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Chapter 6: Digital Modulation 1
Chapter 6: Digital Modulation
Most real communication channels have very poor response in theneighbourhood of zero frequency and hence can be regarded asbandpass channel.
Baseband digital signals can have frequency component as low as DC,so it will be grossly distorted if transmitted over a bandpass channel.
Digital modulation allows a baseband digital signal to be translated toa higher frequency range centering around a carrier frequency.
The higher frequency range is within the passband of the bandpasschannel, hence allowing minimum attenuation transmission.
1 0 1 1
tt
1 0 1 11 0 1 1
x(t)x
s(t)
0 0
X(f)
f f
Xs(f)
UnipolarNRZ
f c
f ccarrier freq f c
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Chapter 6: Digital Modulation 2
There are three basic Digital Modulation techniques :-
Amplitude Shift Keying (ASK)
Frequency Shift Keying (FSK)
Phase Shift Keying (PSK)
Modulation can be binary or Mary (more than twolevels).
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Chapter 6: Digital Modulation 3
x(t)
c(t) = sin 2 f ct
s(t)
x(t) is always a unipolar NRZ waveform for ASK.
0 T b0
V
0 -1 0
1
0 -V
0 V
2T b 3T b 4T b 5T b
1 0 1 1 0
c(t) = sin 2
f ct
x(t)
T b 2T b 3T b 4T b 5T b
s(t) BASK waveform
Generation of Binary Amplitude Shift Keying (BASK) signal
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Chapter 6: Digital Modulation 4
Amplitude Spectrum of BASK Assume x(t) is transmitting a long series of 10101010 .
Assume f c >> r b where r b is the bit rate of x(t).
|S(f)|
(Showing only the +ve freq)
4V
Amplitude spectrum of BASK
0 0 f
2V
2
rb
2rb
2r3
b2
r3b
|X(f)|
*
-f c 0 f c f
|C(f)| 1/2
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Chapter 6: Digital Modulation 5
Generation of a binary FSK (BFSK) signal involves generating twoBASK signals.
The first BASK signal x ask2 (t) is obtained by the unipolar NRZwaveform x(t) modulating a carrier frequency f 2. (Upper branch).
The second BASK signal x ask1 (t) is obtained by passing x(t) through
a NOT gate to obtain , and let modulates another
carrier frequency f 1. (Lower branch).s(t) = x ask2 (t) + x ask1 (t) is the required BFSK waveform.
Given a carrier frequency f c,
f 2 = f c + f d; f 1 = f c f d;where f d is called the frequency deviation.
Generation of Binary Frequency Shift Keying (BFSK) signal
x(t)
x
sin(2 f 2t)
x
sin(2 f 1t)
+x(t) s(t)
x(t)
xask2 (t)
xask1 (t)
x(t)
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Chapter 6: Digital Modulation 6
x
sin(2 f 2t)
x
sin(2 f 1t)
+x(t) s(t)
x(t)
xask2 (t)
xask1 (t)
0 T b0
V
0
0
0
0
0
0
0
2T b 3T b 4T b 5T bV
0
T b
T b
T b
T b
2T b
2T b
2T b
2T b
3T b
3T b
3T b
3T b
4T b
4T b
4T b
4T b
5T b
5T b
5T b
5T b
V
V
V
-V
-V
-V
x(t)
xask2 (t)
xask1 (t)
x(t)
s(t)
1 0 1 1 0
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Chapter 6: Digital Modulation 7
Amplitude Spectrum of BFSK Assume x(t) is transmitting a long series of 10101010 .
Assume f c >> r b where r b is the bit rate of x(t).
0 0 f
2V
2rb
2rb
2r3 b
2
r3b
|X(f)|
f2 -r b /2
|S(f)|
(Showing only the+ve freq)
4V
Amplitude spectrum of BFSK
f1 0 f
f1 +r b /2
f2 f1 -r b /2 f2 +r b /2
f2 -3r b /2 f1 +3r b /2
0 0 f
2V
2rb
2rb
2r3 b
2
r3b
|X(f)|
*
-f 2 -f 1 0 f 1 f 2 f
|C(f)|1/2*
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Chapter 6: Digital Modulation 8
x(t)
c(t) = sin 2 f ct
s(t)
x(t) is always a polar NRZ waveform for PSK.
