ee303: communication systems · chair of communications and array processing imperial college...

EE303: Communication Systems

Professor A. ManikasChair of Communications and Array Processing

Imperial College London

An Overview of Fundamentals: Digital Modulators and Line Codes

Prof. A. Manikas (Imperial College) EE303: Digital Modulators and Line Codes v.18 1 / 54

Table of Contents1 Introduction2 First Modelling of Digital Modulators

Examples of Binary Modulators

3 Second Modelling of Digital ModulatorsSignal ConstellationDistance Between two M-ary signalsConstellation Diagram of Main Modems4ASK and QPSK Block StructuresQPSK: CommentsExamples of QPSK-Rx’s Constellation Diagram

4 Performance Evaluation Criteria5 Line Codes (Wireline Digital Communications)

IntroductionMain Types of Line Codes and their PSDPopular Line CodesExamples of Using Line Codes:Example-1: Connecting Systems with RS232Example-2: PSTN and HDB3 Line Codes

PSD(f) of "line-code" signalsExample: Autocorr. & PSD(f) of a Bipolar Line Code


Introduction

IntroductionGeneral Block Structure of a Digital Communication System


Introduction

Let us focus on the "discrete channel"


Introduction

With reference the previous figures:

at point T : s(t) waveform.I The digital modulator

F takes γcs -bits at a time at some uniform rate rcs = 1Tcs

andF transmits one of M = 2γcs distinct waveforms s1(t), . . . , sM (t)i.e. we have an M -ary communication system.

F If γcs = 1 we have one bit at a time{0 7−→ s11 7−→ s2

i.e. a binary comm. system

I A new waveform (corresponding to a new γcs -bit-sequence) istransmitted every Tcs seconds

at point T̂ : noisy waveform r(t) = βs(t) + n(t).

I The transmitted waveform s(t), affected by the channel, is received at point T̂

at point B̂2 : a binary sequence.I based on the received signal r(t) the digital demodulator has to decidewhich of the M waveforms si (t) has been transmitted in any giventime interval Tcs


Introduction

Binary Comm Systems: use M = 2 possible waveforms

{Tcs←−→s0(t),

Tcs←−→s1(t)};Tcs = Tb (2)

M-ary Comm Systems: use M possible waveforms

{s1(t)←−→Tcs

, s2(t)←−→Tcs

, . . . , sM (t)←−→Tcs

}; Tcs = γcsTb (3)

The M signals (or channel symbols) are characterized by their energyEi

Ei =∫ Tcs

0s2i (t) · dt; ∀i (4)

Furthermore their similarity (or dissimilarity) is characterized by theircross-correlation

ρij =1√EiEj

∫ Tcs

0si (t) · s∗j (t) · dt (5)


Introduction

B2

B̂2


First Modelling of Digital Modulators


Note: Transforming a sequence of 0’s and 1’s to a sequence of ±1’so/p = 1− 2×i/p (6)



Examples of Binary Modulators

A binary digital modulator maps 0’s and 1’s onto two analogue

symbol-waveforms s0(t) and s1(t), that is{0 7→ so (t)1 7→ s1(t)

0 ≤ t ≤ TcsThe three basic binary digital modulators

I ASK (Amplitude Shift-Keyed) 0 7→ so (t) = A0 · cos(2πFc t)1 7→ s1(t) = A1 · cos(2πFc t)for 0 ≤ t ≤ Tcs

I PSK (Phase Shift-Keyed) 0 7→ so (t) = Ac · cos(2πFc t)1 7→ s1(t) = Ac · cos(2πFc t + 180◦)for 0 ≤ t ≤ Tcs

I FSK (Frequency Shift-Keyed) 0 7→ so (t) = Ac · cos(2πF0t)1 7→ s1(t) = Ac · cos(2πF1t)for 0 ≤ t ≤ Tcs



The above binary digital modulators using complex representation:

I ASK (Amplitude Shift-Keyed)

0 7→ so (t) = A0 · exp(j2πFc t)1 7→ s1(t) = A1 · exp(j2πFc t)for 0 ≤ t ≤ Tcs

I PSK (Phase Shift-Keyed)0 7→ so (t) =

=+Ac︷︸︸︷Ac exp(j0◦) · exp(j2πFc t)

