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Demodulation of DPSK

Short answer

DPSK is demodulated using a delay-and-multiply receiver. The received signal is multiplied by a one-bit delayed version, integrated, and the output sign indicates phase change (no change gives positive, 180° change negative), recovering data without requiring a local coherent carrier.

Explain any one method of demodulating DPSK signal.

Differential PSK (DPSK) is the non-coherent form of BPSK. The information is carried by the phase change between consecutive bits, so the receiver uses the previous received bit as its phase reference and needs no locally generated coherent carrier.

DPSK signal (recap)

The data bkb_k is first differentially encoded:

dk=bk⊕dk−1‾d_k=\overline{b_k\oplus d_{k-1}}

i.e. if bk=1b_k=1 the encoded bit is unchanged, if bk=0b_k=0 it toggles. Then dkd_k drives a BPSK modulator (1: phase 0, 0: phase π\pi).

Demodulation method: delay-and-multiply (correlation) receiver

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

Working:

  1. The received signal in bit kk is rk(t)=Acos⁡(ωct+θk)r_k(t)=A\cos(\omega_ct+\theta_k) and in the previous bit, after delay TbT_b, rk−1(t)=Acos⁡(ωct+θk−1)r_{k-1}(t)=A\cos(\omega_ct+\theta_{k-1}).
  2. The multiplier gives
rk rk−1=A22cos⁡(θk−θk−1)+A22cos⁡(2ωct+θk+θk−1)r_k\,r_{k-1}=\frac{A^2}{2}\cos(\theta_k-\theta_{k-1})+\frac{A^2}{2}\cos(2\omega_ct+\theta_k+\theta_{k-1})
  1. The integrator (low-pass action) removes the 2ωc2\omega_c term and keeps y=A2Tb2cos⁡(θk−θk−1)y=\tfrac{A^2T_b}{2}\cos(\theta_k-\theta_{k-1}).
  2. If the phase did not change, y>0y>0, decide bk=1b_k=1; if it changed by π\pi, y<0y<0, decide bk=0b_k=0.

Any unknown fixed carrier phase appears in both θk\theta_k and θk−1\theta_{k-1} and cancels, which is why no carrier recovery is needed.

Example

kk012345678
Data bkb_k10010011
dkd_k (ref. 1)110110111
Phase00π\pi00π\pi000
Phase change?noyesyesnoyesyesnono
Sign of yy+−−+−−++
Output10010011

The output equals the original data.

Remarks

  • Error probability: Pe=12e−Eb/N0P_e=\tfrac12 e^{-E_b/N_0}, about 1 dB worse than coherent BPSK at Pe=10−4P_e=10^{-4}.
  • Errors tend to occur in pairs, because a wrong bit is also the reference for the next bit.
  • The other method is coherent detection of dkd_k (as in BPSK) followed by a differential decoder (bk=dk⊕dk−1‾b_k=\overline{d_k\oplus d_{k-1}}); it gives slightly better PeP_e but needs carrier recovery.

✓ Answer checked by Claude Opus 5.5 (high effort)

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