dc digital communication module iii part1
TRANSCRIPT
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DIGITAL
MODULATION
TECHNIQUES
MODULE 3
PART I
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DIGITAL COMMUNICATION
Module IIIModule III-- Part IPart I
Digital Modulation Techniques: Digital Modulation Techniques: Digital
modulation formats Coherent binary modulation
techniques - PSK, FSK, QPSK, MSK, Non-
coherent binary modulation techniques DPSK -Comparison of binary and quarternary modulation
techniques - M-ary modulation techniques PSK ,
QAM , FSK ( Block level treatment only)
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Binary phase shift keying In binary phase shift keying the transmitted signal is a sinusoid of
fixed frequency. It has one fixed phase when the data is at one level and phase is
changed by 1800 when the data is at another level. If the sinusoid is of amplitude A, it has a power ps = (A/2)2 = A2/2and so A = (2PS) .The transmitted signal is either
In BPSK the data b(t) is a stream of binary digits with voltage levelswhich can be taken as +1 volt for logic high and -1 volt for logic zero.Hence the modulated signal can be represented as
tPtv sBPSK 0cos2)( =
)cos(2)( 0 += tPtv sBPSK tPs 0cos2 =
tPtbtv sBPSK 0cos2)()( =
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Binary phase shift keying
10 1 0 1 0 0 1
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Generation Of BPSK Signal
BPSK signal may be generated by applying a carrier signal to abalanced modulator.
The base band signal b(t) is applied as a modulating signal to thebalanced modulator. A NRZ level encoder converts the binary data
sequence in to bipolar NRZ signal.
Balanced
Modulator
Bipolar NRZ
Level Encoder
Carrier signal
BPSK
Signald(t)b(t)
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Reception Of BPSK Signal The received signal has the form
Where is a nominally fixed phase shift corresponding to the timedelay /0 which depends on the length of path from transmitter toreceiver and the phase shift produced by the amplifiers in the front-end of the receiver preceding the demodulator.
The demodulation technique employed is synchronousdemodulation for which the carrier signal should be reproduced atthe receiver.
tPtbtv sBPSK 0cos2)()( =
)cos(2)(0
+= tPtbs
)/(cos2)(00
+= tPtbs
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Reception Of BPSK Signal
SQUARE LAW
DEVICEBANDPASS
FILTERFREQUENCY
DIVIDER
SYNCHRONOUS
DEMODULATORINTEGRATOR
BIT
SYNCHRONIZER
)cos(2)( 0 +tPtb s )(cos 02
+t )(2cos 0 +t )cos( 0 +t
Recovered
Carrier
)cos(2)( 0 +tPtb s
)(cos2)( 02 +tPtb s
bs TP
tb 2)(Sc
Ss
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Reception Of BPSK Signal
The received signal is squared to generate the signal
The dc component is removed by the band pass filter whose passband is centered around 2f0. At the output we obtain cos2(0 t+ ).
A frequency divider is used to obtain the original carrier cos (0 t+ ).
The received carrier is multiplied with the received signal togenerate which is equal to
It is then applied to an integrator. A bit synchronizer is a device that can recognize precisely the
moment which corresponds to the end of the time interval allocatedto one bit and the beginning of the next.
)(2cos2
1
2
1)(cos 00
2 ++=+ tt
)(cos2)( 02 +tPtb s
++ )(2cos
2
1
2
12)( 0 tPtb s
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Reception Of BPSK Signal At the end of the bit interval it closes switch Sc very briefly to discharge
(dump) the integrator capacitor and leaves the switch Sc open duringthe entire course of the next bit interval.
After this it closes Sc briefly. The output signal at the end of the bit
interval from the integrator (and before closing Sc) is of interest to us. This output signal is made available by switch Ss which samples the
output voltage just prior to dumping of the capacitor. If bit interval Tb is equal to an integral number n of cycles of the carrier
frequency fo such that
the output voltage at the end of a bit interval extending from
time (k-1)Tb to kTb is
bb TnorTn 00
22
==
)(0 bkTv
dttPkTbdtPkTbkTvb
b
b
b
kT
Tk
sb
kT
Tk
sbb )(2cos2
12)(
2
12)()( 0
)1()1(
0 ++=
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Reception Of BPSK Signal The second integral is zero since the integral of a sinusoid over a whole
number of cycles has the value zero.
Thus the system reproduces the bit stream b(t).
bsbb TPkTbkTv2
12)()(0 = b
sb T
PkTb
2
)(=
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Spectrum of BPSK Signal The waveform b(t) is a NRZ binary waveform. The waveform
consists of rectangular pulses of amplitude Vb and each pulse isTb/2 around its centre.
The fourier transform of the single pulse is given
For large number of such positive and negative pulses the powerspectral density S(f) is expressed as S(f) = X(f)2 /Ts where X(f)2denotes the average value of x(f) due to all the pulses in b(t).Ts isthe symbol duration.
b
bbb
fT
fTTVfX
)sin()( =
bV
bV2bT
2bT
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Spectrum of BPSK Signal
In BPSK only one bit is transmitted at a time and so bit duration Tband symbol duration Ts are the same. Tb=Ts
This equation gives the PSD of base band signal b(t). The BPSK signal is generated by modulating a carrier by the base
band signal b(t). Due to modulation by the carrier frequency fc the spectral
components are translated to fc+f and fc-f and the magnitudes aredivided by half.
222)sin(
)(
=
b
b
s
bb
fT
fT
T
TVfS
)1.(..........
)sin(
)(
2
2
= bb
bbfT
fT
TVfS
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Spectrum of BPSK Signal
If the voltage levels of the binary data are vb, it can be expressedin terms of power as Vb= Ps ie, b(t) =Ps . Using this
Equation (1) may be re written as:
The spectrum of NRZ base band signal and the correspondingBPSK signal expressed by equations (2) and (3) are plotted below.
++
+
=22
2
)(
)(sin
2
1
)(
)(sin
2
1)(
bc
bc
bc
bcbBPSK
Tff
Tff
Tff
TffTVfS
)2.........()(
)(sin
)(
)(sin
2)(
22
++
+
=
bc
bc
bc
bcbsBPSK
Tff
Tff
Tff
TffTPfS
)3.(..........)sin(
)(
2
=
b
bbs
fT
fTTPfS
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Spectrum of BPSK Signal)( fSbsTP
2
bsTP
fbf bf2bf-bf2-
cf cf bc ff +bc ff bc ff +bc ff
The spectrum of the BPSK signalis centred on fc and extends fromfc-fb+ fc+fb. So the BW of BPSKsignal is 2fb.
