---
title: "FM numerical: m(t) = 20cos(2π2000t), 200 Hz/V"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:fm-numerical-20cos-50cos-200hz"
source: https://www.bctengineer.com/q/semester_6/communication_systems/fm-numerical-20cos-50cos-200hz
site: BCT Engineering
---

# FM numerical: m(t) = 20cos(2π2000t), 200 Hz/V

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2067 Shrawan (CS I), 2.5×4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/fm-numerical-20cos-50cos-200hz
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

A harmonic signal m(t) = 20 cos(2π 2000 t) Volt is used to frequency modulate the carrier signal c(t) = 50 cos(2π 10⁷ t) Volt. Assuming the frequency sensitivity of the frequency modulator to be 200 Hz/Volt, calculate:
a) peak frequency deviation
b) modulation index
c) bandwidth of modulated signal for over 98% of FM power
d) total modulated signal power

## Short answer

Peak frequency deviation Δf = k_f A_m = 200×20 = 4000 Hz = 4 kHz. Modulation index β = Δf/f_m = 2. Bandwidth over 98% power uses 3 sideband pairs: B = 2×3×2k = 12 kHz. Total power normalised to 1 Ω is A_c²/2 = 1250 W.

## Answer

**Given:** $m(t) = 20\cos(2\pi\, 2000\, t)$ V, $c(t) = 50\cos(2\pi 10^7 t)$ V, $k_f = 200$ Hz/V.

So $A_m = 20$ V, $f_m = 2$ kHz, $A_c = 50$ V, $f_c = 10$ MHz.

#### a) Peak frequency deviation

$$
\Delta f = k_f A_m = 200 \times 20 = 4000\ \text{Hz} = 4\ \text{kHz}
$$

#### b) Modulation index

$$
\beta = \frac{\Delta f}{f_m} = \frac{4000}{2000} = 2
$$

#### c) Bandwidth containing over 98% of power

Bessel coefficients for $\beta = 2$:

| $n$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $J_n(2)$ | 0.2239 | 0.5767 | 0.3528 | 0.1289 | 0.0340 |

Fraction of power in carrier + $n$ sideband pairs $= J_0^2 + 2\sum_{k=1}^{n}J_k^2$:

- up to $n = 2$: $0.0501 + 2(0.3326 + 0.1245) = 0.964$ (96.4%, not enough)
- up to $n = 3$: $0.964 + 2(0.0166) = 0.998$ (99.8%)

So 3 pairs of sidebands are needed:

$$
B = 2 n f_m = 2 \times 3 \times 2\ \text{kHz} = 12\ \text{kHz}
$$

This agrees with Carson's rule: $B_T = 2(\Delta f + f_m) = 2(4 + 2) = 12$ kHz.

#### d) Total modulated signal power

Taking a 1 Ω load (normalised power):

$$
P = \frac{A_c^2}{2} = \frac{50^2}{2} = 1250\ \text{W}
$$

(FM power equals unmodulated carrier power, independent of $\beta$.)

**Answer:** $\Delta f = 4$ kHz, $\beta = 2$, $B = 12$ kHz, $P = 1250$ W (normalised to 1 Ω).

## Check yourself

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

1. With kf = 200 Hz/V and Am = 20 V, what is the peak deviation?
   A. 400 Hz
   B. 4 kHz
   C. 40 kHz
   D. 200 Hz
   Answer: B. delta f = kf Am = 200 x 20 = 4000 Hz. (Key point: Peak deviation)

2. With fm = 2 kHz, what is beta?
   A. 8
   B. 0.5
   C. 2
   D. 4
   Answer: C. beta = 4000/2000 = 2. (Key point: Modulation index)

3. Why are 3 sideband pairs needed for 98% of the power at beta = 2?
   A. Because 2 pairs hold only 96.4% of the power, and 3 pairs hold 99.8%
   B. Because beta + 3 = 5
   C. Because the carrier holds 98%
   D. Because J3 is the largest
   Answer: A. J0^2 + 2(J1^2 + J2^2) = 0.964 is below 98%; adding J3 gives 0.998, so B = 2 x 3 x 2 = 12 kHz. (Key point: 98% power bandwidth)

4. What is the bandwidth and total power (1 Ω)?
   A. 12 kHz and 1250 W
   B. 8 kHz and 2500 W
   C. 12 kHz and 2500 W
   D. 6 kHz and 1250 W
   Answer: A. B = 12 kHz (matches Carson's 2(4 + 2)), and P = Ac^2/2 = 1250 W. (Key point: Bandwidth and power)

## Close variants set in other exams

- [FM numerical: deviation, index, 98% bandwidth, power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-numerical-10cos-80cos-400hz) — Communication Systems, chapter 3, asked 2×

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