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Explain the operating principal, working of Differentially Encoded Phase Shift Keying modulator and demodulator?
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  • The transmitter (modulator) of a DEPSK system is identical to the DPSK transmitter.

  • The differential phase shift keying (DPSK) is a modulation of BPSK.

  • Fig shows the block diagram of DPSK generator and the relevant waveforms.

enter image description here

Operation:

  • d (t) represents the data stream which is to be transmitted .It is applied to one input of an EX-OR logic gate.

  • The EX-OR gate output “b (t)” is delayed by one bit period Tb and applied to the other input of the EX-OR gate. The delayed output is represented by “b(tTb)

  • Depending on the values of “”d (t)” and “b(tTb)” ,the EX-OR gate produces the output sequence “b (t)” .The waveforms for DPSK generator have been drawn by arbitrarily that in the first interval b (0)=0.

  • Output of EX-OR gate is then applied to a bipolar NRZ level shifter which converts “b(t)” to a bipolar level signal b’(t) as shown in the fig.

enter image description here

  • This bipolar NRZ signal b’(t) is then multiplied with the carrier signal to produce the DPSK signal. The DPSK output signal is mathematically expressed as:

    VDPSK(t)=b(t)2PscosωCt

  • So when b(t) =1 , b’(t) =1 hence

    VDPSK(t)=b'(t)2PscosωCt

  • That means no phase shift has been introduced.

    Bt when b(t) =0, b’(t)= -1 hence

enter image description here

  • Thus 180 degree phase shift is introduced to represent b(t) =0

  • The block diagram of DEPSK receiver is shown below. It shows that the signal b (t) is recovered from the received signal, using the synchronous demodulation technique. This is same as the BPSK detection. Once the signal b (t) is recovered, it is applied to one input of an EX-OR gate.

  • The signal b(t) is also applied to a time delay circuit and the delayed signal b(tTb) is applied to the other input of the EX-OR gate as shown below

enter image description here

  • If b(t)= b(tTb) then output of the EX-OR gate will be 0

    d (t) =0 ………if b(t)=b(tTb)

And if b(t) = b(tTb) then output of the EX-OR gate will be 1

d (t) =1 ……… b(t) = b(tTb)

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