Lec_PU_April18

Vector Concept of SignalsVector Concept of Signals

Topics to be CoveredTopics to be CoveredVector space conceptRepresentation of signals using vectorsRepresentation of noise using vectorsRepresentation of modulation schemes

using vector space concept

2

Energy

dttsET

S )(0

2

Signal Space

Digitally modulated signal consists of selection / transmission of one out of a finite set of elements.The signals space concept is similar to vector space.Concept of vector space can be applied to represent digital modulated signals.It allows convenient visualization, mathematical modeling and comparison of various modulation schemes

],...,,[)( 21 nsssSts

Orthogonal Space of Basis Orthogonal Space of Basis FunctionsFunctionsAn N-dim orthogonal space consisting of linearly

independent basis functionsBasis functions are denoted byBasis functions must satisfy the following

conditions:

When Kj are nonzero, the signal space is called Orthogonal; When Kj=1, the signal space is called Orthonormal.

)(tj

Orthogonal Space of Basis Orthogonal Space of Basis FunctionsFunctionsAny finite set of waveforms can be expressed as

a linear combination of N orthogonal waveforms

Also called Generalized Fourier Series

Energy of a Signal Energy of a Signal

For orthonormal functions

Topics to be CoveredTopics to be CoveredMatched filterCorrelator receiversProbability of error

7

OPTIMUM RECEIVER FOR BINARYOPTIMUM RECEIVER FOR BINARY DIGITAL MODULATION SCHEMES DIGITAL MODULATION SCHEMES

The function of a receiver in a digital communication system is to distinguish between all the M transmitted signals sm(t), in the presence of noise

The performance of the receiver is usually measured in terms of the probability of error and the receiver is said to be optimum if it yields the minimum probability of error

4

m=1,2,…,M

Received Signal + Noise Vector Received Signal + Noise Vector

8

Demodulation of Digital SignalsDemodulation of Digital Signals

Matched FilterMatched FilterDetection of signal in presence of additive noiseReceiver knows what pulse shape it is looking for

Channel memory ignored (assumed compensated by other means, e.g. channel equalizer in receiver)

13 - 11

Additive white Gaussian noise (AWGN) with zero mean and

variance N0 /2

s(t)

Transmitted signal

w(t)

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

t = T

y(T)

Matched filter

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

0 tntsthtwthtsty

T is the symbol period

Matched Filter DerivationMatched Filter DerivationDesign of matched filter

◦Maximizes signal power i.e. power at t = T◦Minimizes noise i.e. power of noise

Combined design criteria

13 - 12

power noise averagepower signal ousinstantane

)}({|)(|

SNR pulsepeak is where,max

2

20

tnETs

Matched Filter DerivationMatched Filter Derivation

13 - 13

dffHNdffGtnE N202 |)(|

2 )(} )( {

f2

0N

Noise power spectrum SW(f)

20 |)(|2

)(G )()( fHNffGfG HWN

)(*)()( thtwtn

AWGN Matchedfilter

PSDs are related by


13 - 14

2 2 20 | )(S )(| |)(|

dfeffHTs Tfj

)()( )(0 fSfHfS

dfeffHts tfj )(S )( )( 2 0

)(*)()(0 thtsts

Find h(t) that maximizes pulse peak SNR

dffHN

dfeffH Tfj

20

2 2

|)(|2

| )(S )(|


Schwartz’s inequality

For functions:

upper bound reached if

13 - 15

dffHN

dfeffH Tfj

20

2 2

|)(|2

| )(S )(|

Rkxkx )( )( *21

-

22

-

21

2

2-

1 )( )( )( )( dxxdxxdxxx


13 - 16

)(s )( Hence,

inequality s' Schwartzby )(S )(

whenoccurs which , |)(| 2

|)(| 2 |)(|

2

| )(S )(

|)(| |)(| | )(S )(

)()( and )()(Let

2 *

2

0max

2

020

2 2

222 2

2 21

tTkth

efkfH

dffSN

dffSN

dffHN

dfeffH|

dffSdffHdfeffH|

efSffHf

opt

Tfjopt

-

Tfj

-

Tfj

Tfj

For real s(t)