1 7→ s1(t) = Ac exp(j180◦)︸︷︷︸

=−Ac

· exp(j2πFc t)

for 0 ≤ t ≤ Tcs

I FSK (Frequency Shift-Keyed)

0 7→ so (t) = Ac · exp(j2πF0t)1 7→ s1(t) = Ac · exp(j2πF1t)for 0 ≤ t ≤ Tcs



Communication Systems Classification:1 Coherent (if demodulator is coherent, i.e. it uses a copy of the carrier)2 Non-coherent (if demodulator is non-coherent, e.g. it does not use thecarrier, e.g. envelope detector)

Note: Optimum demodulators are coherent.


Second Modelling of Digital Modulators Signal Constellation

Second Modelling of Digital ModulatorsSignal Constellation

We may represent the signal si (t) by a point w si in a D-dimensionalEuclidean space with D ≤ M.The set of points (vectors) specified by the columns of the matrix

W =[w s1 , w s2 , . . . , w sM

](7)

is known as “signal constellation”.

Ei = wHsi w si =∥∥w si∥∥2 (8)

ρij =1√EiEj

wHsi w sj =wHsi w sj

‖w si ‖∥∥∥w sj ∥∥∥ (9)


Second Modelling of Digital Modulators Distance Between two M -ary signals

Distance Between two M-ary Signals

The distance between two signals si (t) and sj (t) is the Euclideandistance between their associate vectors w si and w sj

i.e. dij =∥∥∥w si − w sj∥∥∥

=√(w si − w sj )H (w si − w sj )

=√wHsi w si + w

Hsj w sj − 2wHsi w sj

=√Ei + Ej − 2ρij

√EiEj

⇒ d2ij = Ei + Ej − 2ρij√EiEj

It is clear from the above that the Euclidean distance dij associatedwith two signals si (t) and sj (t) indicates, like the cross-correlationcoeffi cient, the similarity or dissimilarity of the signals.


Second Modelling of Digital Modulators Constellation Diagram of Main Modems

Constellation Diagram of Main ModemsConsider an M-ary System having the following signals

{s1(t), s2(t), . . . , sM (t)} with 0 ≤ t ≤ Tcs

M-ary ASKI channel symbols: si (t) = Ai · cos(2πFc t) whereAi =given= 2i − 1−M (say)

I dimensionality of signal space = D = 1I if M = 4 then

s t1( )0100 11 10s t2( ) s t3( ) s t4( )

Es1 Es2 Es3 Es4Origin

I if M = 8 then

s t1( )000

Es1

001s t2( )

Es2

011s t3( )

Es3

010s t4( )

Es4Origin

s t5( )110

Es5

s t6( )

Es6

101s t7( )

Es7

100s t8( )

Es8



M-ary PSK

I channel symbols: si (t) = A. cos(2πFc t + 2π

M . (i − 1) + φ)

for i = 1, 2, ..,I dimensionality of signal-space = D = 2

M = 4 & φ = 0◦ M = 4 & φ = 45◦ M = 8 & φ = 0◦

s t1( )

01

0011

s t2( )

s t3( )

Es1

Es2

Es3 Origin

10s t4( )

Es4

s t1( )00Es1

01s t2( )

Es2

11s t3( )

Es3

Origin

10s t4( )

Es4

450Origin

s t1( )000 Es1

001s t2( )

Es2

011s t3( )

Es3

010s t4( )

Es4

s t5( )110Es5

s t6( )

Es6

101s t7( )

Es7

100s t8( )

Es8



M-ary FSK:very diffi cult to be represented using constellation diagram

with

fi − fj = n 12Tcs

fi + fj = m 12Tcs

wheren,m = integers



M-ary Amplitude & Phase - M-ary QAM:1 channel symbols: si (t) = Ai · cos(2πFc t + ϕi + φ) i = 1, 2, . . . ,M

or snm(t) = An · cos(2πFc t + ϕm + φ)

n = 1, 2, . . . ,M1m = 1, 2, . . . ,M2M = M1 ×M2

2 dimensionality of signal space = D = 2PAM-PSK QAM


Second Modelling of Digital Modulators 4ASK and QPSK Block Structures

4ASK and QPSK Block StructuresAn ASK (M=4) Communication System


Second Modelling of Digital Modulators 4ASK and QPSK Block Structures

A QPSK Communication System


Second Modelling of Digital Modulators QPSK: Comments

QPSK: CommentsA QPSK modulator has four “channel-symbols”which are describedby the following equation:

si (t) = A cos(2πFc t +2π

M(i − 1) + φ) for i = 1, 2, 3, 4 (10)

withM = 4 and 0 ≤ t ≤ Tcs

and the modulation process is described by the so called“constellation diagram”.

