--- title: "QAM modulation, demodulation and constellation" subject: "Communication Systems" chapter: 7 key: "semester_6/communication_systems:qam-mod-demod-constellation" source: https://www.bctengineer.com/q/semester_6/communication_systems/qam-mod-demod-constellation site: BCT Engineering --- # QAM modulation, demodulation and constellation Communication Systems · Chapter 7: Digital Modulation Techniques - Asked in: 2081 Bhadra (CS II), 5 marks - Video: [Quadrature Amplitude Modulation - QAM | Basics, Constellation Diagram, Modulator, and Demodulator](https://www.youtube.com/watch?v=rJYreNIlcWM) — 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/qam-mod-demod-constellation - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7 ## Question Explain the working of Quadrature Amplitude Modulation (QAM) with appropriate modulation, demodulation block, and constellation diagram. ## Short answer QAM modulates two quadrature carriers with multi-level amplitudes, achieving high spectral efficiency. Modulator: bits split and mapped to levels, then multiply quadrature carriers and sum. Demodulator: two correlators with local quadrature carriers and slicers recover levels. 16-QAM constellation is a 4×4 Gray-coded grid, each symbol carrying 4 bits. ## Answer **Quadrature Amplitude Modulation (QAM)** is a digital modulation in which each symbol is sent by changing both the **amplitude and phase** of the carrier. It is built from two amplitude-modulated carriers of the same frequency in quadrature, so an $M$-point QAM carries $\log_2 M$ bits per symbol. $$ s_k(t)=\sqrt{\frac{2E_0}{T}}\,a_k\cos 2\pi f_ct-\sqrt{\frac{2E_0}{T}}\,b_k\sin 2\pi f_ct,\quad 0\le t\le T $$ For 16-QAM, $a_k,b_k\in\{-3,-1,+1,+3\}$. #### Modulator ```text +->[2-bit to 4-level]-a_k->(X)<-phi1 bits -->[S/P] | 4 bits/sym | (+)--> s(t) +->[2-bit to 4-level]-b_k->(X)<-phi2 ``` $\phi_1(t)=\sqrt{2/T}\cos 2\pi f_ct$, $\phi_2(t)=-\sqrt{2/T}\sin 2\pi f_ct$. 1. Serial-to-parallel converter sends 2 bits to the I branch and 2 bits to the Q branch. 2. Each 2-bit group is mapped (Gray code) to a level: 00 to -3, 01 to -1, 11 to +1, 10 to +3. 3. The levels amplitude-modulate the quadrature carriers and the outputs are summed. #### Demodulator ```text +->(X)->[int 0..T]->[4-level slicer]--+ | ^ phi1 | r(t) ----+ [P/S]-> bits | v phi2 | +->(X)->[int 0..T]->[4-level slicer]--+ ``` Each correlator recovers one coordinate ($\approx a_k\sqrt{E_0}$ and $b_k\sqrt{E_0}$ plus noise). The slicer compares it with thresholds $0,\pm2\sqrt{E_0}$, picks the nearest level, and the level-to-bits converter plus P/S give the data. Carrier and symbol timing recovery are needed (coherent detection). #### 16-QAM constellation (Gray coded) ```text Q (phi2) 0010 0110 | 1110 1010 +3 | 0011 0111 | 1111 1011 +1 --------------+--------------> I (phi1) 0001 0101 | 1101 1001 -1 | 0000 0100 | 1100 1000 -3 -3 -1 +1 +3 ``` First two bits select the I level, last two the Q level; neighbouring points differ by one bit, so most symbol errors cause only one bit error. The minimum distance between points is $2\sqrt{E_0}$. #### Merits - 4 bits/symbol: bandwidth is 1/4 of BPSK for the same bit rate. - Better error performance than 16-PSK at the same average power. - Not constant envelope, so linear amplifiers are required. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In 16-QAM how are the 4 bits used? A. All 4 bits select one of 16 phases B. 2 bits select the I level and 2 bits select the Q level, each from -3, -1, +1, +3 C. 3 bits for I and 1 bit for Q D. 2 bits select amplitude and 2 bits select frequency Answer: B. S/P sends two bits to each branch, each mapped to one of four levels. (Key point: 16-QAM levels) 2. What are the decision thresholds for the 4-level slicer? A. 0 and +-sqrt(E0) B. +-sqrt(E0) and +-3 sqrt(E0) C. 0 and +-2 sqrt(E0) D. Only 0 Answer: C. Levels are at +-sqrt(E0) and +-3sqrt(E0), so midpoints are 0 and +-2sqrt(E0). (Key point: Slicer thresholds) 3. What is the advantage of Gray coding the constellation? A. It makes the envelope constant B. It increases the minimum distance C. It removes the need for carrier recovery D. Neighbouring points differ by one bit, so most symbol errors cause one bit error Answer: D. A symbol error to a nearest neighbour then flips only one bit. (Key point: Gray coding) 4. Which is true of 16-QAM compared with BPSK at the same bit rate? A. Bandwidth is 1/4 but the envelope is not constant, so linear amplifiers are required B. Bandwidth is 1/2 and the envelope is constant C. Bandwidth is the same but better noise performance D. Bandwidth is 1/4 and constant envelope Answer: A. 4 bits per symbol cuts bandwidth by 4, but varying amplitude needs linear amplification. 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