digital modulation unit 3

Unit 3Unit 3Digital Modulation

Prof A K Ni amProf A K Nigam

Syllabus

• Amplitude Shift Keying (ASK)

• Frequency Shift Keying (FSK), FSK Detection Using PLL

• Binary Phase Shift Keying (PSK)‐ Transmitters, Coherent and non coherent detection, Bit and Baud Rate, Bandwidth and Frequency Spectrum BER and Probability of ErrorFrequency Spectrum. BER and Probability of Error.

• Quadrature Phase Shift Keying (QPSK), QPSK Demodulator, Offset QPSK, Comparison of conventional QPSK and Offset QPSK , BPSK, 8 PSK & 16 PSK

• Quadrature Amplitude Modulation (QAM); 8 QAM & 16 QAM transmitters and receiversQAM transmitters and receivers,

• Band Width efficiency,

• Carrier Recovery; Squaring Loop & Costas Loop, y; q g p p,

• Differential PSK, DBPSK transmitter and receiver, 9/18/2013 2Lt Col A K Nigam, ITM University, Gurgaon

Binary Modulation schemesBinary Modulation schemes

• ASKASK

• FSK

S• PSK

M M d l ti h• QPSK

M‐ary Modulation schemesQPSK

• QAM

9/18/2013 3Lt Col A K Nigam, ITM University, Gurgaon

Need For M‐ary Modulation• Mathematically stated, the Shannon limit for information capacity

• For a standard telephone circuit with a signal to noise power ratio• For a standard telephone circuit with a signal‐to‐noise power ratioof 1000 (30 dB) and a bandwidth of 2.7 kHz, the Shannon limit forinformation capacity is

I = (3 32)(2700) log10 (1 + 1000) = 26 9 kbpsI = (3.32)(2700) log10 (1 + 1000) = 26.9 kbps• Using multilevel signaling, the Nyquist formulation for channel

capacity is I =2B log2 MTh f hi i 26 9 kb h t h• Thus for achieving 26.9 kbps we have to have

26900 =2×2700× log2 M• This gives log2 M=5 or L=32, thus in order to achieve 26.9 kbps we

have to use multilevel signals

Bit and BaudBit and Baud

• Mathematically, baud is the reciprocal of the time of oney, poutput signaling element, and a signaling element mayrepresent several information bits.

• Baud is expressed as baud =1/ts

• In addition, since baud is the encoded rate of change, it alsoequals the bit rate fb divided by the number of bits encodedinto one signaling element. Thus,

AMPLITUDE‐SHIFT KEYING• Mathematically, amplitude‐shift keying is

• Modulating signal [vm(t)] is normalized where + 1 V = logic 1Modulating signal [vm(t)] is normalized where + 1 V = logic 1 and ‐1 V = logic 0.

• Therefore for a logic 1 input, vm(t) = +1 V, Equation reduces to

• And for a logic 0 input, vm(t) = ‐1 V, Equation reduces to

• Thus, the modulated wave is either A cos(wct) or 0.

• Hence, the carrier is either "on“ or "off" which is why amplitude‐shift keying is sometimes referred to as on‐p y goff keying(OOK).

ASK (also called Digital amplitude modulation, DAM)

Performance of BASK/OKK/DAM

• As noise affects only the amplitude of the signal, the noise performance of this is not good

ASK Modulator

Asynchronous DemodulatorAsynchronous Demodulator

Envelope Detector

LPFInput Demodulated output

Synchronous ASK Demodulation

[ ]( / 2 )c b cf f f± ±

( / 2 )c bf f±

[ ]( )c b cf f f

FREQUENCY‐SHIFT KEYING

• FSK is a form of constant‐amplitude anglemodulation similar to standard frequencyq ymodulation (FM) except the modulating signal isa binary signaly g

• FSK is sometimes called binary FSK (BFSK).

