---
title: "AM numerical: s(t) = 5[1 + 0.6cos(6280t)]cos(2π10⁴t)"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:am-5-0-6cos6280t-numerical"
source: https://www.bctengineer.com/q/semester_6/communication_systems/am-5-0-6cos6280t-numerical
site: BCT Engineering
---

# AM numerical: s(t) = 5[1 + 0.6cos(6280t)]cos(2π10⁴t)

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2074 Bhadra (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-5-0-6cos6280t-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 + 0.6 cos(6280t)] cos(2π10⁴t) volts, then find the followings: (a) Modulation Depth, (b) Maximum and Minimum Amplitude of AM wave, (c) Frequency components in modulated signal and their amplitudes, (d) Power dissipated across 1K Ohm resistor.

## Short answer

Modulation depth μ=0.6. Max amplitude 8V, min 2V. Frequency components: carrier 10kHz (5V), USB 11kHz (1.5V), LSB 9kHz (1.5V). Total power across 1kΩ = 14.75mW.

## Answer

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

$A_c = 5$ V, $\mu = 0.6$, $\omega_m = 6280$ rad/s so $f_m = \dfrac{6280}{2\pi} = 999.5 \approx 1$ kHz, and $f_c = 10^4$ Hz = 10 kHz.

#### (a) Modulation depth

$$
\mu = 0.6 \quad (60\%)
$$

#### (b) Maximum and minimum amplitude

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

#### (c) Frequency components and amplitudes

$$
s(t) = 5\cos2\pi(10^4)t + \frac{0.6\times5}{2}\cos2\pi(10^4 + 10^3)t + \frac{0.6\times5}{2}\cos2\pi(10^4 - 10^3)t
$$

| Component | Frequency | Amplitude |
|---|---|---|
| LSB | 9 kHz | 1.5 V |
| Carrier | 10 kHz | 5 V |
| USB | 11 kHz | 1.5 V |

(Using exact $f_m = 999.5$ Hz: 9.0005 kHz and 10.9995 kHz.)

#### (d) Power across 1 kΩ

$$
\begin{aligned}
P_c &= \frac{A_c^2}{2R} = \frac{25}{2000} = 12.5\ \text{mW} \\
P_{USB} = P_{LSB} &= \frac{(1.5)^2}{2\times1000} = 1.125\ \text{mW} \\
P_t &= 12.5 + 2(1.125) = 14.75\ \text{mW}
\end{aligned}
$$

Check: $P_c(1 + \mu^2/2) = 12.5 \times 1.18 = 14.75$ mW.

**Answer:** $\mu = 0.6$; 8 V and 2 V; 10 kHz (5 V), 9 kHz and 11 kHz (1.5 V each); $P_t = 14.75$ mW.

## Check yourself

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

1. From s(t) = 5[1 + 0.6 cos(6280t)] cos(2 pi 10^4 t), the values of mu and fm are:
   A. mu = 0.6, fm about 1 kHz
   B. mu = 3, fm = 6280 Hz
   C. mu = 0.6, fm = 6280 Hz
   D. mu = 5, fm about 1 kHz
   Answer: A. Comparing with Ac[1 + mu cos wm t] cos wc t gives mu = 0.6 and fm = 6280/(2 pi) = about 1 kHz. (Key point: Reading parameters)

2. Amax and Amin are:
   A. 5 V and 0 V
   B. 8 V and 2 V
   C. 3 V and 7 V
   D. 6 V and 4 V
   Answer: B. Amax = Ac(1 + mu) = 5 x 1.6 = 8 V; Amin = Ac(1 - mu) = 5 x 0.4 = 2 V. (Key point: Maximum and minimum amplitude)

3. The side frequencies and their amplitudes are:
   A. 9 kHz and 11 kHz, each 3 V
   B. 9 kHz and 11 kHz, each 0.6 V
   C. 1 kHz and 10 kHz, each 1.5 V
   D. 9 kHz and 11 kHz, each 1.5 V
   Answer: D. Side amplitude = mu Ac/2 = 0.6 x 5/2 = 1.5 V at fc +/- fm = 11 kHz and 9 kHz. (Key point: Frequency components)

4. The total power dissipated in 1 kohm is:
   A. 14.75 mW
   B. 12.5 mW
   C. 25 mW
   D. 29.5 mW
   Answer: A. Pc = 25/2000 = 12.5 mW and Pt = Pc(1 + 0.36/2) = 12.5 x 1.18 = 14.75 mW. (Key point: Total power in 1 kohm)

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