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When a single-tone continuous modulating signal modulates a sinusoidal carrier, isn't the modulated wave periodic? If so, can't we apply fourier series and determine the harmonic frequency components instead of finding the frequency components using fourier transform?

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When a single-tone continuous modulating signal modulates a sinusoidal carrier ... can't we apply fourier series and determine the harmonic frequency components

$\cos(\omega_m t) \cdot \cos(\omega_ct) = \dfrac{\cos[(\omega_m + \omega_c) t] + \cos[(\omega_m - \omega_c) t]}{2}$

Isn't the frequency content evident by inspection?

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  • $\begingroup$ This isn't FM, this is mixing. $\endgroup$ – user6972 Jun 29 '13 at 1:00
  • $\begingroup$ Indeed it isn't FM. The phrase "modulate a carrier", without qualification, most likely means AM and that is what I assumed. Why do you assume FM? $\endgroup$ – Alfred Centauri Jun 29 '13 at 1:18
  • $\begingroup$ @AlfedCentauri mostly because there are 3 forms of modulation AM, FM and PM. FM and PM are similar in spectrum so odds are with me. And no one really does anything with AM anymore because it has poor immunity to noise. If you're doing something involving fourier transforms you're probably well beyond the simplicity of AM. $\endgroup$ – user6972 Jun 29 '13 at 1:27
  • $\begingroup$ @AlfedCentauri I shouldn't say no one uses AM anymore. In conducted communication systems with clean channels AM gets used often. However in any case with a poor channel (radiated for example) AM suffers heavily from noise interference so FM is preferred. $\endgroup$ – user6972 Jul 1 '13 at 1:37
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If the type of modulation you are talking about is FM, then the spectral content depends on its modulation index (M or $\beta$). The full equation for the output is (for a sinusoidal input only):

FM modulation with tones

The term inside the cosine is not easy to isolate.

With some algebra it can be separated out using bessel functions:

enter image description here enter image description here enter image description here

So you can see that this signal has frequency components at $f_c + n*f_m$. Where n is $-\infty$ to $+\infty$. Shocking isn't it? The Bessel Function $J_n\{\beta\}$ tends to approach zero at large values of n, but it is counter-intuitive to see all this frequency content.

So a simple sine wave modulation at frequency Fm can produce a variety of spectra depending on its modulation index (M or $\beta$) which is a ratio of the change in frequency deviation ($\Delta$f) to the maximum frequency of the modulated signal ($f_{m(max)}$). See the plots for different values of M for the same modulating tone at frequency $f_m$ where $M=\frac{\Delta f}{f_m}$

FM spectra

In reality, Carson's Rule is applied to limit the spectral content to about 98% of all the energy. This bandwidth is double the frequency deviation plus the maximum frequency of the input modulation content.

equation

So to answer your question the spectral content is not necessarily directly related to the modulated signal spectra with FM. It's periodic, sure, but there is a lot of content. If you knew the exact modulation parameters you could try to work backwards and estimate the modulated input based on relative amplitudes and spectral frequencies.

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Yes, it's periodic, as long as the frequencies of the modulating signal and the carrier are related by a rational ratio. For example, if the period of the carrier is 3 seconds and that of the modulator is 2 seconds then the signal will be periodic with a period of 6 seconds. In such a case you can use Fourier series to find the spectrum if you want. But if the carrier's period is $\pi$ seconds and the modulator's is 2 seconds then the signal will never quite repeat, and so you can't use Fourier series. I suspect that using Fourier series will generally be messier and more difficult than using a continuous Fourier transform anyway.

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    $\begingroup$ You meant any rational ratio, right? $\endgroup$ – Luboš Motl Jun 28 '13 at 17:17
  • $\begingroup$ @LubošMotl yes - by "integer ratio" I meant "ratio between two integers", but rational ratio is clearly a better way to put it. $\endgroup$ – Nathaniel Jun 29 '13 at 3:37

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