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Geometrical representation of BPSK signal
The BPSK signal can be expressed as
Let cos2fct be the carrier signal represented by
Now bit energy Eb is defined as Eb= Ps Tb and So
tfPtbtv csBPSK 2cos2)()( =
tfTP
PTPtb c
bs
s
bs 2cos2
)(=
tfT
c
b
2cos2 )(1 t
)()()( 1 tTPtbtvThen bsBPSK =
bbs ETP =
tfT
TPtb c
b
bs 2cos2
)(=
)()( 1 tEtvThen bBPSK =
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Geometrical representation of BPSK signal
At the receiver and the point at represents symbol 1 andrepresents symbol 0.
The separation between these two points represents the isolation ofsymbols 1 and 0 from each other. This separation is called the distance d.
In the case of BPSK the distance d is given by
)( 1 tonEb +bE
2 bbb EEEd =+=
2 bEd=
bE bE+
2Distance bE=
0
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Differential Phase Shift Keying
In coherent BPSK, we require expensive circuits toregenerate the carrier at the receiver side.
DPSK is a modification of BPSK which eliminatethe need for a synchronous carrier.
The block diagram of the system that generates
DPSK is shown below.
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Differential Phase Shift Keying
DELAY Tb
BALANCEDMODULATOR
)(tb)(td
)( bTtb
tPs 0cos2
tPtbtv sBPSK 0cos2)()( =tPOr s 0cos2
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Differential Phase Shift Keying
-101111
11-1011
1111-10-10-10-10
VoltageLogiclevel
VoltageLogiclevel
VoltageLogiclevel
b(t)b(t-Tb)d(t)
DELAY Tb
BALANCED
MODULATOR
)(tb)(td
)( bTtb tPs 0cos2
tPtbtvsBPSK 0
cos2)()( =
tPOr s 0cos2
TRUTH TABLE OF EX-OR GATE
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Differential Phase Shift Keying
d(t)
b(t-Tb)
b(t)
0 1 2 3 4 5 6 7 8 9 10 11 12
0 0 1 0 0 1 1 0 0 1 1 1 1 0
0
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Differential Phase Shift Keying
The input to the EXOR gate is the input bit stream d(t)and b(t) delayed by one bit interval given by b (t-Tb).
Because of the feed back involved in the system there is
the some difficulty involved in determining the logiclevels at the start of the waveform We can eliminate this difficulty by assuming that b(t)=0 at
the start of the interval 0. We can also just as well
assume b(t)=1 but ultimately there is no change in thedifferential nature of the wave form.
In the figure waveform of d(t) the bit stream b(t) and b(t-
Tb) is shown. Assuming that b(t)=0 we can complete thewaveform of DPSK by taking b(t) = d(t) b(t-Tb).
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Differential Phase Shift Keying
The response of b(t) to d(t) is that b(t) changes level atthe beginning of each interval in which d(t)=1 and b(t)does not change level when d(t)=0.There is no
correspondence between the levels of d(t) and b(t) asb(t)=1 some times when d(t)=0 and sometimes whend(t)=1
The only invariant feature of the system is that a changein b(t)[up or down] occurs whenever d(t)=1 and nochange in b(t) occurs when d(t)=0
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Differential Phase Shift Keying
Even if we assume that b(t)=1 at the start of the intervalthis particular invariant feature of the system will holdgood.The levels of b(t) will get inverted, but a change in
level occurs whenever b(t)=1 and no change in leveloccurs when b(t)=0 Now b(t) is applied to a balanced modulator to which the
other input is 2Ps Cos0t. The modulator output whichis the transmitted signal is
Altogether when d(t)=0 the phase of the carrier does notchange at the beginning of the bit interval while whend(t)=1 there is a phase change of magnitude 1800
tPtbtv sDPSK 0cos2)()( = tPs 0cos2 =
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Demodulation of DPSK
In the receiver shown above the received signal and the receivedsignal delayed by the bit time Tb are applied to a multiplexer. Themultiplexer output is
))(cos()cos(2)()( 00 ++ bsb TttPTtbtb
)1......(22
2coscos)()( 00
+
+= bbsbT
tTPTtbtb
SYNCHRONOUS
DEMODULATOR
(MULTIPLIER)
DELAY Tb
)cos(2)( 0 +tPtb s
))(cos(2)( 0 + bsb TtPTtb
To Integrator andbit synchronizer
nTb 20 =
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Demodulation of DPSK
This signal is applied to a bit synchronizer and integrator. If the bit interval Tb is equal to an integral number n of
cycles of the carrier frequency f0 such that n.2/ 0 =Tb
or 2n= 0Tb, the output voltage v0(kTb) at the end of abit interval extending from (k-1) Tb to kTb is obtained byintegrating the equation(1).
When 2n = 0Tb the second part of the eq (1) isreduced to zero after integration.
The first part b(t) b(t-Tb) Ps cos 2n is actually a constantmultiplied by b(t) b (t-T
b).
Now we can recover the original data sequence d(t) fromthe product b(t) b(t-Tb),
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Demodulation of DPSK
If d(t)=0 then there was no phase change andb(t) =b(t-Tb), both being +1 volt or -1 volt. In this caseb(t) b(t-Tb)= +1.
If however d(t)=1 then there was a phase change andeither b(t) = 1 or b(t-Tb) = -1volt or vice versa. In eithercase b(t) b (t-Tb)=-1 volt.
( ) ( ) 1 ( ) 0bb t b t T d t = + =
( ) ( ) 1 ( ) 1bb t b t T d t = =
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Advantages and disadvantages
The main advantage of DPSK is that we can avoid theexpensive circuit needed to generate a local carrier atthe receiver .
There is no ambiguity about the polarity of thetransmitted bit sequence.
In DPSK a bit determination is made on the basis of the
signal received in two successive bit intervals. Hence noise in one bit interval may cause errors to two
bit determination. The error rate is DPSK is greater than in PSK. Also there is tendency for bit errors to occur in pairs.
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Quadrature Phase Shift Keying
In the case of BPSK the channel bandwidth is 2fb sinceone bit constitutes a symbol in this case.
But if we can combine two or more bits to form a symbol
and then modulate it the bit rate will be reduced andhence bw requirement is reduced.
In QPSK we bunch together two bits to form a symbol
and then modulate these symbols on quadraturecarriers.
