CS_2011_5_Digital_Modulators.pdf

Digital Modulators & Line Codes
Professor A. Manikas
Imperial College London
EE303 - Communication Systems
An Overview of Fundamental
Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes 2011 1 / 51

Table of Contents
1 Introduction
First Modelling of Digital Modulators
Examples of Binary Modulators
Second Modelling of Digital Modulators: Signal Constellation
Distance Between two M-ary signals
Examples of Signal Constellation Diagram
4ASK and QPSK Block Structures
2 QPSK: Comments
Examples of QPSK-Rx’
s Constellation Diagram
3 Performance Evaluation Criteria
4 Performance of Binary Digital Modulators/Demodulators
5 Line Codes
PSD(f) of "line-code" signals
Main Types of Line Codes and their PSD
Popular Line Codes
HDB3 Line Codes
Example: Autocorr. & PSD(f) of a Bipolar Line Code

Introduction
Introduction
General Block Structure of a Digital Communication System

Introduction
Let us focus on the "discrete channel"

Introduction
With reference the previous …gures:
at point T : s(t) waveform.
I The digital modulator
F takes γcs -bits at a time at some uniform rate rcs = 1
Tcs
and
F 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 ! s1
1 7 ! s2
i.e. a binary comm. system
I A new waveform (corresponding to a new γcs -bit-sequence) is
transmitted every Tcs seconds
at point b
T : noisy waveform r(t) = βs(t) + n(t).
I The transmitted waveform s(t), a¤ected by the channel, is received at point b
T
at point b
B2 : a binary sequence.
I based on the received signal r(t) the digital demodulator has to decide
which of the M waveforms si (t) has been transmitted in any given
time interval Tcs

Introduction
If M = 2 ) Binary Digital Modulator ) Binary Comm. System
If M > 2 ) M-ary Digital Modulator ) M-ary Comm. System
B2 b
B2
F =
2
6
6
6
4
Pr(D1jH1), Pr(D1jH2), . . . , Pr(D1jHM )
Pr(D2jH1), Pr(D2jH2), . . . , Pr(D2jHM )
.
.
.
.
.
. . . . ,
.
.
.
Pr(DK jH1), Pr(DK jH2), . . . , Pr(DK jHM )
3
7
7
7
5
. (1)

Introduction
Binary Com. Systems: use M = 2 possible waveforms
f
Tcs
!
s0(t),
Tcs
!
s1(t)g; Tcs = Tb (2)
M-ary Com. Systems: use M possible waveforms
fs1(t)
!
Tcs
, s2(t)
!
Tcs
, . . . , sM (t)
!
Tcs
g; Tcs = γcs Tb (3)
The M signals (or channel symbols) are characterized by their energy
Ei
Ei =
Z Tcs
0
s2
i (t) dt; 8i (4)
Furthermore their similarity (or dissimilarity) is characterized by their
cross-correlation
ρij =
1
p
Ei Ej
Z Tcs
0
si (t) sj (t) dt (5)

Introduction
B2
b
B2

Introduction First Modelling of Digital Modulators
First Modelling of Digital Modulators
Note: Transforming a sequence of 0’
s and 1’
s to a sequence of 1’
s
o/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 Tcs
The three basic binary digital modulators
I ASK (Amplitude Shift-Keyed)
8
<
:
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)
8
<
:
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)
8
<
:
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)
8
<
:
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)
8
>
>
>
>
>
<
>
>
>
>
>
:
0 7! so (t) =
=+Ac
z }| {
Ac exp(j0 ) exp(j2πFc t)
1 7! s1(t) = Ac exp(j180 )
| {z }
= Ac
exp(j2πFc t)
for 0 t Tcs
I FSK (Frequency Shift-Keyed)
8
<
:
0 7! so (t) = Ac exp(j2πF0t)
1 7! s1(t) = Ac exp(j2πF1t)
for 0 t Tcs

Communication Systems Classi…cation:
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 the
carrier, e.g. envelope detector)
Note: Optimum demodulators are coherent.

