---
title: "AM numerical: S(t) = 20(1 + 0.8cos2π1000t)cos(9424777.96t)"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:am-20-0-8-numerical"
source: https://www.bctengineer.com/q/semester_6/communication_systems/am-20-0-8-numerical
site: BCT Engineering
---

# AM numerical: S(t) = 20(1 + 0.8cos2π1000t)cos(9424777.96t)

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2071 Bhadra (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-20-0-8-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

An AM wave is represented by S_AM(t) = 20(1 + 0.8 cos2π1000t) cos(9424777.96t) volts. Find
i) Amplitude of all frequency components
ii) Modulation index
iii) Maximum and minimum amplitude of AM wave
iv) Frequency of USB and LSB

## Short answer

Carrier amplitude 20V at 1.5MHz; sidebands 8V each at 1.501MHz and 1.499MHz. Modulation index 0.8; maximum amplitude 36V, minimum 4V. USB frequency 1501kHz, LSB 1499kHz.

## Answer

Compare the given wave with the standard single-tone AM equation:

$$
s_{AM}(t) = A_c\,[1 + \mu \cos(2\pi f_m t)]\cos(2\pi f_c t)
$$

So $A_c = 20$ V, $\mu = 0.8$, $f_m = 1000$ Hz and $\omega_c = 9424777.96$ rad/s, which gives

$$
f_c = \frac{\omega_c}{2\pi} = \frac{9424777.96}{2\pi} = 1.5\times10^{6}\ \text{Hz} = 1.5\ \text{MHz}
$$

Expanding the product:

$$
s_{AM}(t) = 20\cos 2\pi f_c t + \frac{\mu A_c}{2}\cos 2\pi(f_c+f_m)t + \frac{\mu A_c}{2}\cos 2\pi(f_c-f_m)t
$$

#### i) Amplitude of all frequency components

$$
\frac{\mu A_c}{2} = \frac{0.8 \times 20}{2} = 8\ \text{V}
$$

| Component | Frequency | Amplitude |
|---|---|---|
| Carrier | 1500 kHz | 20 V |
| Upper sideband | 1501 kHz | 8 V |
| Lower sideband | 1499 kHz | 8 V |

#### ii) Modulation index

The coefficient of the cosine term inside the bracket is the modulation index: $\mu = 0.8$ (80% modulation).

#### iii) Maximum and minimum amplitude

$$
\begin{aligned}
A_{max} &= A_c(1+\mu) = 20(1+0.8) = 36\ \text{V} \\
A_{min} &= A_c(1-\mu) = 20(1-0.8) = 4\ \text{V}
\end{aligned}
$$

Check: $\mu = \dfrac{A_{max}-A_{min}}{A_{max}+A_{min}} = \dfrac{36-4}{36+4} = 0.8$.

#### iv) Frequency of USB and LSB

$$
\begin{aligned}
f_{USB} &= f_c + f_m = 1500 + 1 = 1501\ \text{kHz} \\
f_{LSB} &= f_c - f_m = 1500 - 1 = 1499\ \text{kHz}
\end{aligned}
$$

**Answer:** carrier 20 V at 1.5 MHz, sidebands 8 V each; $\mu = 0.8$; $A_{max} = 36$ V, $A_{min} = 4$ V; USB = 1.501 MHz, LSB = 1.499 MHz.

## Check yourself

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

1. From 20(1 + 0.8 cos 2pi 1000 t) cos(9424777.96 t), what is fc?
   A. 3 MHz
   B. 1 kHz
   C. 9.42 MHz
   D. 1.5 MHz
   Answer: D. fc = 9424777.96/(2 pi) = 1.5 x 10^6 Hz. (Key point: Reading the equation)

2. What is the amplitude of each sideband?
   A. 16 V
   B. 8 V
   C. 20 V
   D. 4 V
   Answer: B. mu Ac/2 = 0.8 x 20/2 = 8 V. (Key point: Sideband amplitude)

3. What are the maximum and minimum amplitudes of the AM wave?
   A. 20 V and 0 V
   B. 40 V and 0 V
   C. 36 V and 4 V
   D. 28 V and 12 V
   Answer: C. Amax = Ac(1+mu) = 36 V, Amin = Ac(1-mu) = 4 V. (Key point: Max and min)

4. What are the USB and LSB frequencies?
   A. 1510 kHz and 1490 kHz
   B. 1501 kHz and 1499 kHz
   C. 1500.8 kHz and 1499.2 kHz
   D. 1520 kHz and 1480 kHz
   Answer: B. fc +/- fm = 1500 +/- 1 kHz. (Key point: Sideband frequencies)

## Related in Communication Systems

- [AM numerical: S(t) = 10(1 + 0.8cos25132.74t)cos(9424777.96t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-10-0-8-25132-numerical) — Communication Systems, chapter 2
- [AM numerical: s(t) = 5[1 + 0.6cos(3140t)]cos(2π10³t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-5-0-6cos3140t-numerical) — Communication Systems, chapter 2
- [AM numerical: s(t) = 5[1 + 0.6cos(6280t)]cos(2π10⁴t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-5-0-6cos6280t-numerical) — Communication Systems, chapter 2
- [AM numerical: u(t) = [20 + 2cos3000πt + 10cos6000πt]cos2πfct](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-20-2cos-10cos-fc-1e5) — Communication Systems, chapter 2
- [AM numerical: S(t) = 10(1 + 0.2cos2π10³t)cos2π10⁶t](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-numerical-10-0-2) — Communication Systems, chapter 2
- [AM wave diagram, modulation index; why near 100%](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-diagram-index-100-percent) — Communication Systems, chapter 2