Question 1 Modulate an input stream of binary bits: m = , 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?

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

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

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

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

Demodulation of m in question 1 k DPSK symbols +1   k -  k-1 00  0  0 mkmk If  k -  k-1 = 0, we decide m k = 0. If  k -  k-1 = , then m k = 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  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:

Integrating

DPSK performance comparison