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QPSK Transmitter
TOGGLE
Flipflop
D Flipflop
D Flipflop
Adder
Even
Clock
Odd
Clock
s 0P cos t
s 0P sin t
Clock se(t)
So(t)
be(t)
bo(t)
)( tvQPSK
b(t)
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QPSK Transmitter Waveforms
1 2 3 4 5 6 7 8 9
0 1 0 1 1 0 0 1 0
1 1 10
0 0 01
Clock
Odd
Clock
Even
Clock
be(t)
bo(t)
b(t)
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QPSK Transmitter
The toggle FF is driven by a clock waveform whose period is the bittime Tb.
The toggle FF generates an odd clock waveform and an even clockwaveform.
These clocks have periods 2Tb.The active edge of one of the clocksand the active edge of the other are separated by bit time Tb.
The bit stream b(t) is applied as the data input to both D flip flops,
one driven by the odd clock and the other driven by the even clock. The FFs register alternate bits in this stream b(t) and hold each
such registered bits for two bit intervals, that is for a time 2Tb.
The bit stream bo(t) registers bits 1,3,5 etc and the bit stream be(t)
registers bits 2,4,6 etc. The bit stream be(t) and bo(t) is either +1volt or -1volt depending on
the data is 1 or 0.
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QPSK Transmitter
The stream be(t) is superimposed on a carrier and bitstream bo(t) is super imposed on a carrier by the use oftwo multipliers to generate the signals se(t) and so(t).
These signals are then added to generate the transmitter output
signal vm(t) which is
When a bit stream with bit time Tb multiplies a carrier, the generated
signal has a bandwidth of 2fb. When b0(t) and be(t) are both of bit duration 2Tb,the frequency is
1/2Tb = 0.5fb so the bw required is 20.5 fb = fb. BW required is halfthat for BPSK
When b0(t) = 1 S0(t) = (Ps sin 0t) and - (Ps sin 0t). When be(t) = 1 Se(t) = (Ps cos 0t) and - (Ps cos 0t)
m s o 0 s e 0v (t) P b (t) sin t P b (t) cos t = +
s 0P cos ts 0P sin t
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QPSK Transmitter
These 4 signals can be represented by phasors as shown below.The phasors are in phase quadrature.
s 0P cos ts 0P cos t
s 0P sin t
s 0P sin t
Q d Ph Shif K i
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Quadrature Phase Shift Keying
The four possible output signals can berepresented as below
)()()( tststv eom +=
s 0P cos ts 0P cos t
s 0P sin t
s 0P sin t
1
1
+=
=
e
o
b
b
1
1
+=
+=
e
o
b
b
1
1
=
+=
e
o
b
b
1
1
=
=
e
o
b
b
{ }1,0
{ }1,1
{ }0,0
{ }0,1
Q d Ph Shif K i
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Quadrature Phase Shift Keying
-451 1
-1351 0
+450 1+1350 0
PHASE
SHIFT
BINARY
INPUT
+135 -45 +45 -135
0 0 1 1 0 1 1 0Dibit input
Phase change
s 0P cos ts 0P cos t
s 0P sin t
s 0P sin t
{ }1,0
{ }1,1
{ }0,0
{ }0,1
( )45cos0
+t( )135cos 0 +t
( )135cos 0 t ( )45cos 0 t
Q d Ph Shif K i
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Quadrature Phase Shift Keying
The QPSK we lump two bits together to form a symbol . The symbol can have any one of four possible values corresponding to
the two bit sequences 00, 01, 10 and11.
We use four distinct signals to represent these symbols.
At the receiver each signal represents one symbol and correspondinglytwo bits .
QPSK Receiver
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QPSK Receiver
Raise input
To fourth
power
Bandpass
Filter
4f0
Frequency
Divider
(by 4)
LATCH
+
b
b
Tk
Tk dt)12(
)12(
+ b
b
Tk
kT d
)22(
2
ttbPttbP esos 00 cos)(sin)(s(t) +=
)(4 ts
t04cos
t0cos
t0sin
tts 0sin)(
tts 0cos)(
g timeat samplinbPT oss
g timeat samplinbPT ess
SamplingSwitch)(tb
QPSK R i
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QPSK Receiver
In QPSK receiver synchronous detection is used and hence it isrequired to locally regenerate the carriers cos0t and sin0t. Carrierregeneration is similar that employed for regeneration . The stepsinvolved in carrier regeneration are :
1. Incoming signal is raised to 4th power.
2. This signal is passed through a bandpass filter of passband 4f0which removes all the frequencies generated during the first step
except 4f0.
3. The signal having frequency 4f0 is passed through a frequencydivider which divides the frequency by 4. The sin40t and
cos40t terms generated in the first step are converted to sin 0tand cos 0t .
QPSK R i
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QPSK Receiver
The regenerated carriers are applied to two synchronousdemodulators, the other input of which is the input signal.
These multipliers or balanced modulators multiplies the incomingsignal with carriers sin 0t and cos 0t
The output of the multipliers are
The first term when expanded gives
) tttbPttbPi esos 000 sincos)(sin)()( +
) tttbPttbPii esos 000 coscos)(sin)()( +
tttbPttbP esos 0002 cossin)(sin)( +
tt
tbP
ttbPes
os 000 cossin22
)(
2cos2
1
2
1
)( +
=
ttbP
ttbPtbP esosos 00 2sin2
)(2cos)(
2
1)(
2
1 +=
QPSK R i
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QPSK Receiver
This signal is then applied to an integrator and bit synchronizer. If the bit duration Tb is an integral number of cycles of the input
signal integration of eq (3) gives zero value for the 2nd and thirdterms.
The integration is performed for Ts =2Tb duration and so the firstterm gives the output Ps b0(t).2Tb = Tb Ps b0(t).
The odd bit stream is thus recovered. Similarly at the output of the
second integrator we get the signal Tb Ps be(t). A bit synchronizer is required to establish the beginnings and endsof the bit intervals of each bit stream so that the time for integration,dumping and sampling can be synchronized .
At the end of each integration time for each individual integratorand just before the accumulation is damped the integrator output issampled.
QPSK R i r
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QPSK Receiver
Samples are taken alternately from one and the other integratoroutput at the end of each bit time Tb and these samples are held inthe latch for the bit time Tb.
Each individual integrator output is sampled at intervals 2Tb. Thelatch output is the recovered bit stream b(t).
