# For the square wave signal, why does a narrower square wave correspond to more spread in the frequency domain?

Consider the square wave.

As we decrease the pulse width $$T_1$$, it seems that the positive Fourier coefficients (located in the central area/lobe) are spreaded out across more frequencies.

In fact, if we decrease $$T_1$$ down to zero, then this is equivalent to an impulse function and and its Fourier transform is a constant which is spreaded across all the frequencies (something like $$1/(2\pi)$$).

Intuitively, why does a narrower signal result in a wider spread in the frequency domain?

• I guess that's just how the math shows it to be. Commented Jun 11 at 15:11
• yeah, to me "narrow in one, wide in the other domain" is what has become intuitive. But that's just because I've seen it that often, and it's a principle that arises directly from the underlying math. If you want to do it for the sinc – the derivation of the sinc shape of the transform of a rectangle is doable, but I personally find integral-sines as you'll need to grasp not too pretty to work with. You can also derive the time-scaling property of the Fourier transform in general – that's honestly easier, because all you need to do is look at what happens to base functions individually. Commented Jun 11 at 16:05
• So, I don't think "why, intuitively…" has an answer. It is like that. Why, intuitively, does the apple fall from the tree to the ground? Why, intuitively, has a polynomial over the complex numbers exactly as many roots as coefficients? There's many things around you that aren't "intuitive" at all, but you can still understand them. Commented Jun 11 at 16:07

Think of what happens when you play back a recording at 2x speed: each frequency of the sound doubles.

When you make the pulse narrower, that's equivalent to speeding it up in the time domain. Therefore you can expect the frequencies to get proportionally higher.

One way to view this concept is from energy conservation. With proper scaling, the Fourier transform is energy conserving. In order to use up its energy in a short window, the signal has to expel its energy through more frequencies, and the converse is true.

I don't know if that's a great answer, but "intuitive" in this case often isn't distinct from the math. The best answer to me personally is from the perspective of the uncertainty principle. When applied to harmonic analysis, it means that signals cannot be localized well in both time and frequency. This makes sense viewing the Fourier transform (or its variants) from a bandpass filter perspective. The bandwidth of each bandpass filter is inversely proportional to the time-domain support. By increasing the duration of the signal, you increase the time-domain support, thereby decreasing the bandwidth of the bandpass filter, which improves the frequency resolution.

For that kind of question ('how to get it'), I'd suggest regarding Fourier transform as a way of determining a level of similarity (or correlation) between the original signal and reference periodic signals - Fourier harmonics.

The narrower the impulse (which may be thought of as a single half-period), the more harmonics (of the possibly infinite set) having yet higher frequencies may be regarded to be 'somewhat resembling' the impulse.

Negative multipliers should be regarded as a measure of 'unlikeness' in a sense they are determined by what part of a complete period of the specific harmonic lies 'outside' the original impulse.

In fact, your second picture doesn't help you in that case. You would be better served by the multi-axes plot of discrete Fourier harmonics, like on the image below (real parts on the left, imaginary parts on the right; stepped lines are for convenience).

Of course, like every analogy or metaphor, this approach can take you only so far.

The "bell curve" is described by the Gaussian function, $$e^{- k x^2}$$, and is its own Fourier transform when $$k = \pi$$: $$\mathcal F(e^{-\pi x^2}) = e^{-\pi x^2}$$. Fourier transforms are their own inverse operation possibly ignoring a constant scaling factor: $$\mathcal F(\mathcal F(f(x))) = \mathcal F(\mathcal F^{-1}(f(x))) = f(x)$$. Suppose we had a similar Gaussian function with $$k=1$$. If the Fourier transform made $$k$$ even smaller than $$1$$, the inverse transform would continue making $$k$$ smaller because the forward transform is identical to the reverse transform (up to a constant). This isn't a proof, but it should make intuitive sense that a narrower time signal transforms to a wider frequency distribution because if it didn't, the attribute that Fourier transforms are their own inverse operation (up to a constant) could no longer apply.