The previous figure (page 20) shows the constellation for φ = 45◦.

From this figure it is clear that the constellation diagram shows themapping of binary digits to QPSK channel symbols (constellationpoints) as well as the square root of the energy

√Es of the channel

symbols. The diagram also indicates a Gray code mapping frombinary digits to channel symbols (constellation points).



Overall, using complex number representation, it is clear from theconstellation diagram that

00 7→ s1(t) =√Es exp(j45◦)︸︷︷︸

=m1

exp(j2πFc t)

01 7→ s2(t) =√Es exp(j135◦)︸︷︷︸

=m2

exp(j2πFc t)

11 7→ s3(t) =√Es exp(j225◦)︸︷︷︸

=m3

exp(j2πFc t)

10 7→ s4(t) =√Es exp(j315◦)︸︷︷︸

=m4

exp(j2πFc t)

(11)

In other words,

si (t) = mi exp(j2πFc t) for i = 1, 2, 3 and 4 (12)



It is important to point out that the four symbols m1,m2,m3 and m4are known as baseband QPSK “channel symbols” and are usedby the “QPSK constellation symbol mapping”block shown in theprevious figure.

For instance, if the bit-pair at the input of the QPSK modulator is“01” then the output is the baseband channel symbol m2.

With reference to the figure given in page 20, the term exp(j2πFc t)(see Equations 11 and 12) is shown at the Transmitter as theup-conversion from baseband to bandpass.

In a similar fashion the down-conversion from baseband to bandpassis shown at the receiver’s front-end using the complex conjugate ofthe transmitter’s carrier, i.e. using exp(−j2πFc t).

Thus, overall, we have exp(j2πFc t) exp(−j2πFc t) = 1.



Based on the above discussion it is clear that the presence of thecarrier does not affect the analysis of the system.

Therefore, it is common practice to ignore the carrier when analyzingcommunication systems, by working on the baseband.

For the rest of this explanatory note the carrier term will be ignoredfrom both Tx and Rx.



Summarising

a QPSK modulator/demodulator is represented by its constellationdiagram and the QPSK symbol mapper transforms the binarysequence to a sequence of QPSK complex channel symbols mi ,forming the baseband QPSK message signal m(t) of bandwidth Bi.e.

m(t) = ∑n

m1,m2,m3,m4︷︸︸︷a[n] · c(t − n · Tcs ); nTcs ≤ t < (n+ 1) · Tcs

where c(t) = rect

{tTcs

}{a[n]} = sequ. of independent data symbols (mi )B = rcs

2 =1

2Tcs



Thus with reference to the binary sequence of bits “001001”(message), by looking at the QPSK constellation diagram it is clearthat we have the following mapping

00 m1 = A exp(j45◦)10 m4 = A exp(j315◦)01 m2 = A exp(j135◦)

whereA =√Es

Note that, for a binary system

m(t) ≡∑n

±1︷︸︸︷a[n] · c(t − n · Tcs ); nTcs ≤ t < (n+ 1) · Tcs

where c(t) =rect

{tTcs

}{a[n]} = sequ. of independent data bits (±1s)B = rcs

2 =1

2Tcs


Second Modelling of Digital Modulators Examples of QPSK-Rx’s Constellation Diagram

Examples of Plots of QPSK- Receiver’s ConstellationDiagram

1.5 1 0.5 0 0.5 1 1.51.5

1

0.5

0

0.5

1

1.5

Re

Im

8 6 4 2 0 2 4 6 8

x 106

8

6

4

2

0

2

4

6

8x 106

ReIm

"good" "bad"


Performance Evaluation Criteria


In general the quality of a digital communication system is expressedin terms of the accuracy with which the binary digits delivered at theoutput of the detector represent the binary digits that were fed intothe digital modulator.