• The general expression for FSK is• The general expression for FSK is

• The modulating signal is a normalized binary waveformwhere a logic 1 = + 1 V and a logic 0 = ‐1 V.g g

• Thus, for a logic l input, we can write

F l i 0 i ( ) 1• For a logic 0 input, vm(t) = ‐1,

• Thus With binary FSK, the carrier center frequency (fc) isshifted (deviated) up and down in the frequency domain bythe binary input signal

• As the binary input signal changes from a logic 0 to a logic 1and vice versa, the output frequency shifts between twofrequencies

(a) mark, or logic 1 frequency (fm), (b) space, or logic 0 frequency (fs).

*The mark and space frequencies are separated from the carrier frequency by the peak frequency ∆f deviation and from each other by 2∆f .

• Frequency deviation is expressed mathematically as

FSK Bit Rate, Baud, and Bandwidth

• The bit time equals the time of an FSK signaling element,and the bit rate equals the baud.

Th b d f bi FSK l b d t i d b• The baud for binary FSK can also be determined bysubstituting N = 1 in

• The minimum bandwidth for FSK is given as

Example 2‐2Determine (a) the peak frequency deviation, (b) minimumbandwidth and (c) baud for a binary FSK signal with a markbandwidth, and (c) baud for a binary FSK signal with a markfrequency of 49 kHz, a space frequency of 51 kHz, and an input bit rate of 2 kbps

Modulation index

• h = FM modulation index called the h‐factor in FSK

f f d t l f f th bi d l ti i l• fo = fundamental frequency of the binary modulating signal

• ∆f = peak frequency deviation (hertz)

FSK Transmitter

FSK Receiver

(a) Noncoherent FSK demodulator

(b) Coherent FSK demodulator

PLL‐FSK demodulator

PLL‐FSK demodulator operation

• As the input to the PLL shifts between the mark and spacep pfrequencies, the dc error voltage at the output of the phasecomparator follows the frequency shift.

B th l t i t f i ( k d• Because there are only two input frequencies (mark andspace), there are also only two output error voltages.

• One represents a logic 1 and the other a logic 0.p g g

PerformancePerformance

• Binary FSK has a poorer error performance than PSK or QAMy p p Qand, consequently, is seldom used for high‐performancedigital radio systems.

• Its use is restricted to low‐performance, low‐cost,asynchronous data modems that are used for dataycommunications over analog, voice‐band telephone lines.

Continuous‐Phase Frequency‐Shift Keying

• Continuous‐phase frequency‐shift keying (CP‐FSK) is binaryFSK except the mark and space frequencies are synchronizedp p q ywith the input binary bit rate.

• This ensures a smooth phase transition in the analog outputsignal when it changes from a mark to a space frequency orvice versa.

• This has effect of limiting the BW after modulation

Non/continuous phase FSK waveformswaveforms

Comparison FSK/CP‐FSKComparison FSK/CP FSK

• CP‐FSK has a better bit‐error performance andCP FSK has a better bit error performance andlower spectral width than conventional binary FSKfor a given signal‐to‐noise ratio.

• The disadvantage of CP‐FSK is that it requiresg qsynchronization circuits and is, therefore, moreexpensive to implement.

PHASE‐SHIFT KEYING• The simplest form of PSK is binary phase‐shift keyingp y p y g(BPSK), where N = 1 and M = 2.

• Therefore, with BPSK, two phases (2^1 = 2) areibl f h ipossible for the carrier.

• One phase represents a logic 1, and the other phaserepresents a logic 0represents a logic 0.

• As the input digital signal changes state (i.e., from a 1to a 0 or from a 0 to a 1), the phase of the output), p pcarrier shifts between two angles that are separated by180°.

h f h l k i ( )• Other names for BPSK are phase reversal keying (PRK)and bi‐phase modulation.BPSK i f f d l ti f• BPSK is a form of square‐wave modulation of acontinuous wave (CW) signal.

