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

# DPSK and its implementation

Communication Systems · Chapter 7: Digital Modulation Techniques

- Asked in: 2070 Asar (CS II), 4 marks
- Video: [Digital Communication 17: DPSK Bandwidth, Modulation, Demodulation, Advantages, Disadvantages](https://www.youtube.com/watch?v=l_mj2u0FOTU) — Study with Dr. Hisham أدرس مع د. هشام
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/dpsk-implementation
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7

## Question

What is DPSK and how it can be implemented?

## Short answer

DPSK conveys information in relative phase changes, avoiding coherent references. Implementation: transmitter uses an XNOR differential encoder followed by a BPSK modulator; receiver multiplies the incoming signal with its one-bit delayed replica, integrates over a bit period, and decides based on sign. It provides simple non-coherent detection at about 1 dB penalty.

## Answer

**DPSK (Differential Phase Shift Keying)** is a non-coherent form of PSK in which a data bit is represented by the **change in carrier phase** relative to the previous bit, not by an absolute phase. Typically bit 1 means no phase change and bit 0 means a $180^\circ$ change. The previous bit acts as the phase reference, so the receiver needs no coherent carrier.

#### Implementation: transmitter

```text
 b_k -->[XNOR]--d_k-->[BPSK modulator]--> DPSK
          ^      |      (phase 0 / pi)
          +-[Tb delay]
```

1. **Differential encoder:** $d_k=\overline{b_k\oplus d_{k-1}}$ with an initial reference bit.
2. **BPSK modulator:** $d_k=1$ gives $A\cos\omega_ct$, $d_k=0$ gives $-A\cos\omega_ct$.

Example: data 1 0 1 1, reference 1 gives $d_k$ = 1 1 0 0 0, phases 0, 0, $\pi$, $\pi$, $\pi$.

#### Implementation: receiver

```text
 r(t)->[BPF]-+------------>(X)->[int 0..Tb]->[sign]-> b_k
             +->[Tb delay]--^
```

The present bit is multiplied by the delayed previous bit and integrated; the result is proportional to $\cos(\theta_k-\theta_{k-1})$. Positive means no phase change (1), negative means a change (0). For the example, changes are no, yes, no, no, giving 1 0 1 1.

DPSK is simpler than coherent PSK and immune to phase ambiguity, with $P_e=\tfrac12e^{-E_b/N_0}$.

## Check yourself

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

1. In DPSK, how is a data bit represented?
   A. By the absolute carrier phase
   B. By the change in carrier phase relative to the previous bit
   C. By the carrier frequency
   D. By the carrier amplitude
   Answer: B. DPSK is non-coherent: bit 1 means no phase change and bit 0 means a 180° change relative to the previous bit. (Key point: What DPSK is)

2. What does the DPSK transmitter consist of?
   A. An XNOR differential encoder with a Tb delay followed by a BPSK modulator
   B. A frequency synthesizer followed by a mixer
   C. A squaring loop and a PLL
   D. Two orthogonal carriers added together
   Answer: A. d_k = XNOR(b_k, d_(k-1)) is fed to a BPSK modulator (phase 0 for d_k = 1, pi for d_k = 0). (Key point: Transmitter structure)

3. For data 1 0 1 1 with reference 1, the phases sent are 0, 0, pi, pi, pi. Where does the single phase change occur?
   A. At the first data bit 1
   B. At the data bit 0
   C. At the last data bit 1
   D. At the reference bit
   Answer: B. Only bit 0 causes a pi change; bits 1 leave the phase unchanged. (Key point: Encoding example)

4. How does the receiver recover the data?
   A. Multiplies the present bit by the Tb-delayed previous bit, integrates and takes the sign (positive = 1)
   B. Locks a PLL to the carrier and compares phase with it
   C. Counts the number of cycles in each bit
   D. Divides the signal by its delayed version
   Answer: A. The result is proportional to cos(theta_k - theta_(k-1)); positive means no change (1), negative means change (0). (Key point: Receiver)

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