR dBb

bb

ESNR eN

N EE E MSN SN MR

ER

=

= =

=

= =

=i

( )( )

( )i

ii

s tts t

=

87


Perform in smldP510.m by using signal model II is misused by signal model I, comparing the signal model ri

and noise variance.2

Be careful to the SNR settings. In text, the author mixed two signal models together and misled the readers a lot.

88


89


1. function [p]=smldP510(snr_in_dB)2. M=4; % quarternary orthogonal signalling3. E=1;4. SNR=exp(snr_in_dB*log(10)/10); % signal to noise ratio per

bit5. sgma=sqrt(E^2/(4*SNR)); % sigma, standard deviation of

noise6. N=10000; % number of symbols being simulated7. % generation of the quarternary data source8. for i=1:N,9. temp=rand; % a uniform random variable over

(0,1)10. if (temp

91

BiorthogonalBiorthogonal SignalsSignals

Signal Constellation

Examples M=2 : Antipodal signal M=4

0 / 2

1 / 2 1

/ 2 1 1

( ) ( ,0,0,...,0), ( ,0,0,...,0)

( ) (0, ,0,...,0), (0, ,0,...,0)

( ) (0,0,0,..., ), (0,0,0,..., )

M

M

M M

s t s

s t s

s t s

+

simulations of communication systemsscholar.fju.edu.tw/課程大綱/upload/049469/content/961... ·...

Documents