--- title: "Line spectrum of a given modulated signal" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:line-spectrum-modulated-signal" source: https://www.bctengineer.com/q/semester_6/communication_systems/line-spectrum-modulated-signal site: BCT Engineering --- # Line spectrum of a given modulated signal Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2080 Chaitra, 3 marks - Video: [Plotting Single sided Spectrum and Double Sided Spectrum](https://www.youtube.com/watch?v=ccQDdO7HZSU) — Prof Manibharathi Deenathayalan - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/line-spectrum-modulated-signal - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question For the given modulated signal draw line spectrums. s(t) = 50cos(2π10⁶t + 60°) − 15cos(2π10⁶t − 60°) + 20sin(4π10⁴t − 190°) + sin(2π10⁴t + 190°). ## Short answer The signal combines to three frequencies: 10 kHz with amplitude 1 and phase 100°, 20 kHz with amplitude 20 and phase 80°, and 1 MHz with amplitude 58.95 and phase 72.73°. These form a one-sided line spectrum. ## Answer **Given:** $s(t)=50\cos(2\pi10^6t+60°)-15\cos(2\pi10^6t-60°)+20\sin(4\pi10^4t-190°)+\sin(2\pi10^4t+190°)$ To draw line spectra, write every term as a cosine with positive amplitude, and combine terms of the same frequency. #### Step 1: 1 MHz terms (phasor addition) $$ \begin{aligned} 50\angle60°&=25+j43.30\\ -15\angle-60°&=-7.5+j12.99\\ \text{Sum}&=17.5+j56.29=58.95\angle72.73° \end{aligned} $$ So $50\cos(\cdot+60°)-15\cos(\cdot-60°)=58.95\cos(2\pi10^6t+72.73°)$. #### Step 2: 20 kHz term ($4\pi10^4t=2\pi\cdot2\times10^4t$) Using $\sin x=\cos(x-90°)$: $$ 20\sin(\omega t-190°)=20\cos(\omega t-280°)=20\cos(2\pi\,20000\,t+80°) $$ #### Step 3: 10 kHz term $$ \sin(\omega t+190°)=\cos(\omega t+100°)=\cos(2\pi\,10000\,t+100°) $$ #### Result $$ s(t)=\cos(2\pi10^4t+100°)+20\cos(2\pi\,2{\times}10^4t+80°)+58.95\cos(2\pi10^6t+72.73°) $$ | $f$ | Amplitude (one-sided) | Phase | |---|---|---| | 10 kHz | 1 | $100°$ | | 20 kHz | 20 | $80°$ | | 1 MHz | 58.95 | $72.73°$ | For the **two-sided** spectrum, each line splits into two of half amplitude at $\pm f$ (0.5, 10, 29.47) with phase $+\theta$ at $+f$ and $-\theta$ at $-f$. ```text One-sided amplitude spectrum 58.95 | | | | 20 | | | 1 | | | | +--+--+------//-----------+--> f 10 20 kHz 1 MHz One-sided phase spectrum (deg) 100 | o 80 | o 72.73 | o +--+--+------//-----------+--> f 10 20 kHz 1 MHz ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What must be done before drawing line spectra? A. Write every term as a cosine with positive amplitude and combine terms of equal frequency B. Square each term C. Differentiate the signal D. Convert to dB Answer: A. Line spectra require standard cosine form with positive amplitudes, combining same-frequency terms by phasors. (Key point: Preparation) 2. Combining 50cos(wt+60) - 15cos(wt-60) gives: A. 35 cos(wt + 0) B. 58.95 cos(wt + 72.73 deg) C. 65 cos(wt + 60) D. 58.95 cos(wt - 72.73 deg) Answer: B. Phasors 25+j43.30 and -7.5+j12.99 sum to 17.5+j56.29 = 58.95 at 72.73 degrees. (Key point: 1 MHz component) 3. In the two-sided spectrum, the 1 MHz line (one-sided 58.95) appears as: A. 58.95 at +f and -f B. 29.47 at +f and -f, with phases +theta and -theta C. 117.9 at +f only D. 29.47 at +f only Answer: B. Each one-sided line splits into two of half amplitude, with phase +theta at +f and -theta at -f. 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