(a) If A and B represent the maximum and minimum amplitudes of an amplitude modulated wave, write the expressi

Question

Q. (a) If A and B represent the maximum and minimum amplitudes of an amplitude modulated wave, write the expression for the modulation index in terms of A & B.

(b) A message signal of frequency 20 kHz and peak voltage 10 V is used to modulate a carrier of frequency 2 MHz and peak voltage of 15 V. Calculate the modulation index. Why the modulation index is generally kept less than one?

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Concept understanding — Modulation Index

Modulation Index: The Depth of the Message

Imagine you are holding a microphone, speaking into it. Your voice is a weak electrical signal — the modulating signal. To send it across a city, you need to "ride" it on a much stronger, high-frequency wave — the carrier wave. Amplitude modulation (AM) is the process of making the carrier's strength (its amplitude) change in step with your voice.

The modulation index is simply a number that tells you how much the carrier's amplitude is being varied by the message. It answers the question: "How deep is the imprint of the message on the carrier?"

The Intuition: A Handshake

Think of the carrier as a tall, steady pillar of height $A_c$ . Your voice signal is a smaller, wavy force of maximum height $A_m$ that pushes and pulls on the top of that pillar. The modulation index $m$ is the ratio:

m = \frac{A_m}{A_c}

If $A_m$ is very small compared to $A_c$ , the pillar barely wobbles — the modulation is shallow ( $m$ is small). If $A_m$ is almost as large as $A_c$ , the pillar wobbles a lot — the modulation is deep ( $m$ is close to 1). If $A_m$ exceeds $A_c$ , the pillar would be pushed so hard that it gets "over-modulated" and the message gets distorted.

Tip

The modulation index is often expressed as a percentage. $m = 0.5$ means 50% modulation. $m = 1$ means 100% modulation — the maximum possible without distortion.

The Precise Statement

For a sinusoidal modulating signal $v_m(t) = A_m \cos(2\pi f_m t)$ and a carrier $v_c(t) = A_c \cos(2\pi f_c t)$ , the amplitude-modulated wave is:

v_{AM}(t) = A_c \left[1 + m \cos(2\pi f_m t)\right] \cos(2\pi f_c t)

where the modulation index $m$ is defined as:

m = \frac{A_m}{A_c}

This is a dimensionless number, always between 0 and 1 for distortion-free AM.

What Happens at Different Values?

$m$ value	What it means	Visual effect
$m = 0$	No modulation	Carrier is a pure sine wave — no message
$0 < m < 1$	Undermodulation	The envelope (the outline of the wave) faithfully follows the message
$m = 1$	100% modulation	The carrier amplitude just touches zero at the troughs — maximum clean swing
$m > 1$	Overmodulation	The envelope crosses zero and inverts — the message is distorted and cannot be recovered cleanly

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The modulation index $m = \frac{A - B}{A + B}$ in terms of envelope extrema; for the given signal, $m = 0.67$ . Keeping $m < 1$ prevents distortion from over-modulation.

Understanding the modulation index

Amplitude modulation superimposes a low-frequency message signal onto a high-frequency carrier. The envelope of the resulting wave varies between a maximum amplitude $A$ and a minimum amplitude $B$ . The modulation index quantifies how deeply the message modulates the carrier—it's the ratio of the message amplitude to the carrier amplitude.

When a carrier $A_c \cos(\omega_c t)$ is modulated by a message $A_m \cos(\omega_m t)$ , the AM wave becomes:

s(t) = A_c[1 + m \cos(\omega_m t)] \cos(\omega_c t)

where $m = \frac{A_m}{A_c}$ is the modulation index. The envelope $A_c[1 + m \cos(\omega_m t)]$ oscillates between:

Maximum: $A = A_c(1 + m)$ when $\cos(\omega_m t) = 1$
Minimum: $B = A_c(1 - m)$ when $\cos(\omega_m t) = -1$

(a) Deriving the modulation index from envelope extrema

Write the envelope bounds:

A = A_c(1 + m), \quad B = A_c(1 - m)

Add the two equations:

A + B = A_c(1 + m) + A_c(1 - m) = 2A_c

So $A_c = \frac{A + B}{2}$ .

Subtract the second from the first:

A - B = A_c(1 + m) - A_c(1 - m) = 2A_c m

So $m = \frac{A - B}{2A_c}$ .

Substitute $A_c$ from step 2:

m = \frac{A - B}{2 \cdot \frac{A + B}{2}} = \frac{A - B}{A + B}

m = \frac{A - B}{A + B}

This expression lets you determine the modulation index directly from the oscilloscope trace of an AM wave.

(b) Calculating the modulation index and the constraint $m < 1$

Given:

Message signal: peak voltage $A_m = 10 \text{ V}$ , frequency $f_m = 20 \text{ kHz}$
Carrier signal: peak voltage $A_c = 15 \text{ V}$ , frequency $f_c = 2 \text{ MHz}$

Apply the definition:

m = \frac{A_m}{A_c} = \frac{10}{15} = \frac{2}{3} \approx 0.67

Why keep $m < 1$ ? …

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Q.(a) If A and B represent the maximum and minimum amplitudes of an amplitude modulated wave, write the expression for the modulation index in terms of A & B.

Understanding the modulation index

(a) Deriving the modulation index from envelope extrema

(b) Calculating the modulation index and the constraint m<1m < 1m<1

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(b) Calculating the modulation index and the constraint $m < 1$

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