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ECNG 6703 - Principles of CommunicationsLinear Modulation - Modulation, Demodulation, Detection &

Synchronization: Part II

Sean Rocke

November 4th, 2013

ECNG 6703 - Principles of Communications 1 / 46
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Outline

1 Mary Baseband Pulse Amplitude Modulation

2 Mary Quadrature Amplitude Modulation

3 Maximum Likelihood Detection

4 Modulation Performance

5 Conclusion

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Mary Baseband Pulse Amplitude Modulation

Mary Baseband Pulse Amplitude Modulation (PAM)

1Dimensional signal set with basis function 0(t) =p(t), wherep(t)is any unit energy pulse.

Consider a symbol sequence with a new symbol occurring every

Tsseconds.The resulting PAM signal is given by,s(t) =

na(n)p(t nTs)

nmay be a finite or infinite symbol sequence.

Minimum Euclidean distance between any two symbols in symbolspace is 2A.

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Baseband PAM Constellation Examples

Question:The average energy of a signal set is given byEavg=

ipiEi, where pi

andEiare the probability of occurrence and energy of symbol i,respectively. Show that, assuming equiprobable symbols, the average

energy of the general PAM constellation is given by

Eavg=

M213 A

2.

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PAM: ContinuousTime Realization

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PAM: ContinuousTime Realization

Received signal: r(t) =s(t) +w(t),w(t)zeromean white

Gaussian noise with PSD N02 W/Hz

MF output:

x(t) =r(t) p(t) = T2+t

T1+t r()p( t)d

x(t) =

ka(k)T2+t

T1+t p( kTs)p( t)d + T2+tT1+t w()p( t)dMF output sampled att=lTs:

x(lTs) =

ka(k)rp((k l)Ts) +v(lTs) =a(l) +v(lTs)

1 rp(mTs) =T2

T1 p(t)p(t mTs)dt= 1, m= 00, m=02 v(lTs)is noise projection in signal space.

Outcome:In signal space the noise term causes the received

signal term to be displaced from the constellation location.

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Example: Binary PAM, CT Realization


M B b d P l A li d M d l i
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PAM: DiscreteTime Realization

Received signal after ADC:r(nT) =s(nT) +w(nT)

MF output:

x(nT) =r(nT) p(nT) = n+ T2Tm=n+

T1T

r(mT)p(mT nT)

x(nT) = n+

T2T

m=n+T1T

(la(l)p(mT lTs)) p(mT nT) +n+ T2T

m=n+T1T

w(mT)p(mT nT)x(nT) = 1T

la(l)rp(lTs nT) +v(nT)

MF output after sample rate conversion n=kTsT:

x(kTs) = 1T

la(l)rp(lTs kTs) +v(kTs) = a(k)T +v(kTs)1 v(kTs) N(0, N02T2 )is noise projection in signal space.


signal term to be displaced from the constellationlocation.


M B b d P l A lit d M d l ti
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PAM: DiscreteTime Realization


M ary Quadrature Amplitude Modulation
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Mary Quadrature Amplitude Modulation

Mary Baseband Quadrature Amplitude Modulation

(QAM)

2Dimensional signal set with basis functions:1 0(t) =

2p(t)cos(0t)

2 1(t) =

2p(t)sin(0t), wherep(t)is any unit energy pulse.

3 Orthonormal basis functions are 900 apart.

At symbol rate,

1

Ts symbols/s, the general MQAM signal is a pulsetrain given by,

s(t) =

2

na0(n)p(t nTs)cos(0t) a1(n)p(t nTs)sin(0t)s(t) =I(t)

2cos(0t)

Q(t)

2sin(0t)

1 Inphase: I(t) =

na0(n)p(t nTs)2 Quadrature:Q(t) =

na1(n)p(t kTs)

Symbol energy for thenth symbol is

T2+nTsT1+nTs

s2(t)dt=a20(n) +a21(n)


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Baseband MQAM Constellation Examples: MPSK

Points equallyspaced around a circle of radiusEavg.Signals have the same energy, and only differ in phase.


