Question

In amplitude modulation, the band with is
$\left( {\text{A}} \right)$ Equal to the signal frequency
$\left( {\text{B}} \right)$ Twice the signal frequency
$\left( {\text{C}} \right)$ Thrice the signal frequency
$\left( {\text{D}} \right)$ Four times the signal frequency

Answer 1

Hint:In modulation, Bandwidth is defined as the difference between the Upper band frequency limit and the lower band frequency limit.

Formula used:
\[\sin A\,\sin B = \dfrac{{\cos (B - A) - \cos (B + A)}}{2}\]

Complete step by step answer:
In modulation, there are two kinds of waves: Modulating or the Input signal and the carrier wave. The carrier wave has a very high frequency as compared to the modulating signal.
We know that a signal can be represented as
\[y(t) = A\sin (\omega t + \phi )\]
Where A is the amplitude of the signal, $\omega $ is the angular frequency of the signal, and is the phase of the signal at the source, and \[y\left( t \right)\] is the signal displacement.
In amplitude modulation, the carrier and modulating signals can be written similarly as: \[{y_m}(t) = {A_m}\sin ({\omega _m}t).....\left( 1 \right)\]
\[{y_c}(t) = {A_c}\sin ({\omega _c}t).....\left( 2 \right)\]
Usually, the carrier signal has a very high frequency than the modulating signal.
Now, as in amplitude modulation, the two signals are superimposed and at the same time the amplitude of the carrier wave is varied by accordance with the modulating signal for transmission, we can write the superimposed signal as:
\[y(t) = ({A_m}\sin ({\omega _m}t) + {A_c})\sin ({\omega _c}t)\]
Now, opening the bracket and multiplying the terms we get,
\[y(t) = {A_m}\sin ({\omega _m}t)\sin ({\omega _c}t) + {A_c}\sin ({\omega _c}t)\]
Using the trigonometric identity, \[\sin A\,\sin B = \dfrac{{\cos (B - A) - \cos (B + A)}}{2}\], we have
\[y(t) = \dfrac{{{A_m}}}{2}\cos ({\omega _m} - {\omega _c})t - \dfrac{{{A_m}}}{2}\cos ({\omega _m} + {\omega _c})t + {A_c}\sin ({\omega _c}t)\]
Here, there are three frequencies, ${\omega _m} - {\omega _c}\;,{\omega _m} + {\omega _c}$ and ${\omega _c}$.
Since the bandwidth is defined as the difference between upper and lower limits, we have
Band width $ = ({\omega _m} + {\omega _c}) - \left( {{\omega _m} - {\omega _c}} \right) = 2{\omega _c} = $ Twice of the signal frequency

Hence, the option B is correct.

Note: Do not confuse amplitude modulation with frequency modulation and also keep in mind that the superimposed signal is not just the vector sum of the two signals.