QPSK Signal Space Representation
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QPSK Signal Space Representation
The QPSK signal can be represented as
Let and be the carriers
ttbPttbP oses 00 sin)(cos)( +
ttbT
TPttbT
TP ob
bse
b
bs 00 sin)(1
cos)(1
+=
0 0
2 2( ) cos ( ) sinb e b o
s s
E b t t E b t t T T
= +
)(1tu
)(2tu
tTtu s01 cos
2
)( =t
Ttu
s02
sin2
)( =
)()()()()( 21 tutbEtutbEts obeb +=
)()( 21 tuEtuE bb =
QPSK Signal Space Representation
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QPSK Signal Space Representation
)(1 tu
bE
1
1
+=
=
e
o
b
b
11
+=+=
e
o
bb
11
=+=
e
o
bb
1
1
=
=
e
o
b
b
)(2 tu
bE
bE
)()( 21 tuEtuE bb +
)()( 21 tuEtuE bb )()( 21 tuEtuE bb
)()( 21 tuEtuE bb +
sE
sE
sE
sE
bEd 2=
QPSK Signal Space Representation
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QPSK Signal Space Representation
The distance of a signal point from origin is
is the symbol energy
Thus = The distance between signal point is given by
Points which differ in a single bit are separated by the distance
The distance specifies the ability of the system to distinguishbetween two signals without error.
bsTP2
bsTP2 sE
bsTP2 sE
bEd 2=
bE2
Spectrum of QPSK Signal
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Spectrum of QPSK Signal
The input sequence b(t) is of bit duration Tb.It is a NRZ bipolarwaveform. It consists of pulses of amplitude vb and each pulses is Tb/2 around its center .
The Fourier transform of such a pulse is given by
For a large number of such positive and negative pulses the powerspectral density is expressed as
In the case of b(t),Ts = Tb so
b
bbb
fT
fTTVfX
)sin()( =
sT
fXfS
2)(
)( =
222)sin(
)(
=
b
b
s
bb
fT
fT
T
TVfS
2
2 )sin()(
=
b
bbbfT
fTTVfS
Spectrum of QPSK Signal
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Spectrum of QPSK Signal Putting
The signal b(t) is divided into be(t) and bo(t) each of bit duration 2Tb = Ts.
If the symbols 1 and 0 are equally likely then we can write powerspectral densities of be(t) and bo(t) as
The base band PSD of QPSK signal equals the sum of individual PSDs
of be(t) and bo(t)
sb PV =2
)(()(
=
s
sbs
fT
fTSinTPfS
2
)(()(
=
s
ssse
fT
fTSinTPfS
2
)(()(
=
s
ssso
fT
fTSinTPfS
)()()( 0 fSfSfS eB +=2
)(2)(
=
s
sssB
fT
fTSinTPfS
Spectrum of QPSK Signal
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Spectrum of QPSK Signal
When this signal is multiplied by a carrier having frequency fo itsspectrum is shifted to f0 and divided by two.
+++
=
2
0
0
2
0
0
)())((
)())(()(
s
s
s
sssQ
TffTffSin
TffTffSinTPfS
Spectrum of QPSK Signal
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Spectrum of QPSK Signal
)( fSssTP2
ssTP
sT1
0f 0f 20bff +
2
bc
ff
20
bff +2
0bff
sT2sT1sT2 22
11 b
bs
f
TT
==
2bf2bf
M ary PSK
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M-ary PSK In BPSK we transmit each bit individually. Depending on whether
b(t) is logic 0 or 1 we transmit one or another of a sinusoid for bittime Tb, the sinusoids differing in phase by 2/2=1800 .
In QPSK we group together two bits and depending on which of thefour two bit word develops we transmit one or another of foursinusoids of duration 2Tb, the sinusoids differing in phase by
2/=900
.
This scheme can be extended . Let as group together N bits so thatin this N bit symbol extending over NTb there are 2N =M possible
symbols. We can represent the symbols by sinusoids of durationNTb = Ts which differ from one another by phase 2/M.
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M ary PSK
In M-ary PSK the wave forms used to identify symbols are
The waveforms are represented by dots in at signal space in whichcoordinate axes are the ortho normal waveforms
( )
mwhere
MmtCosPtv
m
msm
)12(
)1,.....1,0(2)( 0
+=
=+=
sin2
)( 02 tT
tus
=cos2
)( 01 tT
tus
=
[ ]ttptv mmsm 00 sinsincoscos2)( =
tPstP mms 00 sinsin2coscos2 =
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M-ary PSK
mso
mse
oe
Pp
Pp
bypandpDefining
sin2
cos2
=
=
tptptv oem 00 sincos)( =
1( )u t
2( )u t
3
5
7
3
2/M
2/M
2/M
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M ary PSK
SERIAL
TO
PARALLEL
CONVERTER
DIGITAL
TO
ANALOG
CONVERTER
..
......
SINUSOIDAL
SIGNAL
SOURCE
PHASE
CONTROLLED
BY
v(sm)
)(tb
0
12
1N
)( msv output
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M ary PSK The bit stream b(t) is applied to a serial and parallel converter. This converter has facility for storing N bits of a symbol.
When the N bits are presented serially these bits are assembledtogether and is presented all at once of N output lines of theconverter parallelly.
The converter output remains unchanged for the duration NTb of asymbol during which the converter is assembling a new group of N
bits The converter output is applied to a D/A converter. D/A converter
produces an output voltage which assumes one of 2N = M differentvalues in a one to one correspondence to M possible symbols
applied to its input. The D/A output is a voltage v(sm) which depends on the symbol
sm(m =0,1. M-1).
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M ary PSK Finally v(sm) is applied as a control input to a special type of constant
amplitude sinusoidal signal source whose phase m is determinedby v(sm).
The output is a fixed amplitude sinusoidal waveform whose phasehas one to one correspondence to the assembled N bit symbol.
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M ary PSK Receiver
Raise to
Mth
Power
BPF
Frequency
Divider
(by M) t0sin
t0cos
bT
dt0
bT
dt0
ADC
ttvm 0cos)(
ttvm 0sin)(
espT
ospT
tptptv oem 00 sincos)( =
Symbol
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M ary PSK Receiver The carrier recovery circuit requires the following operations
(i) The input signal is raised to the Mth power(ii) The signal is passed through a BP filter of center frequency Mf0(iii) The output of BPF is divided by M.
The recovered carriers are multiplied with the input signal and isapplied to integrators which works with the help of bit synchronizers.
The integrator eliminates all sinusoidal terms and produces theoutputs T
sP
eand T
sP
o.
These signals are applied to a device which reconstructs the digital Nbit symbol which constitutes the transmitted signal .
If serial bit stream output is needed a parallel to serial converter maybe used at the output.
Problems Associated With QPSK
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The BW of QPSK signal is found to be fb as the main lobe of thespectrum has a width of fb.
The main lobe contains 90% of the total signal energy.