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Baseband MQAM Constellation Examples: Square

MQAM

Points on equallyspaced square grid. Only exist for

M=22n, n=1, 2, . . .Signal point projections on0(t) &1(t)axes:

A(M 1),A(M 3), . . . ,A, A, . . . , A(M 3),A(M 1).Average signal energies:

1 M=4 Eavg=2A22 M=16 Eavg=10A23 M=64 Eavg=42A24

GeneralM Eavg= 2

3 (M 1)A2


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Baseband MQAM Constellation Examples: CCITT

More tolerant to phase jitter than equivalent square QAMFallback subset is used when SNR is not high enough to allow

reliable communications.

Can be thought of as APSK constellations.


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Baseband MQAM Constellation Examples: minimum

PeConstellations

Designed to minimize error probability, Pe.

Points constrained to lie on a rectangular grid / concentric circles.Constrained optimization used to minimizePethrough

maximization of normalized Euclidean distances between points.

Much more complex decision regions than square QAM, so not

frequently used in practice.


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y p

QAM Modulator: ContinuousTime Realization


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QAM Demodulator: ContinuousTime Realization


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MQAM: ContinuousTime Realization

Received signal:

r(t) =Ir(t)2cos(0t) Qr(t)2sin(0t) +w(t)MF outputs:

x(t) =T2+t

T1+t Ir()p( t)d +v0(t)

y(t) =T2+t

T1+t Qr()p( t)d +v1(t)v0(t)andv1(t)are MF outputs due to noise.

MF outputs sampled att=lTs:

x(lTs) = ka0(k)rp((k l)Ts) +v0(lTs) =a0(l) +v0(lTs)y(lTs) =ka1(k)rp((k l)Ts) +v1(lTs) =a1(l) +v1(lTs)v0(lTs)andv1(lTs)are noise projections in signal space.




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Example: QPSK Modulator, CT Realization


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Example: QPSK Demodulator, CT Realization


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MQAM: DiscreteTime Realization

Received signal after BPF & ADC:

r(nT) =Ir(nT)2cos(0n) Qr(nT)2sin(0n) +w(nT)After Direct Digital synthesis mixing:

r(nT)

2cos(0n) =Ir(nT) +Ir(nT)cos(20n)

Qr(nT)sin(20n) +w0(nT)

r(nT)2sin(0n) =Qr(nT) +Ir(nT)sin(20n) +Qr(nT)cos(20n) +w1(nT)

MF outputs:

x(nT) =n+ T2

Tm=n+

T1T

Ir(mT)p(mT nT) +v0(nT))

y(nT) =n+ T2

T

m=n+T1T

Qr(mT)p(mT nT) +v1(nT))


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Assuming frequency & phase synchronization, MF outputs

after sample rate conversionn=kTsT:

x(kTs) = 1T la0(l)rp(lTs

kTs) +v0(kTs) = a0(k)

T +v0(kTs)

y(kTs) = 1T

la1(l)rp(lTs kTs) +v1(kTs) = a1(k)T +v1(kTs)

1 v(kTs) N(0, N02T2 )is noise projection in signal space.




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Maximum Likelihood Detection
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Maximum Likelihood (ML) Detection

Scenario:

1 Detector is presented with a series of numbers, r, corresponding toLotransmitted symbols (1 of Mconstellation points),a= [a(0)a(1) . . . a(L0 1)]T.

2 True data sequence,a, unknown (i.e., uncertainty about true datasequence at receiver)

Detection:1 How do we estimate the transmitted sequence?

Detection Problem:

1 Detection Task:estimatingais based on observing samplesr.2 Prior symbol sequence probabilities: P(a)3 Probability transition model: P(r|a)4 Detection objective: Select candidate sequence that maximizes

P(a|r)a= argmaxa{P(a|r)}= argmaxa{P(r|a)P(a)}



M i Lik lih d (ML) D i
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a= argmaxa{P(a|r)}is actually theMaximum a Priori (MAP)detection criterion

Requires knowledge ofP(r

|a)andP(a).

In generalP(a)may be unknown

Designer typically assumes symbol sequences are equally likely.

Thus,P(a) = 1ML0

anda=argmaxa{P(r|a)}(ML decision rule)Requires knowledge ofP(r

|a)only.



