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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 d k = m k d k – 1, where m k is the input bit and d k is the encoded bit at time k, (d 0 = 0). The modulator maps d k = 0 to ‘cos (0) = +1’ and d k = 1 to ‘cos( ) = –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?

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DPSK modulation and encoding 9876543210k 110110100 Encoded bit d k 1 1 1 1 +1 0 1 1 1 1 0 +1 0 1 0 0 0 0 0 DPSK Symbol 1Encoded bit d k-1 0Information bit m k d k = m k d k – 1

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DPSK demodulation (1) We send cos( k = 0) or cos( k = ): But we will receive: k-k-

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DPSK demodulation (2) We form the following variable at the receiver side to determine the information bit that was sent

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DPSK demodulation (3) d k, d k-1 k - k-1 mkmk 0 00 0 1 1 1 0 1 1 0 From d k = m k d k – 1 we conclude that m k = d k d k – 1.

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Demodulation of m in question 1 k0123456789 DPSK symbols +1 +1 +1 k - k-1 00 0 0 mkmk 001110110 If k - k-1 = 0, we decide m k = 0. If k - k-1 = , then m k = 1.

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DPSK performance in AWGN The BER performance of DPSK in AWGN is given by:

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Rayleigh fading channels (1) The received signal model in Rayleigh flat fading channels:

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Rayleigh fading channels (2) We assume that phase distortion has been compensated or is not a deciding factor in the demodulation (as in DPSK):

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Rayleigh fading channels (3) The instantaneous signal to noise ratio in fading channels is affected by and is given by: Therefore the instantaneous BER performance of DPSK signaling is also affected by and is given by: Therefore the average error probability is the given by:

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Integrating

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DPSK performance comparison

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