--- title: "Two-tone AM: spectrum, power, net index (50, 15, 20)" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:am-two-tone-50-15-20" source: https://www.bctengineer.com/q/semester_6/communication_systems/am-two-tone-50-15-20 site: BCT Engineering --- # Two-tone AM: spectrum, power, net index (50, 15, 20) Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2070 Bhadra (CS I), 3+2+2 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-50-15-20 - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question The amplitude modulated signal is given by x(t) = 50 cos(2π × 10⁶ t) + 15 cos(2π × 10⁶ t) cos(2π × 10³ t) + 20 cos(2π × 10⁶ t) cos(4π × 10² t). a) Draw the spectrum. b) Find the total modulated power. c) Find the net modulation index. ## Short answer Rewrite as carrier 50V with modulation indices 0.3 and 0.4. Spectrum: carrier at 1MHz 50V, sidebands at ±1kHz 7.5V and ±200Hz 10V. Total power 1406.25W in 1 ohm. Net modulation index 0.5. ## Answer Rewrite the signal in standard AM form ($f_c = 1$ MHz, $f_1 = 1$ kHz, $f_2 = 200$ Hz, since $4\pi\times10^2 = 2\pi\times200$): $$ x(t) = 50\left[1 + 0.3\cos(2\pi\,10^3t) + 0.4\cos(2\pi\,200t)\right]\cos(2\pi\,10^6t) $$ so $A_c = 50$ V, $\mu_1 = 15/50 = 0.3$, $\mu_2 = 20/50 = 0.4$. #### a) Spectrum Using $\cos A\cos B = \frac{1}{2}[\cos(A+B)+\cos(A-B)]$: $$ x(t) = 50\cos\omega_ct + 7.5\cos(\omega_c \pm \omega_1)t + 10\cos(\omega_c \pm \omega_2)t $$ | Frequency | Amplitude | |---|---| | 999.0 kHz | 7.5 V | | 999.8 kHz | 10 V | | 1000.0 kHz | 50 V | | 1000.2 kHz | 10 V | | 1001.0 kHz | 7.5 V | ```text Amp (V) 50 | | | | 10 | | | | 7.5 | | | | | | +-----+----+----+----+----+--> f (kHz) 999 999.8 1000 1000.2 1001 ``` #### b) Total modulated power Taking a 1 Ω load: $$ \begin{aligned} P_c &= \frac{50^2}{2} = 1250\ \text{W} \\ P_{1\,kHz\ pair} &= 2\times\frac{7.5^2}{2} = 56.25\ \text{W} \\ P_{200\,Hz\ pair} &= 2\times\frac{10^2}{2} = 100\ \text{W} \\ P_t &= 1250 + 56.25 + 100 = 1406.25\ \text{W} \end{aligned} $$ Check: $P_t = P_c\left(1 + \frac{\mu_1^2+\mu_2^2}{2}\right) = 1250(1.125) = 1406.25$ W. #### c) Net modulation index $$ \mu = \sqrt{\mu_1^2 + \mu_2^2} = \sqrt{0.3^2 + 0.4^2} = \sqrt{0.25} = 0.5 $$ **Answer:** spectrum as tabulated; $P_t = 1406.25$ W (1 Ω); net $\mu = 0.5$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are mu1 and mu2? A. 0.15 and 0.2 B. 0.5 and 0.5 C. 15 and 20 D. 0.3 and 0.4 Answer: D. mu = amplitude/Ac: 15/50 and 20/50. (Key point: Standard form) 2. What is the amplitude of each 999 kHz and 1001 kHz component? A. 7.5 V B. 10 V C. 3.75 V D. 15 V Answer: A. 15 cos A cos B gives two terms of 7.5 V. (Key point: Sideband amplitude) 3. What is the total power? A. 1156.25 W B. 1250 W C. 1406.25 W D. 1500 W Answer: C. 1250 + 56.25 + 100. (Key point: Total power) 4. How is net mu found? A. (mu1+mu2)/2 = 0.35 B. mu1 + mu2 = 0.7 C. mu1 x mu2 = 0.12 D. sqrt(mu1^2 + mu2^2) = 0.5 Answer: D. Powers add, so squares add: sqrt(0.09+0.16) = 0.5. (Key point: Net modulation index) ## Close variants set in other exams - [Two-tone AM: net modulation index, power, spectrum](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-two-tone-50-20-12) — Communication Systems, chapter 2, asked 2× ## Related in Communication Systems - [Two-tone AM: spectrum, net index, power, efficiency](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-two-tone-100-30-40) — Communication Systems, chapter 2 - [Two-tone AM: equation, index, power, spectrum](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-two-tone-10cos-15cos-20cos) — Communication Systems, chapter 2 - [Two-tone AM: index, power, spectrum (5cos, 10cos, 15cos)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-two-tone-5cos-10cos-15cos) — Communication Systems, chapter 2 - [AM transmitter 10 kW/11.8 kW: modulation index and power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-10kw-11-8kw) — Communication Systems, chapter 2 - [AM numerical: S(t) = 10(1 + 0.8cos25132.74t)cos(9424777.96t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-10-0-8-25132-numerical) — Communication Systems, chapter 2 - [Double-tone AM expression, bandwidth and modulation index](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adouble-tone-am-bw-index) — Communication Systems, chapter 2 ## Related in other subjects - [AM numerical: 10 sin1000πt on 50 sin 2π×10⁵t](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aam-numeric-10sin1000pi) — Data Communication, chapter 5 - [Modulation; AM and FM modulated waves](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Amodulation-am-fm-wave) — Basic Electronics Engineering, chapter 5 - [Total rainfall, net runoff, W-index of 120-min storm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aphi-w-index-120min-storm) — Engineering Hydrology, chapter 3, asked 2×