0
$\begingroup$

In Wikipedia the definition of the Fourier transform is:

"The Fourier transform of a function of time is a complex-valued function of frequency, whose magnitude (absolute value) represents the amount of that frequency present in the original function, and whose argument is the phase offset of the basic sinusoid in that frequency. "

When it says that represents the amount, I did not understand the amount of what?

$\endgroup$
3
  • 1
    $\begingroup$ that's a vague verbal definition, the mathematically exact one is below. It's really just "amount", i.e. a number without unit $\endgroup$ Commented Nov 10, 2021 at 13:10
  • 1
    $\begingroup$ (think of it as cross-correlation coefficient of an oscillation of every frequency with the signal, cross-correlations have no units, either, and are just "amounts") $\endgroup$ Commented Nov 10, 2021 at 13:38
  • 2
    $\begingroup$ Perhaps "amount of signal" would be less confusing -- perhaps not. I got my head wrapped around the Fourier Transform because my commute to school was an hour and a half on a bicycle, and there were some long straight stretches where I could ponder the math. $\endgroup$
    – TimWescott
    Commented Nov 10, 2021 at 19:51

1 Answer 1

1
$\begingroup$

Wikipedia's statement is best understood by looking at the inverse Fourier transform and murmuring the shibboleth "An integral is just a glorified sum and I understand sums".

If $x(t)$ has Fourier transform $$X(f) = \int_{-\infty}^\infty x(t)\exp(-j2\pi ft) \,\mathrm dt,\tag{1}$$ then the inverse Fourier transform is $$x(t) = \int_{-\infty}^\infty X(f)\exp(j2\pi ft) \,\mathrm df.\tag{2}$$ Note that $\exp(j2\pi ft)$ is a unit-amplitude (complex-valued) oscillator (a.k.a. complex sinusoid) at frequency $f$, $X(f)\exp(j2\pi ft)$ is also an oscillator/sinusoid at frequency $f$ with (complex) amplitude $X(f)$, or as Wikipedia likes to think about it, the sinusoid amplitude (which must be a nonnegative real number according to many folks, none of this socialistic godless communistic "complex-amplitude" nonsense here, thank you) is $|X(f)|$, the magnitude of the Fourier transform $X(f)$ at frequency $f$, while the phase of the sinusoid is $\angle X(f)$. Now, shouting "An integral is just a glorified sum and I understand sums" in a loud ringing voice while looking at $(2)$ might persuade you that $(2)$ is saying that $x(t)$ is the "sum" of the outputs of (infinitely many) complex-valued oscillators where the $f^{\text{th}}$ oscillator has amplitude $|X(f)|$ and phase offset $\angle X(f)$ from the basic complex-valued sinusoid $\exp(j2\pi ft)$. Put another way, the signal $x(t)$ "contains" $|X(f)|$ worth of frequency $f$ at phase offset $\angle X(f)$ and $(2)$ is saying we can reconstruct $x(t)$ by "adding" up all such contributions. You did admit that you understood sums, right?

$\endgroup$
4
  • $\begingroup$ I love that opening paragraph and want to quote you $\endgroup$ Commented Nov 12, 2021 at 14:52
  • $\begingroup$ “Socialistic godless communistic complex amplitude nonsense”… wow that is a bit harsh! $\endgroup$ Commented Nov 12, 2021 at 14:54
  • 2
    $\begingroup$ @DanBoschen Feel free to quote anything I write on dsp.SE (or just copy-and-paste with or without attribution). I don't make any claims about the originality of the sentiments expressed in that paragraph. It is a mishmash of what the instructor in my signals-and-systems class said about 55 years ago. As to the other comment that you find harsh, some people are very firm in their beliefs and unwilling to accept generalizations. Note that I didn't even attempt to address the belief that frequency $f$ must be a nonnegative number (cf. some questions here as to what negative frequency means). $\endgroup$ Commented Nov 13, 2021 at 3:19
  • 2
    $\begingroup$ Just to be clear my last comment about in being harsh was in jest, I found the sentence completely funny and made me laugh out loud when I first read it (and reflecting on my own dependence on the complex representations and "spinning phasors" to visualize things in DSP). Both great quotes. $\endgroup$ Commented Nov 13, 2021 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.