Production Of Amplitude Modulated Wave - Practice Questions & MCQ

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 $x(t)=A_m\sin {\omega }_{m}t+A_c\sin {\omega }_{c}t$ is passed through a square-law device which is a non-linear device which produces an output.
$y(t)=Bx(t)+C{x}^2(t)$ 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 $V_{rms}$ is root mean square value
and $\mathrm{R}=$ Resistance
then Power dissipated in any circuit.

$P=\frac{V_{rms}^2}{R}$

So Carrier Power will be given as

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

$E_c=$ The amplitude of the carrier wave
$R=$ Resistance
Similarly, Total Power of sidebands will be given as

$P_{sb}=\frac{{\left(\frac{m_a{E}_c}{2\sqrt{2}}\right)}^2}{R}+\frac{{\left(\frac{m_a{E}_c}{2\sqrt{2}}\right)}^2}{R}=\frac{m_a^2{E}_c^2}{4R}$

Where 

$m_a=$ modulation index
$E_c=$ the amplitude of carrier waves
$R=$ resistance
And this gives Total power of AM wave as

$\begin{array}{rl}& P_{\text{total }}=P_c+P_{sb}\\ & =\frac{E_c^2}{2R}(1+\frac{m_a^2}{2})\end{array}$

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 $P_t=1.5P=3P_{sb}$