--- title: "Two-tone AM: net modulation index, power, spectrum" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:am-two-tone-50-20-12" source: https://www.bctengineer.com/q/semester_6/communication_systems/am-two-tone-50-20-12 site: BCT Engineering --- # Two-tone AM: net modulation index, power, spectrum Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2080 Chaitra (CS I), 2+2+2 marks - Video: [Calculate modulation index and efficiency of AM signal(Amplitude modulation)](https://www.youtube.com/watch?v=fBsSvkONQ8c) — Signals Systems - 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-50-20-12 - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question For an amplitude modulated signal s(t) = 50cos(2π10⁶t) + 20cos(2π10⁶t)cos(2π10³t) + 12cos(2π10⁶t)cos(4π10²t) a) Calculate the net modulation index. b) Calculate the total modulated power. c) Draw the spectrum of the signal. ## Short answer The standard AM signal has μ₁=0.4 (1kHz) and μ₂=0.24 (200Hz). Net modulation index μ_t = √(0.4²+0.24²) = 0.4665. For R=1Ω, carrier power P_c=1250 W, total modulated power P_t=1250(1+0.2176/2)=1386 W. Spectrum: carrier 50 V at 1MHz, sidebands 10 V at ±1kHz and 6 V at ±200Hz. ## Answer Rewrite the signal in standard two-tone AM form: $$ s(t) = 50\left[1 + 0.4\cos(2\pi10^3t) + 0.24\cos(2\pi\,200\,t)\right]\cos(2\pi10^6t) $$ since $\dfrac{20}{50} = 0.4$, $\dfrac{12}{50} = 0.24$ and $4\pi10^2t = 2\pi(200)t$. So $A_c = 50$ V, $f_c = 1$ MHz, $\mu_1 = 0.4$ at $f_{m1} = 1$ kHz, $\mu_2 = 0.24$ at $f_{m2} = 200$ Hz. #### a) Net modulation index $$ \mu_t = \sqrt{\mu_1^2 + \mu_2^2} = \sqrt{0.16 + 0.0576} = \sqrt{0.2176} = 0.4665 $$ **Answer:** $\mu_t \approx 0.466$ (46.6%). #### b) Total modulated power Assume $R = 1\ \Omega$ (normalised). $$ \begin{aligned} P_c &= \frac{A_c^2}{2R} = \frac{50^2}{2} = 1250\ \text{W} \\ P_t &= P_c\left(1 + \frac{\mu_t^2}{2}\right) = 1250\left(1 + \frac{0.2176}{2}\right) \\ &= 1250 \times 1.1088 = 1386\ \text{W} \end{aligned} $$ **Answer:** $P_t = 1386$ W (sideband power 136 W; efficiency $= 136/1386 = 9.81\%$). #### c) Spectrum Side-frequency amplitudes $= \mu A_c/2$: for 1 kHz, $20/2 = 10$ V; for 200 Hz, $12/2 = 6$ V. | Frequency | Amplitude | |---|---| | 0.999 MHz (999 kHz) | 10 V | | 0.9998 MHz (999.8 kHz) | 6 V | | 1 MHz | 50 V | | 1.0002 MHz (1000.2 kHz) | 6 V | | 1.001 MHz (1001 kHz) | 10 V | ```text A (V) 50 | | | | 10 | | | | 6 | | | | | | +--+-----+--------+--------+-----+---> f (kHz) 999 999.8 1000 1000.2 1001 ``` Bandwidth $= 2 \times 1\text{ kHz} = 2$ kHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the signal written to find the indices? A. mu1 = 20 and mu2 = 12 B. Divide each modulating amplitude by Ac = 50: mu1 = 0.4 and mu2 = 0.24 C. mu1 = 50/20 and mu2 = 50/12 D. mu1 = 0.2 and mu2 = 0.12 Answer: B. mu = Am/Ac, so 20/50 = 0.4 and 12/50 = 0.24. (Key point: Standard form) 2. What is the net modulation index? A. About 0.64 B. About 0.22 C. About 0.93 D. About 0.466 Answer: D. sqrt(0.16 + 0.0576) = sqrt(0.2176) = 0.4665. (Key point: Net index) 3. Taking R = 1 ohm, what is the total modulated power? A. About 1386 W B. About 1250 W C. About 1388 W with Pc = 2500 W D. About 136 W Answer: A. Pc = 50^2/2 = 1250 W and Pt = 1250(1 + 0.2176/2) = 1386 W. (Key point: Total power) 4. What are the side-frequency amplitudes in the spectrum? A. 20 V and 12 V B. 40 V and 24 V C. 10 V at 999 and 1001 kHz, 6 V at 999.8 and 1000.2 kHz D. 5 V and 3 V Answer: C. Amplitude is mu Ac/2, i.e. 20/2 = 10 V and 12/2 = 6 V. (Key point: Spectrum amplitudes) ## Close variants set in other exams - [Two-tone AM: spectrum, power, net index (50, 15, 20)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-two-tone-50-15-20) — 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 - [Modulated signal Z(t) = 50cos{98.6×10⁶×2πt + 15sin(2×10³×2πt)}](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aangle-mod-50cos-98-6mhz) — Communication Systems, chapter 3 - [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 - [Angle modulated signal 20cos(6×10⁸t + 7sin1250t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aangle-mod-20cos-6e8t) — Communication Systems, chapter 3 ## Related in other subjects - [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 140-min storm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aphi-w-index-140min-storm) — Engineering Hydrology, chapter 3, asked 2× - [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× - [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