It is generally taken that it is the fraction of the binary digits thatare delivered back in error that is a measure of the quality of thecommunication system.

This fraction, or rate, is referred to as the bit error probability p e ,or, Bit-Error-Rate BER.



The performance of M-ary communication systems is evaluated bymeans of the average probability of symbol error p e,cs , which, forM > 2, is different than the average probability of bit error (orBit-Error-Rate BER), pe .

That is{pe ,cs 6= pe for M > 2pe ,cs = pe for M = 2 (i.e. Binary Communication Systems)

However because we transmit binary data, the probability of bit errorρe is a more natural parameter for performance evaluation than ρe ,cs .



Although, these two probabilities are relatedi.e.

pe = f{pe ,cs}their relationship depends on the encoding approach which isemployed by the digital modulator for mapping binary digits to M-arysignals (channel symbols)

whereγcs = log2M


Line Codes (Wireline Digital Communications) Introduction

Introduction: Lines Codes (Wireline Digital Comms)Line codes are used for data transmission in a metallic wire, twistedpair, cable.Line coding is baseband signal transmission having:

I a Tx (Digital Modulator1 without carrier)I a cable (communication channel)I a Rx (Digital Demodulator1 without carrier).

Appropriate coding of the transmitted symbol pulse shape canminimise the probability of error by ensuring that the spectralcharacteristics of the digital signal are well matched to thetransmission channel.The line code must also permit the receiver to extract accurate PCMbit and word timing signals directly from the received data, for properdetection and interpretation of the digital pulses.for more info see Chapter 6 in [Ian Glover and Peter Grant, "DigitalCommunications", 3rd edition,Pearson Education Limited, 2010]

1Commonly used names: Tx="Line Encoder" and Rx="Line Decoder".Prof. A. Manikas (Imperial College) EE303: Digital Modulators and Line Codes v.18 31 / 54

Line Codes (Wireline Digital Communications) Main Types of Line Codes and their PSD

Main Types of Line Codes and their PSD



Main Types of Line Codes and their PSD (cont.)


Line Codes (Wireline Digital Communications) Popular Line Codes

Popular Line Codes


Line Codes (Wireline Digital Communications) Popular Line Codes

Popular Line Codes (cont.)


Line Codes (Wireline Digital Communications) Examples of Using Line Codes:

Example-1: Connecting Systems

Example (1)

3rd system: H3(f )

H3(f ) =1

1+ j2πfN.B.: This is a Low Pass Filter



Example-1: Connecting Systems (cont.)

Example (1)

3rd system: H3(f )

H3(f ) =A(f )

1+ A(f ).B(f )NB: This is a Control System




H(f ) = H1(f ).H2(f ).H3(f )

h(t) = h1(t) ? h2(t) ? h3(t)

the dot-symbol denotes "multiplication" and ? denotes "convolution".

These days we have "system-on-chip" (SOC)



Example-1: Connecting SystemsConnecting Systems - Far Apart

Wire Connection? No (due to effects equivalent to "Friction")

Wireline Comm Connection? Yes



Example-1: Connecting Systems (cont.)Connecting Systems with RS232 Data Port

It establishes a two-way (i.e. "full-duplex") data communicationchannel using a single cable of length up to 15m. It uses:

I the pin 2 to Tx dataI the pin 3 to Rx data



Transmitted signal corresponding to the letter "A", using datatransmission over a cable (RS232 data interface/port)

N.B.:one byte of asynchronous transmission




Note: The RS232 is a wireline communication system connecting"smart" devices, e.g. Computers, CPU, PLC (Programmable LogicControllers), etc.:



Example-2: PSTN and HDB3 Line Codes

Example :



Example-2: PSTN and HDB3 Line Codes (cont.)

HDB3 is used in Europe (and many other countries) in PublicSwitched Telephone Networks (PSTN) according to the 2nd CCITTrecommendation (32 channel PCM)

it is used to transmit a frame of duration 1Fs= Ts = 125µs,

containing data from:I 30 users (one 8 bit word per user),I one Frame Alignment 8-bit word, andI one signalling info 8-bit word

using an HDB3 line encoder at the Tx and HDB3 line decoder in theRx.



