--- title: "Multilevel modulation significance; QPSK" subject: "Communication Systems" chapter: 7 key: "semester_6/communication_systems:multilevel-modulation-qpsk" source: https://www.bctengineer.com/q/semester_6/communication_systems/multilevel-modulation-qpsk site: BCT Engineering --- # Multilevel modulation significance; QPSK Communication Systems · Chapter 7: Digital Modulation Techniques - Asked in: 2078 Chaitra, 2+4 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/multilevel-modulation-qpsk - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7 ## Question What are the significances of multilevel modulation? Explain QPSK with its transmitter as well as a receiver block diagram. ## Short answer Multilevel modulation reduces bandwidth by sending log2(M) bits per symbol. QPSK uses four phases each representing a dibit. The transmitter splits bits into I and Q channels and sums after multiplication. The receiver coherently demodulates with two correlators and parallel-to-serial conversion. ## Answer #### Significance of multilevel modulation In **multilevel (M-ary) modulation** each symbol carries $n=\log_2 M$ bits, instead of one bit as in binary schemes. - **Bandwidth saving:** symbol rate is $R_b/\log_2 M$, so bandwidth falls by the factor $\log_2 M$ (QPSK needs half the bandwidth of BPSK). - **Higher data rate** on a band-limited channel (telephone line, radio channel) for the same bandwidth. - **Higher spectral efficiency** (bits/s/Hz); used in modems, Wi-Fi, LTE (16/64-QAM). - **Cost:** for a fixed power, points get closer as $M$ rises, so error probability increases or more power is needed; the receiver is more complex. #### QPSK **Quadrature PSK** sends two bits (a dibit) per symbol using four carrier phases: $$ s_i(t)=\sqrt{\frac{2E}{T}}\cos\left[2\pi f_ct+(2i-1)\frac{\pi}{4}\right],\quad i=1,2,3,4,\ 0\le t\le T $$ with $T=2T_b$ and $E=2E_b$. Expanding, $s_i(t)=\sqrt{E}\cos[(2i-1)\tfrac{\pi}{4}]\phi_1(t)-\sqrt{E}\sin[(2i-1)\tfrac{\pi}{4}]\phi_2(t)$, so QPSK is two BPSK signals on quadrature carriers. | Dibit (Gray) | Phase | Point $(s_{i1},s_{i2})$ | |---|---|---| | 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 | $7\pi/4$ | $(+\sqrt{E/2},+\sqrt{E/2})$ | #### Transmitter ```text bits -> [NRZ level encoder: 1->+1, 0->-1] | [Demux S/P] / \ a1(t) odd a2(t) even | | (X)<-phi1(t) (X)<-phi2(t) | | +----->(+)<-----+ | QPSK ``` $\phi_1(t)=\sqrt{2/T}\cos 2\pi f_ct$, $\phi_2(t)=\sqrt{2/T}\sin 2\pi f_ct$. Each branch is a BPSK signal at half the bit rate. #### Receiver ```text +->(X)->[int 0..T]->[x1>0 ? 1:0]-+ | ^ phi1 | x(t) ---+ [Mux P/S]-> bits | v phi2 | +->(X)->[int 0..T]->[x2>0 ? 1:0]-+ ``` Each correlator output is compared with a zero threshold; the two decisions are multiplexed back into one stream. - Bandwidth: half that of BPSK for the same bit rate. - Bit error probability: $P_e=\tfrac12\text{erfc}\sqrt{E_b/N_0}$, the same as BPSK. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does multilevel modulation save bandwidth? A. The carrier frequency is reduced by M B. Symbol rate is Rb/log2(M), so bandwidth falls by the factor log2(M) C. Each symbol is sent at higher power D. Noise is averaged over M symbols Answer: B. Each symbol carries log2 M bits, so fewer symbols per second are needed. (Key point: Bandwidth saving) 2. What is the cost of increasing M at fixed power? A. Bandwidth increases by M B. The data rate drops C. Points get closer, so error probability rises or more power is needed D. The envelope becomes constant Answer: C. More points in the same signal space reduces the distance between them. (Key point: Cost of M-ary) 3. How is a QPSK transmitter built? A. Four oscillators at different frequencies are switched by the bits B. One BPSK modulator runs at twice the bit rate C. A VCO is driven by the polar NRZ data D. Odd and even bits drive two BPSK modulators on quadrature carriers phi1 and phi2, then are added Answer: D. The demux splits bits into two streams at half rate, each modulating one carrier. (Key point: QPSK structure) 4. What does the QPSK receiver do after the correlators? A. Compares each output with zero, then multiplexes the two decided bits B. Compares the two outputs with each other C. Uses an envelope detector and a comparator D. Divides the output by 2 and squares it Answer: A. Each branch is a BPSK detector with zero threshold; a P/S converter restores the serial stream. 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