---
title: "AM numerical: S(t) = 10(1 + 0.2cos2π10³t)cos2π10⁶t"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:am-numerical-10-0-2"
source: https://www.bctengineer.com/q/semester_6/communication_systems/am-numerical-10-0-2
site: BCT Engineering
---

# AM numerical: S(t) = 10(1 + 0.2cos2π10³t)cos2π10⁶t

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2076 Bhadra (CS I), 1+2+1+2+1 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-numerical-10-0-2
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

An amplitude modulated signal is represented by
S_AM(t) = 10(1 + 0.2 cos2π10³t) cos 2π10⁶t.
a) Identify which type of modulation.
b) Find modulating frequency and carrier frequency
c) Bandwidth of the signal
d) Carrier power, total power, power in side bands
e) Efficiency

## Short answer

DSB-FC single-tone AM with 20% modulation. fm = 1 kHz, fc = 1 MHz, bandwidth = 2 kHz. Carrier power = 50 W, total power = 51 W, sideband power = 1 W. Efficiency = 1.96%.

## Answer

Compare with $s(t) = A_c[1 + \mu\cos2\pi f_mt]\cos2\pi f_ct$: $A_c = 10$ V, $\mu = 0.2$, $f_m = 10^3$ Hz, $f_c = 10^6$ Hz. Take $R = 1\ \Omega$.

#### a) Type of modulation

Carrier term is present and the envelope is $10(1 + 0.2\cos\omega_mt)$: it is **DSB-FC (conventional AM), single-tone, with 20% modulation**.

#### b) Frequencies

$$
f_m = 1\ \text{kHz}, \qquad f_c = 1\ \text{MHz}
$$

#### c) Bandwidth

$$
B = 2f_m = 2 \times 1 = 2\ \text{kHz}
$$

#### d) Powers

$$
\begin{aligned}
P_c &= \frac{A_c^2}{2R} = \frac{10^2}{2} = 50\ \text{W} \\
P_{SB} &= P_c\frac{\mu^2}{2} = 50 \times \frac{0.04}{2} = 1\ \text{W}\ (0.5\ \text{W each}) \\
P_t &= P_c + P_{SB} = 51\ \text{W}
\end{aligned}
$$

#### e) Efficiency

$$
\eta = \frac{P_{SB}}{P_t} = \frac{\mu^2}{2 + \mu^2} = \frac{0.04}{2.04} = 0.0196 = 1.96\%
$$

**Answer:** DSB-FC AM; $f_m$ = 1 kHz, $f_c$ = 1 MHz; $B$ = 2 kHz; $P_c$ = 50 W, $P_t$ = 51 W, $P_{SB}$ = 1 W; $\eta$ = 1.96%.

## Check yourself

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

1. The wave 10(1 + 0.2 cos 2pi 10^3 t)cos 2pi 10^6 t is which type?
   A. DSB-SC
   B. DSB-FC (conventional AM), single tone, 20% modulation
   C. SSB
   D. FM
   Answer: B. The carrier term is present and mu = 0.2. (Key point: Type of modulation)

2. What is the bandwidth?
   A. 1 kHz
   B. 1 MHz
   C. 2 MHz
   D. 2 kHz
   Answer: D. B = 2fm = 2 x 1 kHz. (Key point: Bandwidth)

3. What are Pc, sideband power and Pt?
   A. 50 W, 1 W and 51 W
   B. 100 W, 2 W and 102 W
   C. 50 W, 0.5 W and 50.5 W
   D. 10 W, 0.2 W and 10.2 W
   Answer: A. Pc = 100/2, Psb = Pc mu^2/2 = 1 W, Pt = 51 W. (Key point: Powers)

4. What is the efficiency?
   A. 20%
   B. 4%
   C. 1.96%
   D. 33.3%
   Answer: C. eta = mu^2/(2 + mu^2) = 0.04/2.04. (Key point: Efficiency)

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