---
title: "FM expression after change of modulating signal"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:fm-new-waveform-500hz-2khz"
source: https://www.bctengineer.com/q/semester_6/communication_systems/fm-new-waveform-500hz-2khz
site: BCT Engineering
---

# FM expression after change of modulating signal

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2072 Asoj (CS I), 6 marks
- Video: [Module3_Vid_5_Angle Modulation and Demodulation_FM PM Expressions](https://www.youtube.com/watch?v=X-qlSSqKwN4) — in5minutes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/fm-new-waveform-500hz-2khz
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

A carrier of frequency 10⁶ Hz and amplitude 3 volts is frequency modulated by a sinusoidal modulating signal frequency 500 Hz and peak amplitude 1 volt. The frequency deviation is 1 kHz. The level of the modulating waveform is changed to 5 V peak and the modulating frequency is changed to 2 kHz. Write the expression for the new modulated waveform.

## Short answer

The frequency sensitivity k_f = Δf/A_m = 1 kHz/V. With new amplitude 5 V, new deviation Δf′ = 5 kHz. New modulation index β′ = Δf′/f_m′ = 5 kHz / 2 kHz = 2.5. The FM waveform is s(t) = 3 cos[2π×10^6 t + 2.5 sin(4000π t)] V.

## Answer

**Given:** $f_c = 10^6$ Hz, $A_c = 3$ V, original $A_m = 1$ V, $f_m = 500$ Hz, $\Delta f = 1$ kHz.

**Step 1: Frequency sensitivity of the modulator.** In FM, the deviation is proportional to the modulating amplitude:

$$
k_f = \frac{\Delta f}{A_m} = \frac{1\ \text{kHz}}{1\ \text{V}} = 1\ \text{kHz/V}
$$

**Step 2: New deviation.** New $A_m = 5$ V, new $f_m = 2$ kHz. The deviation depends only on amplitude, not on $f_m$:

$$
\Delta f' = k_f A_m' = 1\ \text{kHz/V} \times 5\ \text{V} = 5\ \text{kHz}
$$

**Step 3: New modulation index.**

$$
\beta' = \frac{\Delta f'}{f_m'} = \frac{5\ \text{kHz}}{2\ \text{kHz}} = 2.5
$$

(The original index was $\beta = 1000/500 = 2$.)

**Step 4: New FM waveform.** Taking the modulating signal as a cosine, $m(t) = 5\cos(2\pi \times 2000\,t)$:

$$
\begin{aligned}
s(t) &= A_c \cos\left[2\pi f_c t + \beta' \sin 2\pi f_m' t\right] \\
&= 3\cos\left[2\pi \times 10^6\, t + 2.5 \sin(2\pi \times 2000\, t)\right]\ \text{V} \\
&= 3\cos\left[6.283\times10^6\, t + 2.5\sin(12566\, t)\right]\ \text{V}
\end{aligned}
$$

(If a sine carrier is used, write $3\sin[\ldots]$ with the same argument.)

**Answer:** $s(t) = 3\cos[2\pi\times10^6 t + 2.5\sin(4000\pi t)]$ V, with $\Delta f = 5$ kHz and $\beta = 2.5$.

## Check yourself

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

1. What is kf?
   A. Delta f x fm
   B. 2 kHz/V
   C. Delta f/fm = 2
   D. Delta f / Am = 1 kHz/V
   Answer: D. Sensitivity is deviation per volt. (Key point: kf)

2. What is the new deviation?
   A. 5 kHz, depends on amplitude not fm
   B. 1 kHz
   C. 2.5 kHz
   D. 10 kHz
   Answer: A. Delta f' = kf x 5 V. (Key point: New deviation)

3. What is the new beta?
   A. 5 x 2 = 10
   B. 5/2 = 2.5
   C. 2
   D. 0.4
   Answer: B. beta = Delta f'/fm'. (Key point: New beta)

4. Which is the new waveform?
   A. 3 cos[2pi x 10^6 t + 2 sin(1000 pi t)]
   B. 5 cos[2pi x 10^6 t + 2.5 sin(4000 pi t)]
   C. 3 cos[2pi x 10^6 t + 2.5 sin(4000 pi t)]
   D. 3 cos[2pi x 10^6 t + 5 sin(4000 pi t)]
   Answer: C. Amplitude stays 3 V. (Key point: Expression)

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