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Question 1 Modulate an input stream of binary bits: m = 0 0 1 1 1 0 1 1 0,with DPSK signaling. Assume that encoder at the transmitter is dk = mk dk – 1, where mk is the input bit and dk is the encoded bit at time k, (d0 = 0). The modulator maps dk = 0 to ‘cos (0) = +1’ and dk = 1 to ‘cos(p) = –1’. Explain how the receiver can demodulate the received signal non-coherently. What is the bit error probability of DPSK signaling in AWGN channel? What is the bit error probability of DPSK signaling in Rayleigh fading channel?
DPSK modulation and encoding dk = mk dk – 1 k 0 1 2 3 4 5 6 7 8 9 Information bit mk 0 0 1 1 1 0 1 1 0 Encoded bit dk-1 0 0 0 1 0 1 1 0 1 1 Encoded bit dk 0 0 1 0 1 1 0 1 1 DPSK Symbol +1 +1 +1 -1 +1 -1 -1 +1 -1 -1
DPSK demodulation (1) • We send cos(qk = 0) or cos(qk = p): • But we will receive: qk-g
DPSK demodulation (2) • We form the following variable at the receiver side to determine the information bit that was sent
DPSK demodulation (3) From dk = mk dk – 1we conclude that mk = dk dk – 1.
Demodulation of m in question 1 If qk - qk-1= 0, we decide mk = 0. If qk - qk-1= p, then mk = 1.
DPSK performance in AWGN • The BER performance of DPSK in AWGN is given by:
Rayleigh fading channels (1) • The received signal model in Rayleigh flat fading channels:
Rayleigh fading channels (2) • We assume that phase distortion has been compensated or is not a deciding factor in the demodulation (as in DPSK):
Rayleigh fading channels (3) • The instantaneous signal to noise ratio in fading channels is affected by a and is given by: • Therefore the instantaneous BER performance of DPSK signaling is also affected by a and is given by: • Therefore the average error • probability is the given by:
