--- title: "QPSK modulation and demodulation" subject: "Communication Systems" chapter: 7 key: "semester_6/communication_systems:qpsk-mod-demod" source: https://www.bctengineer.com/q/semester_6/communication_systems/qpsk-mod-demod site: BCT Engineering --- # QPSK modulation and demodulation Communication Systems · Chapter 7: Digital Modulation Techniques - Asked in: 2075 Asoj (CS II), 8 marks - Video: [QPSK Quadrature Phase Shift Keying (Basics, Modulator, Waveforms, Demodulator & Applications)](https://www.youtube.com/watch?v=ij760lCUtfw) — 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/qpsk-mod-demod - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7 ## Question Explain the modulation and demodulation techniques used in QPSK. ## Short answer QPSK splits the input bit stream into in-phase and quadrature channels, each modulating a carrier at 90° offset, then sums the results. Coherent demodulation uses two correlators with recovered carriers and zero thresholds to recover odd and even bits, recombined by a multiplexer. Signal points are (±√(E/2), ±√(E/2)), doubling bandwidth efficiency to Rb with BPSK-equivalent BER. ## Answer **Quadrature Phase Shift Keying (QPSK)** is a 4-level PSK in which each symbol carries a pair of bits (a dibit) and the carrier takes one of four phases spaced $90^\circ$ apart. $$ s_i(t)=\sqrt{\frac{2E}{T}}\cos\left[2\pi f_ct+(2i-1)\frac{\pi}{4}\right],\quad i=1,\dots,4,\ 0\le t\le T $$ $T=2T_b$ is the symbol duration and $E=2E_b$ the symbol energy. Using $\cos(A+B)$: $$ s_i(t)=\sqrt{E}\cos\left[(2i-1)\tfrac{\pi}{4}\right]\phi_1(t)-\sqrt{E}\sin\left[(2i-1)\tfrac{\pi}{4}\right]\phi_2(t) $$ with $\phi_1(t)=\sqrt{2/T}\cos 2\pi f_ct$ and $\phi_2(t)=\sqrt{2/T}\sin 2\pi f_ct$. So QPSK is the sum of two BPSK signals on quadrature carriers. #### Signal space (Gray coded) ```text phi2 01 * | * 11 | ------------+------------ phi1 | 00 * | * 10 points at (+-sqrt(E/2), +-sqrt(E/2)) ``` #### QPSK modulator ```text bits-->[polar NRZ encoder]-->[Demux S/P] | | a1(t) odd a2(t) even (rate Rb/2) (rate Rb/2) | | phi1 ->(X) (X)<- phi2 | | +->(+)<--+ | QPSK s(t) ``` 1. The data is converted to polar NRZ (+1, -1). 2. The demultiplexer splits it into odd and even bits, each lasting $T=2T_b$. 3. Odd bits modulate the in-phase carrier and even bits the quadrature carrier (two BPSK signals). 4. The adder gives one of four phases; e.g. dibit 11 gives the point $(+\sqrt{E/2},+\sqrt{E/2})$, phase $7\pi/4$. #### QPSK demodulator (coherent) ```text +->(X)->[int 0..T]-x1->[x1>0?1:0]--+ | ^ phi1 | x(t) ----+ [Mux P/S]-> bits | v phi2 | +->(X)->[int 0..T]-x2->[x2>0?1:0]--+ ``` 1. The received signal goes to two correlators fed with the recovered carriers $\phi_1$ and $\phi_2$. 2. Because $\phi_1\perp\phi_2$, each correlator sees only its own BPSK component: $x_1=\pm\sqrt{E/2}$, $x_2=\pm\sqrt{E/2}$ (plus noise). 3. Each output is compared with a zero threshold to give the odd and even bits. 4. The multiplexer recombines them into the original serial stream. #### Properties | Feature | QPSK | BPSK | |---|---|---| | Bits/symbol | 2 | 1 | | Bandwidth | $R_b$ | $2R_b$ | | Bit error probability | $\tfrac12\text{erfc}\sqrt{E_b/N_0}$ | same | | Complexity | Higher | Lower | QPSK doubles bandwidth efficiency without any loss in bit-error performance, which is why it is used in satellite links, DVB-S and CDMA systems. Offset QPSK (Q branch delayed by $T_b$) avoids $180^\circ$ phase jumps. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can QPSK be viewed as the sum of two BPSK signals? A. Two oscillators at different frequencies are added B. The cos(A+B) expansion gives in-phase and quadrature terms on orthogonal carriers C. Two bits are sent at double amplitude D. The carrier is modulated twice in series Answer: B. s_i = sqrt(E)cos(..)phi1 - sqrt(E)sin(..)phi2, each term a BPSK on its own carrier. (Key point: QPSK as two BPSK) 2. What does the demultiplexer in the modulator do? A. Doubles the bit rate B. Converts levels to phases C. Splits the polar NRZ bits into odd and even streams, each lasting T = 2Tb D. Adds the two carriers Answer: C. Odd bits drive the in-phase carrier, even bits the quadrature carrier. (Key point: Modulator steps) 3. Which point corresponds to dibit 11? A. (-sqrt(E/2), -sqrt(E/2)), phase 3pi/4 B. (+sqrt(E/2), -sqrt(E/2)), phase pi/4 C. (-sqrt(E/2), +sqrt(E/2)), phase 5pi/4 D. (+sqrt(E/2), +sqrt(E/2)), phase 7pi/4 Answer: D. From the Gray-coded table, 11 is in the first quadrant of the (x1, x2) plane. (Key point: Dibit example) 4. Why does each correlator see only its own BPSK component? A. Because phi1 and phi2 are orthogonal B. Because the noise is zero C. Because the threshold is zero D. Because the symbol duration is 2Tb Answer: A. Orthogonality kills the cross term after integration. 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