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If a message signal of frequency $f_m$ is amplitude modulated with a carrier signal of frequency $f_c$ and radiated through an antenna, the wavelength of the corresponding signal in air is

ans: $\frac{c}{f_c}$

I know that the amplitude modulated radio wave is given as:

$$R(t) = \left[ 1+ \sin( \omega_s t) \right] C \sin(\omega_c t)$$

By trig identites , we see in the above function contains frequency of $\omega_c, \omega_c + \omega_s, \omega_c - \omega_s$ refer, then why is it that only the carrier frequency is considered for wavelength calculations?

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In the UK AM radio frequencies are around a megahertz and the signal bandwidth is around 5kHz so in effect the frequency is the carrier plus or minus $0.5\%$. To a very good approximation the frequency of the wave is just the carrier frequency.

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  • $\begingroup$ Oh, then the answer shouldn't be exact then $\endgroup$
    – Babu
    Jul 25, 2021 at 5:25
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    $\begingroup$ @Buraian, the velocity of the wave in air is not exactly c either (it's about 0.03% lower). $\endgroup$
    – The Photon
    Jul 25, 2021 at 14:48

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