BPSK ModulatorBPSK Modulator

Balanced Modulator

BPSK modulator: truth table phasor and constellation diagram

Bandwidth considerations of BPSK

.cos (assuming unit amplitude)m cout put cos w t w t=

( )1 [cos( ) cos( )]

neglecting higher frequencies

w w t w w= + + −[cos( ) cos( )]2

2 putting this in aboveeq weget

c m c m

w w t w w

fBut here w π

= putting this in aboveeq.weget2

1 [cos( ) cos( )]

But here w

f ff t f

= + +[cos( ) cos( )]2 2 2

( ) ( )

f fBW f f f

= + + −

+( ) ( )2 2b b

c c bBW f f f= + − − =

Output phase‐versus‐time relationship for a BPSKmodulator

BPSK receiver.

Demodulation

Similarly it can be done for logic 0 (do your self9/18/2013 34Lt Col A K Nigam, ITM University, Gurgaon

Quaternary Phase‐Shift KeyingQuaternary Phase Shift Keying

• QPSK is an M‐ary encoding scheme where N = 2 andQPSK is an M ary encoding scheme where N 2 and M= 4

• Therefore, with QPSK, the binary input data are , Q , y pcombined into groups of two bits, called dibits.

• each dibit code generates one of the four possible g poutput phases (+45°, +135°, ‐45°, and ‐135°).

QPSK transmitter.

• Two bits (di‐bit) are clocked into the bit splitter.

• After both bits have been serially inputted, they areAfter both bits have been serially inputted, they aresimultaneously parallel outputted.

• The I bit modulates a carrier that is in phase withpthe reference oscillator (hence the name "I" for "inphase" channel)

• The Q bit modulate, a carrier that is 90° out ofphase i.e. cosine wave .

l i d l i h ibl h• For a logic 1 = + 1 and a logic 0= ‐ 1 , two phases are possible at theoutput of the –I modulator (+sinwct and ‐ sinwct)

Si il l t h ibl t th t t f th Q b l d• Similarly two phases are possible at the output of the Q balancedmodulator (+coswct), and (‐coswct).

• For input of Q =I= 1 the two inputs to the I balanced modulator are• For input of Q =I= 1, the two inputs to the I balanced modulator are +1 and sinwct, and The two inputs to the Q balanced modulator are +1 and coswct.

• Outputs are(a) I balanced modulator =(+1)(sinwct) = +1 sinwct(b) Q balanced modulator =(+1)(coswct) = +1 coswct(b) Q balanced modulator (+1)(coswct) +1 coswct

Output of the linear summer for 1,1 input dibits is

cosc csin w t w t= +

2 90 90

c csin w t sin w t

⎧ ⎫⎛ ⎞ ⎛ ⎞⎪ ⎪0 02 90 902sin cos2 2

cw t⎧ ⎫⎛ ⎞ ⎛ ⎞+⎪ ⎪= ⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠⎝ ⎠⎩ ⎭

02 902 sin2

cw t⎛ ⎞+= ⎜ ⎟

⎝ ⎠2

1.414sin( 45 }ocw t

⎝ ⎠= +

(Similarly it can be calculated for all other input combinations)9/18/2013 39Lt Col A K Nigam, ITM University, Gurgaon

Truth tableTruth table

Phasor diagram

Constellation diagram

(How to remember: For 1 +ve will go and for 0 –ve will go)I is for sine and Q for cosine

Salient featuresSalient features

• Each of the four possible output phasors has exactly theac o t e ou poss b e output p aso s as e act y t esame amplitude. Therefore, the binary informationmust be encoded entirely in the phase of the outputi lsignal.

• The angular separation between any two adjacentphasors in QPSK is 90°phasors in QPSK is 90 .

• Thus a QPSK signal can undergo almost a +45° or ‐45°shift in phase during transmission and still retain thep gcorrect encoded information when demodulated at thereceiver.

Output phase‐versus‐time relationship for a QPSKmodulatormodulator.

Bandwidth considerations of QPSK

The highest fundamental frequency at the input and fastest rate of change at the output of the balance modulators is equal to one‐fourth of the binary input bit rate.

