---
title: "AM numerical: s(t) = 5[1 + 0.6cos(3140t)]cos(2π10³t)"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:am-5-0-6cos3140t-numerical"
source: https://www.bctengineer.com/q/semester_6/communication_systems/am-5-0-6cos3140t-numerical
site: BCT Engineering
---

# AM numerical: s(t) = 5[1 + 0.6cos(3140t)]cos(2π10³t)

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2077 Chaitra (CS I), 2+2+2+2 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-5-0-6cos3140t-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

An AM wave is represented by s(t) = 5[1 + 0.6cos(3140t)].cos(2π10³t) volts, then find the followings:
a) Modulation percentage,
b) Maximum and minimum amplitude of AM wave,
c) Power dissipated across 1K ohm resistor and
d) Frequency of USB and LSB.

## Short answer

Given s(t)=5[1+0.6cos(3140t)]cos(2π10³t), modulation index μ=0.6 → 60%. Max amplitude = Ac(1+μ)=8V, min = Ac(1-μ)=2V. Carrier power Pc=Ac²/(2R)=12.5mW, total power Pt=Pc(1+μ²/2)=14.75mW across 1kΩ. fm=3140/(2π)≈500Hz, fc=1000Hz; USB=fc+fm≈1.5kHz, LSB=fc-fm≈500Hz.

## Answer

Compare with $s(t) = A_c[1 + \mu\cos\omega_mt]\cos\omega_ct$:

$A_c = 5$ V, $\mu = 0.6$, $\omega_m = 3140$ rad/s, $\omega_c = 2\pi\times10^3$ rad/s.

$$
f_m = \frac{3140}{2\pi} = 499.75\ \text{Hz} \approx 500\ \text{Hz}, \qquad f_c = 1000\ \text{Hz}
$$

#### a) Modulation percentage

$$
\%\,\text{modulation} = \mu \times 100 = 0.6 \times 100 = 60\%
$$

#### b) Maximum and minimum amplitude

$$
\begin{aligned}
A_{max} &= A_c(1 + \mu) = 5(1.6) = 8\ \text{V} \\
A_{min} &= A_c(1 - \mu) = 5(0.4) = 2\ \text{V}
\end{aligned}
$$

Check: $\mu = \dfrac{8-2}{8+2} = 0.6$.

#### c) Power across 1 kΩ

$$
\begin{aligned}
P_c &= \frac{A_c^2}{2R} = \frac{5^2}{2\times1000} = 12.5\ \text{mW} \\
P_t &= P_c\left(1 + \frac{\mu^2}{2}\right) = 12.5(1 + 0.18) = 14.75\ \text{mW}
\end{aligned}
$$

Each sideband: $\frac{\mu^2}{4}P_c = 0.09 \times 12.5 = 1.125$ mW.

#### d) USB and LSB frequencies

$$
\begin{aligned}
f_{USB} &= f_c + f_m = 1000 + 499.75 = 1499.75\ \text{Hz} \approx 1.5\ \text{kHz} \\
f_{LSB} &= f_c - f_m = 1000 - 499.75 = 500.25\ \text{Hz} \approx 0.5\ \text{kHz}
\end{aligned}
$$

**Answer:** 60%; $A_{max} = 8$ V, $A_{min} = 2$ V; $P_t = 14.75$ mW; USB ≈ 1.5 kHz, LSB ≈ 500 Hz.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the modulation percentage of s(t) = 5[1 + 0.6cos(3140t)]cos(2pi 10^3 t)?
   A. 6%
   B. 60%
   C. 30%
   D. 83%
   Answer: B. mu = 0.6, and percentage modulation = mu x 100. (Key point: Modulation percentage)

2. What are the maximum and minimum amplitudes?
   A. 5.6 V and 4.4 V
   B. 3 V and 0 V
   C. 8 V and 3 V
   D. 8 V and 2 V
   Answer: D. Amax = 5(1.6) = 8 V and Amin = 5(0.4) = 2 V. (Key point: Envelope extremes)

3. What is the total power in a 1 kohm load?
   A. 14.75 mW
   B. 12.5 mW
   C. 25 mW
   D. 14.75 W
   Answer: A. Pc = 25/2000 = 12.5 mW and Pt = 12.5(1 + 0.18) = 14.75 mW. (Key point: Total power)

4. With fc = 1000 Hz and fm = 3140/2pi = 500 Hz, what are the USB and LSB?
   A. 3.14 kHz and 0
   B. 1.0005 kHz and 0.9995 kHz
   C. About 1.5 kHz and 0.5 kHz
   D. 3140 Hz and 1000 Hz
   Answer: C. fUSB = fc + fm and fLSB = fc - fm. (Key point: Sideband frequencies)

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