# A signal that can be transformed by Fourier transform and its frequency

A signal A, which is in time domain, can be transformed by Fourier transform into its frequency contents.

Then, is the signal A's frequency the highest frequency part of its Fourier transform? (i.e. is the signal itself oscillating at the frequency of its highest Fourier transform frequency part?)

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You asking: what is the frequency of the signal ? Or what is the frequency of spectrum of signal ? Or what is frequency of signal with particular spectrum ? Broadly, you are trying to apply single charateristic to object with multiples of same charateristics. Try use the wording like "average" or "maximum", then you will get better question –  user924 May 27 '12 at 15:34
We need further information from you on this question. Are there a specific subset of signals you are thinking about or any possible signal? Please try to answer @RocketSurgeon also. –  Kortuk May 29 '12 at 14:35

Like The Photon says, it's the lowest non-zero frequency, it's called the fundamental, and the other harmonics are integer multiples of it. That means the fundamental is the frequency with the longest period in the signal.

This AM signal is the product of two frequencies, a low baseband signal frequency and a higher modulating frequency, which in this case is exactly 10 times the baseband frequency. The signal's period is the lower frequency's period, and its inverse is the fundamental's frequency.
The function is (3 + sin($\omega_0$)) $\times$ sin($\omega_m$). Since

$sin(x) \times sin(y) = \dfrac{cos(x – y) – cos (x + y)}{2}$

we have

$V_t = 3 \cdot sin(\omega_m t) + \dfrac{cos(\omega_m t - \omega_0 t)}{2} - \dfrac{cos(\omega_m t + \omega_0 t)}{2}$

and, with $\omega_m$ = 10 $\times$ $\omega_0$

$V_t = 3 \cdot sin(10 \cdot \omega_0 t) + \dfrac{cos(9 \cdot \omega_0 t)}{2} - \dfrac{cos(11 \cdot \omega_0 t)}{2}$

which can be written in the standard Fourier series form:

$V_t = -\dfrac{1}{2}sin(9 \cdot \omega_0 t - \dfrac{\pi}{2}) + 3 \mbox{ } sin(10 \cdot \omega_0 t) + \dfrac{1}{2}sin(11 \cdot \omega_0 t - \dfrac{\pi}{2})$

For a repeating signal the frequency of the fundamental is greater than zero, and the harmonics show in the spectrum as equal-spaced lines.
For a non-repeating signal the limit of the signal's period goes to $\infty$ so that the frequency of the fundamental goes to $\displaystyle \lim_{f \to 0}$, and the series of harmonics forms a continuous spectrum.

"Sometimes it's difficult to see the fundamental sine in it. Take for instance the sum of a 3Hz sine and a 4Hz sine. The resulting waveform will repeat once every second, that's 1Hz. The 1Hz is the fundamental, even if its amplitude is zero. The series can be written as

$V_t = 0 \cdot sin(\omega_0 t) + 0 \cdot sin(2 \omega_0 t) + sin(3 \omega_0 t) + sin(4 \omega_0 t)$

All the following terms also have zero amplitude.

Why is the fundamental frequency 1Hz,and not 0.5Hz, for instance? 3Hz and 4Hz are also multiples of that. The fundamental is the greatest common divider of the composing harmonics, and the GCD of 3 and 4 is 1. If you would choose a lower frequency its period will show a repetition of the signal, twice in the case of 0.5Hz.

A note on GCD
It has been suggested that GCD only applies to integers, like in the given example. GCD can also be applied to the rationals, however. I found that the definition $GCD\left(\dfrac{a}{b}, \dfrac{c}{d} \right) = \dfrac{GCD(a\cdot d, c \cdot b)}{b \cdot d}$ seems to work, and has been confirmed to be the correct method.

Note that also in the AM example the amplitude of the fundamental is zero. The modulated signal only consists of the 9th, 10th and 11th harmonic. GCD(9 $\omega_0$, 10 $\omega_0$, 11 $\omega_0$) = $\omega_0$.

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I downvoted :) Fourier transform is specifically made for non-periodic signals, while Fourier series are for periodic ones. The spectrum can be an arbitrary function, as well as the signal in the time domain. And periodic in one domain means discrete in the other, and vice versa, while the signal can easily be none of them. –  clabacchio May 29 '12 at 9:44
@clabacchio - I still don't know what's wrong in my answer. –  stevenvh May 29 '12 at 9:50
I may have misread it, in that case I'll revert and apologize, but I understand that it's about harmonics and multiple frequencies, while the question is about Fourier transform (I assume OP has the same concept), which in my experience is for non-periodic signals, then doesn't produce harmonics but something like the first row –  clabacchio May 29 '12 at 9:55
Really, nothing personal, but I was trying to push the asnwers I like more to the top. I've upvoted you so many times, you can forgive me :) –  clabacchio May 29 '12 at 9:57
@clabacchio - OP asks about "the signal's frequency" and "the highest frequency part". Those hint at discrete values, and my guess was that OP meant series where he said transform. To me this absolutely points to the concept of fundamental frequency (and harmonics). –  stevenvh May 29 '12 at 10:00

The Fourier transform gives, like you said, the "frequency contents" of the signal. The signal has content at all of the frequencies where the transform is non-zero.

