Timing andsynch-

ronization

noise

gi(t) si(t)

y(T) r(t)

Digitalbaseband/bandpasswaveform

mi

Detect

To other destinations

FormatD/A

Information sink

Demod-ulate &sample

Block Diagram of a DCS in 411

Informationsource

FormatA/D

XMT

Multi-plex

From other sources

Pulsemodu.

Band-pass

modu.

RCV

h(t):Wave-formchan-nel

(band-width

limited)

ISI

Digitalinput

im

Digitaloutput

im

Digitaloutput

Demul-tiplex

h(t):channelimpulseresponse

@G. Gong 2

Chapter 3 Waveform Coding Technique (Text book, Chapter 5)

Purpose: to study baseband transmission system with A/D and D/A converters.

1. Review of Introduction to PCM2. Binary Signal Detection in AWGN and Error

Probability of PCM System3. Multiplexing of PCM Signals4. Uniform Quantization and Signal-to-Noise Ratio 5. Differential Pulse-Code Modulation (DPCM) 6. Delta Modulation (DM)

1. Review of Introduction to PCM

)(ta )(taδ { }iasample quantize PCM

encode

binarycodeword

010 011 ..

Analog-to-Digital (A/D) conversion :

{ } signal. digital amplitude-discrete time,-discrete :signal discrete amplitude-continuous time,-discrete :)(signal analog amplitude-continuous time,- continuous :)(

iata

ta

δ

∑ −=n

siTttata )()()( δδ , Ts is the sampling interval.where

@G. Gong 4

Discretize: To discretize the amplitude of , the most convenient way is quantize the amplitude to L = 2n levels.

bits 2loglog 22 nnL ==

)( taδ

Then, each signal level can be represented by an n-bit codeword (a bit stream is called a codeword).

For example, an n = 3 bits digital signal can be transmitted to an L = 2n = 8 level PAM signal or be represented as a 3-bit codeword and transmitted using a 2 level PAM signal.

Level L 4-bit codeword0123456789101112121415

0000000100100011010001010110011110001001101010111100110111101111

01234567

000001010011100101110111

Level L 3-bit codeword

@G. Gong 6

The transmission of an L-level signal, obtained from the quantized signals, as an n-bit codeword, 2-level signal is referred as pulse-code modulation (PCM).

Pulse Code Modulation (PCM)

Procedure of PCM at Transmitter:

Step 1. The source information is sampled.

Step 2. Each sample is quantized to one of L levels.

Step 3. Each quantized amplitude is encoded into an n-bitcodeword.

Step 4. Each bit is represented by a baseband waveform for transmission.

1.3 3.6 2.3 0.7 -0.7 -2.4 -3.4Natural sample value

1.5Quantized sample value 3.5 2.5 0.5 -0.5 -2.5 -3.5Code number 5 7 6 4 3 1 0

PCM sequence 101 111 110 100 011 001 000

0Ts

2Ts 3Ts 4Ts 5Ts 6Ts 7Ts

Pulse waveform

4

Quantizationlever

00.5 11.5 2

34

2.53.5

-1-0.5-1.5

-2-2.533.5

Ts 2Ts 3Ts 4Ts 5Ts 6Ts 7Ts

Codenumber x(t) (v)

x(t)76543210

Example 1.

Transmission of PCM signals

Matchedfilter

PCMdecoder

Reconstructionfilter

Decisiondevice

snf=rate snf sf

XR

)(ˆ ta{ }ib { }iar(t)

a(t)δ(t) +

channel

r(t)

n(t)

Pulsegeneratorquantizer

PCMencodersampler

PCM sequence PCM waveform

Wf s 2≥ sf snf

{ }ia { }ib)(taδ

XT

ai: an L-level digital signal with L = 2n;bi: a binary sequence.Note. One data sample is represented by n bits. Thus the bit rate = nfs.

@G. Gong 9

Operation at Tx (transmitter):(1) Sampling (2) Quantizing(3) Encoding into binary codewords(4) Waveform representation of each bit of the

codeword.

Operation at Rx (receiver):(1) Matched filtering and equalization (2) Synchronization for correct sampling(3) Threshold decision to produce binary codeword sequence(4) Decoding to convert PCM to PAM (decoder-output)(5) Reconstruction the analog signal by pass PAM signal through

a low-pass filter with cutoff frequency equal to the message bandwidth.