Generation of Binary Phase Shift Keying (BPSK) signal
0 T b 2T b 3T b 4T b 5T b
1 0 1 1 0
c(t) = sin 2 f ct
x(t)
s(t) BPSK waveform
0 V
0 -1
0
1
0
0 T b 2T b 3T b 4T b 5T b
V
-V
-V
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Chapter 6: Digital Modulation 9
Amplitude Spectrum of BPSK Assume x(t) is transmitting a long series of 10101010 .
Assume f c >> r b where r b is the bit rate of x(t).
|S(f)|
(Showing only the +ve freq) Amplitude spectrum of BPSK
0 0 f
2
rb2rb
2r3 b
2
r3b
|X(f)|
*
-f c 0 f c f
|C(f)| 1/2
V
2
V
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Chapter 6: Digital Modulation 10
Optimal Receiver for Binary Digital Modulation Systems
The diagram below shows the overall diagram of a binary datatransmission over a bandpass channel.
The source is generating a binary bit stream b k , and representedelectrically by a line code.
The line code waveform modulates a carrier to produce a digitallymodulated signal (e.g. ASK / FSK / PSK) which is transmittedacross the bandpass channel.
Ideally, the demodulator reproduces a replica of the b k bit stream, butbecause of the additive white Gaussian noise (AWGN) introduced atthe channel, some of the bits will be different (bit error).
The function of the matched filter is to minimise these errors.
Modulator BandpassChannel
ThresholdDevice
Sampleevery T b sec
Demodulator
MatchedFilter
Carrier Clock
Carrier Clock
k b k b
AWGN
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Chapter 6: Digital Modulation 11
How to Design a Matched Filter for Digital Modulation Signal ?Example 1.1
Design a matched filter for the casewhen its input is an ASK waveform asshown below.
Solution:
Inputwaveform
of matchedfilter
0
V
-V
1 0 1 1 0
Basic waveforms:
s2(t)
0 V
-V
s1(t)
0
0 V
-V
0 V
-V
s2(Tb-t) s1(Tb-t)
s2(t)
s1(t)
s2(t) s1(t)
s2(-t) s1(-t)
{Impulse response h(t)0 T b
0
V
-V
0
V
0 -V
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Chapter 6: Digital Modulation 12
Therefore if the output of the bandpass channel feeding thematched filter is a BASK waveform of amplitude V volt,then the matched filter must have a characteristic where itsimpulse response is as shown:
MatchedFilter
(t)
0 t
Impulseresponse
h(t)
Once you know the impulse response characteristic of thematched filter, you can implement the matched filter circuitfrom the h(t) waveform using signal processing technique.
V0
-VTb t
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Chapter 6: Digital Modulation 13
Coherent BPSK System
Matched Filter
bk is a polar NRZ waveform.
The bandpass filter (BPF) limits the noise power entering the matched
filter.The matched filter is implemented by the Integrate and DumpCorrelation receiver.
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Chapter 6: Digital Modulation 14
Example 1.2
An integrate and dump correlation receiver is shown in Fig 1.2.1.
If its input is a BPSK waveform of amplitude V volt and f c = 2 r b,where f c is the carrier frequency and r b the bit rate, sketch thewaveforms at A to E for a 1101 sequence. Explain the operationsof SW1 and SW2. Assume an ideal noiseless bandpass channelwith infinite bandwidth. Assume r b = 1200 b/s.
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Chapter 6: Digital Modulation 15
Solution:
For binary sequence { 1 1 0 1}, BPSK waveform at A is:
1 1 0 1
t
-V
V
0
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Chapter 6: Digital Modulation 16
Waveform at B is based on the equation: s 2(t) s1(t) for each bit frame:
To sketch waveform B, repeats2(t) s1(t) waveform pattern for everybit frame.
s2(t) (binary 1)
s1(t) (binary 0)
s2(t) s1(t) :
V-V
2V
Bit
Duration
V-V
-2V
2Vt
1 1 0 1
Waveform B :-2V
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Chapter 6: Digital Modulation 17
Waveform C is the multiplication of Waveform A and Waveform B
A:
B:
-2V2
2V2
tC:
1 1 0 1
-V
Vt0
2Vt
-2V
0
0
Consider binary 1 frame:
t V V
t V
t V
c
c
c
2cos2
2cos12
sin2BA x
22
2
22
Consider binary 0 frame:
t V V
t V
t V
c
c
c
2cos2
2cos12
sin2BA x
22
2
22
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Chapter 6: Digital Modulation 18
Waveform D is the integration of Waveform C.
