--- title: "AM wave diagram, modulation index; why near 100%" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:am-diagram-index-100-percent" source: https://www.bctengineer.com/q/semester_6/communication_systems/am-diagram-index-100-percent site: BCT Engineering --- # AM wave diagram, modulation index; why near 100% Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2072 Asoj (CS I), 6+2 marks - Video: [Amplitude Modulation- AM Waveform- Draw Modulating Signal, Carrier Wave, AM wave- Modulation index](https://www.youtube.com/watch?v=19q2X6Prr7s) — Engineering Made Easy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/am-diagram-index-100-percent - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question Draw a neat diagram of amplitude-modulated wave and derive an expression for modulation index. Justify why AM transmitters are generally operated with the modulation index as close to 100% as possible. ## Short answer Modulation index μ = (Amax−Amin)/(Amax+Amin). Power efficiency η = μ²/(2+μ²) increases from 1.96% at μ=0.2 to 33.3% at μ=1. Operating close to 100% puts maximum power in sidebands, improving signal strength and S/N, but μ must not exceed 1 to avoid envelope distortion and spectral splatter. ## Answer #### AM wave and modulation index ```text v(t) Amax - __ __ / \ envelope / \ Ac - -/----\--------------/----\--- Amin - | \__ __/ | |||||||||||||||||||||||||||| carrier 0 ---+--------------------------------> t |||||||||||||||||||||||||||| -Amin- \ /\ /\ / -Amax- \/ envelope (mirror) |<----- 1/fm ------>| Peak-to-peak at crest = 2Amax, at trough = 2Amin ``` AM wave: $s(t) = A_c[1 + \mu\cos\omega_mt]\cos\omega_ct$, with $\mu = A_m/A_c$. **Derivation of $\mu$ from the waveform.** The envelope is $A_c(1 + \mu\cos\omega_mt)$. It is largest when $\cos\omega_mt = 1$ and smallest when $\cos\omega_mt = -1$: $$ \begin{aligned} A_{max} &= A_c(1 + \mu) \\ A_{min} &= A_c(1 - \mu) \end{aligned} $$ Adding and subtracting: $$ \begin{aligned} A_{max} + A_{min} &= 2A_c \Rightarrow A_c = \frac{A_{max} + A_{min}}{2} \\ A_{max} - A_{min} &= 2\mu A_c \Rightarrow \mu = \frac{A_{max} - A_{min}}{2A_c} \end{aligned} $$ $$ \mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}} $$ Also $A_m = \dfrac{A_{max} - A_{min}}{2}$, so $\mu = A_m/A_c$. In terms of peak-to-peak values read on an oscilloscope, $\mu = \dfrac{V_{pp,max} - V_{pp,min}}{V_{pp,max} + V_{pp,min}}$. #### Why operate close to 100% - **Power efficiency:** $\eta = \dfrac{\mu^2}{2 + \mu^2}$ rises with $\mu$: 1.96% at $\mu = 0.2$, 15.3% at $\mu = 0.6$, 33.3% at $\mu = 1$. More of the transmitted power carries information. - **Better signal at the receiver:** detected audio amplitude is proportional to $\mu A_c$, so higher $\mu$ gives a stronger output and higher S/N for the same carrier power. - But $\mu$ must **not exceed 1**: over-modulation distorts the envelope, generates spurious sidebands (splatter) and causes adjacent-channel interference. So transmitters are run as close to 100% as possible without exceeding it. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For s(t) = Ac[1 + mu cos(wm t)] cos(wc t), Amax and Amin are: A. Ac mu and Ac/mu B. Ac(1 + mu) and Ac(1 - mu) C. Ac and 0 D. 2Ac and Ac Answer: B. The envelope is largest when cos = 1 and smallest when cos = -1. (Key point: Envelope extremes) 2. Solving the two envelope equations gives mu equal to: A. (Amax + Amin)/(Amax - Amin) B. Amax/Amin C. (Amax - Amin)/(Amax + Amin) D. (Amax - Amin)/2 Answer: C. Subtracting gives 2 mu Ac and adding gives 2 Ac; the ratio is mu. (Key point: Derived formula) 3. Why are AM transmitters run with mu as close to 100% as possible? A. Bandwidth shrinks B. The carrier power is eliminated C. The receiver needs no detector D. Efficiency mu^2/(2 + mu^2) and detected signal strength increase Answer: D. A larger mu puts more power into the sidebands and gives a stronger detected audio signal. (Key point: Why mu near 100%) 4. Why must mu not exceed 1? A. Over-modulation distorts the envelope, creating spurious sidebands and adjacent-channel interference B. Efficiency would fall C. The carrier would vanish D. The bandwidth would halve Answer: A. Beyond 100% the envelope no longer follows the message, causing splatter. 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