Mathematical Analysis

.cos (assuming unit amplitude)( )

m cout put cos w t w tneglecting higher frequencies=

1 [cos( ) cos( )]2 c m c mw w t w w= + + −

2 putting this in aboveeq.weget4

fBut here w π=

1 [cos( ) cos( )]2 4 4

b bc c

f ff t f= + + −

( ) ( )4 4 2b b b

c cf f fBW f f= + − − =

QPSK receiver

Output of the I product detectorOutput of the I product detector

(For input I=0 and Q=1)

Output of the Q product detector

Offset QPSK.I difi d f f QPSK h h bi h I d Q• Is a modified form of QPSK where the bits on the I and Qchannels are offset in phase from each other by one‐half of a bittime.

• Because changes in the i channel occur at the midpoints of the qchannel bits and vice versa, there is never more than a single bitchange in the dibitchange in the dibit

• Therefore, there is never more than a 90° shift in the outputphase compared to conventional QPSK in which a change in thep p ginput dibit from 00 to 11 or 01 to 10 causes a corresponding180° shift in the output phase.

Ad t f OQPSK i th li it d h hift th t t b• Advantage of OQPSK is the limited phase shift that must beimparted during modulation.

• disadvantage of OQPSK is that changes in the output phase g Q g p poccur at twice the data rate in either the I or Q channel".

Offset keyed (OQPSK)

Baud and minimum bandwidthBaud and minimum bandwidth

• OQPSK the baud and minimum bandwidthOQPSK the baud and minimum bandwidth are twice that of conventional QPSK for a given transmission bit rategiven transmission bit rate.

• OQPSK is sometimes called OKQPSK (offset‐keyed QPSK)keyed QPSK).

8‐PSK Modulation• With 8 PSK three bits are encoded forming tribits and• With 8‐PSK, three bits are encoded, forming tribits and

producing eight different output phases.

• To encode eight different phases, the incoming bits areencoded in groups of three, called tribits (2^3 = 8)

8‐PSK transmitter.

How to find anglesFor 111 inputFor 111 inputOutput= 1.307sinwt+0.541 coswtAngle= tan^‐1(.541/1.307) in 1st quadrangle=67.5°9/18/2013 54Lt Col A K Nigam, ITM University, Gurgaon

Output Phases

NotePhases are +/ (22 5°+45°)Phases are +/‐(22.5°+45°)

Output phase‐versus‐time relationship for an 8‐PSK modulatorPSK modulator

Phasor DiagramPhasor Diagram

Constellation Diagram: 8 PSK

Note: Format is QIC

I ‐ve (0) I +ve (1)Q +ve (1) Q +ve (1)Q ve ( ) Q ve ( )

I ‐ve (0) I +veQ +ve (1) Q ‐veQ ( ) Q

Bandwidth considerations of 8‐PSK• With 8‐PSK, the data are divided into three channels,• the bit rate in the I, Q, or C channel is equal to one‐third of the binary input

data rate (fb /3).

A l iAnalysiscos . ( )

( )c mout put w t cos w t amplitudeassumed tobeunity

neglecting higher frequencies=( )

1 [cos ( ) cos ( )]2 c m c m

g g g f q

w w t w w= + + −

2 putting this in aboveeq.weget6

fBut here w π=

1 [cos ( ) cos ( )]2 6 6

b bc c

f ff t f

= + + −

( ) ( )6 6 3b b b

c cf f fBW f t f= + − − =

8‐PSK receiver.

8 PSK Performance8 PSK Performance

• With 8‐PSK, the angular separation between adjacent output , g p j pphases is only 45° (360 / 8 ).

• Therefore, 8‐PSK can undergo only a 22.5° phase shift during transmission and still retain its integrity.

16‐PSK

Truth table

Constellation diagram

16 PSK Performance16 PSK Performance

• With 16‐PSK, the angular separation between adjacent , g p joutput phases is only 22.5° (360 / 16).

• Therefore, 16‐PSK can undergo only a 11.25° phase shift during transmission and still retain its integrity.

QUADRATURE – AMPLITUDE MODULATION

8‐QAM

• 8‐QAM is an M‐ary encoding techniquewhere M = 8.

• Unlike 8‐PSK, the output signal from an 8‐QAM modulator is not a constant‐amplitudesignal.

