--- title: "Design goals of digital modulation; coherent BPSK" subject: "Communication Systems" chapter: 7 key: "semester_6/communication_systems:design-goals-bpsk" source: https://www.bctengineer.com/q/semester_6/communication_systems/design-goals-bpsk site: BCT Engineering --- # Design goals of digital modulation; coherent BPSK Communication Systems · Chapter 7: Digital Modulation Techniques - Asked in: 2073 Chaitra (CS II), 2+8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/design-goals-bpsk - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7 ## Question What are the design goals of digital modulation techniques? Explain coherent binary PSK modulation technique with its signal space diagram, modulator and demodulator. ## Short answer Key design goals are maximizing data rate and interference robustness while minimizing error probability, power, bandwidth, and circuit complexity. Coherent BPSK uses antipodal signals ±√(E_b) φ1(t), represented by points ±√E_b on one axis. The modulator applies polar NRZ to a product modulator; the demodulator correlates with a local carrier and decides with a zero threshold. ## Answer #### Design goals of digital modulation A good digital modulation scheme aims to: 1. **Maximise data rate** for the available channel. 2. **Minimise probability of symbol/bit error** for a given $E_b/N_0$. 3. **Minimise transmitted power** (important for battery and satellite systems). 4. **Minimise channel bandwidth** (high spectral efficiency, bits/s/Hz). 5. **Maximise resistance to interference** and fading. 6. **Minimise circuit complexity and cost.** These goals conflict (e.g. higher-level schemes save bandwidth but need more power), so the choice is a trade-off. #### Coherent binary PSK In BPSK the carrier phase is shifted by $180^\circ$ between the two symbols: $$ s_1(t)=\sqrt{\frac{2E_b}{T_b}}\cos 2\pi f_ct\ \ (\text{symbol }1),\qquad s_2(t)=-\sqrt{\frac{2E_b}{T_b}}\cos 2\pi f_ct\ \ (\text{symbol }0) $$ for $0\le t\le T_b$, where $E_b$ is the bit energy and $f_c=n_c/T_b$. The two signals are antipodal. #### Signal space diagram Only one orthonormal basis function is required: $$ \phi_1(t)=\sqrt{\frac{2}{T_b}}\cos 2\pi f_ct,\quad 0\le t\le T_b $$ so $s_1(t)=\sqrt{E_b}\,\phi_1(t)$ and $s_2(t)=-\sqrt{E_b}\,\phi_1(t)$. The coordinates are $$ s_{11}=\int_0^{T_b}s_1(t)\phi_1(t)dt=+\sqrt{E_b},\qquad s_{21}=-\sqrt{E_b} $$ ```text Z2 (decide 0) | Z1 (decide 1) s2 | s1 ---*-------------------+-------------------*---> phi1 -sqrt(Eb) 0 +sqrt(Eb) decision boundary ``` Distance $d_{12}=2\sqrt{E_b}$. The decision boundary is the midpoint, $x_1=0$. #### Modulator ```text binary data -->[Polar NRZ level encoder]--> b(t) 1 -> +sqrt(Eb), 0 -> -sqrt(Eb) | (X)<-- phi1(t) | BPSK s(t) ``` The level encoder converts 1 and 0 to $+\sqrt{E_b}$ and $-\sqrt{E_b}$; the product modulator multiplies by $\phi_1(t)$. The result is a constant-envelope carrier whose phase flips with the data. #### Demodulator (coherent detector) ```text x(t) -->(X)-->[ integral 0..Tb ]--x1--> [Decision device] ^ x1 > 0 -> 1 phi1(t) x1 < 0 -> 0 (from carrier and bit-timing recovery) ``` 1. The received signal $x(t)=s_i(t)+w(t)$ is multiplied by a locally generated $\phi_1(t)$, exactly in phase with the transmitted carrier. 2. The integrator over one bit gives $x_1=\pm\sqrt{E_b}+n$, where $n$ is Gaussian with variance $N_0/2$. 3. The decision device compares $x_1$ with zero. #### Performance $$ P_e=\frac12\,\text{erfc}\left(\sqrt{\frac{E_b}{N_0}}\right) $$ BPSK has the lowest error probability of all binary schemes (3 dB better than coherent BFSK). Its null-to-null bandwidth is $2/T_b$. Its drawback is the need for an accurate coherent reference (Costas or squaring loop). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the choice of a digital modulation scheme described as a trade-off? A. Because the design goals conflict, e.g. higher-level schemes save bandwidth but need more power B. Because every scheme gives the same bit error rate C. Because bandwidth and power are always reduced together D. Because only one goal can ever be measured Answer: A. The goals (data rate, power, bandwidth, error rate, complexity) pull against each other, so improving one usually costs another. (Key point: Design trade-off) 2. In coherent BPSK, how many orthonormal basis functions are needed and how are the two symbols related? A. Two basis functions; the symbols are orthogonal B. One basis function; the symbols are antipodal (phase differs by 180°) C. One basis function; the symbols differ in frequency D. Two basis functions; the symbols differ by 90° Answer: B. s1 = +sqrt(Eb)phi1 and s2 = -sqrt(Eb)phi1, so a single cosine basis function is enough and the signals are antipodal. (Key point: BPSK signals and basis) 3. What are the distance between the two BPSK message points and the decision boundary? A. d = sqrt(Eb), boundary at sqrt(Eb) B. d = 2sqrt(Eb), boundary at x1 = 0 C. d = 2sqrt(Eb), boundary at x1 = sqrt(Eb) D. d = sqrt(2Eb), boundary at x1 = 0 Answer: B. The points are at +sqrt(Eb) and -sqrt(Eb), so they are 2sqrt(Eb) apart and the midpoint boundary is x1 = 0. (Key point: Signal space and decision) 4. Which statement about coherent BPSK is correct? A. Pe = (1/2)exp(-Eb/N0) and no carrier reference is needed B. Pe = (1/2)erfc(sqrt(Eb/N0)); it needs an accurate coherent reference such as a Costas loop C. It is 3 dB worse than coherent BFSK D. Its null-to-null bandwidth is 1/Tb Answer: B. BPSK has the lowest error probability of binary schemes (3 dB better than coherent BFSK) but needs carrier recovery; bandwidth is 2/Tb. 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