--- title: "AM numerical: 15 sin2π(1500t) audio on 60 sin2π(100,000t)" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:am-audio-15sin-60sin" source: https://www.bctengineer.com/q/semester_6/communication_systems/am-audio-15sin-60sin site: BCT Engineering --- # AM numerical: 15 sin2π(1500t) audio on 60 sin2π(100,000t) Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2072 Magh (CS I), 2+2+2+2 marks; 2071 Magh (old course), 2×4 marks (asked 2 times) - 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-audio-15sin-60sin - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question An audio signal given as 15 sin2π(1500t) amplitude modulates a carrier given as 60 sin2π(100,000t). Determine the following: i) Construct the modulated wave ii) Determine the modulation index and percent modulation iii) What frequencies would present in a spectrum analysis of the modulated wave? iv) Sketch audio and carrier wave ## Short answer Modulation index 0.25 (25%). Modulated wave: s(t)=60[1+0.25 sin(2π1500t)] sin(2π100000t). Spectrum contains carrier at 100 kHz (60 V) and sidebands at 98.5 kHz and 101.5 kHz (7.5 V each). ## Answer **Given:** message $m(t)=15\sin2\pi(1500t)$, so $A_m=15$ V, $f_m=1.5$ kHz. Carrier $c(t)=60\sin2\pi(100{,}000t)$, so $A_c=60$ V, $f_c=100$ kHz. #### i) Modulated wave For standard AM, the carrier amplitude varies as $A_c+m(t)$: $$ \begin{aligned} s(t)&=[A_c+m(t)]\sin2\pi f_ct\\ &=[60+15\sin2\pi(1500t)]\sin2\pi(100{,}000t)\\ &=60[1+0.25\sin2\pi(1500t)]\sin2\pi(100{,}000t) \end{aligned} $$ Expanding with $\sin A\sin B=\frac{1}{2}[\cos(A-B)-\cos(A+B)]$: $$ s(t)=60\sin2\pi(10^5t)+7.5\cos2\pi(98{,}500t)-7.5\cos2\pi(101{,}500t) $$ The envelope varies between $60+15=75$ V and $60-15=45$ V. #### ii) Modulation index and percent modulation $$ \begin{aligned} \mu&=\frac{A_m}{A_c}=\frac{15}{60}=0.25\\ \text{Percent modulation}&=0.25\times100=25\% \end{aligned} $$ Check: $\mu=\frac{E_{max}-E_{min}}{E_{max}+E_{min}}=\frac{75-45}{75+45}=0.25$. #### iii) Frequencies in the spectrum | Component | Frequency | Amplitude | |---|---|---| | Lower side frequency $f_c-f_m$ | 98.5 kHz | 7.5 V | | Carrier $f_c$ | 100 kHz | 60 V | | Upper side frequency $f_c+f_m$ | 101.5 kHz | 7.5 V | Side-frequency amplitude $=\mu A_c/2=0.25\times60/2=7.5$ V. Bandwidth $=2f_m=3$ kHz. ```text 60 V | 7.5 V | 7.5 V | | | ----+-----+-----+-----> f (kHz) 98.5 100 101.5 ``` (Extra: on a 1 $\Omega$ load, $P_c=60^2/2=1800$ W, each sideband $28.125$ W, total $1856.25$ W.) #### iv) Sketch of audio and carrier waves Audio period $T_m=1/1500=0.667$ ms; carrier period $T_c=1/100{,}000=10\ \mu$s, so about 66.7 carrier cycles fit in one audio cycle. ```text Audio: 15 sin(2*pi*1500 t), peak 15 V 15 | .--. | / \ 0 +-'--------'\--------/----> t | \ / -15 | '--' |<---- 0.667 ms ---->| Carrier: 60 sin(2*pi*1e5 t), peak 60 V 60 | /\ /\ /\ /\ /\ /\ 0 +/--\/--\/--\/--\/--\/--\--> t -60 | |<>| 10 us AM wave: envelope between 45 V and 75 V 75 | /\/\/\ 45 |/\/ \/\/\ /\/ 0 +-------------------------> t -45 |\/\ /\/\/\ /\/\ -75 | \/\/\/ \/\/ ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For m = 15 sin(2pi 1500t) and carrier 60 sin(2pi 100000t), the modulation index is: A. 4 B. 0.25 C. 0.5 D. 15 Answer: B. mu = Am/Ac = 15/60 = 0.25, i.e. 25 percent. (Key point: Modulation index) 2. Which frequencies are present in the AM spectrum? A. 1.5 kHz and 100 kHz B. 98.5 kHz, 100 kHz and 101.5 kHz C. 98.5 kHz and 101.5 kHz only D. 100 kHz and 101.5 kHz Answer: B. A single-tone AM wave has carrier fc and side frequencies fc +/- fm. (Key point: Spectrum) 3. What is the amplitude of each side frequency? A. 15 V B. 7.5 V C. 3.75 V D. 60 V Answer: B. mu Ac/2 = 0.25 x 60/2 = 7.5 V. (Key point: Side amplitude) 4. Between what values does the AM envelope vary? A. 0 and 60 V B. 45 V and 75 V C. 15 V and 60 V D. 30 V and 90 V Answer: B. Envelope is 60 +/- 15 V, i.e. 45 to 75 V; mu = (75-45)/(75+45) = 0.25. 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