@G. Gong 10

2. Binary Signal Detection in AWGN and Error Probability of PCM System

2. Binary Signal Detection in AWGN and Error Probability of PCM System

Three major sources of noise that effects the performance of PCM systems:

Intersymbol interference (in Chapter 4)

Channel noise, modeled as AWGN (in this section)

Quantization noise (in the next section)

@G. Gong 11

The effect of channel noise is to introduce transmission errors in reconstruction of the original PCM waveform at the receiver output. In other words,

1 mistaken for a symbol 00 mistaken for a symbol 1

Note. The following treatment is general, it can be applied to any binary waveform transmission. Objective: to compute the probability of transmission error for the optimal receiver. (In which sense is it optimal?)

Waveform selection: the possible waveforms for binary symbols 0 and 1.

where s(t) is a real function with duration T , i.e.,

This is referred to as the on-off signaling (or non-return-zero (NRZ) signaling).

⎩⎨⎧ ≤≤

=otherwise00

)( TtA

ts

)()(s10)(s0

1

0

tstt=↔=↔

(1) Examples of Signal Sets for Binary Data Transmission

Antipodal signal set:

Orthogonal signal set:

)()(s1)()(s0

1

0

tsttst

=↔−=↔

⎩⎨⎧ ≤<

=↔

⎩⎨⎧ ≤≤

=↔

otherwise ,02/2/ ,

)(s1

otherwise ,02/0 ,

)(s0

1

0

TtTAt

TtAt

Structure of Receiver: the above figure, LTI, which is the matched filter, is followed by a sampler and threshold comparator.

Receiver structure for detecting binary signals in AWGN

modulator si(t) +

decisiondevice

T

2)( with )( 0NfStn N =

)()()( tntstr i +=binary sequence

binary sequence

)()()(ariabledecision v

000 TnTsTY oio +=

detector:matched filter

h(t)

AWGN:

)(tY

(2) General Model

Channel noise process: the signal s only corrupted by the AWGN process n(t) with psd N0/2. (Here noise includes the thermal noise originated in the receiver itself).

Description of the Receiver Structure

2)( psd with noise additive is )( where

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

0NfStn

itntsn(t)str

n

ii

=

=+=+∗= δ

Received signal: due to the memoryless channel, the received signal

This is the input to the LTI filter.

The output of the filter (it will be the matched filter in the optimal case) is denoted by

Y(t) = sio(t) + no(t), i = 1, 2

where sio(t) is the output of the matched filter to the input si(t)no(t) is the output of the matched filter to the input n(t)

h(t)n(t)WSS

no(t)WSS

h(t)si(t) sio(t)

The filter is followed by a sampler. The output of the sampler is the random variable

)()()( 0000 TnTsTY io +=

The threshold device: Y(T0) is compared with a threshold value α.

Decision rule: if , decide “1” is sent

if , decide “0” is sent

α≥)( 0TYα<)( 0TY

@G. Gong 17

202)(

2)()()( fHNfHfSfS nn o

=⋅=

h (t)n(t)WSS

no(t)WSS

∫+∞

∞−== )]([)()( 0

22 TnEdffHfS on

The psd of the noise component no(t) is given by

The average power of no(t) is

∫+∞

∞−== dffStnEP

oo non )()]([ 2

Noise Component of the Filter Output

∫∫+∞

∞−

+∞

∞−== dtthNdffHN )(

2)(

2 2020

@G. Gong 18

Note. A random process X(t) is called a Gaussian process if at time t, X(t) is a Gaussian random variable, and for any (t1, t2, …, tn), the random variables X(t1), X(t2), …, X(tn) are jointly Gaussian.

Property 1: Both Y(t) and no(t) are gaussian processes (why?).

Decision variable: The decision is made based on the output of the detector Y(t) at time t = T0. We define it as a decision variable, i.e., the decision variable is given by

)()()( 000 TnTsTY oio +=

Question: From Property 1, both Y(T0) and no(T0) are gaussian random variables, what are their means and variances?

(If X is a gaussian random variable with mean μ and variance σ2, we denote it as ). ),(~ 2σμNX

Input to the Threshold Device

Case 1: suppose that 0 is sent. The decision variable is given by

We define a random process:

Then

The mean of

The variance of

)()()( 11 tntstY oo +=

)()()( 0000 TnTsTY oo +=

)()()( 00 tntstY oo +=

Case 2: suppose that 1 is sent. The decision variable is given by

We define a random process:

Then

The mean of

The variance of

)( 00 TY

)()]([)( 000000 TsTYET o==μ

)( 00 TY

)0()]([))(( 02000 onRTnETYVar ==

)()()( 0010 TnTsTY oo +=

ttYtY allfor ),()( 0= ttYtY allfor ),()( 1=

)( 01 TY

)()]([)( 010101 TsTYET o==μ

)( 01 TY

)0()]([))(( 02001 onRTnETYVar ==

Observation: The two decision variables have the same variance which is independent of which signal is sent. Conclusion: The only difference between two decision variables is the mean. It is expected that the ability of the decision device to discriminate between these two cases should depend on the difference of the means .)()( 0001 TT μμ −

