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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$\begingroup$ Are you talking FM? Are you trying to compute the modulated signal directly from the frequency content using a fourier series? I don't think this will be easy. $\endgroup$– user6972Commented Jun 29, 2013 at 0:55
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1$\begingroup$ just curious, If you have lost control of your account see I accidentally created two accounts; how do I merge them?. $\endgroup$– dmckee --- ex-moderator kittenCommented Jul 3, 2013 at 15:01
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$\begingroup$ Would dsp.stackexchange.com be a better home for this question? $\endgroup$– Qmechanic ♦Commented Jul 3, 2013 at 15:27
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3 Answers
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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?
answered Jun 28, 2013 at 16:48
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$\begingroup$ 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? $\endgroup$ Commented Jun 29, 2013 at 1:18
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$\begingroup$ @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. $\endgroup$– user6972Commented Jun 29, 2013 at 1:27
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$\begingroup$ @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. $\endgroup$– user6972Commented Jul 1, 2013 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):
The term inside the cosine is not easy to isolate.
With some algebra it can be separated out using bessel functions:
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 $M=\frac{\Delta f}{f_m}$
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.
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.
answered Jun 28, 2013 at 16:54
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1$\begingroup$ You meant any rational ratio, right? $\endgroup$ Commented Jun 28, 2013 at 17:17
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$\begingroup$ @LubošMotl yes - by "integer ratio" I meant "ratio between two integers", but rational ratio is clearly a better way to put it. $\endgroup$– N. VirgoCommented Jun 29, 2013 at 3:37