Production Of Amplitude Modulated Wave

Production of Amplitude Modulated Wave

Any signal that is generated from a source and needs to be sent over large distances from the source to the receiver, needs to be modified. This can be done by superimposition with a carrier signal to ensure the signal can be transmitted in a suitable bandwidth. Amplitude modulation can be produced by a variety of methods. A conceptually simple method is shown in the block diagram below

Here the modulating signal $A_m \sin \omega_m t$ is added to the carrier signal $A_c \sin \omega_c t$ to produce the signal x (t). This signal $x(t)=A_m \sin \omega_m t+A_c \sin \ omega_ t$ is passed through a square-law device which is a non-linear device which produces an output.

$y(t)=B x(t)+C x^2(t)$ where B and C are constants.

This square waveform passes through a bandpass filter. The bandpass filter is a device which filters out the noise that is the unwanted frequencies. For example, if the frequencies of the system differ from those including ‘ω’ and ω±ω’, then the bandpass filter automatically rejects them.

Yet, the process is incomplete. The modulated signal generated is quite weak and cannot sustain attenuation over a large distance. This demands strengthening the signal. We get this by amplification of the signal using an amplifier diode. The quality of the signal does not change only its strength increases by the amplifier which forms the second last part of the circuit.

Finally, the amplified and modulated signal goes to a transmitter or antenna for radiation at a particular bandwidth frequency. This antenna transmits the signal over large distances using radiation. But this alone does not ensure the signal will reach its destination.

Power in AM Waves

If $V_{r m s}$ is the root mean square value

and R = Resistance

then Power dissipated in any circuit.
$
P=\frac{V_{r m s}^2}{R}
$
So Carrier Power will be given as

$
P_c=\frac{\left(\frac{E_{\mathrm{c}}}{\sqrt{2}}\right)^2}{R}=\frac{E_c^2}{2 R}
$

$E_c=$ The amplitude of the carrier wave

$
\mathrm{R}=\text { Resistance }
$
Similarly, the Total Power of sidebands will be given as

$
P_{s b}=\frac{\left(\frac{m_a E_c}{2 \sqrt{2}}\right)^2}{R}+\frac{\left(\frac{m_a E_c}{2 \sqrt{2}}\right)^2}{R}=\frac{m_a^2 E_c^2}{4 R}
$

Where

$m_a=$ modulation index
$E_c=$ the amplitude of carrier waves
$\mathrm{R}=$ resistance
This gives the Total power of the AM wave as

$
\begin{aligned}
& P_{\text {total }}=P_c+P_{s b} \\
& =\frac{E_c^2}{2 R}\left(1+\frac{m_a^2}{2}\right)
\end{aligned}
$

where
$m_a=$ modulation index
$E_c=$ the amplitude of carrier waves
$R=$ Resistance
Note-maximum power in the AM wave without distortion Occurs when $m_a=1$
I.e $P_t=1.5 P=3 P_{s b}$

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Solved Examples Based on the Production of Amplitude Modulated Wave

Example 1: A certain transmitter radiates 9 W power while transmitting an unmodulated carrier wave. If the amplitude of carrier wave voltage varies sinusoidally, the peak voltage of carrier wave would be (take antenna resistance = $50 \Omega$)

1) 15 V

2) 30 V

3) $30 / \sqrt{2} V$

4) 900 V

Solution:

Carrier Power
$
P_c=\frac{\left(\frac{E_{\mathrm{c}}}{\sqrt{2}}\right)^2}{R}=\frac{E_c^2}{2 R}
$

wherein
$E_c=$ The amplitude of carrier wave
$R=$ Resistance
We know that the power of carrier wave is given by $P_c=\frac{E_c^2}{2 R} \quad$ where $E_c=$ peak amplitude of carrier wave
Now $P=9 W$

$
\begin{aligned}
& R=50 \Omega \\
& E_c=\sqrt{2 R P_c} \text { volt } \\
& E_c=\sqrt{2 \times 9 \times 50} \mathrm{volt} \\
& E_c=30 \mathrm{volt}
\end{aligned}
$

Hence, the correct answer is the option (2).

