Does anyone have a mechanism to understand intuitively (and automatically) why the Fourier Transform of certain functions have certain shapes (at least for some functions, not necessarily for all)? I know what kind of operator the Fourier Transform is and what it does to a function but somehow I can't see intuitively and automatically why why the Fourier Transform of certain functions have certain shapes. For example, is there a intuitive reason why the Fourier Transform of a pulse (box function) is in the $\mathbb{sinc}$?

Thanks in advance.

  • 1
    $\begingroup$ If you look at lots and lots of transforms, some forms of them (especially linear combinations of the ones with which you've already become extremely familiar) might start becoming intuitive. $\endgroup$
    – hotpaw2
    Jun 22, 2014 at 21:30
  • $\begingroup$ Don't look for intuition. Do everything abstractly using math symbols and you'll be doing the same math as everyone else, without struggling to find familiarities and intuition. In theory. $\endgroup$ Jun 22, 2014 at 22:50
  • $\begingroup$ @EnjoysMath I do the exact opposite. I can't understand an equation until I've mentally visualized it in a more intuitive graphical form. commons.wikimedia.org/wiki/… commons.wikimedia.org/wiki/… $\endgroup$
    – endolith
    Jun 23, 2014 at 13:59

2 Answers 2


What might seem intuitive differs greatly between individuals. But let's start with a few basic things about a Fourier transform that people who have studied it might know about it. Intuitively or not.

Basic concept 1: Something symmetric around t=0 in the time domain is strictly real in the frequency domain (as only cosine functions and DC are purely symmetric in the time domain).

Basic concept 2: The duality principle of the FT implies that if any transform looks intuitive in one direction, the transform in the opposite direction should also.

Transform 1: A DC level in the time domain corresponds to an impulse at 0 in the frequency domain.

Transform !B: Getting away from impulses and infinities, but staying with the same picture, a strictly real and really skinny Gaussian pulse in the frequency domain at DC must convert to a really really wide Gaussian envelope in the time domain, also symmetric around t=0. And vice versa. Going in between those two (and depending on your graph scaling), a medium width Gaussian in one domain should correspond to a medium width Gaussian in its transform domain.

Possible intuitive transform concept 2 (if you have studied signal modulation concepts): Adding strictly real and symmetric sidebands to a carrier in the frequency domain corresponds to AM modulation with a cosine sinusoid in the time domain. The farther away the sidebands from the carrier center, the higher the frequency of the modulation. The closer the sidebands, the lower the frequency of the modulating sinusoid.

Thus if we add symmetric wide sidebands to a wide Gaussian, that will correspond to a narrow Gaussian that is high frequency AM modulated in the other domain. If we add only narrow sidebands close in to a narrow Gaussian, that will correspond to a really wide Gaussian with a low frequency AM modulation in the other domain.

Possible intuitive transform concept 3 (if you have done much filter design): If you have had to design FIR filters, you might know that the sharper the cut-off and the flatter the pass-band desired in the frequency domain, the longer the FIR filter becomes to meet those requirements, corresponding to a longer and wider impulse response in the time domain. Or, in the other direction, the sharper-edged and flatter-topped a pulse becomes in the time domain, the wider the bulk of its frequency response spans in the frequency domain.

So, let's take that Gaussian pulse, and add some similar shaped side-bands within the bulk of its main lobe so that its top become flatter. The Gaussian in the other domain will get amplitude (AM) modulated with sinusoidal modulators, mostly of frequencies where the tops of the added sidebands are much higher that the original Gaussian roll-off. (To make it even flatter, perhaps remove the some or all of carrier (original Gaussian) to produce a modulation over 100%, e.g. that crosses zero.)

Let's also chop off the tails of the Gaussian with a brick wall filter and increase the flatness of the pass-band top even more to “square it up” in one domain. In the other domain, that will cause lots of added ripples extending wider than before, or a “fattening up” of the tails of the original Gaussian.

So as we “square up” a Gaussian in one domain, it gets amplitude modulated plus fatter tails in the other domain.

Thus, a skinny “squared up” Gaussian pulse in one domain might corresponds to a wide slowly AC modulated pulse with fat tails in the other domain, And a wide but “squared up” Gaussian might correspond to something like a narrow pulse with fast AC modulation, but also with fat tails, in the other domain.

An AC modulated hump with fat tails starts to approach looking more similar to a Sinc waveform than the original Gaussian waveform, and with a width inversely proportional to the width of the "squared up" looking waveform in the other domain.

Intuitive? Who knows…

  • $\begingroup$ +1 for the effort! ;) If you would add any plots that would benefit OP. $\endgroup$
    – jojeck
    Jun 23, 2014 at 12:29
  • $\begingroup$ Of course, this answer may still beg the question regarding intuitiveness: Why does AM produce a pair of sidebands? Why does a sharper FIR filter have a longer transient response? etc. $\endgroup$
    – hotpaw2
    Jun 23, 2014 at 19:13
  • $\begingroup$ I think you can even make it a wiki entry at some point. $\endgroup$
    – jojeck
    Jun 23, 2014 at 19:43

functions in time that wiggle a lot will have a greater portion of its energy in the frequency domain at higher frequencies. because of duality, you can switch the roles of time and frequency. a function in time that has the greater portion of its energy at times much earlier and later than $t=0$ will wiggle a lot in the frequency domain.

  • $\begingroup$ I understand the first sentence but not the second. Will a pulse, say rect$(t)$, have less variability in the frequency domain that the delayed version rect$(t-\tau)$ because the latter has "greater portion of its energy at times much later that $t=0$"? $\endgroup$ Jun 24, 2014 at 3:00
  • $\begingroup$ sure, if $\tau$ is big, there's gonna be an $e^{-j 2 \pi \tau f}$ factor multiplying the spectrum that isn't there when $\tau=0$. that's gonna make the spectrum wiggle a bit, no? $\endgroup$ Jun 24, 2014 at 3:27
  • $\begingroup$ Since $|e^{j2\pi f\tau}| = 1$, I wouldn't think of the high-frequency spectrum as "wiggling", at least not in the same sense as a time function wiggling and thus having more energy at higher frequencies than a smoother, more nearly constant, function. $\endgroup$ Jun 24, 2014 at 11:47
  • $\begingroup$ well, since "spectrum" $X(f)$ is a complex function with real and imaginary parts, i wouldn't think of it as only $|X(f)|$. but such as it is, then i'll just change "earlier or later" to "earlier and later". then you get to change your example from "$\mathrm{rect}(t-\tau)$" to "$\mathrm{rect}(t-\tau) + \mathrm{rect}(t+\tau)$" and you'll see all the wiggling your heart desires. $\endgroup$ Jun 24, 2014 at 14:35

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