--- title: "BPSK modulation and signal space diagram" subject: "Communication Systems" chapter: 7 key: "semester_6/communication_systems:bpsk-signal-space" source: https://www.bctengineer.com/q/semester_6/communication_systems/bpsk-signal-space site: BCT Engineering --- # BPSK modulation and signal space diagram Communication Systems · Chapter 7: Digital Modulation Techniques - Asked in: 2078 Chaitra, 5 marks - Video: [Binary Phase Shift Keying (BPSK) Modulation and Demodulation Technique](https://www.youtube.com/watch?v=Pq-TGHRBMsw) — WIT Solapur - Professional Learning Community - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/bpsk-signal-space - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7 ## Question Explain BPSK modulation technique with its relevant diagram and signal space diagram. ## Short answer BPSK uses antipodal signals (0° and 180° phase) for bits 1 and 0. The modulator multiplies the NRZ polar signal with the carrier. The coherent demodulator integrates and compares to zero. Signal space is one-dimensional with points at ±√Eb, distance 2√Eb. ## Answer **Binary Phase Shift Keying (BPSK)** sends a bit by shifting the carrier phase by $180^\circ$: phase $0$ for symbol 1 and phase $\pi$ for symbol 0. Amplitude and frequency stay constant. $$ s_1(t)=\sqrt{\frac{2E_b}{T_b}}\cos 2\pi f_ct,\qquad s_2(t)=-\sqrt{\frac{2E_b}{T_b}}\cos 2\pi f_ct,\qquad 0\le t\le T_b $$ $f_c=n_c/T_b$ so each bit holds a whole number of cycles. $s_1$ and $s_2$ are **antipodal** signals. #### Waveform (data 1 0 1) ```text data: 1 0 1 /\ /\ \ /\ / /\ /\ / \/ \ \/ \/ / \/ \ phase flips at 1->0 and 0->1 ``` #### Modulator ```text binary --> [NRZ polar encoder] --> b(t) = +-1 | (X) <-- sqrt(2/Tb) cos 2pi fc t | BPSK s(t) ``` The polar NRZ signal (+1 for 1, -1 for 0) multiplies the carrier, so the product modulator flips the phase whenever the data sign changes. #### Coherent demodulator ```text x(t) -->(X)--> [integrate 0..Tb] --> x1 --> [decision] ^ x1>0 -> 1 phi1(t) from carrier recovery x1<0 -> 0 ``` The correlator gives $x_1=\int_0^{T_b}x(t)\phi_1(t)dt$, which equals $\pm\sqrt{E_b}$ plus noise. #### Signal space diagram One basis function is enough ($N=1$): $$ \phi_1(t)=\sqrt{\frac{2}{T_b}}\cos 2\pi f_ct,\qquad s_1(t)=\sqrt{E_b}\,\phi_1(t),\quad s_2(t)=-\sqrt{E_b}\,\phi_1(t) $$ ```text region Z2 | region Z1 s2 | s1 -----*-------------+-------------*-----> phi1 -sqrt(Eb) 0 +sqrt(Eb) boundary ``` - Distance between points: $d=2\sqrt{E_b}$, the largest possible for a given energy. - Decision threshold at zero. - $P_e=\tfrac12\,\text{erfc}\sqrt{E_b/N_0}$, the best among binary schemes. - Bandwidth $\approx 2/T_b=2R_b$ (null to null). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are bits represented in BPSK? A. Amplitude A for 1 and 0 for 0 B. Phase 0 for 1 and phase pi for 0 with constant amplitude and frequency C. Frequency f1 for 1 and f2 for 0 D. Phase pi/2 for 1 and 3pi/2 for 0 Answer: B. The carrier is shifted by 180 degrees, giving antipodal signals +-sqrt(2Eb/Tb)cos(2 pi fc t). (Key point: BPSK signal) 2. How does the BPSK modulator work? A. Unipolar on-off data switches the carrier on and off B. Data drives a VCO between two frequencies C. Polar NRZ (+-1) data multiplies the carrier, flipping phase when the sign changes D. Data is added to the carrier Answer: C. A product modulator with +-1 data flips the sign, i.e. a 180-degree phase shift. (Key point: Modulator) 3. Where are the BPSK signal points in signal space? A. At 0 and sqrt(2Eb) B. On two axes at (sqrt(Eb), 0) and (0, sqrt(Eb)) C. At four points on a circle D. On one axis at +sqrt(Eb) and -sqrt(Eb) Answer: D. One basis function is enough (N = 1) and the signals are antipodal. (Key point: Signal space) 4. What is the distance between the BPSK points and why is it important? A. 2 sqrt(Eb), the largest for given energy, giving the best binary Pe B. sqrt(2Eb), giving 3 dB worse Pe C. sqrt(Eb), giving the same Pe as ASK D. 2Eb, giving zero error Answer: A. Antipodal signals maximize distance, so Pe = (1/2)erfc sqrt(Eb/N0) is the best among binary schemes. 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