Introduction Second Modelling of Digital Modulators: Signal Constellation
Second Modelling of Digital Modulators: Signal
Constellation
We may represent the signal si (t) by a point (wsi
) in a
D-dimensional Euclidean space with D M.
The set of points (vectors) speci…ed by the columns of the matrix
W = ws1
, ws2
, . . . , wsM
(7)
is known as “signal constellation”.
Ei = wH
si
wsi
= wsi
2
(8)
ρij = 1
p
Ei Ej
wH
si
wsj
=
w H
si
w sj
kw si
k w sj
(9)

Distance Between two M-ary Signals
The distance between two signals si (t) and sj (t) is the Euclidean
distance between their associate vectors wsi
and wsj
i.e. dij = wsi
wsj
=
q
(wsi
wsj
)H (wsi
wsj
)
=
q
wH
si
wsi
+ wH
sj
wsj
2wH
si
wsj
=
q
Ei + Ej 2ρij
p
Ei Ej
) d2
ij = Ei + Ej 2ρij
p
Ei Ej
It is clear from the above that the Euclidean distance dij associated
with two signals si (t) and sj (t) indicates, like the cross-correlation
coe¢ cient, the similarity or dissimilarity of the signals.

Examples of Signal Constellation Diagram
Consider an M-ary System having the following signals
fs1(t), s2(t), . . . , sM (t)g with 0 t Tcs
M-ary ASK
I channel symbols: si (t) = Ai cos(2πFc t) where
Ai =given= 2i 1 M (say)
I dimensionality of signal space = D = 1
I if M = 4 then
s t
1( )
01
00 11 10
s t
2( ) s t
3( ) s t
4( )
Es1 Es2 Es3 Es4
Origin
I if M = 8 then
s t
1( )
000
Es1
001
s t
2( )
Es2
011
s t
3( )
Es3
010
s t
4( )
Es4
Origin
s t
5( )
110
Es5
s t
6( )
Es6
101
s t
7( )
Es7
100
s t
8( )
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 t
1( )
01
00
11
s t
2( )
s t
3( )
Es1
Es2
Es3 Origin
10
s t
4( )
Es4
s t
1( )
00
Es1
01
s t
2( )
Es2
11
s t
3( )
Es3
Origin
10
s t
4( )
Es4
450
Origin
s t
1( )
000
Es1
001
s t
2( )
Es2
011
s t
3( )
Es3
010
s t
4( )
Es4
s t
5( )
110
Es5
s t
6( )
Es6
101
s t
7( )
Es7
100
s t
8( )
Es8

M-ary FSK:
very di¢ cult to be represented using constellation diagram
with
fi fj = n 1
2Tcs
fi + fj = m 1
2Tcs
where
n, 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)
8
<
:
n = 1, 2, . . . , M1
m = 1, 2, . . . , M2
M = M1 M2
2 dimensionality of signal space = D = 2
PAM-PSK QAM

4ASK and QPSK Block Structures
An ASK (M=4) Communication System

A QPSK Communication System

QPSK: Comments
QPSK: Comments
A QPSK modulator has four “channel-symbols” which are described
by the following equation:
si (t) = A cos(2πFc t +
2π
M
(i 1) + φ) for i = 1, 2, 3, 4 (10)
with
M = 4 and 0 t Tcs
and the modulation process is described by the so called
“constellation diagram”.
The previous …gure (page 20) shows the constellation for φ = 45 .
From this …gure it is clear that the constellation diagram shows the
mapping of binary digits to QPSK channel symbols (constellation
points) as well as the square root of the energy
p
Es of the channel
symbols. The diagram also indicates a Gray code mapping from
binary digits to channel symbols (constellation points).

