---
title: "Two-tone AM: equation, index, power, spectrum"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:am-two-tone-10cos-15cos-20cos"
source: https://www.bctengineer.com/q/semester_6/communication_systems/am-two-tone-10cos-15cos-20cos
site: BCT Engineering
---

# Two-tone AM: equation, index, power, spectrum

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2076 Baisakh (CS I), 2×4 marks
- Video: [AM examples on Multi tone AM signals in Communication Engineering by Engineering Funda](https://www.youtube.com/watch?v=0oX236SrXQo) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/am-two-tone-10cos-15cos-20cos
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

The signals m(t) = 10cos2×10³πt + 15cos1×10³πt, c(t) = 20cos3×10⁶πt are applied to AM. Find
a) The equation of the resulting signal.
b) Modulation index
c) Total power
d) Draw spectrum

## Short answer

AM signal: s(t)=[20+10cos(2π1000t)+15cos(π1000t)]cos(3π10⁶t). Modulation indices μ1=0.5, μ2=0.75; total μt=√(0.5²+0.75²)=0.901. Carrier power Pc=200W (1Ω), total power Pt=Pc(1+μt²/2)=281.25W. Spectrum: carrier at 1.5MHz, sidebands at ±1kHz (5V) and ±500Hz (7.5V).

## Answer

Given:

$$
m(t) = 10\cos(2\pi\,1000\,t) + 15\cos(2\pi\,500\,t), \quad c(t) = 20\cos(2\pi\times1.5\times10^6t)
$$

So $A_{m1} = 10$ V at $f_{m1} = 1$ kHz, $A_{m2} = 15$ V at $f_{m2} = 500$ Hz, $A_c = 20$ V, $f_c = 1.5$ MHz. Assume standard AM with $k_a = 1$: $s(t) = [A_c + m(t)]\cos\omega_ct$, and $R = 1\ \Omega$.

#### a) Equation of AM signal

$$
\begin{aligned}
s(t) &= [20 + 10\cos(2\pi10^3t) + 15\cos(\pi10^3t)]\cos(3\pi10^6t) \\
&= 20[1 + 0.5\cos(2\pi10^3t) + 0.75\cos(\pi10^3t)]\cos(3\pi10^6t)
\end{aligned}
$$

Expanded:

$$
\begin{aligned}
s(t) = {} & 20\cos 2\pi(1.5\text{M})t \\
& + 5\cos2\pi(1.501\text{M})t + 5\cos2\pi(1.499\text{M})t \\
& + 7.5\cos2\pi(1.5005\text{M})t + 7.5\cos2\pi(1.4995\text{M})t
\end{aligned}
$$

#### b) Modulation index

$$
\mu_1 = \frac{10}{20} = 0.5, \quad \mu_2 = \frac{15}{20} = 0.75
$$

$$
\mu_t = \sqrt{0.5^2 + 0.75^2} = \sqrt{0.8125} = 0.9014
$$

#### c) Total power

$$
\begin{aligned}
P_c &= \frac{A_c^2}{2} = \frac{400}{2} = 200\ \text{W} \\
P_t &= P_c\left(1 + \frac{\mu_t^2}{2}\right) = 200\left(1 + \frac{0.8125}{2}\right) = 281.25\ \text{W}
\end{aligned}
$$

(Efficiency $= 81.25/281.25 = 28.9\%$.)

#### d) Spectrum

```text
 A (V)
 20 |                  |
    |                  |
7.5 |         |        |        |
  5 |   |     |        |        |     |
    +---+-----+--------+--------+-----+---> f (MHz)
     1.499 1.4995    1.5     1.5005 1.501
```

Bandwidth $= 2 \times 1$ kHz $= 2$ kHz.

**Answer:** $\mu_t = 0.901$, $P_t = 281.25$ W.

## Check yourself

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

1. What are fm1, fm2 and fc?
   A. 2 kHz, 1 kHz and 3 MHz
   B. 1 kHz, 500 Hz and 1.5 MHz
   C. 1 MHz, 500 kHz and 1.5 kHz
   D. 1 kHz, 500 Hz and 3 MHz
   Answer: B. 2 pi f = 2x10^3 pi gives f = 1 kHz, 10^3 pi gives 500 Hz, and 3x10^6 pi gives 1.5 MHz. (Key point: Frequencies from the equation)

2. What are mu1 and mu2?
   A. 10 and 15
   B. 2 and 1.33
   C. 0.2 and 0.75
   D. 0.5 and 0.75
   Answer: D. mu = Am/Ac = 10/20 and 15/20. (Key point: Individual indices)

3. What is mu_t, and is the wave over-modulated?
   A. 0.901, so no over-modulation
   B. 1.25, so over-modulated
   C. 0.25, so under-modulated
   D. 0.5, so no over-modulation
   Answer: A. sqrt(0.25 + 0.5625) = 0.9014 which is below 1. (Key point: Net index)

4. What is the total power?
   A. 200 W
   B. 81.25 W
   C. 281.25 W
   D. 240 W
   Answer: C. Pc = 400/2 = 200 W and Pt = 200(1 + 0.8125/2) = 281.25 W. (Key point: Total power)

## Related in Communication Systems

- [Two-tone AM: net modulation index, power, spectrum](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-two-tone-50-20-12) — Communication Systems, chapter 2, asked 2×
- [Two-tone AM: index, power, spectrum (5cos, 10cos, 15cos)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-two-tone-5cos-10cos-15cos) — Communication Systems, chapter 2
- [Two-tone AM: spectrum, power, net index (50, 15, 20)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-two-tone-50-15-20) — Communication Systems, chapter 2, asked 2×
- [Two-tone AM: spectrum, net index, power, efficiency](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-two-tone-100-30-40) — Communication Systems, chapter 2
- [AM transmitter 10 kW/11.8 kW: modulation index and power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-10kw-11-8kw) — 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

## Related in other subjects

- [AM numerical: 10 sin1000πt on 50 sin 2π×10⁵t](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aam-numeric-10sin1000pi) — Data Communication, chapter 5
- [AM numerical: 20 sin2π500t on 60 sin2π10⁶t](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aam-numeric-20sin500) — Data Communication, chapter 5