--- title: "AM numerical: u(t) = [20 + 2cos3000πt + 10cos6000πt]cos2πfct" subject: "Communication Systems" chapter: 2 key: "semester_6/communication_systems:am-20-2cos-10cos-fc-1e5" source: https://www.bctengineer.com/q/semester_6/communication_systems/am-20-2cos-10cos-fc-1e5 site: BCT Engineering --- # AM numerical: u(t) = [20 + 2cos3000πt + 10cos6000πt]cos2πfct Communication Systems · Chapter 2: Amplitude Modulation - Asked in: 2070 Magh (CS I), 2×4 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/am-20-2cos-10cos-fc-1e5 - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2 ## Question Given the modulated wave, u(t) = [20 + 2cos3000πt + 10cos6000πt] cos2πfct where fc = 10⁵ Hz. i) Sketch the spectrum of the signal ii) Find power contained in each frequency component iii) Calculate efficiency iv) Transmission bandwidth of the system ## Short answer Carrier: 20V (200W) at 100kHz; sidebands: 1V (0.5W) at 98.5 & 101.5kHz, 5V (12.5W) at 97 & 103kHz. Total power 226W; efficiency 11.5%. Bandwidth 6kHz (97-103kHz). ## Answer Expand the given wave ($f_c = 10^5$ Hz = 100 kHz; message tones $f_1 = 1500$ Hz and $f_2 = 3000$ Hz): $$ \begin{aligned} u(t) &= 20\cos 2\pi f_c t + 2\cos(2\pi\,1500t)\cos 2\pi f_ct + 10\cos(2\pi\,3000t)\cos 2\pi f_ct \\ &= 20\cos 2\pi f_ct + \cos 2\pi(f_c \pm 1500)t + 5\cos 2\pi(f_c \pm 3000)t \end{aligned} $$ (each $\pm$ means two terms, one with + and one with −). #### i) Spectrum (one-sided amplitude spectrum) ```text Amp (V) 20 | | | | | | 5 | | | | 1 | | | | | | +----+----+----+----+----+----> f (kHz) 97 98.5 100 101.5 103 ``` | Frequency | Amplitude | |---|---| | 97 kHz | 5 V | | 98.5 kHz | 1 V | | 100 kHz | 20 V | | 101.5 kHz | 1 V | | 103 kHz | 5 V | #### ii) Power in each component Taking a 1 Ω load, power of a cosine of amplitude $A$ is $A^2/2$. $$ \begin{aligned} P_c &= \frac{20^2}{2} = 200\ \text{W} \\ P_{98.5} = P_{101.5} &= \frac{1^2}{2} = 0.5\ \text{W each} \\ P_{97} = P_{103} &= \frac{5^2}{2} = 12.5\ \text{W each} \\ P_t &= 200 + 2(0.5) + 2(12.5) = 226\ \text{W} \end{aligned} $$ #### iii) Efficiency $$ \eta = \frac{P_{sidebands}}{P_t} = \frac{26}{226} = 0.115 = 11.5\% $$ (Check with $\mu_1 = 2/20 = 0.1$, $\mu_2 = 10/20 = 0.5$: $\mu^2 = 0.26$, $\eta = \frac{0.26}{2+0.26} = 11.5\%$.) #### iv) Transmission bandwidth The highest message frequency is 3 kHz, so $$ B_T = 2f_{max} = 2 \times 3 = 6\ \text{kHz} \quad (97\ \text{to}\ 103\ \text{kHz}) $$ **Answer:** carrier 200 W, 1.5 kHz sidebands 0.5 W each, 3 kHz sidebands 12.5 W each (total 226 W); efficiency 11.5%; bandwidth 6 kHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the amplitude of the 98.5 kHz and 101.5 kHz components? A. 2 V each B. 10 V each C. 1 V each D. 5 V each Answer: C. 2cos(...)cos(...) gives two terms of amplitude 1 V. (Key point: Sideband amplitudes) 2. What is the power in the carrier (1 ohm)? A. 400 W B. 100 W C. 200 W D. 20 W Answer: C. P = A^2/2 = 400/2 = 200 W. (Key point: Power) 3. What is the efficiency? A. 13% B. 11.5% C. 88.5% D. 26% Answer: B. Sideband power 26 W over total 226 W = 11.5%. (Key point: Efficiency) 4. What is the transmission bandwidth? A. 6 kHz B. 200 kHz C. 1.5 kHz D. 3 kHz Answer: A. B = 2 x highest message frequency = 2 x 3 kHz. 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