Still a small amount of power lies outside this main lobe in the formof side lobes.
When QPSK is used for multi channel communication on adjacentcarriers this side lobe power poses many problems.
The wide spectrum of QPSK signal is due to the nature of the baseband signal.
This signal consists of abrupt changes and abrupt changes give riseto spectral components at high frequencies.
When this base band spectra is translated to a high carrierfrequency as a result of multiplication it retains its shape.
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Q
One solution is to pass the base band signal through a LPF toeliminate the abrupt changes and to band limit it.
But the band limiting give rise to a pulse that is not time limitedand hence inter symbol interference results.
Another alternative is to pass the modulated signal through aband pass filter to suppress the side lobes.
When a signal like QPSK where there are abrupt phase changes
is passed through a band pass filter substantial changes in theamplitude of the signal occurs at times of abrupt phase changesdue to the effect of filters.
Such amplitude variations can cause problems in QPSK that
employ repeaters.
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Q So a method of modulation called Minimum Shift Keying isdeveloped where the major changes are:
(i) The base band wave form that multiplies the quadrature
carriers are much smoother than the abrupt rectangular wave
form of QPSK.(ii) The wave form of MSK exhibits phase continuity. There are
no abrupt phase changes as in QPSK. So inter symbol
interference is also avoided.
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y g In MSK also we start with the original bit stream b(t). It is divided in to an odd and even bit stream bo(t) and be(t) as in fig b and
fig c . The odd stream consists of alternate bits b1,b3,b5 etc and the even
stream consists of b2,b4,b6 etc.
Each bit in both streams is held for two bit intervals 2Tb=Ts, the symboltime.
The MSK transmitter also generates two sinusoidal waveformssin (2t/4Tb) and cos (2t/4Tb).
It is essential that sin (2t/4Tb) passes through zero precisely at the endof symbol time be(t) and cos (2t/4Tb) passes through zero at the end ofthe symbol time bo(t).
Now the products of be(t)sin(2t/4Tb) and bo(t) cos (2t/4Tb) is generatedwhich are shown in fig e and fig f.
In MSK the transmitted signal is
)1.....(sin4
2cos)(2cos
4
2sin)(2)( 00 t
T
ttbPt
T
ttbPtv
b
os
b
esMSK
+
=
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y g
In the case of QPSK the quadrature carries are multiplied by therectangular abruptly changing odd and even bit streams. But in MSKthe carriers are multiplied by the smoother waveforms shown in fig eand f.
The side lobes generated by this smoother waveforms will be smallerthan those associated with the rectangular waveform and have easierto suppress.
Equation(1) can be re-written as
+
+
+= tTtT
tbPtv
bb
esMSK 00
4
2sin4
2sin2
)(2)(
+
+ t
Tt
T
tbP
bb
os
4
2sin
4
2sin
2
)(2 00
+
+
+= t
T
tbtbt
T
tbtbPtv
b
eo
b
oesMSK
4
2sin
2
)()(
4
2sin
2
)()(2)( 00
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y g
Now bo = 1 and, be = 1
If bo = be than CL = 0 and CH =1If bo = -be than CH = 0 and CL =1
( ) ( )
++
+= t
tbtbt
tbtbPtv eooesMSK 00 sin
2
)()(sin
2
)()(2)(
2
2 eoL
eoH
bbCandbbCDefining =+=
=+= 00 LH and
)2.......(sin)(2sin)(2)( ttCPttCPtv LLsHHsMSK +=
Minimum Shift Keying
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y g Thus depending on the value of bits be and bo in each bit interval the
transmitted signal is at angular frequency H or at L as in the case ofFSK and the magnitude is always. sP2
ShiftedH-1-1
0L-11
Shifted
L1-1
0H11
PhaseTransmittedFrequency
bebo
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y g In MSK the two frequencies fH and fL are chosen to ensure that the two
possible signals are orthogonal over the bit interval Tb.
0sinsin
0
= tt LT
H
b
( ) ( ) 0coscos2
1..
0
=+ ttei LHT
LH
b
( ) ( ) 02cos2cos21
..0
=+ tfftffei LHT
LH
b
( )
( )
( )
( )
0
2
2sin
2
2sin
2
1..
0
=
+
+
bT
LH
LH
LH
LH
ff
tff
ff
tffei
( )( )
( )( )
)3......(02
2sin
2
2sin
2
1.. =
++
LH
bLH
LH
bLH
ff
Tff
ff
Tffei
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Equation(3) will be satisfied if
Now we know that
)
( )
2(4)
2
H L b
H L b
f f T n
f f T m
=
+ =
422
42
4
2000
bb
b
H
ff
f
T+=+=+=
( )54
.. 0 +=b
Hfffei
4
22
4
2
4
2 000
bb
b
L
ff
f
T
Similarly ===
( )64
.. 0 =b
L
fffei
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Substituting (5) and (6) in equations (4)
Similarly
Altogether we have
Since n=1 fH and fL are close together as possible for orthogonalityto prevail.
So this system is called Minimum shift keying.
nTf
ff
f bbb =
+
442 00 nT
fi.e. b
b =2
2 1 =ni.e.
mTf
ff
f bbb =
++
442 00 mTfi.e. bo =22
mffi.e. bo =
14 m
ffi.e. bo 4 =
1=n mf
f bo4
=
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The carrier frequency f0 should be an integral multiple of fb/4
4)1(
444 0
bbbbH
fm
ffm
fff +=+=+=
4)1(
444 0 bbbbL fmffmfff ===
Minimum Shift Keying Transmitter
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BPF
BPF
ADD
SUB
ADD
)( 0
+
+
+
-
+
+
t0sin
tcos
t)sin(21
0 +
t)sin(21
0
)(2 tbP os
)(2 tbP es
tttbP os 0sincos)(2
tttbP es 0cossin)(2
tt )sin()sin( 00 + ttei 0cossin..
tt )sin()sin( 00 ++ ttei 0sincos..
)(tvMSK
)(0
+
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First and are multiplied resulting in the productwhich is equal to
Two band pass filters are used to separate the two terms of the
above equation and These components are then added to produce
and i.e.