Consider an integrator circuit as shown in Fig (a) below during binary 1 frame:
)(t voD =
D varieslinearly with t
over 0 t Tb
Valueof D at
t = T b
Fig (a)
C = t V V t v ci 2cos)(22
K = circuit constant
D =
bT T
c
T
T
cc
T T
c
T
o
T KV t KV t KV t KV
t KV
t KV dt t V K dt V K t v
bbb
bbbb
2
0
2
0
25-
0
2
0
2
0
2
0
2
0
2
2sin10x3.3
2sin2
2cos)(
2V2
-2V2 KV2Tb
t
t
0C
D 0
-KV2Tb
1 1 0 1
Tb
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Chapter 6: Digital Modulation 19
SW1 is closed at the end of each bit duration for a very short duration tosample the D waveform. After sampling D, SW1 is opened again followed
by the short closure of SW2 to discharge the capacitor so that Dwaveform drops to zero to initialize the start of the next bit-framewaveform of D.
t
tD
= SW1 operation
= SW2 operation1 0 11
0E
KV2Tb
-KV2Tb
0
V
-V
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Chapter 6: Digital Modulation 20
Waveform A to E:
A:
B:
-2V2
2V2tC:
1 1 0 1
-V
Vt0
2Vt
-2V
0
0
t
tD
= SW1 operation
= SW2 operation1 0 11
0E
KV2Tb
-KV2Tb
0
V
-V
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Chapter 6: Digital Modulation 22
Solution :
Probability of bit error for a matched filter in general is given by:
8erfc
2
1
22erfc
2
1Pe
wherebT
dt t st s0
212
2 )]()([2
}Given in Formula ListDoc
22
21
c
b
b
b
c
c
T
T
T
r
T
f
Note:
From Example 1.2
s2(t) s1(t)
waveform is :-
2V
Tb
-2V
s2(t) s1(t) = 2V sin ct
Tc
b
cb
c
b
c
b
T
c
c
T
T T
c
T c
T
c
T
T V T
V T
T T
V t t
V
dt t dt V
dt t V
dt t V dt t st s
bb
b bb
bb
22
2
00
2
0 0
2
0
2
0
22
0
212
2
48sin
214
4sin
2
142sin
2
14
2cos4
22cos18
sin42
)]()([2
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Chapter 6: Digital Modulation 23
8erfc
21
22erfc
21
Pe
(from Formula List)
Substitute b
T V 24
into above equation:
2erfc
21
84
erfc21
8 / 4
erfc21
P
2
22
e
b
bb
T V
T V T V
i.e.
2erfc
21P
2
ebT V
for matched filter with BPSK input.
bb
T V T V 222 44
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Chapter 6: Digital Modulation 24
Differential Phase Shift Keying (DPSK)
In the coherent BPSK system mentioned earlier, it wasassumed that the carrier signal s 2(t) s1(t) at the integrateand dumped correlation receiver was synchronised with the
carrier of the incoming BPSK waveform in terms of frequencyand phase.
Also the two switches at the receiver that samples every T b second must synchronise with the bit frame of the BPSK waveform.
Synchronisation of carrier frequency and bit clock requiresextra overhead signalling if transmitted together with thetransmission signal or if generated locally would requirecomplex hardware at the receiver.
A DPSK system gets around the need for a coherent reference
signal by encoding the bit stream at the source before generatingthe BPSK signal. At the receiver, it uses the previous BPSK bit-framewave to correlate with the present BPSK bit-frame wave.
