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In frequency modulation
A) The amplitude of the modulated wave varies as the frequency of the carrier wave.
B) The frequency of the modulated wave varies as the amplitude of the modulating wave.
C) The amplitude of the modulated wave varies as the amplitude of the carrier wave.
D) The frequency of the modulated wave varies as the frequency of the modulating wave.

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Last updated date: 20th Sep 2024
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Answer
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Hint: The modulation is the method of changing the frequency, amplitude, and the phase of the carrier wave is changed according to the signal wave. We can say the carrier wave as a modulated wave as it is changed for the signal wave. And the signal wave can be called as the modulating wave because the parameter of the carrier wave is changed according to it.

Complete step by step answer:
The modulation is used for transmitting the signal wave to higher distances without any change. There are three types of modulation according to the parameters we change. They are:
Amplitude modulation
(i) Frequency modulation
(ii) Phase modulation
When the frequency of the carrier wave (modulated wave) is changed for the traveling of the signal wave. This is due to the low frequency of the signal or modulating wave. Hence the frequency of the modulated wave varies as the frequency of the modulating wave.

Hence the correct option is D.

Note:
The frequency modulation is very important for radio transmission etc. After we modulate the signal, the antennae radiate the modulated signal out and then the receiver receives it. That antenna needs to be in order to the wavelength of the modulated signal.
Length of the antennae = wavelength of the modulated signal.
$ \Rightarrow L = \lambda $
$ \Rightarrow L = \dfrac{c}{\nu }$
$ \Rightarrow L = \dfrac{{3 \times {{10}^8}}}{\nu }$
If the modulated signal is in the audible frequency \[20Hz - 20KHz\], then
\[ \Rightarrow L = \dfrac{{3 \times {{10}^8}}}{{20 \times {{10}^3}}}\]
\[ \Rightarrow L = 15000m\].
We cannot build the antennae of this much longer. So we choose the very high-frequency carrier wave which is more than the audible frequency. Suppose, if we choose the carrier wave of frequency \[3000KHz\]. Then the length would be
\[ \Rightarrow L = \dfrac{{3 \times {{10}^8}}}{{1000 \times {{10}^3}}}\]
\[L = 100m\]. It is possible to build the antennae of this length. Hence, we should use a very high-frequency carrier wave.