---
title: "AM numerical: S(t) = 10(1 + 0.8cos25132.74t)cos(9424777.96t)"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:am-10-0-8-25132-numerical"
source: https://www.bctengineer.com/q/semester_6/communication_systems/am-10-0-8-25132-numerical
site: BCT Engineering
---

# AM numerical: S(t) = 10(1 + 0.8cos25132.74t)cos(9424777.96t)

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2069 Bhadra (CS I), 2×5 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-10-0-8-25132-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

An AM wave is represented by S_AM(t) = 10(1 + 0.8cos 25132.74t) cos(9424777.96t) volts. Find:
i) Amplitude of all frequency components
ii) Modulation index
iii) Maximum and minimum amplitude of AM wave
iv) Bandwidth of the signal
v) Power spectrum of the modulated signal.

## Short answer

From s(t)=10[1+0.8cos(25132t)]cos(9424778t), carrier amplitude 10V, sidebands 4V each, modulation index 0.8, Amax=18V, Amin=2V, bandwidth 8kHz. Power: carrier 50W, sidebands 8W each, total 66W.

## Answer

Compare with $s_{AM}(t) = A_c[1 + \mu\cos\omega_mt]\cos\omega_ct$:

- $A_c = 10$ V, $\mu = 0.8$
- $f_m = \dfrac{25132.74}{2\pi} = 4000$ Hz = 4 kHz
- $f_c = \dfrac{9424777.96}{2\pi} = 1.5\times10^6$ Hz = 1.5 MHz

Expanding:

$$
s_{AM}(t) = 10\cos\omega_ct + 4\cos(\omega_c+\omega_m)t + 4\cos(\omega_c-\omega_m)t
$$

#### i) Amplitude of all frequency components

Sideband amplitude $= \mu A_c/2 = 0.8\times10/2 = 4$ V.

| Component | Frequency | Amplitude |
|---|---|---|
| LSB | 1496 kHz | 4 V |
| Carrier | 1500 kHz | 10 V |
| USB | 1504 kHz | 4 V |

#### ii) Modulation index

$\mu = 0.8$ (80%).

#### iii) Maximum and minimum amplitude

$$
\begin{aligned}
A_{max} &= A_c(1+\mu) = 10(1.8) = 18\ \text{V} \\
A_{min} &= A_c(1-\mu) = 10(0.2) = 2\ \text{V}
\end{aligned}
$$

#### iv) Bandwidth

$$
B = 2f_m = 2\times4 = 8\ \text{kHz}
$$

#### v) Power spectrum

Normalized power (1 Ω load), $P = A^2/2$ for each cosine:

$$
\begin{aligned}
P_c &= \frac{10^2}{2} = 50\ \text{W} \\
P_{USB} = P_{LSB} &= \frac{4^2}{2} = 8\ \text{W} \\
P_t &= 50 + 8 + 8 = 66\ \text{W}
\end{aligned}
$$

Check: $P_t = P_c(1+\mu^2/2) = 50(1.32) = 66$ W.

In two-sided form, each power is split equally between $+f$ and $-f$: impulses of weight 25 W at $\pm1500$ kHz and 4 W at $\pm1496$ and $\pm1504$ kHz.

```text
 P (W)
  50 |          |
     |          |
   8 |     |    |    |
     +-----+----+----+------> f (kHz)
         1496 1500 1504
 (one-sided power spectrum)
```

**Answer:** carrier 10 V, sidebands 4 V; $\mu = 0.8$; $A_{max} = 18$ V, $A_{min} = 2$ V; $B = 8$ kHz; powers 50 W, 8 W, 8 W (total 66 W, 1 Ω).

## Check yourself

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

1. What is fm for 25132.74 rad/s?
   A. 4 kHz
   B. 25 kHz
   C. 8 kHz
   D. 2 kHz
   Answer: A. 25132.74/(2 pi) = 4000 Hz. (Key point: Message frequency)

2. What is the bandwidth?
   A. 8 kHz
   B. 1.5 MHz
   C. 4 kHz
   D. 16 kHz
   Answer: A. B = 2 fm. (Key point: Bandwidth)

3. What are the carrier and each sideband powers (1 ohm)?
   A. 100 W and 16 W
   B. 10 W and 4 W
   C. 50 W and 4 W
   D. 50 W and 8 W
   Answer: D. A^2/2: 100/2 and 16/2. (Key point: Powers)

4. Amax and Amin are
   A. 18 V and 2 V
   B. 10 V and 0
   C. 8 V and 2 V
   D. 18 V and 8 V
   Answer: A. 10(1.8) and 10(0.2). (Key point: Max/min)

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