--- title: "QPSK modulator, demodulator and signal space" subject: "Communication Systems" chapter: 7 key: "semester_6/communication_systems:qpsk-mod-demod-signal-space" source: https://www.bctengineer.com/q/semester_6/communication_systems/qpsk-mod-demod-signal-space site: BCT Engineering --- # QPSK modulator, demodulator and signal space Communication Systems · Chapter 7: Digital Modulation Techniques - Asked in: 2080 Baisakh (CS II), 10 marks; 2070 Chaitra (CS II), 8 marks (asked 2 times) - 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-signal-space - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7 ## Question Explain the Modulator, Demodulator, and Signal Space Diagram for QPSK modulation with relevant derivations. ## Short answer QPSK uses four phases (π/4, 3π/4, 5π/4, 7π/4) each carrying two bits. The modulator splits bits into I and Q channels multiplied by orthogonal carriers and summed. The coherent demodulator correlates with the same carriers and uses zero thresholds. Signal space has four points at ±√(E/2). ## Answer **Quadriphase-shift keying (QPSK)** is a digital modulation in which the carrier takes one of four phases, each carrying a pair of bits (a **dibit**). It sends 2 bits per symbol, so it needs half the bandwidth of BPSK for the same bit rate, with the same bit error probability. #### Signal set With symbol duration $T = 2T_b$ and symbol energy $E$: $$ s_i(t) = \sqrt{\frac{2E}{T}}\cos\left[2\pi f_ct + (2i-1)\frac{\pi}{4}\right], \quad 0 \le t \le T,\ i = 1, 2, 3, 4 $$ Expanding the cosine: $$ s_i(t) = \sqrt{E}\cos\left[(2i-1)\frac{\pi}{4}\right]\phi_1(t) - \sqrt{E}\sin\left[(2i-1)\frac{\pi}{4}\right]\phi_2(t) $$ with two orthonormal basis functions $$ \phi_1(t) = \sqrt{\frac{2}{T}}\cos(2\pi f_ct), \qquad \phi_2(t) = \sqrt{\frac{2}{T}}\sin(2\pi f_ct), \quad 0 \le t \le T $$ ($f_c$ is a multiple of $1/T$, so $\int_0^T\phi_1\phi_2\,dt = 0$ and each has unit energy.) Coordinates $s_{i1} = \sqrt{E}\cos[(2i-1)\pi/4]$, $s_{i2} = -\sqrt{E}\sin[(2i-1)\pi/4]$: | Dibit (Gray) | Phase | $s_{i1}$ | $s_{i2}$ | |---|---|---|---| | 10 | $\pi/4$ | $+\sqrt{E/2}$ | $-\sqrt{E/2}$ | | 00 | $3\pi/4$ | $-\sqrt{E/2}$ | $-\sqrt{E/2}$ | | 01 | $5\pi/4$ | $-\sqrt{E/2}$ | $+\sqrt{E/2}$ | | 11 | $7\pi/4$ | $+\sqrt{E/2}$ | $+\sqrt{E/2}$ | #### Signal space diagram ```text phi2 | 01 * | * 11 (-a,+a) | (+a,+a) | --------------+--------------- phi1 | 00 * | * 10 (-a,-a) | (+a,-a) | a = sqrt(E/2) ``` The four points lie on a circle of radius $\sqrt{E}$. Gray coding makes adjacent points differ by one bit. The distance between adjacent points is $d = \sqrt{2E} = 2\sqrt{E_b}$ (with $E = 2E_b$), the same as BPSK, so the bit error probability is the same: $$ P_e(\text{bit}) = \frac12\,\text{erfc}\sqrt{\frac{E_b}{N_0}} $$ #### QPSK modulator ```text +-> a1(t) -->(x)----+ binary | (odd bits) ^ | data --[serial to] | | b(t) [parallel ] phi1(t) (+)--> QPSK s(t) | (even bits) | | +-> a2(t) -->(x)----+ ^ phi2(t) ``` 1. The input bit stream $b(t)$ is converted to polar NRZ form ($1 \to +\sqrt{E_b}$, $0 \to -\sqrt{E_b}$). 2. A **demultiplexer** (serial-to-parallel converter) splits it into odd bits $a_1(t)$ and even bits $a_2(t)$, each of duration $T = 2T_b$. 3. $a_1(t)$ multiplies the in-phase carrier $\phi_1(t)$ and $a_2(t)$ multiplies the quadrature carrier $\phi_2(t)$, giving two BPSK signals. 