QPSK: Comments
Overall, using complex number representation, it is clear from the
constellation diagram that
00 7! s1(t) =
p
Es exp(j45 )
| {z }
=m1
exp(j2πFc t)
01 7! s2(t) =
p
Es exp(j135 )
| {z }
=m2
exp(j2πFc t)
11 7! s3(t) =
p
Es exp(j225 )
| {z }
=m3
exp(j2πFc t)
10 7! s4(t) =
p
Es exp(j315 )
| {z }
=m4
exp(j2πFc t)
(11)
In other words,
si (t) = mi exp(j2πFc t) for i = 1, 2, 3 and 4 (12)

QPSK: Comments
It is important to point out that the four symbols m1, m2, m3 and m4
are known as baseband QPSK “channel symbols” and are used
by the “QPSK constellation symbol mapping” block shown in the
previous …gure.
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 …gure given in page 20, the term exp(j2πFc t)
(see Equations 11 and 12) is shown at the Transmitter as the
up-conversion from baseband to bandpass.
In a similar fashion the down-conversion from baseband to bandpass
is shown at the receiver’
s front-end using the complex conjugate of
the transmitter’
s carrier, i.e. using exp( j2πFc t).
Thus, overall, we have exp(j2πFc t) exp( j2πFc t) = 1.

QPSK: Comments
Based on the above discussion it is clear that the presence of the
carrier does not a¤ect the analysis of the system.
Therefore, it is common practice to ignore the carrier when analyzing
communication systems, by working on the baseband.
For the rest of this explanatory note the carrier term will be ignored
from both Tx and Rx.

QPSK: Comments
Summarising
a QPSK modulator/demodulator is represented by its constellation
diagram and the QPSK symbol mapper transforms the binary
sequence to a sequence of QPSK complex channel symbols mi ,
forming the baseband QPSK message signal m(t) of bandwidth B
i.e.
m(t) = ∑
n
m1,m2,m3,m4
z}|{
a[n] c(t n Tcs ); nTcs t < (n + 1) Tcs
where
8
>
<
>
:
c(t) = rect
n
t
Tcs
o
fa[n]g = sequ. of independent data symbols (mi )
B = rcs
2 = 1
2Tcs

QPSK: Comments
Thus with reference to the binary sequence of bits “001001”
(message), by looking at the QPSK constellation diagram it is clear
that we have the following mapping
00 m1 = A exp(j45 )
10 m4 = A exp(j315 )
01 m2 = A exp(j135 )
where
A =
p
Es
Note that, for a binary system
m(t) ∑
n
1
z}|{
a[n] c(t n Tcs ); nTcs t < (n + 1) Tcs
where
8
>
<
>
:
c(t) =rect
n
t
Tcs
o
fa[n]g = sequ. of independent data bits ( 1s)
B = rcs
2 = 1
2Tcs

QPSK: Comments Examples of QPSK-Rx’
s Constellation Diagram
Examples of Plots of QPSK- Receiver’
s Constellation
Diagram
-1.5 -1 -0.5 0 0.5 1 1.5
-1.5
-1
-0.5
0
0.5
1
1.5
Re
Im
-8 -6 -4 -2 0 2 4 6 8
x 10
6
-8
-6
-4
-2
0
2
4
6
8
x 10
6
Re
Im
"good" "bad"

Performance Evaluation Criteria
In general the quality of a digital communication system is expressed
in terms of the accuracy with which the binary digits delivered at the
output of the detector represent the binary digits that were fed into
the digital modulator.
It is generally taken that it is the fraction of the binary digits that
are delivered back in error that is a measure of the quality of the
communication 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 by
means of the average probability of symbol error p e,cs , which, for
M > 2, is di¤erent than the average probability of bit error (or
Bit-Error-Rate BER), pe .
That is
pe,cs 6= pe for M > 2
pe,cs = pe for M = 2 (i.e. Binary Communication Systems)
However because we transmit binary data, the probability of bit error
ρe is a more natural parameter for performance evaluation than ρe,cs .