The outputs of adder and subtractor are then multiplied by the oddand even bit streams and resulting in the
products and
When these products are added we get
which is the required MSK wave form.
t0sin tcostt cossin 0 ( )tt )sin()sin(2
100 ++
t)sin(2
10 + t)sin(2
10
( )tt )sin()sin( 00 ++
tt )sin()sin( 00 + tt sincos 0
)(2 tbP os )(2 tbP estttbP os 0sincos)(2 tttbP es 0cossin)(2
0 02 ( ) sin cos 2 ( ) cos sin s e s oP b t t t P b t t t +
MSK Waveforms
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b1=1 b2=1 b3=0 b4=0 b5=0 b6=0 b7=0 b8=0
b1
b3
b5
b7
b2 b4 b6 b8
b(t)
bo
(t)
be(t)
bT
t
4
2cos
bT
t
4
2sin
b
eT
ttb 4
2sin)(
b
oT
ttb
4
2cos)(
)(a
)(b
)(c
)(d
)(e
)( f
2b 4b 6b
8b
1b 3b
5b 7b
MSK Waveforms
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b
eT
ttb
4
2sin)(
b
oT
ttb42cos)(
2b 4b 6b
8b
1b 3b5b 7b
(1,1) (-1,1) (-1,-1) (-1,-1) (-1,1) (1,1) (1,1)
bH ff 5.1=
bL ff 1=
)(tvMSK
eo bb = eo bb = eo bb = eo bb = eo bb = eo bb = eo bb =
)(g
)(h
)(i
Minimum Shift Keying Receiver
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Storesample
Store
sample
+
b
b
Tk
Tkdt
)12(
)12(
+ b
b
Tk
kT dt
)22(
2
tt = cossinx(t) 0
tt = sincosy(t) 0
bTkatSample )12(t +=
bTkatSample )22(t +=
bkTatSwitch =t
)(tvMSK
Minimum Shift Keying Receiver2222
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ADD
SUB
)(tvMSKSquaring
Circuit
BPF
BPF
Filter &Amp
H2
L2
2
2
tP Hs 2sinsin)(sin)(
2222
ttCttC LLHH +
tP Ls 2sin
tHsin2
1
tLsin2
1
tt = cossinx(t) 0
tt = sincosy(t) 0
tk scos
+
+
+
_
Minimum Shift Keying Receiver
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Synchronous detection is used in MSK receiver. First the carrier is recovered by a carrier recovery circuit as in fig(2).
The input signal
is first squared and the output of the squarer consists , among otherterms and which consists of double
frequency terms and .
When it is passed through two BPFs of central frequencies and
and assuming phase modification the output signal is
and .
The frequency is divided by 2 to obtain and
The signals are then added to obtain which is
equal to
sin)(2sin)(2 ttCPttCP LLsHHs +
sin)( 22
ttC LL ttC HH 22 sin)(
tH2cos tL2cos
H2
L2 tP Hs 2sin
tP Ls 2sin
tH
sin tL
sin
tt LH sinsin +tt LHLH
+
2cos
2sin
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So the expression becomes Similarly the signals are subtracted to obtain
The subtracted signal is thus Thus the carriers are recovered.
Also the signals and are multiplied and passed
through a LP to generate signal at symbol frequency
When passed through a LPF the output is
Once the carriers are regenerated the incoming MSK signal ismultiplied by these carriers x(t) and y(t).
tt = cossinx(t) 0
tttt LHLHLH
+
=2
cos2
sinsinsin
tt = sincosy(t) 0
tHsin tLsin
s( ) ( )tttttz LHLHLH ++== coscossinsin)(( ) ttLH = 2coscos
t
fb
= 422cos tfb
= 22cos
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The outputs of the multipliers are passed to an integrate and dumpcircuit.
The integrator integrates staggered overlapping intervals of symboltime Ts=2Tb.
The first integrator integrates over (2k-1)Tb to (2k+1)Tb and thesecond one over 2kTb to (2k+2)Tb .
The sinusoidal terms are eliminated by the integrator.
At the end of each integration time the integrator output is storedand then it is dumped.
The switch at the output swings back and forth at the bit rate so thatthe final output waveform is the original data bit stream.
Signal space representation of MSK
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2( ) sinL Ls
u t tT
=
tT
tu Hs
H sin2)( =0
1
=
+=
L
H
C
C
1
0
=
=
L
H
C
C
0
1
==
L
H
C
C
1
0
+=
=
L
H
C
C
sss ETP =2 sd E=
Signal space representation of MSK
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Here CH(t) and CL(t) can assume 1.
When CH(t) = 1, CL(t)=0 and CL(t) = 1, CH(t)=0
The ortho normal unit vectors are given by
sin)(2
sin)(2
ttCT
TPttCT
TP LLs
ssHH
s
ss +=
)()()()( tutCEtutCE LLsHHs +=
sin
2
)( tTtuwhere HsH = sin
2
)(nd tTtua LsL =
tT
tu cs
L sin2
)( =tT
tu Hs
H sin2
)( =
sin)(2sin)(2)( ttCPttCPtv LLsHHsMSK +=
Signal space representation of MSK
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The end points of 4 possible signal vectors are indicated by dots. The smallest distance between signal points is bs EEd 42 ==
Binary Frequency Shift Keying
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In BFSK two frequencies are used to represent binary data 0 and 1 The binary data waveform d(t) generates a signal
where d(t) is +1 or -1 corresponding to the logic levels 1 and 0 of the
data waveform. The transmitted signal is of amplitude and is either
Where 0 is the nominal carrier frequency and d(t) is represented bya constant offset (0+) or (0-) from the carrier frequency.
(0+) is the higher frequency and (0-) is the low frequency and isrepresented by H and L
[ ]ttdtPtv sBFSK += )(cos2)( 0
sP2
tPts sH )cos(2)( 0 +=
tPtsor sL )cos(2)( 0 =
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A scheme for generating BFSK is shown below. Two balanced modulators are used, one with carrier frequency H
and the other with L Voltage values of pH(t) and pL(t) are related to the voltage values of
d(t) as given below.
When pH(t)=1, pL(t)=0 generated signal is
When pH(t)=0, pL(t)=1 generated signal is At the output of the adder, corresponding to d(t) one of these
waveforms occur, but not both
+1V0V0-1V
0V+1V1+1V
pL(t)pH(t)b(t)d(t)
tP Hs cos2
tP Ls cos2
BFSK-Complete Block Diagram2
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ADDER
)(tpTP Hbs
)(tpTP Lbs
tTH
bcos
tT
L
b
cos2
)(tvBFSK
LEVEL
SHIFTER
LEVEL
SHIFTERINVERTER
POLAR TONONPOLAR
CONVERTOR
)(td
)(tb
)(tpH
)(tpL
BFSK- Transmitter
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The input signal b(t) is either 1 or 0. The upper level shifter produces corresponding to 1 and no
signal corresponding to 0.
