---
title: "FM numerical: s(t) = 9cos(70000t + 16sin2000t)"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:fm-numerical-3cos-9cos-16sin"
source: https://www.bctengineer.com/q/semester_6/communication_systems/fm-numerical-3cos-9cos-16sin
site: BCT Engineering
---

# FM numerical: s(t) = 9cos(70000t + 16sin2000t)

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2065 Kartik (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/fm-numerical-3cos-9cos-16sin
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

A modulating signal m(t) = 3cos(2000t) modulates the carrier signal c(t) = 9cos(70000t) to produce the modulated signal s(t) = 9cos(70000t + 16 sin2000t). Calculate: the total modulated signal power, modulation index, peak frequency deviation and the bandwidth of modulated signal.

## Short answer

From the signal, modulation index β = 16. Peak frequency deviation Δf = β f_m = 16×(2000/(2π)) ≈ 5.093 kHz. Bandwidth by Carson’s rule B_T = 2(Δf + f_m) ≈ 10.82 kHz. Total power P = A_c²/2 = 40.5 W.

## Answer

**Given:** $m(t) = 3\cos(2000t)$, $c(t) = 9\cos(70000t)$, $s(t) = 9\cos(70000t + 16\sin 2000t)$.

So $A_c = 9$ V, $\omega_m = 2000$ rad/s, $\beta = 16$.

$$
f_m = \frac{2000}{2\pi} = 318.31\ \text{Hz}, \qquad f_c = \frac{70000}{2\pi} = 11.14\ \text{kHz}
$$

#### Total modulated signal power

FM has constant amplitude, so (normalised to 1 Ω):

$$
P = \frac{A_c^2}{2} = \frac{9^2}{2} = 40.5\ \text{W}
$$

#### Modulation index

From the expression directly:

$$
\beta = 16
$$

#### Peak frequency deviation

$$
\Delta f = \beta f_m = 16 \times 318.31 = 5092.96\ \text{Hz} \approx 5.093\ \text{kHz}
$$

(In rad/s: $\Delta\omega = 16 \times 2000 = 32000$ rad/s. The frequency sensitivity is $k_f = \Delta f/A_m = 5092.96/3 = 1697.65$ Hz/V.)

#### Bandwidth (Carson's rule)

$$
B_T = 2(\Delta f + f_m) = 2(5092.96 + 318.31) = 10822.5\ \text{Hz} \approx 10.82\ \text{kHz}
$$

(Equivalently $2(\beta+1)\omega_m = 2 \times 17 \times 2000 = 68000$ rad/s.)

**Answer:** $P = 40.5$ W, $\beta = 16$, $\Delta f = 5.093$ kHz, $B_T \approx 10.82$ kHz.

## Check yourself

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

1. For s(t) = 9 cos(70000 t + 16 sin 2000 t), what is the modulation index?
   A. 16
   B. 9
   C. 2000
   D. 70000
   Answer: A. The coefficient of the sine inside the argument is beta = 16. (Key point: Reading beta)

2. What is the modulating frequency fm?
   A. 2000 Hz
   B. 318.3 Hz
   C. 11.14 kHz
   D. 3 Hz
   Answer: B. fm = wm/(2π) = 2000/(2π) = 318.31 Hz. (Key point: fm from omega)

3. What is the peak frequency deviation?
   A. 32 kHz
   B. 16 Hz
   C. 5.09 kHz
   D. 318 Hz
   Answer: C. delta f = beta fm = 16 x 318.31 = 5093 Hz (32000 rad/s). (Key point: Peak deviation)

4. What are the total power (1 Ω) and Carson bandwidth?
   A. 81 W and 5.09 kHz
   B. 40.5 W and 5.09 kHz
   C. 81 W and 10.82 kHz
   D. 40.5 W and 10.82 kHz
   Answer: D. P = 81/2 = 40.5 W; BT = 2(5093 + 318.3) = 10.82 kHz. (Key point: Power and bandwidth)

## Close variants set in other exams

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