JEE Mains Physics · Communication Systems

Modulation Index, Sidebands and Bandwidth

In amplitude modulation the carrier's amplitude follows the message; the modulation index μ = Aₘ/A_c fixes the largest and smallest amplitudes, and the wave carries f_c and f_c ± fₘ, a bandwidth of 2fₘ.

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Why this matters

Twenty-nine PYQs, six of them asking for a number: nine from 2021, ten from 2022 and ten from 2023. Six are about the modulation index itself, eleven read the largest and smallest amplitudes of the modulated wave, and twelve ask which frequencies the wave carries, its bandwidth, or how many stations fit in a band. This is the largest page of the chapter, and the work is short arithmetic once the two formulas are fixed.

Concept 1 of 3: Amplitude modulation and the modulation index

In amplitude modulation the carrier keeps its frequency, but its amplitude rises and falls with the message. The modulation index says how deep that rise and fall is, as a fraction of the carrier's amplitude. If the message is bigger than the carrier, the amplitude would have to go below zero, and the shape of the message is lost.

Definition

Carrier $c(t) = A_c \sin \omega_c t$ ; message $m(t) = A_m \sin \omega_m t$ .
AM wave: $c_m(t) = (A_c + A_m \sin \omega_m t) \sin \omega_c t$ . The amplitude of the carrier changes with the message; its frequency does not.
Modulation index $\mu = \dfrac{A_m}{A_c}$ , often given as a percentage.
$\mu \le 1$ is needed to avoid distortion. With $\mu > 1$ (over-modulation) the envelope no longer follows the message.
From a graph of the message, $A_m$ is half its peak-to-peak swing: a square wave between +a and −a has $A_m = a$ .

Modulation index

\mu = \frac{A_m}{A_c}, \qquad \mu \le 1

Worked example

A message of amplitude 3 V modulates a carrier of amplitude 12 V. Find the modulation index. What message amplitude would give

\mu = 0.75

with the same carrier?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q103Moderate

Example 1 · Communication Systems · Modulation Index, Sidebands and Bandwidth

In an amplitude modulation, a modulating signal having amplitude of

XV

is superimposed with a carrier signal of amplitude Y V in first case. Then, in second case, the same modulating signal is superimposed with different carrier signal of amplitude

2YV

. The ratio of modulation index in the two cases respectively will be:

Message over carrier, not the other way

\mu = A_m/A_c

. Writing

A_c/A_m

gives a value above 1 for any normal AM wave, which should itself be a warning.

Read the amplitude, not the swing

A message drawn between +a and −a has amplitude a, not 2a. Taking the full swing doubles

\mu

It is the carrier whose amplitude changes

Statements say the amplitude of the 'modulating' or 'modulated' signal is varied. In AM it is the amplitude of the carrier that is varied, in step with the message.

Drill 5 more on amplitude modulation and the modulation index

Concept 2 of 3: Maximum and minimum amplitude of an AM wave

The amplitude of an AM wave swings between the carrier plus the message and the carrier minus the message. So the largest and smallest amplitudes hold both pieces of information: their average is the carrier, and half their difference is the message. Divide one by the other and you have the modulation index.

Definition

$A_{\max} = A_c + A_m$ and $A_{\min} = A_c - A_m$ .
So $A_c = \dfrac{A_{\max} + A_{\min}}{2}$ and $A_m = \dfrac{A_{\max} - A_{\min}}{2}$ .
$\mu = \dfrac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}}$ . Peak-to-peak values are both doubled, so they give the same $\mu$ .
In terms of $\mu$ : $A_{\max} = A_c(1 + \mu)$ , $A_{\min} = A_c(1 - \mu)$ , and $\dfrac{A_{\max}}{A_{\min}} = \dfrac{1 + \mu}{1 - \mu}$ .
Each side band has amplitude $\dfrac{\mu A_c}{2} = \dfrac{A_m}{2}$ .

Modulation index from the envelope

\mu = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}}, \qquad A_{\text{side band}} = \frac{\mu A_c}{2} = \frac{A_m}{2}

Worked example

An AM wave has a largest amplitude of 10 V and a smallest amplitude of 4 V. Find the carrier amplitude, the message amplitude, the modulation index and the amplitude of each side band.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q109Moderate

Example 2 · Communication Systems · Modulation Index, Sidebands and Bandwidth

For an amplitude modulated wave the minimum amplitude is

3\text{ }V

, while the modulation index is

60\%

. The maximum amplitude of the modulated wave is:

μ is not the ratio of the largest to the smallest

\mu = (A_{\max} - A_{\min})/(A_{\max} + A_{\min})

. Dividing

A_{\max}

A_{\min}

gives

(1 + \mu)/(1 - \mu)

, a different number that the options also carry.

A side band carries half the message amplitude

Each side band has amplitude

\mu A_c/2 = A_m/2

. Using

\mu A_c

gives twice the right value.