Although, these two probabilities are related
i.e
pe = ffpe,cs g
their relationship depends on the encoding approach which is
employed by the digital modulator for mapping binary digits to M-ary
signals (channel symbols)
where
γcs = log2 M
.

An Important Bound involving dij (distance between the ith and jth
constellation point):
The symbol error probability pe,cs is bounded as follows:
pe,cs
M
∑
i=1
Pr(si )
M
∑
j=1
j6=i
T
n
dij
p
2N0
o
(13)
de…nition of the “minimum distance”: dmin = min
8i,j
fdij g

Performance of Binary Digital Modulators/Demodulators
Performance of Binary (equiprobable) Digital
Modulators/Demodulators
Consider a Binary Communication system
0 7! s1
1 7! s2
or a more popular notation:
0 7! s0
1 7! s1
Constellation diagram:
0 1
s0 s1
p
E0
p
E1
pe =T
n
d
p
2N0
o
= . . . =T
np
(1 ρ)EUE
o

Thus, at the output of an optimum digital demodulator the probability
of error can be calculated by using the following expression:
pe = T
q
(1 ρ)EUE (14)
where EUE= Eb
N0
and PSDni (f ) = N0
2
with
8
>
>
>
>
>
<
>
>
>
>
>
:
Eb = 1
2
R Tcs
0
(s0(t)2 + s1(t)2)dt
= average signal energy
ρ = 1
Eb
R Tcs
0
s0(t)s1(t)dt
= the time cross-correlation between signals

N.B.:
(1 ρ)Eb
N0
= " =) pe = #
if Eb
N0
= …xed then the optimum system is that for which the
correlation coe¤ is -1
i.e. ρ = 1 =) s0(t) = s1(t)
This is known as optimum, or
ideal binary Communication System

Examples
Baseband MODEMS:
Antipodal!
0 7! s0(t) = Ac
1 7! s1(t) = Ac
0 t Tcs
s(t) = Ac m(t) (fa[n]g = sequ. of independent data bits ( 1s))
pe =T
n
Ac
σ
o
COHERENT MODEMS:
1. Amplitude Shift-Keyed (ASK) or On-O¤ Keying (OOK)
I (ASK or OOK) !
0 7! s0(t) = 0
1 7! s1(t) = Ac cos(2πFc t)
0 t Tcs
I s(t) = Ac m(t) cos(2πFc t)
I pe =T
q
Es1
2N0

2. Biphase Shift-Keyed:
I general !
0 7! s0(t) = Ac cos(2πFc t ∆θ)
1 7! s1(t) = Ac cos(2πFc t + ∆θ)
0 t Tcs
pe =T
q
2 EUE sin2(∆θ)
I Phase-Reversal Keying (RSK)
(RSK) !
0 7! s0(t) = Ac sin(2πFc t)
1 7! s1(t) = Ac sin(2πFc t)
0 t Tcs
pe =T
np
2 EUE
o

N.B.: for ∆θ = π
2 then general = RSK and it is called BPSK
BPSK s(t) = Ac sin(2πFc t + m(t) π
2 ) (15)
Equation 15 can be written as follows:
BPSK s(t) = Ac m(t) cos(2πFc t) (16)
) BPSK can be considered as
PM
AM
The PSD(f )’
s of m(t) and s(t) are shown below:

3. Frequency Shift-Keyed (FSK)
(FSK) :
0 7! s0(t) = Ac cos(2πFc t)
1 7! s1(t) = Ac cos(2π(Fc + ∆f )t)
0 t Tcs , ∆f = m
2Tcs
coherent
z }| {
CFSK : pe =T
np
EUE
o
discontinuous FSC

