--- title: "Multi-tone AM: spectrum, bandwidth, power efficiency" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:am-50-0-3-0-2-spectrum" source: https://www.bctengineer.com/q/semester_6/communication_systems/am-50-0-3-0-2-spectrum site: BCT Engineering --- # Multi-tone AM: spectrum, bandwidth, power efficiency Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2073 Bhadra (CS I), 4+3+3 marks - Video: [AM examples on Multi tone AM signals in Communication Engineering by Engineering Funda](https://www.youtube.com/watch?v=0oX236SrXQo) — 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/am-50-0-3-0-2-spectrum - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question An amplitude modulated wave is given by s(t) = 50(1 + 0.3cos3141.60t + 0.2cos2513.28t)cos10⁶t i) Draw the amplitude spectrum of s(t) ii) Determine the bandwidth of s(t) iii) Calculate the power efficiency ## Short answer Spectrum: carrier 50 V at fc = 159.155 kHz, sideband pairs at ±500 Hz (7.5 V) and ±400 Hz (5 V). Bandwidth = 2 × 500 Hz = 1 kHz. Power efficiency η = (0.3²+0.2²)/(2+0.3²+0.2²) = 0.13/2.13 = 6.10%. ## Answer Compare with $s(t) = A_c[1 + \mu_1\cos\omega_1t + \mu_2\cos\omega_2t]\cos\omega_ct$: | Quantity | Value | |---|---| | $A_c$ | 50 V | | $\mu_1$, $\omega_1$ | 0.3, 3141.60 rad/s → $f_1 = 500$ Hz | | $\mu_2$, $\omega_2$ | 0.2, 2513.28 rad/s → $f_2 = 400$ Hz | | $\omega_c$ | $10^6$ rad/s → $f_c = 159.155$ kHz | #### i) Amplitude spectrum Expanding: $$ \begin{aligned} s(t) = {} & 50\cos\omega_ct + 7.5\cos(\omega_c \pm 3141.6)t + 5\cos(\omega_c \pm 2513.28)t \end{aligned} $$ (Side amplitudes $\mu A_c/2$: $0.3 \times 50/2 = 7.5$ V; $0.2 \times 50/2 = 5$ V.) | ω (rad/s) | f | Amplitude | |---|---|---| | 996 858.4 | 158.655 kHz | 7.5 V | | 997 486.7 | 158.755 kHz | 5 V | | 1 000 000 | 159.155 kHz | 50 V | | 1 002 513.3 | 159.555 kHz | 5 V | | 1 003 141.6 | 159.655 kHz | 7.5 V | ```text A(V) 50 | | | | 7.5 | | | | 5 | | | | | | +---+-----+--------+--------+-----+---> f fc-500 fc-400 fc fc+400 fc+500 (Hz) fc = 159.155 kHz ``` #### ii) Bandwidth $$ B = 2f_{max} = 2 \times 500 = 1000\ \text{Hz}\ (= 6283.2\ \text{rad/s}) $$ #### iii) Power efficiency $$ \mu_t^2 = \mu_1^2 + \mu_2^2 = 0.09 + 0.04 = 0.13 $$ $$ \eta = \frac{\mu_t^2}{2 + \mu_t^2} = \frac{0.13}{2.13} = 0.0610 = 6.10\% $$ Check with powers (1 Ω): $P_c = 1250$ W, $P_{SB} = 1250 \times 0.13/2 = 81.25$ W, $P_t = 1331.25$ W, $\eta = 81.25/1331.25 = 6.10\%$. **Answer:** $B = 1$ kHz; $\eta = 6.10\%$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the amplitudes of the side frequencies for mu1 = 0.3 and mu2 = 0.2 with Ac = 50 V? A. 15 V and 10 V B. 0.3 V and 0.2 V C. 7.5 V and 5 V D. 3.75 V and 2.5 V Answer: C. Each side amplitude is mu Ac/2: 0.3 x 50/2 = 7.5 V and 0.2 x 50/2 = 5 V. (Key point: Side amplitudes) 2. The message tones have frequencies f1 and f2 of: A. 3141.6 Hz and 2513.28 Hz B. 159 kHz and 500 Hz C. 500 Hz and 400 Hz D. 1000 Hz and 800 Hz Answer: C. Dividing the angular frequencies by 2 pi gives 500 Hz and 400 Hz. (Key point: Frequencies of tones) 3. The bandwidth of this multi-tone AM wave is: A. 900 Hz B. 500 Hz C. 100 Hz D. 1000 Hz Answer: D. B = 2 f_max = 2 x 500 = 1000 Hz. (Key point: Bandwidth) 4. The total modulation index squared and the efficiency are: A. 0.13 and 6.10% B. 0.5 and 20% C. 0.13 and 13% D. 0.36 and 15.3% Answer: A. mu_t^2 = 0.09 + 0.04 = 0.13, and eta = 0.13/(2 + 0.13) = 6.10%. 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