Fourier Transform as limit case of Fourier Series in $sin$ and $rect$ functions

Question

I have a weird doubt about Fourier Series and Fourier Transform.

I know a periodic function can be expressed by the sum of simple sinusoidal functions multiplied by some integer coefficients using the Fourier Series. The sinusoidal functions used to "compose" the original function are called the Harmonics of the original function and if we look at function's Spectrum (i.e. Frequency/Amplitude diagram) we see only the frequencies of the harmonics have Amplitude values $\neq0$, i.e. the frequencies of the $\sin$ and $\cos$ functions which compose the original function

I also know that if the function is not periodic you can transform it in Frequency Domain using the Fourier Transform, which is basically the "limit case" of the Fourier series, using an Integral instead of the discrete sums used by the Fourier Series.

knowing this, I always thought it made sense that the Fourier Transform of a $\sin$ of frequency $A$ is:

$$ \mathscr{F}\{sin({2\pi At})\} = \frac{1}{2i} [\delta(f-A) - \delta(f+A)]$$

In fact I thought that since the sin function has only one frequency (his own actually), it is perfectly natural that its spectrum just has an impulse (all information "concentrated" in one point) right at the point with that frequency. Actually it presents two impulses, but still this is the intuitive explanation I gave to myself...

But then I was expecting something similar - for example - for the $rect$ function. In fact (I thought) we can also "periodicize" a rect function obtaing a succession of positive halves of a Square Wave. So, for the same reason explained before, I was expecting the $rect$ function's Forurier Transform to have an impulse in its own frequency (inverse of its period); yet the Fourier Transform of $rect(t/T)$ (with $T$ its period) is:

$$ \mathscr{F}\{rect(\frac{t}{T})\} = T sinc(fT)$$ where $$sinc(fT) = \frac{sin(fT)}{fT}$$

which is definitely not an impulse, but is actually a continuous function...

What am I missing here? I'm looking for an intuitive grasp of the relation between Fourier Series and Fourier Transform (e.g. frequency of $sin$ function I stated before) rather that formal demonstrations.
Maybe the link between these two operators as I thought it is completely wrong? Should associate/think about Fourier Series and Fourier Transform in some other way?

Answer 1

Functions may be similarly periodic but that does not mean they will have similar Fourier series/transforms. We can come up with all sorts of functions with equal frequencies but entirely different Fourier series, as in the image below.

Recall that Fourier series decompose functions into harmonics, sin/cos waves of different frequencies. And many Fourier series will continue to add higher frequency components as the series progresses. On the other hand, the only functions that do not, and are entirely encapsulated by the first component, are pure sine waves.

But harmonics of higher frequencies may still contribute to the shape of the function, despite having no effect on its period. Observe that all harmonics will repeat when the first does, since all of $\sin 2x, \sin 3x, \ldots$ all repeat when $\sin x$ does. They all have $2\pi$ as a frequency, even if not necessarily a fundamental frequency.

A Fourier series/transform is therefore not an indication of a function's frequency alone, but specifically how much it has of a sin/cos wave with each frequency. Or more correctly, it decomposes the function into a basis of complex exponentials. A sine wave will therefore be entirely encapsulated by few terms. But a square wave will not, as no sum of sin/cos waves will exactly equal it, so it has high frequency components.