Check which ratio is asked

Questions ask for maximum to minimum or minimum to maximum, and as a ratio like 50 : x. Write both amplitudes first, then set them in the order the stem gives.

Drill 10 more on maximum and minimum amplitude of an am wave

Concept 3 of 3: Side-band frequencies and the bandwidth of an AM wave

Multiply out the AM wave and it splits into three plain sine waves: the carrier itself and two waves just above and just below it, at the carrier frequency plus and minus the message frequency. The message frequency on its own is not in the wave. The band from the lower side band to the upper one is 2fₘ wide, and that is the space each station needs.

Definition

Expanding with $\sin a \sin b = \tfrac{1}{2}[\cos(a - b) - \cos(a + b)]$ : $c_m(t) = A_c \sin \omega_c t + \dfrac{\mu A_c}{2}\cos(\omega_c - \omega_m)t - \dfrac{\mu A_c}{2}\cos(\omega_c + \omega_m)t$ .
Frequencies present: $f_c$ , $f_c - f_m$ (lower side band) and $f_c + f_m$ (upper side band).
Bandwidth $= (f_c + f_m) - (f_c - f_m) = 2f_m$ . Find f from $\omega$ first: $f = \omega/2\pi$ .
Stations that fit in a band without overlapping: $N = \dfrac{\text{band}}{2f_m}$ , rounded down.
A square-law modulator gives $f_m$ , $2f_m$ , $f_c$ , $2f_c$ and $f_c \pm f_m$ ; a band-pass filter keeps only $f_c$ and $f_c \pm f_m$ , so the output bandwidth is again $2f_m$ . At the receiver a rectifier and an envelope detector recover the message.
Frequency modulation, for contrast: the deviation ratio is $\Delta f/f_m$ , and by Carson's rule the bandwidth is $2(\Delta f + f_m)$ .

Side bands and bandwidth

f_{\text{LSB}} = f_c - f_m, \quad f_{\text{USB}} = f_c + f_m, \qquad \text{BW} = 2f_m

Worked example

The carrier

12\sin(1.6\pi \times 10^{6} t)

V is amplitude modulated by

4\sin(8\pi \times 10^{3} t)

V. Which frequencies does the AM wave contain? Find its bandwidth and the amplitude of each side band.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q6Moderate

Example 3 · Communication Systems · Modulation Index, Sidebands and Bandwidth

The amplitude of

15\sin(1000\pi t)

is modulated by

10\sin(4\pi t)

signal. The amplitude modulated signal contains frequency (ies) of A.

500\text{ }Hz

2\text{ }Hz

250\text{ }Hz

498\text{ }Hz

502\text{ }Hz

Choose the correct answer from the options given below:

Convert ω to f first

\sin(\omega t)

the number in front of t is

\omega = 2\pi f

. Reading it as f makes every frequency, and the bandwidth,

2\pi

times too large.

The message frequency is not in the AM wave

An AM wave contains

f_c

and

f_c \pm f_m

only. Listing

f_m

on its own among the frequencies present is wrong.

Bandwidth is twice the message frequency

The bandwidth is

2f_m

, set by the message alone. The carrier frequency does not enter it, and

f_m

alone is half the answer.

Each station needs 2fₘ, not fₘ

Dividing the band by

f_m

counts twice as many stations as can really fit. Divide by

2f_m

and round down.

Drill 11 more on side-band frequencies and the bandwidth of an am wave

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Amplitude modulation and the modulation index
Modulation index
$\mu = \frac{A_m}{A_c}, \qquad \mu \le 1$
Maximum and minimum amplitude of an AM wave
Modulation index from the envelope
$\mu = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}}, \qquad A_{\text{side band}} = \frac{\mu A_c}{2} = \frac{A_m}{2}$
Side-band frequencies and the bandwidth of an AM wave
Side bands and bandwidth
$f_{\text{LSB}} = f_c - f_m, \quad f_{\text{USB}} = f_c + f_m, \qquad \text{BW} = 2f_m$

Watch out for (10)

Message over carrier, not the other way
→ Amplitude modulation and the modulation index
Read the amplitude, not the swing
→ Amplitude modulation and the modulation index
It is the carrier whose amplitude changes
→ Amplitude modulation and the modulation index
μ is not the ratio of the largest to the smallest
→ Maximum and minimum amplitude of an AM wave
A side band carries half the message amplitude
→ Maximum and minimum amplitude of an AM wave
Check which ratio is asked
→ Maximum and minimum amplitude of an AM wave
Convert ω to f first
→ Side-band frequencies and the bandwidth of an AM wave
The message frequency is not in the AM wave
→ Side-band frequencies and the bandwidth of an AM wave
Bandwidth is twice the message frequency
→ Side-band frequencies and the bandwidth of an AM wave
Each station needs 2fₘ, not fₘ
→ Side-band frequencies and the bandwidth of an AM wave

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