4. The **sum** of the two BPSK signals is the QPSK signal: $$ s(t) = a_1(t)\phi_1(t) + a_2(t)\phi_2(t) $$ Since each branch runs at $R_b/2$, the null-to-null bandwidth is $B = 2/T = R_b$ (BPSK needs $2R_b$). #### QPSK demodulator (coherent receiver) ```text +->(x)->[integrate 0..T]->[decide]->a1^-+ x(t) | ^ thresh 0 | --------+ phi1(t) [parallel]-> b^ | [to serial] +->(x)->[integrate 0..T]->[decide]->a2^-+ ^ thresh 0 phi2(t) ``` The received signal is $x(t) = s_i(t) + w(t)$, with $w(t)$ white Gaussian noise of PSD $N_0/2$. 1. Two correlators multiply $x(t)$ by locally generated coherent references $\phi_1(t)$ and $\phi_2(t)$ and integrate over one symbol: $$ x_1 = \int_0^T x(t)\phi_1(t)\,dt = s_{i1} + w_1, \qquad x_2 = \int_0^T x(t)\phi_2(t)\,dt = s_{i2} + w_2 $$ Because of orthogonality, the in-phase branch sees only $s_{i1}$ and the quadrature branch only $s_{i2}$; $w_1, w_2$ are independent Gaussian with variance $N_0/2$. 2. Each output is compared with a **threshold of zero**: $x_1 > 0 \Rightarrow$ 1, else 0 (same for $x_2$). This is the minimum-distance decision in the signal space; the decision regions are the four quadrants. 3. A **multiplexer** (parallel-to-serial converter) combines the two decided bit streams into the original sequence. #### Error performance Each branch is a BPSK detector with $E_b = E/2$: $$ P_e(\text{bit}) = \frac12\,\text{erfc}\sqrt{\frac{E_b}{N_0}}, \qquad P_e(\text{symbol}) \approx \text{erfc}\sqrt{\frac{E}{2N_0}} $$ So QPSK gives the error performance of BPSK with twice the bandwidth efficiency (2 bits/s/Hz in theory). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does QPSK need half the bandwidth of BPSK for the same bit rate? A. Each symbol carries 2 bits, so the symbol duration is T = 2Tb B. Its carrier frequency is halved C. It uses only one basis function D. It transmits at half the energy per bit Answer: A. Two bits per symbol halves the symbol rate, so null-to-null bandwidth is Rb instead of 2Rb. (Key point: Dibit and bandwidth) 2. Which pair of orthonormal basis functions is used for QPSK? A. cos(2 pi fc t) and cos(4 pi fc t) B. sqrt(2/T)cos(2 pi fc t) and sqrt(2/T)sin(2 pi fc t) C. Two cosines at different frequencies fc and fc + 1/T D. A single cosine and its negative Answer: B. Cosine and sine of the same carrier are orthogonal over T and have unit energy, giving a two-dimensional signal space. (Key point: Basis functions) 3. Where do the four QPSK message points lie in the signal space? A. On the phi1 axis at +-sqrt(E) and +-sqrt(E)/2 B. At (+-sqrt(E), 0) and (0, +-sqrt(E)) C. At (+-sqrt(E/2), +-sqrt(E/2)), on a circle of radius sqrt(E) D. At (+-E, +-E) in the four quadrants Answer: C. Phases pi/4 odd multiples give coordinates +-sqrt(E)cos45 = +-sqrt(E/2), all at distance sqrt(E) from the origin. (Key point: Signal space geometry) 4. How does the coherent QPSK receiver decide each bit? A. The two correlator outputs are compared with each other B. The envelope of the signal is compared with a threshold of half its peak C. The phase of the carrier is measured against 90 degrees D. Each correlator output is compared with a threshold of zero Answer: D. Decision regions are the four quadrants, so the sign of x1 and x2 gives the two bits. (Key point: Demodulator decision) 5. How does the bit error probability of QPSK compare with BPSK? A. The same, (1/2)erfc sqrt(Eb/N0), since each branch is a BPSK detector B. Twice as large, because two bits are sent together C. Lower, because the bandwidth is halved D. 3 dB worse, because the points are closer Answer: A. Each quadrature branch is an independent BPSK with Eb = E/2 per bit, so QPSK matches BPSK performance with half the bandwidth. 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