Intuitive way to understand the square wave spectrum?

Question

What is a intuitive way to understand that a transform of a square wave can result into something like this?

Answer 1

Is it intuitive to switch the time and frequency axes? We can do it because the Fourier transform is essentially its own inverse. So, the spectrum of a square wave looks the same way as the time profile of a signal that contains all frequencies up to $\omega$, with equal amplitude (because rectangular wave has constant modulus) and phase (because rectangular wave is a positive function).

Let's simplify further and replace continuous superposition of frequencies by discrete. So, at time $0$ we strike $n$ strings that vibrate with angular frequencies $$ \frac{1}{N}\omega,\frac{2}{N}\omega,\frac{3}{N}\omega,\dots, \frac{N-1}{N}\omega, \omega$$ The first moment is the loudest, when all strings are in phase. (This is the peak on your picture). But that quickly changes. Half of all strings have frequencies between $\omega/2$ and $\omega$. They form a packet that stays together for a while; certainly for more than half of one period $T=2\pi/\omega$. Indeed, at time $t\approx 2T/3$ most of high-frequency contributions are negative. On the graph below this happens around $0.4$.

This is why we see a negative minimum when adding these frequencies together:

After that the pattern becomes less pronounced. But the basic idea remains: the behavior of the sum is driven by its high-frequency components, simply because there are more of them (which becomes apparent if one groups the frequencies on logarithmic scale).