M i Lik lih d (ML) D t ti
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Consider observing received signal,r(t), consisting ofL0 symbols,each of durationTsseconds

Received signal,r(t) =s(t) +w(t), wherew(t)is a zeromeanwhite Gaussian RP with PSD N02 W/Hz

Signal sampled everyTseconds to get

r(nT) =s(nT) +w(nT), n=0,1, . . . , NL0 1Assumptions:

1 Phase/frequency/symbol timing synchronization2 ExactlyNsamples/symbol interval

Observation/sample vectors:1 r= [r(0)r(T) . . . r((NL0 1)T)]T2 s= [s(0)s(T) . . . s((NL0 1)T)]T3 w= [w(0)w(T) . . . w((NL0 1)T)]T



M i Lik lih d (ML) D t ti
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IID noise signal samples,w(nT) N(0, N02T)Hence,P(w) = 1

(22)L0N

2

e 1

22NL

0

1

n=0 w2(nT)

Symbol vector,a= [a(0)a(1) . . . a(L0 1)]T, where1 a(k) = [a0(k)a1(k) . . . aK1(k)]

T

2

a(k) S={s0, s1, . . . , sM1}The likelihood function,

P(r|a) = 1(22)

L0N

2

e 1

22

NL01n=0 |r(nT)s(nT;a)|

2

The loglikelihood function,(a) =log

{P(r

|a)

}ML estimates1 a= argmaxaSL0 {(a)}

a= argmaxaSL0 {L01

k=0 |x(k) a(k)|2}(for symbol sequence)2 a(k) =argmaxaS

{|x(k)

a(k)

|2

}(for single symbol)


Modulation Performance

E l PAM B d idth P f
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Example: PAM Bandwidth Performance

Consider independent & equallylikely symbols in pulse train

s(t) =

na(n)p(t nTs)PSD given byPs(f) =

EavgTs

|P(f)|2, where1 P(f)- continuoustime fourier transform of pulse shape p(t)

Pusle shape examples:

1 NRZ - nonreturn to zero2 RZ - Return to zero3 MAN - Manchester4 HS - Half Sine5 SRRC - SquareRoot Raised Cosine

Pulse shape BW typically of the formBW = BTs =

BRblog2M, where

1 B- constant depending on pulse shape & BW definition adopted2 Rs- symbol rate (symbols/s)3 Rb- bit rate (bits/s)4 M- constellation size (symbols)



Pulse Shape Examples
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Babs:|P(f)|2 =0 forf BabsB%:

B%0 |P(f)|2df = 100

0 |P(f)|2df

-60dB BW:

|P(f)

|2


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Pulse Shape Examples: RC
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Pulse Shape Examples: RC



Pulse Shape Examples: SRRC
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Pulse Shape Examples: SRRC



Calculating Error Probability
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P(E) =

M1m=0P(E|a=sm)P(a=sm)

Typical assumption: equiprobable symbol occurrences

P(a= sm) = 1M, m=0,1, . . . , M 1

P(E) = 1MM1m=0P(E|a=sm)

How to calculateP(E|a=sm)?



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P(E|a(k) = +A) =1P( 2ATx(kTs)) =1P( ATv(kTs) AT)P(E|a(k) = +3A) =1P(0x(kTs) 2AT) =1P( ATv(kTs))P(E) = 32 Q

2Eavg5N0

Pb(E) = 34 Q4Eb5N0



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For Mary PAM:

P(E) =2 M1M Q

6Eavg(M21)N0

Pb(E) =

2(M1)Mlog2M

Q6log2MEb(M21)N0



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Calculating Error Probability: Union Bound


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Ca cu at g o obab ty U o ou d

Useful approximation when decision regions are irregularlyshaped, making the previous approach extremely complicated

Assumings0 was transmitted,

P(E|s0) =P([s=s1] [s=s2] . . . [s= sM1]|s0)

Since probability of union of events is upper bounded by the sumof the probabilities of the events, P(E|s0)M1

n=1 P(s= sn|s0)Assuming equallylikely symbols,

P(E) 1M

M1m=0

M1n=0 P(s= sn|sm)

Each pairwise error probability given by,

P(s= sn|sm) =Q

dm,n2Tv

=Q

d2m,n

2Eavg

EavgN0



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g y



Error Probability Comparison
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y p



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y p



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y p



Spectral Efficiency Comparison
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Conclusion

Conclusion
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We covered:

Modulation & Demodulation Realization

Modulation & Demodulation Performance Evaluation

Your goals for next class:

Continue ramping up your MATLAB & Simulink skills

Work on your CW exercises

Work on HW 4for submission next week


Q & A

Thank You
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Questions????

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