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