Example-2: PSTN and HDB3 Line Codes (cont.)Single-Channel Path of 2nd CCITT rec. (32-channels PCM)


Line Codes (Wireline Digital Communications) PSD(f) of "line-code" signals

PSD(f) of Line Code SignalsSignals - Definition:

s(t) = ∑na[n]c(t − nTcs ); nTcs < t < (n+ 1)Tcs (13)

whereI c(t) is an energy signal of duration Tcs . This implies that c(t − nTcs )is defined in the interval nTcs < t < (n+ 1)Tcs

I {a[n]} is a binary sequenceAutocorrelation function of a binary sequence {a[n]}

Raa[k ] = E {a[n]a[n+ k ]} (14)

=I

∑i=1(a[n]a[n+ k ])i th-pair .Pr(i

th-pair) (15)

I where I denotes the total number of "pairs"



PSDs (f ) =|FT(c (t))|2

Tcs·

Raa[0] ++∞∑

k=−∞k 6=0

Raa[k ] exp(−j2πfkTcs )

(16)

Note that if Raa[k ] ={E{

a2n}for k = 0

E {an} · E {an+k} for k 6= 0

i.e. Raa[k ] ={

µ2a + σ2a for k = 0 where µa = mean and σa = stdµ2a for k 6= 0

then

(16)⇒PSDs (f ) = σ2a|FT(c(t))|2

Tcs︸︷︷︸Continuous Spectrum

+µ2aT 2cs· comb 1

Tcs(|FT(c(t))|2)︸︷︷︸

Discrete Spectrum



EXAMPLE. . . FOR YOU. . .if an = 0, 1 with Pr {an = 0} = Pr {an = 1} = 1

2 find the spectrumof “unipolar RZ” line-code signal.


Line Codes (Wireline Digital Communications) Example: Autocorr. & PSD(f) of a Bipolar Line Code

Example: Autocorr. & PSD(f) of a Bipolar Line Code

1 Show that for a bipolar line code the autocorrelation function of thecode sequence {a[n]} is as follows:

Raa[k ] =

1/2 if k = 0−1/4 if k = 10 if k ≥ 2

(17)

2 If Pr(0) = Pr(±1) = 0.5 and c(t) =rect{

tTcs

}, derive an expression

for the power spectral density PSD(f ) for the bipolar line codewaveform

m(t) =∞

∑n=−∞

a[n] · c(t − nTcs ) (18)



Solution: Autocorrelation R[0], R[1]

Raa[0] = E{

a[n]2}=

{0× 0 = 0→ 1

2(±1)× (±1) = +1→ 1

2

}= 0× 1

2+ 1× 1

2

=12

(19)

Raa[1] = E {a[n]a[n+ 1]} =

1st 2nd 1st×2rd Pr

0, 0, = 0 1/4

0, ±1, = 0 1/4

±1, 0, = 0 1/4

±1, ∓1, = −1 1/4

= 0× 1

4+ 0× 1

4+ 0× 1

4+ (−1)× 1

4

= −14

(20)



Solution: Autocorrelation R[k],k>1

Raa[k ]with k>1

=

1st 2nd 3rd 1st×3rd Pr

0, 0, 0 = 0 1/8

0, 0, ±1 = 0 1/8

0, ±1, 0 = 0 1/8

0, ±1, ∓1 = 0 1/8

±1, 0, 0 = 0 1/8

±1, 0, ∓1 = −1 1/8

±1, ∓1, 0 = 0 1/8

±1, ∓1, ±1 = +1 1/8

= (ignoring the 2nd column and multiplying 1st with 3rd we have)

= 0× 68+ 1× 1

8+ (−1)× 1

8= 0 (21)



Solution: PSD(f)

PSD(f ) =

∣∣∣FT(rect{ tTcs

})∣∣∣2

Tcs{R [0] + 2R [k ] cos(2πkTcs )}

=T 2cs sinc

2(fTcs )Tcs

{12− 2× 1

4cos(2πTcs )

}

= Tcs sinc2(fTcs ){12− 12cos(2πTcs )

}

= Tcs sinc2(fTcs ) · sin2(2πTcs2

)(22)


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