--- title: "Why DPSK over PSK; DPSK modulator, demodulator, signal space" subject: "Communication Systems" chapter: 7 key: "semester_6/communication_systems:dpsk-vs-psk-mod-demod" source: https://www.bctengineer.com/q/semester_6/communication_systems/dpsk-vs-psk-mod-demod site: BCT Engineering --- # Why DPSK over PSK; DPSK modulator, demodulator, signal space Communication Systems · Chapter 7: Digital Modulation Techniques - Asked in: 2072 Kartik (CS II), 1+2+2+1 marks - Video: [Differential Phase Shift Keying DPSK (Definition, Waveform, DPSK transmitter & DPSK receiver)](https://www.youtube.com/watch?v=ZTKDZ6QPMlE) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/dpsk-vs-psk-mod-demod - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7 ## Question Why DPSK is preferred than PSK? Explain the Modulator, Demodulator and Signal Space Diagram for DPSK system. ## Short answer DPSK is preferred because it eliminates the need for coherent carrier recovery, avoiding phase ambiguity with only ~1 dB penalty. The modulator differentially encodes data (1=no phase change, 0=180° change) and applies BPSK. The demodulator multiplies the signal by a one-bit delayed version, integrates, and decides sign. Signal space points are antipodal ±√E_b, with detection based on phase difference. ## Answer #### 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 ```text 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 $d_0=1$: | $b_k$ | | 1 | 0 | 1 | 1 | 0 | |---|---|---|---|---|---|---| | $d_k$ | 1 | 1 | 0 | 0 | 0 | 1 | | Phase | 0 | 0 | $\pi$ | $\pi$ | $\pi$ | 0 | #### Demodulator ```text 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. ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is DPSK preferred over coherent PSK? A. It needs about 1 dB less Eb/N0 B. It needs no coherent carrier and is not affected by 180° phase ambiguity C. It uses a wider bandwidth for better accuracy D. It uses two orthogonal carriers Answer: B. DPSK uses the previous bit as phase reference, so carrier recovery is unnecessary and phase ambiguity does not matter. (Key point: Why DPSK is preferred) 2. What is the price paid for using DPSK instead of coherent PSK? A. About 1 dB more Eb/N0 for the same error rate B. Twice the bandwidth C. A need for carrier recovery D. A lower data rate Answer: A. DPSK has Pe = (1/2)exp(-Eb/N0), roughly 1 dB worse than coherent BPSK. (Key point: Cost of DPSK) 3. With d_k = XNOR(b_k, d_(k-1)) and d_(k-1) = 1, what is d_k and the resulting phase for data bit b_k = 0? A. d_k = 1, phase 0 B. d_k = 0, phase pi C. d_k = 1, phase pi D. d_k = 0, phase 0 Answer: B. XNOR of 0 and 1 is 0, so d_k = 0 and the carrier is sent with phase pi (a change); bit 1 means no change. (Key point: Differential encoder) 4. How does the DPSK demodulator decide a bit? A. It compares x with a locally generated coherent carrier B. It multiplies the signal by its one-bit-delayed version, integrates over Tb and takes y > 0 as 1, y < 0 as 0 C. It counts zero crossings in each bit D. It squares the signal and filters the second harmonic Answer: B. The integrated product is proportional to cos(theta_k - theta_(k-1)): positive for no phase change (1), negative for a pi change (0). (Key point: DPSK demodulator) ## Related in Communication Systems - [QPSK modulator, demodulator and signal space](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aqpsk-mod-demod-signal-space) — Communication Systems, chapter 7, asked 2× - [FSK modulator, demodulator and signal space](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afsk-mod-demod-signal-space) — Communication Systems, chapter 7, asked 2× - [BPSK modulation and signal space diagram](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abpsk-signal-space) — Communication Systems, chapter 7 - [BFSK generation, non-coherent detection and signal space](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abfsk-noncoherent-signal-space) — Communication Systems, chapter 7 ## Related in other subjects - [MSK, GMSK and OFDM modulator/demodulator](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Amsk-gmsk-ofdm-modem) — Wireless Communication, chapter 4 - [Why serial transfer is preferred over parallel](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Aserial-vs-parallel-preferred) — Instrumentation II, chapter 3 - [T/F: ACSR preferred over AAAC](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atf-acsr-vs-aaac) — Transmission and Distribution System Design, chapter 4 - [Why NOS preferred over DOS](https://www.bctengineer.com/question.md?key=semester_7%2Fdistributed_system%3Anos-vs-dos-realization) — Distributed System, chapter 3, asked 2×