The lower level shifter on the other hand produces
corresponding to 0 and no signal corresponding to 1. When these signals are multiplied and added the output is
corresponding to b(t)=1 and corresponding to b(t)=0
bsTP
bsTP
tP Hs cos2
tP Ls cos2
BFSK Receiver
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ENVELOPE
DETECTOR
ENVELOPE
DETECTOR
BPF
BPF
COMPARATOR
[ ]ttdtPs + )(cos2 0
)(td
bf2
bf2
bH fff += 0
bL fff = 0
BFSK Receiver
BPSK signal is applied to two bandpass filters with centre
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g pp pfrequencies f0+fb and f0-fb and with bandwidths 2fb.
Also fH-fL=2fb. The filter frequency ranges selected do not overlapand each filter has a pass band wide enough to pass the main lobein the spectrum.
Hence the filter will pass nearly all energy in the transmission at fHwhere as it will be blocked by the other filter.
The second filter will pass nearly all energy in the transmission at fLwhere as it will be blocked by the first filter.
The filter outputs are applied to two envelope detectors and theoutputs of envelope detectors are applied to a comparator.
If the transmitted signal is at fH the output of the upper envelopedetector will produce a higher voltage when compared to the lower
detector. The comparator compares the envelope detector outputs and a
decision is made in favour of high.
BFSK Receiver
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If the transmitted signal is at fL the output of the lower envelopedetector will produce a higher voltage when compared to the upperdetector.
The comparator compares the envelope detector outputs and a
decision is made in favour of low.
BFSK Receiver (Synchronous detection)BFSK SIGNAL
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COMPARATOR
CARRIERRECOVERY
CIRCUITtLcos
tHcos
OUTPUT
bT
dt0
bT
dt0
[ ]ttdtPs + )(cos2 0
BFSK Receiver
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In the presence of noise and other interference the demodulationmethod using envelope detection produces wrong results.
So synchronous detection method is employed in thesecircumstances.
The carrier recovery circuit regenerates the two carriers cosHt andcosLt .
The incoming signal is multiplied by these two carriers and ispassed through an integrate and dump circuit synchronized by bit
synchronizers. The sinusoidal terms are eliminated and a term containing the bit
information is obtained at the output.
This information is applied to a comparator circuit which comparesthe voltage levels of the integrator outputs and makes a decisionregarding the transmitted bit i.e. high or low.
This type of demodulation is called coherent detection.
Spectrum of BFSK Signal
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The BFSK signal may be written as
The expression for BPSK signal is given by
When we compare equations (1) and (2) we can see that the two
terms of eq(1) are identical to eq(2) with the difference that b(t) is abipolar waveform assuming +1 and -1 where as pH(t) and pL(t) arenon polar signals assuming values of +1 and 0.
If we can convert eq(1) in the same form as eq(2) we can easily
obtain the spectrum by comparing with the spectrum of BPSKsignal.
)1......(cos2)(cos2)()( tPtptPtpts LsLHsH +=
)2......(cos2)()( 0tPtbts sBPSK =
Spectrum of BFSK Signal
)('
11
)( tt HH 1)(1)(' tth HH
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Substituting in eq(1)
The first term of eq(3) has a spectrum that is a single impulse
located at fH and the second term of has a spectrum that is a singleimpulse located at fL.
)('22)( tptpHH +=
)('2
1
2
1)( tptp LL +=
1)(1)(' =+= tptpwhen HH0)(1)(' == tptpwhen HH1)(1)(' +=+= tptpwhen LL
0)(1)(' == tptpwhen LL
cos)('2
1
2
12cos)('
2
1
2
12)( ttpPttpPts LLsHHs
++
+=
cos ' ( )cos cos ( ) ' ( )cos .........( 3 )2 2 2 2
s s s s H H H L L L
P P P P t p t t t t p t t = + + +
Spectrum of BFSK Signal
PSD f bit t i i b)(' tPs
2
)sin()( bfTTPfS
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PSD of bit stream is given by
When it is multiplied by its spectrum is translated to fHand divided by 2.
Similarly the spectrum of is given by
Complete spectra is obtained by adding all the 4 terms.
)('2
tpP Hs )sin()(
=
b
bbs
fTfTTPfS
tHcos
++
+
=22
)(
)(sin
)(
)(sin
2)(
bH
bH
bH
bHbsq
Tff
Tff
Tff
TffTPfS
ttpP LLs cos)('2
++
+
=22
)(
)(sin
)(
)(sin
2)(
bL
bL
bL
bLbsp
Tff
Tff
Tff
TffTPfS
Spectrum of BFSK Signal
[ ])()(1)( fffffS
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The spectra is plotted for fH-fL=2fb With such a selection the BW may be made minimum without
interference between spectra of the signals.
[ ] )()(21)( LHBPSK fffffS +
=
+
+
22
)(
)(sin
)(
)(sin
2 bL
bL
bH
bHbs
Tff
Tff
Tff
TffTP
Spectrum of BFSK Signal
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HfHf Lf Lf
bL ff bL ff + bH ff bH ff +
HfLfHf Lf
bf2
bf4
Geometrical Representation of BFSK
In the general case a modulated signal can be represented in terms
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In the general case a modulated signal can be represented in termsof ortho normal vectors u1(t) and u2(t) in signal space.
In the case of PSK the orthogonality of the vectors u1(t) and u2(t)
results from their phase quadrature. But in the case of BFSK we should select the frequencies carefully
to ensure orthogonality.
Let m and n be integers. Then
)()()( 2211 tuCtuCts +=
1 2( ) cos Hs
u t tT
=
1
2( ) cos 2 b
b
u t m f t T
= 22
( ) cos 2 bb
u t n f t T
=
bitsymbolforTT bs == 1 2( ) cos Ls
u t tT
=
Geometrical Representation of BFSK
The vectors u1(t) and u2(t) are the mth
and nth
harmonics of thef f f
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The vectors u1(t) and u2(t) are the m and n harmonics of thefundamental frequency fb.
Different harmonics are orthogonal over the interval of thefundamental period.
If the frequencies fH and fL in a BFSK system is selected to be
the corresponding signal vectors are
bH mff =
L b f n f =nmwhere >
1
2( ) cos 2 ( ) H s b b b
b
s t P T m f t E u t T
= =
2
2( ) cos 2 ( ) L s b b b
b
s t P T n f t E u t T
= =
Geometrical Representation of BFSK
Now sH(t) and sL(t) are orthogonal and can be represented in signali b l
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Now sH(t) and sL(t) are orthogonal and can be represented in signalspace as given below.
The distance between signal points is which is less thanthat of BPSK signals.