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Chapter 6: Digital Modulation 25
DPSK System
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Chapter 6: Digital Modulation 26
Example 1.4
Fig 1.4.1 shows the block diagram of a DPSK system. The binaryinput is in unipolar NRZ format of amplitude 2V volt, at a bit rateof 1200 b/s. The carrier is sin ct where c = 4800 rad/s.
Assume that the input is a long series of 10101010 . Assumethe receiver bandpass filter (BPF) has negligible effect on theshape of the received signal waveform. The receiver lowpassfilter is assumed ideal and has a cut-off frequency f co = r b /2 =600 Hz. Sketch the waveforms at points A to G as indicated in Fig1.4.1 for a 1010 frame. Assume distortionless transmission path.
Also assume that the encoder output is binary 1 prior to the 1010frame.
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Chapter 6: Digital Modulation 27
Fig 1.4.1
Transmitter
Receiver
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Chapter 6: Digital Modulation 28
Solution:
A has unipolar format; bit rate r b = 1200 b/s
sec 2400
1 PeriodCarrierHz240048002
rad/s48002
ccc
cc
T f f
f
sec 1200
1bT
Bdelay
One bit duration cb
b T r T 224001
x2120011
= two carrier periods
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Chapter 6: Digital Modulation 29
X Y Z(o/p)
0 0 1
0 1 0
1 0 0
1 1 1
X-NOR
Bdelay
-T b 0 T b0
2V
0 0
2V
0 0
2V
0
-V 0
V
0 -1
0 1
0 -V
0 V
-T b
-T b
-T b
-T b
-T b
T b
T b
T b
T b
T b
2T b
2T b
2T b
2T b
2T b
2T b
3T b
3T b
3T b
3T b
3T b
3T b
4T b
4T b
4T b
4T b
4T b
4T b
1 0 1 0 A
Bdelay
B
C
sinct
D
1 01
1 1
0
0 0
1
Transmitter:
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Chapter 6: Digital Modulation 30
Receiver:
Ddelay
If D and D delay has same polarity:
t V V
t V
t V
c
c
c
2cos22
2
2cos1
sinDxDE
22
2
22delay
If D and D delay has differentpolarity:
t V V
t V
t V
c
c
c
2cos22
22cos1
sinDxDE
22
2
22delay
-T b 0 -V
0
V
0
0
V
0
-V
V 2
-V 2
0
T b
T b
T b
2T b
2T b
2T b
3T b
3T b
3T b
4T b
4T b
4T b
D
Ddelay
E
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Chapter 6: Digital Modulation 31
0
0
V 2
0
0
K
-1 0
0
2V
-V 2
-K
Tb 2Tb 3Tb 4Tb
2Tb
23Tb
25Tb
27Tb
2Tb
23Tb
25Tb
2
7Tb
1 0 1 0
E
F
G
E waveform consists of a squarewave (cyan) and high freq (2 f c)phase-switching ripple added to it.
E passes through a LPF of cut-off freq r b /2 Hz which retains thefundamental freq of the cyansquare wave, but remove all otherhigher freqs.
Hence F is a sine wave of freq = r b /2 = 1200/2 = 600 Hz .
F is sampled at its highestexcursion value (centre of each bitinterval).
If F > 0, G = binary 1 (2V volt)else G = binary 0 ( 0 volt).
G bit stream is hence = A bit stream.
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Chapter 6: Digital Modulation 32
Probability of bit error for DPSK is:
2
TVexp
21
P b2
e
where is the single-sided power spectral density of thewhite channel noise. P e for DPSK is hence higher (worst) than
BPSK.Disadvantages of DPSK:
Asynchronous transmission is not possible because of theneed to synchronise the previous bit frame with the currentbit frame at the receiver correlator to allow multiplication .
Another minor error is that an error will propagate to theadjacent bit.
Example
Case 1: No error Case 2: Error
Received phase (D) 0 0 0 0 0 0 0
Correlation output (F) + + - + - - - - + -
Output sequence (G) 1 1 0 1 0 0 0 0 1 0
Red = Error ; Green = Propagated error
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Chapter 6: Digital Modulation 33
Quadrature Phase Shift Keying (QPSK)
In binary phase shift keying (BPSK):
Binary 0
carrier with 180 degree ( rad) phase shift.Binary 1 carrier with 0 degree (0 rad) phase shift.
