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Amplitude Modulation

Amplitude Modulation

Edited By Vishal kumar | Updated on Sep 09, 2024 10:20 PM IST

Amplitude Modulation (AM) is a technique used in electronic communication, most commonly for transmitting information via a radio carrier wave. In AM, the amplitude of the carrier wave is varied in proportion to that of the message signal, such as an audio signal. This allows the signal to be encoded and transmitted over long distances.

A common real-life example of amplitude modulation is AM radio broadcasting. In this system, radio stations encode sound signals onto a high-frequency carrier wave, which is then transmitted through the air. When you tune your AM radio to a specific frequency, the radio receiver demodulates the signal, extracting the audio content so you can hear the broadcast.

Amplitude Modulation

The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency (AF) signal is known as amplitude modulation (AM). Carrier wave remains unchanged in AM frequency. The amplitude of a modulated wave is varied in accordance with the amplitude of the modulating wave.

Modulation Index

The ratio of change of amplitude of the carrier wave to the amplitude of the original carrier wave is called the modulation factor or degree of modulation or modulation index (m).
$\begin{aligned} & \mu_a=\frac{\text { Change in amplitude of carrier wave }}{\text { Amplitude of original carrier wave }}=\frac{E_m}{E_c} \\ & \text { where } \quad \mu_a=\frac{E_m}{E_c}=\frac{E_{\max }-E_{\min }}{E_{\max }+E_{\min }}\end{aligned}$
If a carrier wave is modulated by several sine waves the total modulated index m is given by

$m_t=\sqrt{m_1^2+m_2^2+m_3^2+\ldots \ldots}$

Voluage Equation for AM Wave

Suppose voltage equations for carrier wave and modulating wave are

$e_c=E_c \cos \omega_c t$ and
$e_m=E_m \sin \omega_m t=m E_c \sin \omega_m t$
where,

$e_c=$ Instantaneous voltage of carrier wave,
$\mathrm{E}_{\mathrm{c}}=$ Amplitude of the carrier wave,
$\omega_c=2 \pi f_c=$ Angular velocity at the carrier frequency $f_c$
$e_m=$ The instantaneous voltage of modulating.
$E_m=$ The amplitude of the modulating wave,
$\omega_m=2 \pi f_m=$ Angular velocity of modulating frequency 'f'
The voltage equation for AM wave is
$e=E \sin \omega_c t=\left(E_c+e_m\right) \sin \omega_c t=\left(E_c+e_m \sin \omega_m t\right) \sin \omega_c t$
$=E_c \sin \omega_c t+\frac{m_a E_c}{2} \cos \left(\omega_c-\omega_m\right) t-\frac{m_a E_c}{2} \cos \left(\omega_c+\omega_m\right) t$
The above AM wave indicated that the AM wave is equivalent to the summation of three sinusoidal waves, one having amplitude 'E' and the other two having amplitude $\frac{m_o E}{2}$.

Sideband Frequencies

The AM wave contains three frequencies, $f_c,\left(f_c+f_m\right)$ and $\left(f_c-f_m\right) . f_c$ is called carrier frequencies, $\left(f_c+f_m\right)$ and $\left(f_c-f_m\right)$ are called sideband frequencies.
$\left(f_c+f_m\right):$ Upper sideband (USB) frequency
$\left(f_c-f_m\right):$ Lower sideband (LSB) frequency

In general sideband frequencies are close to the carrier frequency.

Bandwidth

The two sidebands lie on either side of the carrier frequency at equal frequency interval fm.

So, bandwidth $=\left(f_c+f_m\right)-\left(f_c-f_m\right)=2 f_m$

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Solved Example Based On Amplitude Modulation

Example 1: Choose the correct statement :

1) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

2)In amplitude modulation, the frequency of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal

3)In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

4)In frequency modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the frequency of the audio signal

Solution:

The amplitude of a modulated wave is varied in accordance with the amplitude of the modulating wave

In amplitude modulation, the amplitude of the high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

Hence, the correct answer is the option (1).