8‐QAM modulator

• The incoming data are divided into groups of three bits (tribits): the

8‐QAM modulator • The incoming data are divided into groups of three bits (tribits): the

I, Q, and C bit streams.

E h t h bit t l t thi d f th i i d t• Each stream has a bit rate equal to one‐third of the incoming datarate.

• The I and Q bits determine the polarity of the PAM signal at theoutput of the 2‐to‐4‐level converters

• The C channel determines the magnitude.

• Because the c bit is fed un‐inverted to both the i and the q channelBecause the c bit is fed un inverted to both the i and the q channel2‐to‐4‐level converters, the magnitudes of the I and Q PAM signalsare always equal.

• Their polarities depend on the logic condition of the i and q bitsand, therefore, may be different.9/18/2013 67Lt Col A K Nigam, ITM University, Gurgaon

Truth TableTruth Table

Phasor and constellation diagramg

Output phase and amplitude‐versus‐timerelationship for 8 QAMrelationship for 8‐QAM

Bandwidth considerations of 8‐QAM.

• N=3N=3

• Thus the minimum bandwidth required for 8‐QAM is fb / 3 the same as in 8 PSKQAM is fb / 3, the same as in 8‐PSK.

8‐QAM receiver.

• An 8‐QAM receiver is almost identical to theAn 8 QAM receiver is almost identical to the 8‐PSK receiver

16‐QAM

• As with the 16‐PSK 16‐QAM is an M‐aryAs with the 16 PSK, 16 QAM is an M arysystem where M =16.

• The input data are acted on in groups of four• The input data are acted on in groups of four(2^4 = 16).

A i h 8 QAM b h h h d h• As with 8‐QAM, both the phase and theamplitude of the transmit carrier are varied.

QAM transmitter

• The input binary data are divided into four channels: I, I', Q, and Q'.

• The bit rate in each channel is equal to one‐fourth of the input bit rateThe bit rate in each channel is equal to one fourth of the input bit rate (fb/4).

• The I and Q bits determine the polarity at the output of the 2 to 4 level• The I and Q bits determine the polarity at the output of the 2‐to‐4‐level converters

• logic 1 = positive• logic 0 =negative

• The I' and Q' bits determine the magnitudeThe I and Q bits determine the magnitude • logic 1 = 0.821V • logic 0 = 0.22 V

QAMtransmitter

I and Q, 1 = +ve logic 0 =‐ve

I’ and Q’ logic 1 = 0.821V, logic 0 = 0.22 V

Computation of values of amplitude and phasesComputation of values of amplitude and phases

• The outputs from the I and Q channel product modulators arecombined in the linear summer and produce a modulated output

• For a quadbit input of I= 1, I' = 0, Q = 1, and Q' = 0 i.e.(1010) logic 1q p , , Q , Q ( ) gfor I and Q gives +ve value and logic 0 for I’ and Q’ gives 0.22 V

•• Thus summer output = 0.22 sin wc t + 0.22cos wc tThus summer output 0.22 sin wc t + 0.22cos wc t

=0.22{sinwct + sin(90 + wct)}=0.22[2{sin(wct + 45). cos45}=0.311sin(wct+45)

• Similarly other values of amplitude and phase can be computed

Values of amplitude and phases

Phasor and constellation diagramg

1 0 2 3

How to remember9/18/2013 79Lt Col A K Nigam, ITM University, Gurgaon

BANDWIDTH EFFICIENCY

• Bandwidth efficiency isBandwidth efficiency is

Example

Qu. For 16‐PSK and a transmission system with a 10 kHz bandwidth, determine the maximum ,bit rate.

Solutionh b d id h ffi i f i hi h• The bandwidth efficiency for 16‐PSK is 4, which means that four bits can be propagated through the system for each hertz of bandwidth.the system for each hert of bandwidth.

