---
title: "Demodulation of DPSK"
subject: "Communication Systems"
chapter: 7
key: "semester_6/communication_systems:dpsk-demodulation"
source: https://www.bctengineer.com/q/semester_6/communication_systems/dpsk-demodulation
site: BCT Engineering
---

# Demodulation of DPSK

Communication Systems · Chapter 7: Digital Modulation Techniques

- Asked in: 2068 Bhadra (CS I), 8 marks
- Video: [UNIT-3 L5.1 | example of DPSK |  Generation and Detection of DPSK | DPSK signal](https://www.youtube.com/watch?v=4XKHtkoeVZY) — Techno Tutorials ( e-Learning)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/dpsk-demodulation
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7

## Question

Explain any one method of demodulating DPSK signal.

## 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.

## Answer

**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 $b_k$ is first differentially encoded:

$$
d_k=\overline{b_k\oplus d_{k-1}}
$$

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

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

```text
 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 $k$ is $r_k(t)=A\cos(\omega_ct+\theta_k)$ and in the previous bit, after delay $T_b$, $r_{k-1}(t)=A\cos(\omega_ct+\theta_{k-1})$.
2. The multiplier gives

$$
r_k\,r_{k-1}=\frac{A^2}{2}\cos(\theta_k-\theta_{k-1})+\frac{A^2}{2}\cos(2\omega_ct+\theta_k+\theta_{k-1})
$$

3. The integrator (low-pass action) removes the $2\omega_c$ term and keeps $y=\tfrac{A^2T_b}{2}\cos(\theta_k-\theta_{k-1})$.
4. If the phase did not change, $y>0$, decide $b_k=1$; if it changed by $\pi$, $y<0$, decide $b_k=0$.

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

#### Example

| $k$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| Data $b_k$ | | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 |
| $d_k$ (ref. 1) | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
| Phase | 0 | 0 | $\pi$ | 0 | 0 | $\pi$ | 0 | 0 | 0 |
| Phase change? | | no | yes | yes | no | yes | yes | no | no |
| Sign of $y$ | | + | − | − | + | − | − | + | + |
| Output | | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 |

The output equals the original data.

#### Remarks

- Error probability: $P_e=\tfrac12 e^{-E_b/N_0}$, about 1 dB worse than coherent BPSK at $P_e=10^{-4}$.
- 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 $d_k$ (as in BPSK) followed by a differential decoder ($b_k=\overline{d_k\oplus d_{k-1}}$); it gives slightly better $P_e$ but needs carrier recovery.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does DPSK carry information?
   A. In the absolute phase of each bit
   B. In the phase change between consecutive bits
   C. In the amplitude of each bit
   D. In the frequency of each bit
   Answer: B. The previous bit acts as the phase reference, so no coherent carrier is needed. (Key point: DPSK idea)

2. With d_k = NOT(b_k XOR d_(k-1)), what happens when b_k = 0?
   A. The encoded bit stays the same
   B. The encoded bit becomes 0
   C. The encoded bit toggles
   D. The encoded bit becomes 1
   Answer: C. If b_k = 1 the bit is unchanged; if b_k = 0 it toggles, producing a phase change. (Key point: Encoding rule)

3. After multiplying r_k by the delayed r_(k-1) and integrating, y is proportional to
   A. cos(theta_k + theta_(k-1))
   B. cos(2 wc t)
   C. sin(theta_k - theta_(k-1))
   D. cos(theta_k - theta_(k-1))
   Answer: D. The 2wc term is removed; y = (A^2 Tb/2)cos(change), positive for no change (bit 1), negative for pi change (bit 0). (Key point: Delay-and-multiply)

4. Why is carrier recovery not needed in the delay-and-multiply receiver?
   A. An unknown fixed phase appears in both theta_k and theta_(k-1) and cancels
   B. The signal has constant envelope
   C. The delay removes the noise
   D. The integrator restores the carrier
   Answer: A. Only the difference of phases matters. (Key point: Unknown carrier phase)

## Related in Communication Systems

- [Why DPSK over PSK; DPSK modulator, demodulator, signal space](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adpsk-vs-psk-mod-demod) — Communication Systems, chapter 7
- [Demodulation of DSB-FC AM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbfc-demodulation-method) — Communication Systems, chapter 2
- [DSB-SC vs SSB; synchronous demodulation of DSB-SC](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbsc-vs-ssb-synchronous-demod) — Communication Systems, chapter 2
- [PLL components and AM demodulation using PLL](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apll-components-am-demod) — Communication Systems, chapter 2, asked 3×