Example 2: A signal $A$ cos $\omega t$ is transmitted using $v_0 \sin \omega_0 t$ as carrier wave. The correct amplitude-modulated (AM) signal is :

1) $v_0 \sin \omega_0 t+\frac{A}{2} \sin \left(\omega_0-\omega\right) t+\frac{A}{2} \sin \left(\omega_0+\omega\right) t$
2) $v_0 \sin \left[\omega_0(1+0.01 A \sin \omega t) t\right]$
3) $\left(v_0+A\right) \cos \omega t \sin \omega_0 t$
4) NoneSolution:

Amplitude modulated signal

$\begin{aligned} & =E_c \sin \omega_c t+\frac{m E_c}{2} \cos \left(\omega_c-\omega_m\right) t \quad-\frac{m E_c}{2} \cos \left(\omega_c+\omega_m\right) t \\ & E_c=\text { amplitude of carrier wave }=V_0 \\ & m=\frac{E_m}{E_c}=\frac{A}{V_0}\end{aligned}$

therefore

The AM signal is given by

$V_0 \sin \omega_0 t+\frac{A}{2} \sin \left(\omega_0-\omega\right) t+\frac{A}{2} \sin \left(\omega_0+\omega\right) t$

Hence, the correct answer is the option (1).

Example 3: A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index?

1) 0.6

2) 0.5

3) 0.4

4) 0.3

Solution:

$
\begin{aligned}
& m_a=\frac{E_m}{E_c} \\
& A_c=100 \mathrm{~V} \\
& A_c+A_m=160 \mathrm{~V}=A_{\text {max }} \\
& A_c-A_m=40=A_{\text {min }}
\end{aligned}
$

From (1) \& (2)
$
A_c=100 \mathrm{~V} \& A_m=60 \mathrm{~m}
$

So,
$
\mu=\frac{A_m}{A_c}=0.6
$

Hence, the correct answer is the option (1).

Example 4: An amplitude-modulated signal is plotted below :

Which one of the following best describes the above signal?

1) $\left(9+\sin \left(4 \pi \times 10^4 t\right)\right) \sin \left(5 \pi \times 10^5 t\right) V$
2) $\left(1+9 \sin \left(2 \pi \times 10^4 t\right)\right) \sin \left(2.5 \pi \times 10^5 t\right) \mathrm{V}$
3) $\left(9+\sin \left(2.5 \pi \times 10^5 t\right)\right) \sin \left(2 \pi \times 10^4 t\right) \mathrm{V}$
4) $\left(9+\sin \left(2 \pi \times 10^4 t\right)\right) \sin \left(2.5 \pi \times 10^5 t\right) V$

Solution:

Voltage equation for AM wave -

$
\begin{aligned}
& e_c=E_c \cos \omega_c t \\
& e_m=E_m \sin \omega_m t
\end{aligned}
$

Resultant Modulated wave
$
e=\left(E_c+e_m \sin \omega_m t\right) \cdot \sin \omega_c t
$

From the graph
$
\begin{aligned}
& E_{\min }=8 v \text { and } E_{\max }=10 v \\
& E_C=\frac{E_{M A X}+E_{M I N}}{2}
\end{aligned}
$
and
$
\begin{aligned}
& T_s=\text { Time period of signal wave }=100 \mu \mathrm{s} \\
& T_c=\text { Time period of carrier wave }=8 \mu \mathrm{s}
\end{aligned}
$

So signal equation is
$
\begin{aligned}
& =\left[9 \pm 1 \sin \left(\frac{2 \pi t}{T_s}\right) \sin \left(\frac{2 \pi t}{T_c}\right)\right] \\
& =\left[9 \pm \sin \left(2 \pi \times 10^4 t\right) \sin \left(2.5 \pi \times 10^5 t\right)\right]
\end{aligned}
$

Hence, the correct answer is the option (4).

Example 5: An audio signal $v_m=20 \sin 2 \pi(1500 t)$ amplitude modulates a carrier $v_c=80 \sin 2 \pi(100,000 t)$. The value of per cent modulation is $\qquad$ ___.

1) 25

2) 40

3) 15

4) 50

Solution:

The percentage change in the modulated wave and the carrier wave is given by :

$\begin{aligned} & \% \text { modulation }=\frac{\mathrm{Am}}{\mathrm{Ac}} \times 100 \\ & \% \text { modulation }=\frac{20}{80} \times 100 \\ & \% \text { modulation }=25 \%\end{aligned}$

Hence, the correct answer is the option (1).

Summary

Amplitude Modulation (AM) is a communication technique where the amplitude of a high-frequency carrier wave is varied in proportion to the amplitude of an audio signal. This method is widely used in AM radio broadcasting, where sound signals are encoded onto a carrier wave for transmission and later demodulated by receivers. Key concepts in AM include modulation index, sideband frequencies, and bandwidth, which determine the efficiency and clarity of the transmitted signal.

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