Matched FilterMatched Filter Given transmitted signal s(t) of duration T, matched filter is

given by hopt(t) = k s(T-t) for any k Duration and shape of impulse response of the optimal filter

is determined by pulse shape s(t) hopt(t) is scaled, time-reversed, and shifted version of s(t) Optimal filter maximizes peak pulse SNR

SNR Does not depend on pulse shape s(t) Directly proportional to signal energy (energy per bit) E Inversely proportional to power spectral density of noise

13 - 17

SNR2|)(| 2 |)(| 2

0

2

0

2

0max

NEdtts

NdffS

Nb

Matched Filter ReceiverMatched Filter Receiver

13 - 18

Matched filter ExampleMatched filter ExampleReceived SNR is maximized at time T

T

example:

transmit filter receive filter

t

)()( tstTs

T t

)( ts

T t

)(ts

Matched Filter: optimal receive filter for maximizedNS

(matched)

Correlation Filter Correlation Filter

At t=T

At t=T, matched filter becomes a correlator

Correlator and Matched Filter Correlator and Matched Filter Outputs for a Sine Wave InputOutputs for a Sine Wave Input

Matched Filter ReceiverMatched Filter Receiver

13 - 22

Matched filter ExampleMatched filter ExampleReceived SNR is maximized at time T

T

example:

transmit filter receive filter

t

)()( tstTs

T t

)( ts

T t

)(ts

Matched Filter: optimal receive filter for maximizedNS

(matched)

Correlation Filter Correlation Filter

At t=T

At t=T, matched filter becomes a correlator

Correlator and Matched Filter Correlator and Matched Filter Outputs for a Sine Wave InputOutputs for a Sine Wave Input

Channel ModelChannel ModelAWGN channel model:

◦Signal vector is deterministic.◦Elements of noise vector are i.i.d

(independent and identically distributed) Gaussian random variables with zero-mean and variance . The noise vector pdf is

◦The elements of observed vector are independent Gaussian random variables. Its pdf is

26

),...,,( 21 iNiii aaas

),...,,( 21 Nrrrr

),...,,( 21 Nnnnn

nsr i

2/0N

0

2

2/10

exp1)(NN

pn

nn

0

2

2/10

exp1)|(NN

p ii

srsrr

Correlation Receiver Correlation Receiver for BPSK/BASKfor BPSK/BASK

One basis function, so one branchThe received signal is input to a correlator The correlator output is compared with a

threshold to decide which symbol was transmitted.

27

T

0

)(1 t)(tr Decision

CircuitCompare r

with threshold.

m̂

Sample at T

Probability of Error Probability of Error BPSK/BASKBPSK/BASK

29

a1(T) is the correlator output when s1(t) is senta2(T) is the correlator output when s2(t) is sentn0(T) is the correlator output when AWGN is present at the input

Output of the correlator:

Hypotheses: H1: s1 is assumed to have been transmitted H2: s2 is assumed to have been transmitted

Probability of Error:Probability of Error: BPSK/BASK BPSK/BASK

30

Probability of Error: BPSK/BASKProbability of Error: BPSK/BASK

31

Probability of ErrorProbability of Error BPSK/BASK BPSK/BASK

32

Probability of ErrorProbability of Error BPSK/BASK BPSK/BASK

33

Where Q(x) is the Complementary Error Function

And for BPSK

b

b

Ea

Ea

2

1

2/020 NNow

)2(0NEQP b

B

Coherent detection of BFSKCoherent detection of BFSK

For correct detection, the carriers must be orthogonal.

Error ProbabilityError Probability BPSK and BFSK BPSK and BFSK

BPSK and BFSK with coherent detection:

35

)(1 t

2s1sbE

“0” “1”

bE

)(2 t

bE221 ss

)(1 t

2s

1s

)(2 tbE

bEbE221 ss“0”

“1”

BPSK BFSK

2

0

NEQP b

B

2/

2/

0

21

NQPB

ss

0

NEQP b

B

Assignment#4Assignment#4Due Before start of next class on Due Before start of next class on

Saturday 23 April, 2011Saturday 23 April, 2011

◦ Q 1: Show that the probability of error for coherently detected BFSK is given by

◦ Q 2: Derive the value of decision threshold if the binary transmitted symbols in BPSK are not equiprobable.