@G. Gong 20

Property 2. We set . Then

and the decision variable Y(T0) has the following distributions:

)0(2onR=σ

1 and 0both for ,))(( 20 === iiTYVar i σ

( ) ed transmittis 1 if ),(~)()( 201010 σTsNTYTY o=

( ) ed transmittis 0 if ),(~)()( 20000 σTsNTYTY o=

A

t

s(t)

T

A

tT

)()( tTsth −=

⎪⎩

⎪⎨

⎧

≤<−≤≤

=otherwise0

2)2(0,

)( 2

2

1 TtTtTATttA

ts o

tts o allfor ,0)(0 =

Example. Let h(t) = s(T-t). For the on-off (or nonreturn-to-zero (NRZ) unipolar ) signaling, we have

T 2T

)()()(1 thtsts o ∗=

A2

t

The means and variances of the decision variables are given by

0)()]([)( 000000 === TsTYET oμ

⎩⎨⎧

<≤−<≤

=

==

TTTTTATTTA

TsTYET

2 ),(0 ,

)()]([)(

002

002

0010101μ

dtthNTnETYVar i

∫∞

∞−=

==

)(2

)]([))((

20

0200

2σ

TAN 20

2=

The difference of the means is given by )()()( 010001 TTT μμμ =−

We denote by

the probability that decision made by the receiver is wrong when 0 is sent

the probability that decision made by the receiver is wrong when 1 is sent

)0|(eP

)1|(eP

The receiver is wrong when 1 is sent if and only if

Thusα≤= )()( 010 TYTY

}1|)({ )1|( 0 α≤= TYPeP

})({ 01 α≤= TYP

]/))([( 01 σα Ts o−Φ=

]/))([( 01 σα−= TsQ o

The Error Probabilities

)1|(eP ]/))([( 01 σα−= TsQ o

The receiver is wrong when 0 is sent if and only if

Thusα>= )()( 000 TYTY

}0|)({ )0|( 0 α>= TYPeP

})({ 00 α>= TYP

]/))([(1 00 σα Ts o−Φ−=

})({1 00 α≤−= TYP

]/))([( 00 σα TsQ o−=

)0|(eP ]/))([( 00 σα TsQ o−=

@G. Gong 23

Let X = Y(T0) and be the pdf of X given that symbol i is transmitted, i = 0, 1. Then

)|( ixf X

( ) ),(ith variablerandom theof pdf theequals )1|( 201 σTsNwxf oX

( ) ),( with variablerandom theof pdf theequals )0|( 200 σTsNxf oX

Alternative way: Using conditional densities

@G. Gong 24

α

)1|(xf X

t

)0|(xf X

Error occurs in the following two cases:

(1) X ≥ α if 0 is transmitted

(2) X < α if 1 is transmitted

)( 01 Ts o)( 00 Ts o

TATsTtTtTATttA

ts oo2

12

2

1 )( otherwise0

2)2(0,

)( =⇒⎪⎩

⎪⎨

⎧

≤<−≤≤

=

TATT 210 )( and 0)( == μμ

Example 2. Continue the on-off signaling and h(t) = s(T-t) (called the matched filter). Determine the error probabilities for sampling time T0 = T.

TAdtth 22 )( =∫∞

∞−Note that

the energy of the signal, denoted as E. Thus the variance

ENTANRon 0

20

2

21

21)0( ===σ

]/)[()1|( 2 σα−= TAQeP

( )ENEQeP 0)2/1(/)()1|( α−=

]/[)0|( σαQeP =

( )ENQeP 0)2/1(/)0|( α=

2/2/)( 2 ETA ==αIf we set

then the expressions becomes

⎟⎟⎠

⎞⎜⎜⎝

⎛==

02)1|()0|(

NEQePeP

Solution.