)(1 tu
)(2 tu
)( tsL
)( tsH
bE
bE
2 bd E=
2 bd E=
1cos2)( fortPts HsH =
0cos2)( fortPts LsL =
M-ary FSK Transmitter
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SERIAL
TO
PARALLEL
CONVERTER
N- BIT
D/A
CONVERTER
FREQUENCY
MODULATOR.
)(td
0d
1Nd
M-ary FSKoutput
M-ary FSK Receiver
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BPF(f0)
BPF(f1)
BPF(fM-1)
ENVELOPE
DETECTOR
ENVELOPE
DETECTOR
ENVELOPE
DETECTOR
LARGEST
SIGNAL
OUTPUT
N-BIT
ADC
M-aryFSK
M-ary FSK At the transmitter a serial to parallel converter converts the serial bit
stream to parallel formS ti th d t i l d i ll l f d d t
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stream to parallel form. Sometimes the data is already in parallel form and we need not use
a serial to parallel converter. The converter output is applied to a frequency converter that
generates a carrier waveform whose frequency is determined bythe modulating waveform. The transmitted signal for the duration of the symbol interval is of
frequency f0, f1, .,fM-1 where M=2N.
At the receiver the incoming signal is applied to M parallel BPFseach followed by an envelope detector. The BPFs have centre frequencies f0, f1, .,fM-1 The envelope detector apply their outputs to a device which
determines which of the detector outputs is the largest. Largest output is applied to an N-bit DAC whose output is theoriginal transmitted signal. This parallel data may be converted toserial form if required.
M-ary FSK
The probability of error is minimized by selecting frequencies f0
, f1
,fM 1 mutually orthogonal
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.,fM-1 mutually orthogonal. For this the carrier frequency is selected to be successive even
harmonics of the symbol frequency fs=1/Ts
Thus the lowest frequency is f0=kfs, f1=(k+2)fs, f3=(k+4)fs, . In this case the spectral density patterns of the individual possible
transmitted signals overlap as shown below.
In order to pass M-ary FSK the required BW is B=2Mfs.
Since fs=fb/N and M=2N, B=2N+1 fb/N
M-ary FSK
kff fkf )( fkf )4(
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2fs
BW = 2Mfs
skff =1 sfkf )2(2 += sfkf )4(3 +=
Quadrature Amplitude Shift Keying (QASK)
QASK is an amplitude and phase shift keying system in which thephase as well as amplitude of the carrier is varied according to the
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p p gmodulating bit stream.
In the case of BPSK,QPSK and M-ary PSK we transmit in any
symbol interval one signal or another which are distinguished fromone another in phase but are all of the same amplitude.
In each of these systems the end points of the signal vectors insignal space falls on the circumference of a circle.
The ability to distinguish one signal vector from another in thepresence of noise will depend on the distance between vector endpoints.
So the noise immunity can be improved if the signal vectors are
varied in phase amplitude. Such a type of modulation system is called QAPSK or QASK.
QASK Example
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Quadrature Amplitude Shift Keying (QASK)
Consider a system in which a symbol is transmitted for every 4 bitcombinations. There are thn 24=16 possible symbols.
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p y Now we have to generate 16 distinguishable signals.
One possible geometrical representation of these 16 signals is
shown in figure(1). Each signal point is equally distant from its neighbouring points, the
distance being d=2a.
When all the 16 signals are equally likely the average energy
associated with a signal is
( ) ( ) ( ) ( )[ ] 222222222 1099994
1aaaaaaaaaEs =+++++++=
s
Ea 1.0=
sEad 1.022 ==
QASK Each symbol represents 4 bits and the normalised symbol energy
Es=4Eb where Eb is the normalised bit energy.
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s b b gy
A typical signal can be represented by
where k1 and k2 are each equal to 1 or 3
bs EEa 1.01.0 == bEd 4.02=
)1().........()()( 2211 tauktauktvQASK +=
tT
tus
01 cos2
)( =
tTtu s02 sin
2
)( =
sEa 1.0=)3(
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QASK Constellation
a3D(a, 3a) C(3a, 3a)
2a )(2 tu
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a
a2
a
a2
a3
a a2 a3aa2a3
A(a, a) B(3a, a)
O )(1 tu
QASK Generator
D Qkb )(tAe
tPs 0cos
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D Q
D Q
D Q
DAC
DAC
ADDER
CLOCK GENERATOR (TS)
1+kb
2+kb
3+kb
)(tAo
tPs 0sin
QASK Generator The 4 bit symbol bk+3 bk+2 bk+1 bk is stored in the 4 bit register made
up of 4 flip flops. A new symbol is presented once per interval T =4T and the
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A new symbol is presented once per interval Ts=4Tb and thecontents are updated at each active edge of the clock which has aperiod Ts.
Two bits are presented to one D/A converter and two bits are to thesecond converter.
The converter output Ae(t) modulates the balanced modulator whoseinput carrier is the even function and Ao(t) modulatesthe modulator with odd function carrier.
The transmitted signal is then
Comparing eq(3) with the general equation
tPs 0cos
)3.......(sin)(cos)()( 00 ttAPttAPtv osesQASK +=
tPktPktv ssQASK 0201 sin2.0cos2.0)( += 2.032.0)(),( = ortAtA oe
QASK Receiver
T
)(tAe
ttAPttAPoses 00
sin)(cos)( +
0b
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Raise to
4th
Power
BPF(4f0)
FrequencyDivider
(by 4) t0sin
t0cos
sT
dt0
sT
dt0
ADC
ADC
)(tAo
0
1b
2b
3b
QASK Receiver
As in the case of QPSK carriers are regenerated at the receiver.
The incoming signal is raised to the 4th power.
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A band pass filter extracts the component at 4f0.
It is divided by 4 to get
Neglecting all terms which are not in the frequency range 4f0
( )40024 sin)(cos)()( ttAttAPtv oesQASK +=
ttAtAtAtA
Ptv oeoesQASK 04444
44cos
8
)()(6)()()(
+=
t
tAtAtAtA
Poeoe
s 0
22
4sin2
))()()(()(
+
QASK Receiver
When the signal is divided by 4 we obtain the quadrature carriers
andt0sin t0cos
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Two balanced modulators are used to multiply the incoming signalwith the retrieved carriers.
After that it is applied to an integrator circuit operating insynchronism with a symbol time synchronizer.
At the output of the integrator all sinusoidal terms are reduced tozero and we get Ae(t) and Ao(t).
These voltage levels are applied to a D/A converter which producesthe original bit stream.