◦ Hint: See Sec. 3.2 and App B.2.

36

0

NEQP b

B

Quiz#4Quiz#4◦Q : (a) A BASK signal transmits 2000 bps. If the

energy per bit is 4 Joules and the noise PSD is 1 W/Hz, how many erroneous bits will be received in 1 second?

◦(b) Also sketch the PDFs of the received signal for both the possible transmitted signals and indicate the area that corresponds to the probability of correct decision.

37

Coherent Detection Coherent Detection of MPSK of MPSK

38

Compute Choose smallest1

2arctanrr ̂

|ˆ| i

T

0

)(1 t

T

0

)(2 t)(tr

1r

2r

m̂

3s

7s

“110”)(1 t

4s 2ssE

“000”

)(2 t

6s 8s

1s5s

“001”“011”“010”

“101”

“111” “100”

8-PSK

Decision variable r̂z

Information resides in the phase, which needs to be detected

Coherent Detection of M-QAM Coherent Detection of M-QAM (Amplitude Phase Shift Keying)(Amplitude Phase Shift Keying)

Coherent detection of M-QAM

39

)(1 t

)(2 t

2s1s 3s 4s0000 0001 0011 0010

6s5s 7s 8s

10s9s 11s 12s

14s13s 15s 16s

1000 1001 1011 1010

1100 1101 1111 1110

0100 0101 0111 0110

16-QAM

©2000 Bijan Mobasseri 41

Signal space dimensionSignal space dimensionHow many basis functions does it take to

express a signal? It depends on the dimensionality of the signal

Some need just 1 some need an infinite number.

The number of dimensions is N and is always less than the number of signals in the set

N<=M


Example: Fourier seriesExample: Fourier series

Remember Fouirer series? A signal was expanded as a linear sum of sines and cosines of different frequencies. Sounds familiar?

Sines and cosines are the basis functions and are in fact orthogonal to each other

cos 2nfot To cos 2mfot dt 0,m n

fo 1/To


InterpretationInterpretations1(t) is now condensed into just two

numbers. We can “reconstruct” s1(t) like this

s1(t)=(1)1(t)+(-0.5)2(t)

Another way of looking at it is this

1

-0.5

1

2


Signal constellationSignal constellationFinding individual components of each

signal along the two dimensions gets us the constellation

s4

s1

s2

s3

1

2

-0.5

-0.5 0.5


Energy in simple languageEnergy in simple language

What we just saw says that the energy of a signal is simply the square of the length of its corresponding constellation vector

3

2 E=9+4=13E


Constrained energy signalsConstrained energy signals

Let’s say you are under peak energy Ep constraint in your application. Just make sure all your signals are inside a circle of radius sqrt(Ep )

Ep


Correlation of two signalsCorrelation of two signals

A very desirable situation in is to have signals that are mutually orthogonal. How do we test this? Find the angle between them

s1

s2 cos 12 s

1

Ts2

s1 s2

transpose


Find the angle between sFind the angle between s11 and s and s22

Given that s1=(1,2)T and s2=(2,1)T, what is the angle between the two?

s1T s2 1 2

21

2 2 4

s1 1 4 5

s2 4 1 5

cos 12 4

5 5

45

12 36.9o


Distance between two signalsDistance between two signals

The closer signals are together the more chances of detection error. Here is how we can find their separation

d12

2 s1 s22 s1 j s2 j 2

j1

N

(1)2 (1)2 2 d12 2

1 2

1

2


Constellation building using Constellation building using correlator bankscorrelator banks

We can decompose the signal into its components as follows

s(t)

