Why DPSK over PSK; DPSK modulator, demodulator, signal space

Why DPSK is preferred over PSK

Coherent PSK needs a local carrier exactly in phase with the received carrier; carrier recovery circuits are complex and suffer from a $180^\circ$ phase ambiguity. DPSK sends information in the phase difference between successive bits, so the previous bit serves as the reference. No coherent carrier is needed, the receiver is simpler, and phase ambiguity does not matter. The cost is about 1 dB more $E_b/N_0$ ($P_e=\tfrac12e^{-E_b/N_0}$).

Modulator

 b_k -->[XNOR]---d_k--->[level shift]-->(X)--> DPSK
          ^       |      1->+1,0->-1     ^
          |   [delay Tb]           A cos wc t
          +--d_(k-1)--+

Differential encoding: $d_k=\overline{b_k\oplus d_{k-1}}$ (bit 1: no change, bit 0: change), starting from an arbitrary reference bit. The encoded sequence then drives a BPSK modulator.

Example, reference $d_0=1$:

Demodulator

 r(t)-->[BPF]--+-------------->(X)-->[int 0..Tb]-->[y>0?1:0]
               |                ^
               +-->[delay Tb]---+

The product of the present and one-bit-delayed signals, integrated over $T_b$, gives $y\propto\cos(\theta_k-\theta_{k-1})$. No phase change gives $y>0$ (bit 1); a $\pi$ change gives $y<0$ (bit 0). For the example, phase changes no, yes, no, no, yes give 1, 0, 1, 1, 0.

Signal space diagram

Over one bit, the transmitted points are the same as BPSK, $\pm\sqrt{E_b}$ on $\phi_1$; what matters is whether the point stays or moves.

      phase pi          phase 0
 -------*-------+-------*-------> phi1
     -sqrt(Eb)  0   +sqrt(Eb)
   bit 1: stay on same point
   bit 0: jump to other point

Viewed over two bit intervals ($2T_b$), DPSK is binary orthogonal signalling: "no change" $(\sqrt{E_b},\sqrt{E_b})$ and "change" $(\sqrt{E_b},-\sqrt{E_b})$, energy $2E_b$, detected non-coherently.