Example 2: A carrier wave is amplitude-modulated to a depth of 50 %. The ratio between the total power of side bonds to the power of unmodulated carrier wave would be

1)1 :1

2)1: 2

3)1: 4

4)1: 8

Solution

Total Power of side bands
$
P_{s b}=\frac{\left(\frac{m_a E_c}{2 \sqrt{2}}\right)^2}{R}+\frac{\left(\frac{m_a E_c}{2 \sqrt{2}}\right)^2}{R}=\frac{m_a^2 E_c^2}{4 R}
$

wherein
$m_a=$ modulation index
$E_c=$ the amplitude of carrier waves
$R=$ resistance
Depth of modulation $=50 \%$
which means modulation index $\left(m_a\right)=1 / 2$
Now we know
Total power of side bonds $=\frac{m_a^2 E_c^2}{4 R}$
Total power of unmodulated carries $=\frac{E_c^2}{2 R}$

Ratio b/w power of sidebands and carriers $=\frac{m_a^2 E_c^2}{4 R} / \frac{E_c^2}{2 R}$

$
\begin{aligned}
& \qquad=\frac{m_a^2}{2} \\
& \text { ratio }=\frac{(1 / 2)^2}{2}\left(m_a=1 / 2\right) \\
& \text { Ratio }=1 / 8
\end{aligned}
$
Hence, the correct answer is the option (4).

Example 3: A signal of 0.1 kW is transmitted in a cable. The Attenuation of cable is -5 dB per km and cable length is 20 km. The power received at the receiver is 10 ^-x W. The value of x is ______.

[Gain in $d B=10 \log _{10}\left(\frac{P_o}{P_i}\right)_{}$]

1) 2

2) 4

3) 6

4) 8

Solution:

$\begin{aligned} & \text { Power of signal transmitted: } \mathrm{P}_{\mathrm{i}}=0.1 \mathrm{KW}=100 \mathrm{~W} \\ & \text { Rate of attenuation }=-5 \frac{d B}{K m} \\ & \text { The total length of path }=20 \mathrm{Km} \\ & \text { Total loss suffered }=-5 \times 20=-100 \mathrm{~dB} \\ & \text { Gain in } \mathrm{dB}=10 \log _{10}\left(\mathrm{P}_0 / \mathrm{P}_{\mathrm{i}}\right) \\ & \Rightarrow-100=10 \log _{10}\left(\mathrm{P}_0 / \mathrm{P}_{\mathrm{i}}\right) \\ & \Rightarrow \log _{10}\left(\mathrm{P}_{\mathrm{i}} / \mathrm{P}_0\right)=10 \\ & \Rightarrow \log _{10}\left(\mathrm{P}_{\mathrm{i}} / \mathrm{P}_0\right)=\log _{10} 10^{10} \\ & \Rightarrow \frac{100}{\mathrm{P}_0}=10^{10} \\ & \Rightarrow \mathrm{P}_0=\frac{1}{10^8}=10^{-8} \\ & \Rightarrow \mathrm{x}=8\end{aligned}$

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

Example 4: A signal of 100 THz frequency can be transmitted with maximum efficiency by :

1) Coaxial cable

2) Optical fibre

3) Twisted pair of copper wires
4) Water:

Solution

$\rightarrow$ A signal of THz frequency can be transmitted with maximum efficiency by optical fibre.

$\rightarrow$ Since optical communication using fibres is performed in the frequency range of 1 THz to 100 THz

Hence, the correct answer is the option (2).

Example 5: The maximum and minimum voltage of an amplitude-modulated signal are 60 V and 20 V respectively. The percentage modulation index will be :

1) $0.5 \%$
2) $50 \%$
3) $2 \%$
4) $30 \%$

Solution:
$
\begin{aligned}
& \mathrm{V}_{\max }=60 \mathrm{~V} \\
& \mathrm{~V}_{\min }=20 \mathrm{~V}
\end{aligned}
$
The percentage modulation index is

$
\begin{aligned}
& \mu \%=\frac{V_{\max }-V_{\min }}{V_{\max }+V_{\min }} \times 100 \\
& =\frac{40}{80} \times 100 \\
& =50 \%
\end{aligned}
$

Hence, the correct answer is the option (2).

Summary

The process in which an amplitude-modulated (AM) wave is created is the carrier wave, high in frequency, combined with a kind of information signal, which is usually audio. This is done using a modulator circuit such that the amplitude of the carrier wave varies directly in proportion to the instantaneous amplitude of the information signal. The AM wave so formed contains both the carrier frequency and the variations which augur the nature of the original audio wave that was modulated. It's further transmitted thousands of miles using radio waves and intercepted by an AM radio, which would demodulate that signal back into its original audio content. This simplicity and high effectiveness are both what carry forward AM into a fundamental technique in communication.