1

2

N

dt0

T

dt0

T

dt0

T

s1

s2

sN

N components


Detection in the constellation Detection in the constellation spacespace

Received signal is put through the filter bank below and mapped to a point

s(t)

1

2

N

dt0

T

dt0

T

dt0

T

s1

s2

sN

componentsmapped to a single point


Constellation recovery in noiseConstellation recovery in noise

Assume signal is contaminated with noise. All N components will also be affected. The original position of si(t) will be disturbed


Actual exampleActual exampleHere is a 16-level constellation which is

reconstructed in the presence of noise

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2Eb/No=5 dB


Detection in signal spaceDetection in signal space

One of the M allowable signals is transmitted, processed through the bank of correlators and mapped onto constellation question is based on what we see , what was the transmitted signal?

received signalwhich of the four did it come from


Minimum distance decision ruleMinimum distance decision ruleIt can be shown that the optimum

decision, in the sense of lowest BER, is to pick the signal that is closest to the received vector. This is called maximum likelihood decision making

this is the most likely transmitted signal

received


Defining decision regionsDefining decision regionsAn easy detection method, is to compute

“decision regions” offline. Here are a few examples

decide s1

decide s2

s1s2

measurement

decide s1decide s2

decide s3 decide s4

s1s2

s3 s4

decide s1

s1


More formally...More formally...Partition the decision space into M

decision regions Zi, i=1,…,M. Let X be the measurement vector extracted from the received signal. Then

if XZi si was transmitted


Error probabilityError probability

we can write an expression for error like thisP{error|si}=P{X does not lie in Zi|si was

transmitted}Generally

Pe P X Zi | si P{i1

M si}


Example: BPSKExample: BPSK((BBinary inary PPhase hase SShift hift KKeying)eying)

BPSK is a well known digital modulation obtained by carrier modulating a polar NRZ signal. The rule is

1: s1=Acos(2πfct)0:s2= - Acos(2πfct)

1’s and 0’s are identified by 180 degree phase reversal at bit transitions


Signal space for BPSKSignal space for BPSK

Look at s1 and s2. What is the basis function for them? Both signals can be uniquely written as a scalar multiple of a cosine. So a single cosine is the sole basis function. We have a 1-D constellation

A-Acos(2pifct)


Bringing in EBringing in Ebb

We want each bit to have an energy Eb. Bits in BPSK are RF pulses of amplitude A and duration Tb. Their energy is A2Tb/2 . Therefore

Eb= A2Tb/2 --->A=sqrt(2Eb/Tb)We can write the two bits as follows

s1 t 2EbTb

cos 2fct

s2 t 2EbTb

cos 2fct


BPSK basis functionBPSK basis functionAs a 1-D signal, there is one basis

function. We also know that basis functions must have unit energy. Using a normalization factor

E=1

1 t 2Tb

cos 2fct


Formulating BERFormulating BER

BPSK constellation looks like this

√Eb-√Eb

X|1=[√Eb+n,n]

transmitted

received

noise

nois

e

Pe1 P Eb n 0 |1 is transmitted

if noise is negative enough, it will pushX to the left of the boundary, deciding 0instead


Finding BERFinding BERLet’s rewrite BERBut n is gaussian with mean 0 and

variance No/2

Pe1 P Eb n 0 |1 P n Eb

-sqrt(Eb)


BER for BPSKBER for BPSKUsing the trick to find the area under a

gaussian density(after normalization with respect to variance)

BER=Q[(2Eb/No)0.5]


BPSK ExampleBPSK ExampleData is transmitted at Rb=106 b/s. Noise

PSD is 10-6 and pulses are rectangular with amplitude 0.2 volt. What is the BER?