If the original signal is periodic, it's Fourier transform will have characteristic spikes or peaks at the oscillation frequency of the signal and its harmonics. For example, if the signal is repetitive with frequency f, the Fourier transform will have peaks at f, 2*f, 3*f, etc. Depending on the nature of the signal, some of these peaks could be missing (for example, a pure sine wave will just show the fundamental frequency, a square wave will only have odd harmonics, etc.)

So for a periodic signal you might say that the oscillation frequency of the signal is the lowest frequency of the Fourier transform, not the highest.

Edit: As Stevenvh and Teleclavo point out, it's possible that the missing peaks include the fundamental. It's even possible for there to be many many missing peaks below the first one observed in the spectrum. For example, take a 1 Hz square wave with extremely fast rising edges (say 10 ps). Now apply a high-pass filter with cut-off at 1 GHz. Depending on how sharp is the filter, you might see no output below 10 MHz and a series of peaks at 1 Hz intervals from there up to 10 GHz, meaning the first 10 million harmonics are absent. but the repetition period remains 1 s.

And its also possible to have an aperiodic signal that has a spectrum composed of multiple peaks. My answer relates to cases where you have independent information that tells you the signal is periodic.

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The oscillation frequency is at the frequency of every frequency it is oscillating at would be more correct to me. The fact we are having a large number of users arguing adamantly about this is a sign that we are trying to define a quantity that cannot be defined, you make the point of specifying to look at harmonics and such it must be period and I find your answer the best fit. –  Kortuk May 28 '12 at 10:00

No. For instance the Fourier transform of a Gaussian pulse is another Gaussian pulse, neither of which appears to oscillate.

However, the opposite is true. If a signal is a sinusoidal-like oscillator, then the FT or FFT will show a spike or peak.

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It is neither the lowest frequency component (of the Fourier transform) nor the highest. The Fourier transform exists even for non-periodic signals, and those signals (obviously) do not oscillate at any frequency.

I'll say even more. Even for periodic signals, it is still neither the lowest nor the highest frequency component, what -in general- determines its period. If you add two sinusoids of 19 kHz and 20 kHz (as it is commonly done in intermodulation tests for audio equipment), the spectrum has one delta at 19 kHz, and one delta at 20 kHz. However, the resulting signal has a period of 1 kHz. Neither 19 kHz, nor 20 kHz.

If you mention Fourier, forget that one signal may "oscillate" at only one frequency, and you try to see what determines the frequency of that oscillation. In general, a signal "oscillates" at infinite frequencies. All those frequencies which have nonzero component, at their Fourier transform.

Added: Even for a periodic signal, the fundamental frequency is not the one of the lowest frequency component (invisible or not) in the Fourier transform.

Take this simple example:

$S(t)=1+sin(2\pi f_1·t)$

with $f_1=$1 kHz, which looks like this:

Notice that its period is 1 ms.

Now look at the Fourier transform, S(f), of S(t):

What is the frequency of its lowest frequency component? Zero. Is f=0 the fundamental frequency of S(t)? No. The fundamental frequency of S(t) is 1 kHz.

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@FedericoRusso But my answer is not different because of that. It is different because it clearly states that it is neither the lowest nor the highest, and that -in general- signals oscillate at infinite frequencies (if decomposed into sinusoids). It is misleading to say, as stevenh did, that a non-repeating signal has a fundamental frequency. –  Telaclavo May 27 '12 at 15:30
@FedericoRusso I know what you mean, and you know what I mean. The difference is that the definition is on my side. You show me a reputable reference that defines or even talks about the fundamental of a non-periodic signal. I can show you infinite, that define it for periodic signals: home.iitk.ac.in/~kundu/paper86.pdf home.iitk.ac.in/~kundu/paper88.pdf ... –  Telaclavo May 27 '12 at 16:19
@FedericoRusso Hopefully the added example will put an end to this discussion. –  Telaclavo May 27 '12 at 17:21
OP asks about "the signal's frequency", that's asking for an explanation about fundamental (definition) and harmonics (definition). I should downvote you for not answering OP's question. –  Federico Russo May 27 '12 at 19:00
@FedericoRusso. Sure. The fundamental frequency is only defined for periodic signals, and it is equal to the inverse of the period with which they repeat, in time. // The OP did NOT mention neither "fundamental" nor "harmonics", in his question. Why should I mention those? To tell him that a signal does not have, in general, "a frequency", but instead, infinite ones (the ones that show up in its transform)? // Downvote whoever you want. // You are starting to be tiresome, and no, there's nothing wrong with me. –  Telaclavo May 27 '12 at 19:51