--- title: "Two-tone AM: index, power, spectrum (5cos, 10cos, 15cos)" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:am-two-tone-5cos-10cos-15cos" source: https://www.bctengineer.com/q/semester_6/communication_systems/am-two-tone-5cos-10cos-15cos site: BCT Engineering --- # Two-tone AM: index, power, spectrum (5cos, 10cos, 15cos) Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2072 Asoj (CS I), 2+2+3 marks - Video: [AMPLITUDE MODULATION (SOLVED PROBLEMS)](https://www.youtube.com/watch?v=7f3Y2dvfnqI) — Prof. Barapate's Tutorials - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/am-two-tone-5cos-10cos-15cos - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question The signals m(t) = 5cos 2×10³πt + 10cos 1×10³πt, c(t) = 15cos 2×10⁶πt are applied to AM. i) Find modulation index ii) Find total power iii) Draw spectrum ## Short answer Modulation indices: μ1=5/15=0.333, μ2=10/15=0.667, total μt = √(0.111+0.444)=0.745. Total power Pt = (15²/2)(1+0.5556/2)=112.5×1.2778=143.75 W. Spectrum: carrier 15 V at 1 MHz, sidebands at ±500 Hz (5 V) and ±1 kHz (2.5 V); bandwidth 2 kHz. ## Answer Given $m(t) = 5\cos(2\pi\,1000\,t) + 10\cos(2\pi\,500\,t)$ and $c(t) = 15\cos(2\pi\times10^6t)$. So $A_{m1} = 5$ V at 1 kHz, $A_{m2} = 10$ V at 500 Hz, $A_c = 15$ V, $f_c = 1$ MHz. Assume standard AM $s(t) = [A_c + m(t)]\cos\omega_ct$ and $R = 1\ \Omega$. $$ s(t) = 15\left[1 + \tfrac13\cos(2\pi10^3t) + \tfrac23\cos(\pi10^3t)\right]\cos(2\pi10^6t) $$ #### i) Modulation index $$ \mu_1 = \frac{5}{15} = 0.333, \quad \mu_2 = \frac{10}{15} = 0.667 $$ $$ \mu_t = \sqrt{\mu_1^2 + \mu_2^2} = \sqrt{\frac19 + \frac49} = \sqrt{0.5556} = 0.7454 $$ #### ii) Total power $$ \begin{aligned} P_c &= \frac{A_c^2}{2R} = \frac{15^2}{2} = 112.5\ \text{W} \\ P_t &= P_c\left(1 + \frac{\mu_t^2}{2}\right) = 112.5\left(1 + \frac{0.5556}{2}\right) = 143.75\ \text{W} \end{aligned} $$ #### iii) Spectrum Side amplitudes $A_m/2$: 2.5 V (1 kHz tone), 5 V (500 Hz tone). | Frequency | Amplitude | |---|---| | 999 kHz | 2.5 V | | 999.5 kHz | 5 V | | 1000 kHz | 15 V | | 1000.5 kHz | 5 V | | 1001 kHz | 2.5 V | ```text A(V) 15 | | | | 5 | | | | 2.5 | | | | | | +----+----+--------+--------+----+---> f (kHz) 999 999.5 1000 1000.5 1001 ``` **Answer:** $\mu_t = 0.745$; $P_t = 143.75$ W; bandwidth 2 kHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With Ac = 15 V and tone amplitudes 5 V and 10 V, mu1 and mu2 are: A. 0.333 and 0.667 B. 5 and 10 C. 0.5 and 1 D. 3 and 1.5 Answer: A. mu = Am/Ac: 5/15 = 0.333 and 10/15 = 0.667. (Key point: Individual indices) 2. The total modulation index is: A. 1.0 B. 0.745 C. 0.556 D. 0.333 Answer: B. mu_t = sqrt(1/9 + 4/9) = sqrt(0.5556) = 0.7454. (Key point: Total index) 3. Taking R = 1 ohm, the total power is: A. 112.5 W B. 225 W C. 143.75 W D. 150 W Answer: C. Pc = 15^2/2 = 112.5 W, and Pt = 112.5(1 + 0.5556/2) = 143.75 W. (Key point: Total power) 4. The 500 Hz tone gives side frequencies at what frequencies and amplitude? A. 999 and 1001 kHz, 2.5 V B. 999.5 and 1000.5 kHz, 10 V C. 999.5 and 1000.5 kHz, 2.5 V D. 999.5 and 1000.5 kHz, 5 V Answer: D. Side amplitude is Am/2 = 10/2 = 5 V at fc +/- 500 Hz. 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