1. TELE4653 Digital Modulation & Coding
Detection Theory
Wei Zhang
w.zhang@unsw.edu.au
School of Electrical Engineering and Telecommunications
The University of New South Wales
April 19th, 2010

2. Short Review
Last lecture
MAP v.s. ML receivers
Decision region
Error probability
Correlation and MF receiver
Today’s
Error probability analysis for the memoryless modulation schemes
PAM
PSK
QAM

3. ASK
ASK Constellation
M-ary PAM signals are represented by
Eg
sm = Am
2
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 .

4. ASK - Symbol Error Probability
Average Energy
Assuming equally probable signals
M M
1 d 2 Eg
Eav = Em = (2m − 1 − M)2
M 2M
m=1 m=1
(M 2 − 1)d 2 E g
= . (Recall Tutorial 1)
6
Symbol Error Probability
On the basis that all amplitude levels are equally likely
M −1 Eg
PM = Pr{|r − sm | > d }
M 2
2(M − 1) d 2 Eg
= Q
M N0

5. ASK - Symbol Error Probability
Pe for PAM
In plotting, it is preferred to use the
SNR per bit rather than per symbol,
E
i.e. Eb av = log avM , and
2
2(M − 1) (6 log2 M)Eb av
PM = Q
M (M 2 − 1)N0

6. Signal Representation
PSK
Recall the signal waveform
2π
sm (t) = g (t) cos[2πfc t + (m − 1)], m = 1, 2, . . . , M, 0 ≤ t ≤ T
M
and have the vector representation
2π 2π
sm = [ Es cos (m − 1) Es sin (m − 1)]
M M
Recall the optimum detector for AWGN channel, the correlation
metrics
C (r , sm ) = r • sm r = [r1 r2 ]

7. Distribution of Phase
Phase of Received Signal
Due to the symmetry of the signal constellation and the noise, we
√
consider the transmitted signal phase is Θ = 0, i.e. s1 = [ Es 0],
and the received signal vector
r1 = Es + n1
r 2 = n2 .
As a result, the pdf. function is
√ 2 +r 2
1 − (r1 − 2σs2)
E 2
pr (r1 , r2 ) = 2
e r .
2πσr

8. Distribution of Phase
Phase Θr of Received Signal
Denote
V = 2
r1 + r22
r2
Θr = tan−1
r1
which yields
2 +E −2√E V cos Θ
V −V s
2
s r
pV ,Θr (V , Θr ) = 2
e 2σr
2πσr
and take the marginal probability is
∞ √
(V − 2γs cos Θr )2
1 −γs sin2 Θr
pΘr (Θr ) = e Ve − 2 dV .
2π 0

9. Error Probability for PSK
π
M
PM = 1 − pΘr (Θr )dΘr
π
−M

10. QAM
Signal Representation
QAM signal waveform
sm (t) = Amc g (t) cos(2πfc t) − Ams g (t) sin(2πfc t)
Eg Eg
⇒ sm = [Amc Ams ]
2 2
4-QAM
The error probability is dominated by the minimum distance between pairs of
signal points. Therefore, the scheme performance can be compared through the
(a) (b)
average transmitter power Pav = Pav = 2A2 (T1) where dmin = 2A.

11. QAM
8-QAM

12. QAM
16-QAM and above
Rectangular QAM:
easily generated - two PAM signal
easily demodulated
average power slightly greater than the optimal
Therefore, it is most frequently used in practice, and its error
probability is given as
PM = 1 − (1 − P√M )2
where
1 3Eav
P√M = 2(1 − √ )Q
M (M − 1)N0
if M = 2k and k is even.

13. QAM
When k is odd
The symbol error probability is tightly upper-bounded by
3Eav 2 3Eav
PM ≤ 1 − 1 − 2Q ≤ 4Q
(M − 1)N0 (M − 1)N0