---
title: "DSB-FC single tone expression and modulation index"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:dsbfc-single-tone-index"
source: https://www.bctengineer.com/q/semester_6/communication_systems/dsbfc-single-tone-index
site: BCT Engineering
---

# DSB-FC single tone expression and modulation index

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2068 Bhadra (CS I), 6+2 marks
- Video: [Understanding Single Tone AM: Expression And Spectrum | GATE Communications](https://www.youtube.com/watch?v=6W2vrTczIok) — Ekeeda GATE & ESE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/dsbfc-single-tone-index
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

Derive the expression for double side band full carrier amplitude wave where the message contains a single tone frequency component. Also find the expression for modulation index.

## Short answer

For message Am cos(omega_m t) and carrier Ac cos(omega_c t), the DSB-FC wave is s(t) = Ac[1 + mu cos(omega_m t)] cos(omega_c t) with mu = ka Am / Ac. Modulation index is mu = (Amax - Amin) / (Amax + Amin).

## Answer

**DSB-FC (standard AM)** is the AM wave in which the carrier amplitude varies linearly with the message, and the carrier and both sidebands are transmitted.

#### Derivation

Let the message and carrier be

$$
m(t) = A_m\cos\omega_mt, \qquad c(t) = A_c\cos\omega_ct, \qquad \omega_c \gg \omega_m
$$

In AM the instantaneous amplitude of the carrier is

$$
A(t) = A_c + k_am(t) = A_c + k_aA_m\cos\omega_mt
$$

where $k_a$ is the amplitude sensitivity of the modulator. The AM wave is

$$
\begin{aligned}
s(t) &= [A_c + k_aA_m\cos\omega_mt]\cos\omega_ct \\
&= A_c\left[1 + \frac{k_aA_m}{A_c}\cos\omega_mt\right]\cos\omega_ct \\
&= A_c[1 + \mu\cos\omega_mt]\cos\omega_ct
\end{aligned}
$$

with $\mu = k_aA_m/A_c$. Using $\cos A\cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$:

$$
s(t) = \underbrace{A_c\cos\omega_ct}_{\text{carrier}} + \underbrace{\frac{\mu A_c}{2}\cos(\omega_c+\omega_m)t}_{\text{USB}} + \underbrace{\frac{\mu A_c}{2}\cos(\omega_c-\omega_m)t}_{\text{LSB}}
$$

So the wave has three components: the carrier at $f_c$ and two sidebands at $f_c \pm f_m$, each of amplitude $\mu A_c/2$. The bandwidth is $2f_m$.

```text
 Amp
  Ac  |         |
      |         |
 uAc/2|    |    |    |
      +----+----+----+---> f
        fc-fm  fc  fc+fm
```

Power (1 Ω): $P_t = \dfrac{A_c^2}{2}\left(1 + \dfrac{\mu^2}{2}\right)$.

#### Modulation index

$$
\mu = \frac{k_aA_m}{A_c} = \frac{\text{peak amplitude change}}{\text{carrier amplitude}}
$$

From the waveform, $A_{max} = A_c(1+\mu)$ and $A_{min} = A_c(1-\mu)$. Solving:

$$
\mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}}
$$

For distortion-free envelope detection, $0 < \mu \le 1$; $\mu > 1$ is over-modulation.

## Check yourself

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

1. For a single-tone message, which expression gives the DSB-FC wave?
   A. s(t) = Ac cos(ωc t + μ cos ωm t)
   B. s(t) = Ac[1 + μ cos ωm t] cos ωc t
   C. s(t) = Ac μ cos ωm t cos ωc t
   D. s(t) = Ac cos ωc t + μ cos ωm t
   Answer: B. The carrier amplitude is Ac + ka Am cos ωm t = Ac[1 + μ cos ωm t], and this multiplies cos ωc t. (Key point: Amplitude of AM wave)

2. Expanding s(t) with the product-to-sum identity, which components appear and with what amplitudes?
   A. Carrier Ac only, with sidebands at fc ± fm each of amplitude μAc
   B. Two sidebands only, each of amplitude μAc/2
   C. Carrier Ac and sidebands at fc ± fm, each of amplitude μAc/2
   D. Carrier Ac and one sideband at fc + fm of amplitude μAc
   Answer: C. The product gives the carrier Ac cos ωc t plus USB and LSB, each of amplitude μAc/2, so bandwidth is 2fm. (Key point: Spectral components)

3. What is the modulation index μ in terms of the circuit quantities?
   A. μ = ka Am / Ac
   B. μ = Ac / (ka Am)
   C. μ = ka Ac / Am
   D. μ = fm / fc
   Answer: A. μ = ka Am / Ac is the peak amplitude change divided by the carrier amplitude. (Key point: Definition of modulation index)

4. With Amax = Ac(1+μ) and Amin = Ac(1−μ), what is μ?
   A. (Amax + Amin)/(Amax − Amin)
   B. (Amax − Amin)/Amax
   C. Amax/Amin
   D. (Amax − Amin)/(Amax + Amin)
   Answer: D. Subtracting gives 2μAc and adding gives 2Ac, so μ = (Amax − Amin)/(Amax + Amin). (Key point: Modulation index from the waveform)

## Related in Communication Systems

- [Single tone DSB-FC AM in time and frequency domain](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asingle-tone-dsbfc-time-freq) — Communication Systems, chapter 2
- [Double-tone AM expression, bandwidth and modulation index](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adouble-tone-am-bw-index) — Communication Systems, chapter 2
- [Angle modulation; single tone FM time and frequency domain](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aangle-mod-single-tone-fm-time-freq) — Communication Systems, chapter 3
- [Need for modulation; single tone AM expressions](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amodulation-need-am-single-tone) — Communication Systems, chapter 2, asked 2×
- [Double-tone AM; DSB vs SSB](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adouble-tone-am-dsb-vs-ssb) — Communication Systems, chapter 2
- [USB-SSB expression for a random band limited message](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ausb-ssb-random-message) — Communication Systems, chapter 2

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- [Single tone FM parameters](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afm-single-tone-numeric) — Data Communication, chapter 5
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