• Therefore, the maximum bit rate is simply the product of the bandwidth and the bandwidth ffi iefficiency, or

• bit rate = 4 x 10,000 = 40,000 bps

DIFFERENTIAL PHASE‐SHIFT KEYING

Is an alternative form of digitalgmodulation where the binary inputinformation is contained in the differencebetween two successive signalingelements rather than the absolute phase.p

OP of mod is same for input as ‘1’ and is shifted by 180° for input ‘0’

Input Data

(initial reference bit is assumed a logic 0, If the initialreference bit is assumed a logic 1, the output from theXNOR circuit is simply the complement of that shown)

DemodulationDemodulation

Change of phase indicates 0 same phase indicates 1Change of phase indicates 0, same phase indicates 1

PROBABILITY OF ERROR AND BIT ERROR RATE• Probability of error P(e) and bit error rate (BER) are often used• Probability of error P(e) and bit error rate (BER) are often used

interchangeably

• It is a function of the carrier‐to‐noise power ratio (or, morespecifically, the average energy per bit‐to‐noise power densityratio) and the number of possible encoding conditions used (M‐ary)ary).

• Energy per bit is simply the energy of a single bit of information.Mathematically,

Energy per bit is Eb = C.Tb=C/fb.............................1

• Noise power density is the thermal noise power normalized to a1 H b d idth (i th i t i 1 H1‐ Hz bandwidth (i.e., the noise power present in a 1‐Hzbandwidth). Mathematically, noise power density is

No = N/B N= Noise, B= BW…………………………2/ ,

From eq 1 and 2 we can writeFrom eq 1 and 2 we can write

b bE C fN N B

C BN f

⎛ ⎞⎛ ⎞= ×⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠B

N fStated in dB

⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞⎛ ⎞ +⎜ ⎟ ⎜ ⎟⎜ ⎟0

dB B dBdBN N f

⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

dBB dB dB

EB Cf N N

⎛ ⎞⎛ ⎞ ⎛ ⎞= −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

PSK Error Performance• For PSK systems, the phase

difference between two consecutivei li i t i 2 / Msignaling points is α =2π / M

• For error free transmission themaximum shift of angle permissible α =2 π/2Mg pis α= 2π /2M either way fromsignaling point…………………………….1

If d di b i

α =2 π/2Md

• If d= distance between consecutivetwo signaling points and D=Peaksignal amplitude then

sin α=(d/2)/D or

• Sin(2π /2M)=d/2D from Eq. 1

or d/2 =D.sin(π/M)…………………….2

Computation of Bit Error Probability

• If no. of states =M, then no. of bits required to encode these=log2M

If E /N i th bit t i d it ti th i• If Eb/N0 is the bit energy to noise density ratio then energy in all bits

log bED M⎛ ⎞

= × ⎜ ⎟

• Putting this in eq. 2 we get

logD MN

= × ⎜ ⎟⎝ ⎠

/ 2 log ................3bEd Sin MM Nπ ⎛ ⎞⎛ ⎞= × ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠• Now the probability that the noise will be greater than z is

given by error function

2 ∞22( ) z

P z e dzπ

∞−= ∫

• Now maximum error can be d/2, thus error will be obtained / ,by putting z=d/2 from eq. 3

• As there are log2M bits per signal, the bit error probability ld bwould be

1( ) ( )log

P e erfc zM

• Where Z is given by

2log M

/ 2 log ................3bEz d Sin MM Nπ ⎛ ⎞⎛ ⎞= = × ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

• And M is no. of levels

Example of BPSKExample of BPSK

M=2, thus from eq. 3 we get

/ 2 log 22

bEd SinN

π ⎛ ⎞⎛ ⎞= × ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

1( )l 2

b bE EP e erfc erfcN N

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠2 0 0log 2 N N⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Example of QPSKExample of QPSK

