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Differential coding; DPSK modulation and detection

Short answer

Differential coding represents each data bit by a change or no change in the encoded sequence. DPSK modulation uses an XNOR gate to encode bits (1=keep, 0=toggle) then drives a BPSK modulator. Detection multiplies the signal by a one-bit delayed copy and integrates; positive output means no phase change (bit 1), negative means a change (bit 0).

What do you understand by differential coding? Explain differential phase shift keying modulation and detection with example and diagrams.

Differential coding

Differential coding represents each data bit by a change or no change in the encoded sequence, rather than by an absolute level. The encoder compares the present data bit with the previous encoded bit:

dk=bk⊕dk−1‾(1: keep previous symbol, 0: toggle)d_k=\overline{b_k\oplus d_{k-1}}\quad(\text{1: keep previous symbol, 0: toggle})

An initial reference bit is chosen arbitrarily. Decoding uses only two neighbouring received bits, bk=dk⊕dk−1‾b_k=\overline{d_k\oplus d_{k-1}}, so a complete inversion of the received stream (e.g. 180∘180^\circ phase ambiguity) does not cause errors.

DPSK modulation

Differential PSK = differential encoding + BPSK. The encoded bit dkd_k selects carrier phase 0 (dk=1d_k=1) or π\pi (dk=0d_k=0).

 b_k -->[XNOR]--d_k-->[polar level]-->(X)--> DPSK s(t)
          ^       |                    ^
          |  [delay Tb]           A cos wc t
          +--------+

Example, data 1 0 0 1 0 0 1 1 with reference 1:

kk012345678
bkb_k10010011
dkd_k110110111
Phase00π\pi00π\pi000

Each 0 in the data causes a π\pi phase change; each 1 leaves the phase unchanged.

DPSK detection

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

The received bit is multiplied by the previous bit and integrated:

y=∫0TbAcos⁡(ωct+θk) Acos⁡(ωct+θk−1) dt≈A2Tb2cos⁡(θk−θk−1)y=\int_0^{T_b}A\cos(\omega_ct+\theta_k)\,A\cos(\omega_ct+\theta_{k-1})\,dt\approx\frac{A^2T_b}{2}\cos(\theta_k-\theta_{k-1})
kk12345678
θk−θk−1\theta_k-\theta_{k-1}0π\piπ\pi0π\piπ\pi00
Sign of yy+−−+−−++
Output10010011

The output equals the data. No carrier recovery is needed because any constant phase offset cancels in the difference.

  • Pe=12e−Eb/N0P_e=\tfrac12e^{-E_b/N_0} (about 1 dB worse than coherent BPSK).
  • Errors tend to come in pairs since each bit is the reference for the next.

✓ Answer checked by Claude Opus 5.5 (high effort)

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