Using the complementary error function,

⎟⎟⎠

⎞⎜⎜⎝

⎛==

021

21)1|()0|(

NEerfcePeP

@G. Gong 26

)0|( sent} is 0 {)1|( sent} is 1 {)( ePPePPeP +=

The average probability of error is defined as

The Average Probability of Error

If two signals are transmitted equally likely, i.e.,

21 sent} is 0 { sent} is 1 { == PP

then the average probability of error is given by

)(eP { ]/))([()2/1( 01 σα−= TsQ o }]/))([( 00 σα TsQ o−+

For the on-off signaling, the matched filter, the sampling time T, and the threshold 2/2/2/)( 2

1 ETATs o ===α

)(eP ⎟⎟⎠

⎞⎜⎜⎝

⎛=

02NEQ ⎟

⎟⎠

⎞⎜⎜⎝

⎛=

021

21

NEerfc

@G. Gong 27

(3) Optimization of the Threshold

For an AWGN channel, the optimal threshold in the sense of minimizing the error probabilities is given by

This threshold value is called the minimax threshold in the literature.For this threshold, the expression of error probability becomes

2/)]()([ 0100 TT μμα +=

)]2/())()([()0|()1|()( 0001 σμμ TTQePePePm −===

@G. Gong 28

(4) Matched Filter for the AWGN Channel

We define a signal-to-noise ratio as

Pm(e) becomes

)2/()]()([)( 0001 σμμ TTSNR o −=

])[()( om SNRQeP =

Note that

⇒

where

||||2))(())(()(

0

0001

hNThsThsSNR o

∗−∗=

))(()()( 000000 ThsTsT o ∗==μ))(()()( 010101 ThsTsT o ∗==μ

Goal: Find the optimum filter such that the error probability Pm(e) is minimized.

{ } { } 2/1

-

22/1

-

2 |)(|)(|||| ∫∫∞+

∞

∞+

∞== dffHdtthh

@G. Gong 29

Let us define

∫+∞

∞−= dfefHfSTs fTj

iio02

0 )()()( π

20 ||)(|| Thg ∗

Then

)()()( 01 tststg −=

02* )()( fTjefHfV π=

Then maximizing (SNR)o is equivalent to maximize

Note that

2*

222

0 )()()()(||)(*|| 0 ∫∫∞+

∞−

∞+

∞−== dffVfGdfefHfGThg fTj π

@G. Gong 30

∫∫∫∞+

∞−

∞+

∞−

∞+

∞≤ dffVdffGdffVfG

2*22

-

* )()()()(

Using the Schwarz’s inequality (Appendix G),

where equality holds if and only if G(f) = kV(f) where k is some constant. Thus, to maximum (SNR)o means that G(f) = kV(f) , i.e.,

02* )( )( fTjefGkfH π−=

Note that the value k does not change (SNR)o, so it can be free to choose. We conclude that the optimum filter has impulse response

This is called the match filter for the AWGN channel.

)]()([)()( 00010 tTstTstTgth −−−=−= λλ

@G. Gong 31

By a proper choice of the constant, we can write

We assume that

We now express the signal-to-noise ratio for the matched filter in terms of more fundamental parameters of the signal set.

2/||2/)(||)(

0

0

NtTgSNR o

−=

)]()([212/)()( 000100

tTstTstTgtgT −−−=−=

@G. Gong 32

We use the fact

where Ei is the energy of si, and ρ is the integral in the second term. We write

, The parameter E called the average energy for the signal set and ris the correlation coefficient (or normalized correlation) for the two signals s0 and s1.Thus

The signal-to-noise ratio for a receiver with the matched filter is therefore given by

∫+∞

∞−= dttgtg TT

22 )]([||)(|| 00 ∫

+∞

∞−−−−= dttTstTs 2

0001 )]()([41

[ ]∫ ∫+∞

∞−

∞

∞−−+= duususduusus )()(

21)()(

41

0120

21

ρ21)(

41 10 −+= EE

2/)1(||)(|| 20

rEtgT −=

2/)( 10 EEE += Er /ρ=

2/10}/)1({)( NrESNR o −=

@G. Gong 33

Observations:

Signal-to-noise ration does not depend on the sampling time T0 if the matched filter is used. The intuitive reason for this is that the matched filter, as defined by

automatically compensates for any changes in the sampling time. It should also noted that, because the noise is WSS, the variance of the output noise is independent of the sampling time. (This is true for any LTI filter.)

Signal-to-noise ratio and hence the error probabilities depend on two signal parameters only. These are the average energy in the two signals s0 and s1 and their inner product. The detailed structure of the signals are unimportant.

)]()([)( 0001 tTstTsth −−−= λ

@G. Gong 34

Example 3. Determine the signal-to-noise ratio and the error probability of the antipodal signal set for the matched filter receiver.

Solution.

The resulting signal-to-noise ratio is

and the error probabilities for the minimax threshold are given by

Question: How about the SNR and the error probabilities of the orthogonal signal set?

10 EEE ==

∫∞

∞−= dttsts )()( 01ρ 1

21 )]([ Edtts −=−= ∫

∞

∞−1 −=⇒ r

0/2)( NESNR o =

( )0/2)0|()1|( NEQePeP ==