/ 2 log 4 bEd Si π ⎛ ⎞⎛ ⎞ × ⎜ ⎟⎜ ⎟ 20

/ 2 log 44

d SinN

= × ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟

22 N N⎜ ⎟

⎝ ⎠

1( )log 2

b bE EP e erfc erfcN N

= =2 0 0g

Error rates ofError rates of PSK modulationmodulation systems

QAM Error Performance

1 1( ) ( )l

LP e erfc zL L

−⎛ ⎞= ⎜ ⎟⎝ ⎠2

( ) ( )log

Where z is givenby

⎜ ⎟⎝ ⎠

bL EzL N

= ×− 0

FSK Error Performance

• The probability of error for coherent FSK is

1( )2 2

bEP e erfcN

=02 2N

ASK Error Performance

• The probability of error for coherent ASK isp y

1( ) bEP e erfc⎛ ⎞

= ⎜ ⎟⎜ ⎟0

( )2 2

P e erfcN⎜ ⎟⎜ ⎟

⎝ ⎠

• The probability of error for non‐coherent FSK is

1 E⎛ ⎞

1( )2 2

bEP e erfcN

⎛ ⎞−= ⎜ ⎟

⎝ ⎠

DPSK Error PerformanceDPSK Error Performance

bENP e e−

P e e=

Finding Error function values usingFinding Error function values using approximate formula

( )xef

( )erfc xx π

For exampleperfc(.7)=.493erfc(.5)=.878

BANDWIDTH EFFICIENCY

• Also called information density or spectral efficiency, often y p y,used to compare the performance of one digital modulation technique to another.

B d idth ffi i i d fi d• Bandwidth efficiency is defined as

Clock RecoveryClock Recovery

• Squaring LoopSquaring Loop

• Costas Loop

d l• Remodulator

Squaring Loop• The incoming modulated signal is squared and band‐pass

filtered to extract the carrier component at 2 times itsoriginal frequencyoriginal frequency.

• This signal is then fed into a phase locked loop whose otherinput comes from a VCO.

• The error output of the phase locked loop is converted into aDC voltage which is fed back into the VCO to cause it tooscillate at a frequency which is almost same as the carrieroscillate at a frequency which is almost same as the carrierfrequency such that the error output reduces to nearly zero.

• This is then divided by two to give the in phase carrierfrequency

BPF DividerPLLSquarer

i ( )I t t t PSK i t±

BPF DividerPLLSquarer

sin ( )

cInput to squarer w t PSK inputOutput of squarer

2sin1 (1 cos 2 )2

= + −( )2

This is filtered to give1 cos 22

This is divided by to givein phasecarrier frequency

2This is divided by to givein phasecarrier frequency

COSTAS LOOP

• It consists of two coherent detectors supplied with the samereceived PSK inputreceived PSK input

• Carrier is generated locally by using a VCO which is having aphase difference of φ for simplicity we assume that it hasamplitude =1 volt.

• This carrier is given as it is to I product modulator

O h d fi i li d h l ll d i• Other product figure is applied the locally generated carrierwith phase of 90° shift of as shown.

• Both the outputs of I and Q channel are passed through aBoth the outputs of I and Q channel are passed through aLPF and are fed to a phase discriminator which is consistingof a multiplier followed by a low pass filter

f h f l l h h l• Output of the final LPF is error voltage which is proportionalto sin2φ and it corrects VCO frequency to carrier frequencyby bringing to φ to 0y g g φ

. c o s . ( ) c o s ( )c c c

O u tp u t o f I M o d u la to r isA w t m t w tA

φ= × +

{ }( ) c o s ( 2 ) c o s2

( ) c o s2

A m t w t

AA fte r L P F it is m t

= → →2

. c o s . ( ) s in ( )c c c

O u tp u t o f Q M o d u la to r isA w t m t w t φ= × +

( ) s in ( 2 ) s in ( )2

( ) s in ( 2 ) s in

A m t w t

= + − −

= + +{ }( ) s in ( 2 ) s in2 cm t w t

A f te r L P

φ φ+ +

( ) s in2

cAF i t is m t φ=

m in1( ) c o s . ( ) s in s in 2 ( )

2 2 4 2c c c

O u tp u t o f p h a s e d is c r i a to r isA A Am t m t m tφ φ φ⎡ ⎤= = ⎢ ⎥⎣ ⎦

0 0a s e r r o r v o l ta g e g o e s toa n d i t lo c k s to th e c a r r ie r fr e q u e n c y

φ⎣ ⎦

→9/18/2013 110Lt Col A K Nigam, ITM University, Gurgaon

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