1. 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?

2. 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

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

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

5. 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.

6. 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.

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

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

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

10. 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:

11. Integrating

12. DPSK performance comparison