In quadrature phase shift keying (QPSK), 2 bits are used to form4 possible phase shifts:
Binary-pair 00 carrier with 0 degree (0 rad) phase shift.
Binary-pair 01 carrier with 90 degree ( /2 rad) phase shift.
Binary-pair 11 carrier with 180 degree ( rad) phase shift.
Binary-pair 10 carrier with 270 degree (3 /2 rad) phase shift.
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Chapter 6: Digital Modulation 34
If the carrier frequency of the QPSK system is a cosine wave, the QPSK modulated signal can be expressed as:
"10"270
"11"180
"01"90"00"0
0)2cos(
)(
for
for
for for
T t t f A
t si
bic
i
where each dibit (2 bits) forms a phase-shifted carrier wavesymbol.
Hence the symbol rate r s for QPSK is half the bit rate (r b);i.e. r s = r b /2
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Chapter 6: Digital Modulation 35
Example 1.5
A QPSK modulation scheme can be described by the equation below:
"10"270
"11"180
"01"90"00"0
0)2cos(
)(
for
for
for for
T t t f A
t si
bic
i
If the input to the modulator is a binary bit stream: 00011011sketch the QPSK modulated waveform.
Solution:
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Chapter 6: Digital Modulation 36
Probability of Bit Error (P e) for QPSK
Consider the QPSK coding scheme shown earlier:
"10"270
"11"180
"01"90
"00"0 0)2cos(
)(
for
for
for
for T t t f A
t si
bic
i
It is highly unlikely for the channel noise to cause the phase toshift more than 90 degree.
So if the channel noise cause a phase transition of 90 degree, itwill incur only one bit error.
Therefore the bit error rate (BER) is similar to BPSK scheme.Hence P e for QPSK is similar to BPSK, as shown below:
} Dibit coded usingGray code
2
TVerfc
21
P b2
e
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Chapter 6: Digital Modulation 37
Comparison of QPSK with BPSK
As mentioned earlier, the symbol rate for QPSK is half that of the bit rate of BPSK.
A PSK signal bandwidth depends on the rate of phase change,the higher the phase change rate the larger the signalbandwidth.
Since QPSK phase change rate is half that of BPSK, it occupieshalf the bandwidth of BPSK signal.
If Gray code is used in coding QPSK dibits, the BER of QPSK issimilar to that of BPSK.
Important goals of digital communication systems is to make
signal with as small a bandwidth as possible so that more suchsignals can go through the same transmission channel throughmultiplexing. The signal modulation scheme should also allowlow BER.
Hence QPSK is preferred over BPSK in many applications.
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Chapter 6: Digital Modulation 38
Mary PSK (MPSK) In MPSK, the phase of the carrier takes one of M possible values,namely, i = 360i/M degree, where i = 0, 1, 2, , M -1 .
Accordingly, during each symbol interval of duration T s, one of Mpossible symbols:
1,.. .,2,1,0360
2cos)(
M i
M i
t f At s ci
For example for QPSK, M = 4 so that
3,2,1,0902cos)( iit f At s ci
For large M, although the transmission signal bandwidth issmall, P e is larger unless the signal increases in power.
More complex equipment is needed for larger M.
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Chapter 6: Digital Modulation 39
Quadrature Amplitude Modulation (QAM)This scheme is a combination of ASK and PSK, resulting in moreefficient (smaller) signal bandwidth and better P e if compared
with Mary PSK (M > 4). A constellation diagram for 16-QAM is shown below.Each symbol (a point in the diagram) uses 4 bits.The in-phase axis (x-axis) and the quadrature-phase axis (y-axis) each has 4 amplitude levels resulting in a combination of 16 symbols (indicated by 16 points in the diagram).
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Chapter 6: Digital Modulation 40
Comparison of Digital Modulation Systems
Choice of digital modulation methods is dependent mainly on errorperformance, bandwidth efficiency (in bps/Hz) and equipmentcomplexity.
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Chapter 6: Digital Modulation 41
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Chapter 6: Digital Modulation 42
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Applications
Cable Data ModemsDigital RadioDigital Communications by Satellite
Refer to your Bound Notes.