Upcoming SlideShare
×

# Tele4653 l7

398 views

Published on

Published in: Design
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
398
On SlideShare
0
From Embeds
0
Number of Embeds
6
Actions
Shares
0
11
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Tele4653 l7

1. 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. 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. 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. 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. 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. 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. 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. 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. 9. Error Probability for PSK π M PM = 1 − pΘr (Θr )dΘr π −M
10. 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. 11. QAM 8-QAM
12. 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. 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