TELE4653 Digital Modulation & CodingDetection Theory

Wei Zhang

w.zhang@unsw.edu.au

School of Electrical Engineering and TelecommunicationsThe University of New South Wales

April 19th, 2010

Short Review

Last lecture

I MAP v.s. ML receivers

I Decision region

I Error probability

I Correlation and MF receiver

Today’s

Error probability analysis for the memoryless modulation schemes

I PAM

I PSK

I QAM

ASK

ASK ConstellationM-ary PAM signals are represented by

sm =

√Eg2Am

where Eg is the energy of the basic signal pulse. The amplitude is

Am = (2m − 1−M)d , m = 1, 2, . . . ,M

and the distance between adjacent signals is dmin = d√

2Eg .

ASK - Symbol Error Probability

Average Energy

Assuming equally probable signals

Eav =1

M

M∑m=1

Em =d2Eg2M

M∑m=1

(2m − 1−M)2

=(M2 − 1)d2Eg

6. (Recall Tutorial 1)

Symbol Error Probability

On the basis that all amplitude levels are equally likely

PM =M − 1

MPr{|r − sm| > d

√Eg2}

=2(M − 1)

MQ(√d2Eg

N0

)

ASK - Symbol Error Probability

Pe for PAMIn plotting, it is preferred to use theSNR per bit rather than per symbol,i.e. Eb av = Eav

log2 M, and

PM =2(M − 1)

MQ(√

(6 log2 M)Eb av

(M2 − 1)N0

)

Signal Representation

PSKRecall the signal waveform

sm(t) = g(t) cos[2πfct +2π

M(m − 1)], m = 1, 2, . . . ,M, 0 ≤ t ≤ T

and have the vector representation

sm = [√Es cos

2π

M(m − 1)

√Es sin

2π

M(m − 1)]

Recall the optimum detector for AWGN channel, the correlationmetrics

C (r , sm) = r • sm r = [r1 r2]

Distribution of Phase

Phase of Received Signal

Due to the symmetry of the signal constellation and the noise, weconsider the transmitted signal phase is Θ = 0, i.e. s1 = [

√Es 0],

and the received signal vector

r1 =√Es + n1

r2 = n2.

As a result, the pdf. function is

pr (r1, r2) =1

2πσ2r

e− (r1−

√Es )2+r2

22σ2

r .

Distribution of Phase

Phase Θr of Received Signal

Denote

V =√r21 + r2

2

Θr = tan−1 r2r1

which yields

pV ,Θr (V ,Θr ) =V

2πσ2r

e−V 2+Es−2

√EsV cos Θr

2σ2r

and take the marginal probability is

pΘr (Θr ) =1

2πe−γs sin2 Θr

∫ ∞0

Ve−(V−√

2γs cos Θr )2

2 dV .

Error Probability for PSK

PM = 1−∫ π

M

− πM

pΘr (Θr )dΘr

QAM

Signal RepresentationQAM signal waveform

sm(t) = Amcg(t) cos(2πfct)− Amsg(t) sin(2πfct)

⇒ sm = [Amc

√Eg2

Ams

√Eg2]

4-QAM

The error probability is dominated by the minimum distance between pairs ofsignal points. Therefore, the scheme performance can be compared through theaverage transmitter power P

(a)av = P

(b)av = 2A2 (T1) where dmin = 2A.

QAM

8-QAM

QAM

16-QAM and aboveRectangular QAM:

I easily generated - two PAM signal

I easily demodulated

I average power slightly greater than the optimal

Therefore, it is most frequently used in practice, and its errorprobability is given as

PM = 1− (1− P√M)2

where

P√M = 2(1− 1√M

)Q(√ 3Eav

(M − 1)N0

)if M = 2k and k is even.

QAMWhen k is oddThe symbol error probability is tightly upper-bounded by

PM ≤ 1−[1− 2Q

(√ 3Eav

(M − 1)N0

)]2

≤ 4Q(√ 3Eav

(M − 1)N0

)