Question

A signal of frequency $20kHz$ and the peak voltage of $5V$ is used to modulate a carrier wave of frequency $1.2MHz$ and the peak voltage $25V$. Choose the correct statement.
(A) $Modulation{\text{ }}index = 5$, side frequency bands are at $1400kHz$ and $1000kHz$.
(B) $Modulation{\text{ }}index = 0.8$, side frequency bands are at $1180kHz$ and $1220kHz$.
(C) $Modulation{\text{ }}index = 5$, side frequency bands are at $21.2kHz$ and $18.8kHz$.
(D) $Modulation{\text{ }}index = 0.2$, side frequency bands are at $1220kHz$ and $1180kHz$.

Answer 1

We are given with the signal frequency and its peak voltage and also with the frequency of the carrier wave and its peak voltage. We are then asked to find the modulation index and the side frequency band of the resultant. Thus, we will use the formula of modulation index. Also, we will use the side frequency band formula.

Formulae used:
$Modulation{\text{ }}Index = \frac{{Signal{\text{ Peak }}Voltage}}{{{\text{Carrier Peak }}Voltage}}$
And,
Side frequency band:
$Upper{\text{ }}band{\text{ }}limit = Carrier{\text{ }}Frequency{\text{ }} + {\text{ }}Signal{\text{ }}Frequency$
$Lower{\text{ }}band{\text{ }}limit = Carrier{\text{ }}Frequency{\text{ - }}Signal{\text{ }}Frequency$

Complete step by step solution:
Here, $Signal{\text{ }}Peak{\text{ }}Voltage = 5V$
$Carrier{\text{ }}Peak{\text{ }}Voltage = 25V$
Thus, $Modulation{\text{ }}Index = \frac{5}{{25}} = \frac{1}{5} = 0.2$
Now, $Signal{\text{ }}Frequency = 20kHz$
$Carrier{\text{ }}Frequency = 1.2MHz = 1.2 \times {10^3}kHz = 1200kHz$
Thus, The side frequency band:
$Upper{\text{ }}Band{\text{ }}Limit = \left( {1200 + 20} \right)kHz = 1220kHz$
And, $Lower{\text{ }}Band{\text{ }}Limit = \left( {1200 - 20} \right)kHz = 1180kHz$
Hence, the correct option is (D).

Additional information:
Modulation can be simply understood as the manipulation of the basic parameter of a wave (frequency or amplitude) in order to enhance the performance of the wave. In other words, a manipulated or a modulated wave will have a better performance to a great extent over the original wave.
Here, we calculated the modulation index which physically signifies the quality of the transmitted signal. For instance, if the value of the index is small, then the amount of variation in the carrier amplitudes is small and so on.

Note:
After all the required calculations, we got that the carrier amplitude variation will be up to an amount of $0.2$. Also, the variation of the wave will range between $1220kHz$ and $1180kHz$. This will imply that the amplitude of the carrier wave will be varied by an amount of $0.2$ and the frequency of the same will fluctuate between the values having a lower range of $1180kHz$ and a higher range of $1220kHz$. Thus, this is a benefit of mathematical calculations which assists us to have an idea of the implications of these variations.