First we need energy per bit, Eb. 1’s and 0’s are sent by

2Eb

Tbcos(2fct)

2EbTb

0.2


Solving for ESolving for Ebb

Since bit rate is 106, bit length must be 1/Rb=10-6

Therefore, Eb=20x10-6=20 w-sec


Solving for BERSolving for BERNoise PSD is No/2 =10-6. We know for BPSK

BER=0.5erfc[(Eb/No)0.5]What we have is then

Finish this using erf tablesBER

12erfc Eb

No

12erfc 2 10 7

2 10 6

12erfc( 0.1)

12erfc(0.316)


M-ary signalingM-ary signalingBinary communications sends one of only 2 levels; 0

or 1There is another way: combine several bits into

symbols1 0 1 1 0 1 1 0 1 1 1 0 0 1 1

Combining two bits at a time gives rise to 4 symbols; a 4-ary signaling


QPSK constellationQPSK constellation

45o

0001

11 10

√E

1 t 2T

cos2fct

2 t 2T

sin2fctBasis functions S=[0.7 √E,- 0.7 √E]


QPSK decision regionsQPSK decision regions

0001

11 10

Decision regions re color-coded


QPSK error rateQPSK error rate

Symbol error rate for QPSK is given by

This brings up the distinction between symbol error and bit error. They are not the same!

Pe erfc(E

2No

)


Symbol errorSymbol errorSymbol error occurs when received vector is

assigned to the wrong partition in the constellation

When s1 is mistaken for s2, 00 is mistaken for 11

0011s1s2


Symbol error vs. bit errorSymbol error vs. bit errorWhen a symbol error occurs, we might

suffer more than one bit error such as mistaking 00 for 11.

It is however unlikely to have more than one bit error when a symbol error occurs

10 10 11 1000

11 10 11 1000

10 symbols = 20 bits

Sym.error=1/10Bit error=1/20


QPSK vs. BPSKQPSK vs. BPSK

Let’s compare the two based on BER and bandwidth

BER BandwidthBPSK QPSK BPSK QPSK

12erfc

EbNo

1

2erfc

EbNo

Rb Rb/2

EQUAL


M-phase PSK (MPSK)M-phase PSK (MPSK)

If you combine 3 bits into one symbol, we have to realize 23=8 states. We can accomplish this with a single RF pulse taking 8 different phases 45o apart

si(t) 2ET

cos 2fct (i 1)4

, i 1,...,8

0

,0 t T


8-PSK constellation8-PSK constellation

Distribute 8 phasors uniformly around a circle of radius √E

45o

Decision region


Quadrature Amplitude Modulation Quadrature Amplitude Modulation (QAM)(QAM)

MPSK was a phase modulation scheme. All amplitudes are the same

QAM is described by a constellation consisting of combination of phase and amplitudes

The rule governing bits-to-symbols are the same, i.e. n bits are mapped to M=2n symbols


16-QAM constellation using Gray 16-QAM constellation using Gray codingcoding

16-QAM has the following constellationNote gray coding where adjacent symbolsdiffer by only 1 bit

0010001100010000

1010

1110

0110

1011

1111

0111

1001

1101

0101

1000

1100

0100


Vector representation Vector representation of 16-QAMof 16-QAM

There are 16 vectors, each defined by a pair of coordinates. The following 4x4 matrix describes the 16-QAM constellation

[ai ,bi ]

3,3 1,3 1,3 3,3 3,1 1,1 1,1 3,1

3, 1 1, 1 1, 1 3, 1 3, 3 1, 3 1, 3 3, 3


What is energy per symbol in What is energy per symbol in QAM?QAM?

We had no trouble defining energy per symbol E for MPSK. For QAM, there is no single symbol energy. There are many

We therefore need to define average symbol energy Eavg

Eavg 1M

ai2 bi

2 i1

M


EEavgavg for 16-QAM for 16-QAM

Using the [ai,bi] matrix and using E=ai^2+bi^2 we get one energy per signal

E

18 10 10 1810 2 2 1010 2 2 1018 10 10 18

Eavg=10

Relation between ERelation between Ebb/N/N00 & S/N & S/N

With R=Rb

83

In M-ary schemes

In most cases e.g. FSK

Time Division MultiplexingTime Division Multiplexing

Lec_PU_April18

Documents

f s f h f s

t h t s t s

t h t w t

df e f f h t st f

h t w t h t s t v

df f hndf f g t

df f hndf e f f ht f

dt t s ets