chapter 6 digital modulation

7/31/2019 Chapter 6 Digital Modulation

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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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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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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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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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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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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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Amplitude Spectrum of BFSK Assume x(t) is transmitting a long series of 10101010 .


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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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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Amplitude Spectrum of BPSK Assume x(t) is transmitting a long series of 10101010 .


|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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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DPSK System


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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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Fig 1.4.1

Transmitter

Receiver


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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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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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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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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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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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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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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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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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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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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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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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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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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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Applications

Cable Data ModemsDigital RadioDigital Communications by Satellite

Refer to your Bound Notes.

chapter 6 digital modulation

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