continuous FSC

1 ASK pe =T
nq
1
2 EUE
o
2 non-coherent ASK pe = 0.5 exp(
Es1
4N0
) + 0.5T
q
Es1
2N0
3 FSK pe =T
np
EUE
o
4 FSK (non-coherent) pe = 1
2 exp 1
2 EUE
5 BPSK pe =T
np
2EUE
o
6 BPSK (di¤erential) pe = 1
2 exp f EUEg
7 QPSK pe =T
np
2EUE
o
8 MSK pe =T
np
1.7EUE
o
9 Gaussian MSK pe T
np
1.36EUE
o
10 M-ary PSK (coherent) pe 2T
np
4EUE sin( π
2M )
o
11 M-ary QAM pe 4(1 1
p
M
)T
nq
3
M 1 EUE
o
12 DS-BPSK and DS-QPSK SSS pe =T
p
2EUEequ

Line Codes PSD(f) of "line-code" signals
Line Codes
PSD(f) of Line Code Signals
de…nitions:
s(t) = ∑
n
a[n]c(t nTcs ); nTcs < t < (n + 1)Tcs (17)
where
I c(t) is an energy signal of duration Tcs . This implies that c(t nTcs )
is de…ned in the interval nTcs < t < (n + 1)Tcs
I fa[n]g is a binary sequence having an autocorrelation function
Raa[k] = E fa[n]a[n + k]g
=
I
∑
i=1
(a[n]a[n + k])ith-pair . Pr(ith
-pair)
where I denotes the total number of "pairs"
I Insert Figure

PSDs (f ) = jFT(c(t))j2
Tcs
2
4Raa[0] +
+∞
∑
k= ∞
k6=0
Raa[k] exp( j2πfkTcs )
3
5 (18)
Note that if Raa[k] =
E a2
n for k = 0
E fang E fan+k g for k 6= 0
i.e. Raa[k] =
µ2
a + σ2
a for k = 0 where µa = mean and σa = std
µ2
a for k 6= 0
then
(18) )PSDs (f ) = σ2
a
jFT(c(t))j2
Tcs
| {z }
Continuous Spectrum
+
µ2
a
T2
cs
comb 1
Tcs
(jFT(c(t))j2
)
| {z }
Discrete Spectrum

EXAMPLE. . . FOR YOU. . .
if an = 0, 1 with Pr fan = 0g = Pr fan = 1g = 1
2 …nd the spectrum
of “unipolar RZ” line-code signal.

Line Codes Main Types of Line Codes and their PSD
Main Types of Line Codes and their PSD

Line Codes Popular Line Codes
Popular Line Codes

Line Codes HDB3 Line Codes
HDB3 Line Codes

Line Codes 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 the
code sequence fa[n]g is as follows:
Raa[k] =
8
<
:
1/2 if k = 0
1/4 if k = 1
0 if k 2
(19)
2 If Pr(0) = Pr( 1) = 0.5 and c(t) =rect
n
t
Tcs
o
, derive an expression
for the power spectral density PSD(f ) for the bipolar line code
waveform
m(t) =
∞
∑
n= ∞
a[n] c(t nTcs ) (20)

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
=
1
2
(21)
Raa[1] = E fa[n]a[n + 1]g =
8
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
:
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
9
>
>
>
>
>
>
>
=
>
>
>
>
>
>
>
;
= 0
1
4
+ 0
1
4
+ 0
1
4
+ ( 1)
1
4
=
1
4
(22)

Solution: Autocorrelation R[k],k>1
Raa[k]
with k>1
=
8
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
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
9
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
=
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
;
= (ignoring the 2nd column and multiplying 1st with 3rd we have)
= 0
6
8
+ 1
1
8
+ ( 1)
1
8
= 0 (23)

Solution: PSD(f)
PSD(f ) =
FT(rect
n
t
Tcs
o
)
2
Tcs
fR[0] + 2R[k] cos(2πkTcs )g
=
T2
cs sinc2
(fTcs )
Tcs
1
2
2
1
4
cos(2πTcs )
= Tcs sinc2
(fTcs )
1
2
1
2
cos(2πTcs )
= Tcs sinc2
(fTcs ) sin2 2πTcs
2
(24)

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