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The Fourier transform of $x(t)=\operatorname{rect}(t)$ is $X(f)=\operatorname{sinc}(f)$

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The Fourier transform of a periodic train of rectangular pulses $x(t)=\sum\limits_{n=-\infty}^{\infty}\operatorname{rect}(t-n)$ is a sampled sinc:

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Picture taken from here.

From a math point of view it is fine because:

  1. The periodic train of rects is a periodic signal that can be expressed as a discrete sum of sine waves according to Fourier Series Expansion.

  2. The periodic train of rects is the convolution of a rect and a train of dirac deltas. The product of their Fourier Transforms is a sampled sinc since they are a sinc and again train of dirac deltas in frequency domain.

So, my question is a about the physical meaning of this. A single rect has a continuous spectrum, that means that it has a certain amout of power (and energy) at each possible frequency (from $-\infty$ to $+\infty$). Why, if I repeat the rect infinite times, do I remove some frequencies from its spectrum and leave only a discrete set of frequencies? Why does repeating periodically a signal means removing some frequency intervals from the single pulse spectrum?

In time domain, frequency may be seen as the speed of variation of the signal. Well, it is the same for a single rect and a train of rects (if we suppose the same fall and rising times and same rect duration).

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    $\begingroup$ Have you tried sketching using pencil and paper what the functions $\operatorname{rect}(t), \operatorname{rect}\left(t-\frac 12\right), \operatorname{rect}\left(t+\frac 12\right)$ corresponding to $n=0,1,-1$ in your infinite sum look like? All other rect functions in your infinite sum have value $0$n in the interval $\left(-\frac 12, +\frac 12\right)$, and so adding out these three will show you what $x(t) $ looks like in $\left(-\frac 12, +\frac 12\right)$, and hopefully explain why the sampled sinc function you are showing is totally inapplicable to your question as asked. $\endgroup$ May 27 at 15:06
  • $\begingroup$ @DilipSarwate Thanks for your observation. I've made a mistake in writing the train of rects $\endgroup$
    – Kinka-Byo
    May 27 at 19:41
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    $\begingroup$ Your correction to the definition of $x(t)$ doesn't help very much. Now, the rect functions $\operatorname{rect}(t), \operatorname{rect}(t-1), \operatorname{rect}(t+1)$ have non-overlapping supports $\left(-\frac 12,+\frac 12\right), \left(+\frac 12,+\frac 32\right),\left(-\frac 32,-\frac 12\right)$ respectively, but the sum of these three rects is $1$ everywhere in $\left(-\frac 32, \frac 32\right)$ except at $t=\pm \frac 12$. $\endgroup$ May 27 at 19:56
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    $\begingroup$ @DilipSarwate , perhaps the definition of the $\operatorname{rect}(x)$ should include the midpoints of the steps: $\operatorname{rect}(\pm \tfrac12)=\tfrac12$. $\endgroup$ May 27 at 22:11
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Why does repeating periodically a signal means removing some frequency intervals from the single pulse spectrum?

Because your signal is periodic. A periodic signal only contains frequencies that are integer multiples of the signal's repetition frequency. In other words, if the signal is periodic all of it's constituent parts must be periodic as well.

In time domain, frequency may be seen as the speed of variation of the signal.

Not really. "speed of variation" would be the first derivate of the signal. It's somewhat related to higher frequencies since the Fourier Transform of a differentiator is proportional to the frequency itself but it's not the same thing. If you look at a square wave you also need high frequency to keep the flat parts nice and flat.

So, my question is a about the physical meaning of this.

That's always a tricky proposition. One possible answer is "there is no physical meaning since it's physically impossible to have a signal that's infinite in time". Sine waves are a useful mathematical concept but they DO NOT EXIST in reality. Of course, for practical purposes you can always generate something that's "close enough" for your specific application.

Another explanation

Let's look at what happens if you repeat the rect just once. In the time domain you can represent this as a convolution with two pulses, i.e.

$$h(t) = \delta(t) + \delta(t-T_0) $$

So you add a delayed copy to the original signal. In the frequency domain this looks like

$$H(\omega) = 1 + e^{-j\omega T_0}$$

The delay turns into a phase shift so you add add a phase shifted copy of the spectrum the original one. If the phase shift at a certain frequency is 0 degree, they will just add. But if the phase shift at a different frequency is 180 degrees they will actually cancel. So some frequencies will be amplified others will be cancelled that's called a "comb filter" a graph of the amplitude with these notches looks like the teeth of a comb.

The more you repeat the original rectangle, the more "combing" you get and eventually all frequencies are gone except the ones that line up exactly with the period of the repetition.

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    $\begingroup$ +1 Nice answer, but note that the OP's function $x(t)=\sum\limits_{n=-\infty}^{\infty}\operatorname{rect}(t-\frac{n}{2})$ has value $2$ almost everywhere, and its Fourier transform is $2\delta(f)$, the $2\operatorname{sinc}(f)$ sampled exactly once at $f=0)$, all other frequencies having been eliminated! $\endgroup$ May 27 at 15:14
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Okay, this function

$$x(t)=\sum\limits_{n=-\infty}^{\infty}\operatorname{rect}(t-n)$$

is identically equal to one: $x(t) = 1 \quad \forall t\in\mathbb{R}$

because

$$ \operatorname{rect}(x) \triangleq \begin{cases} 1,\quad & |x|<\tfrac12 \\ \tfrac12, \quad & |x|=\tfrac12 \\ 0, \quad & |x|>\tfrac12 \\ \end{cases} $$

Now, if these rectangular pulse functions were spaced further apart (or closer with overlap-add) then

$$x(t)=\sum\limits_{n=-\infty}^{\infty}\operatorname{rect}(t-nT)$$

Now the $\operatorname{sinc}(f)$ function is sampled at integer multiples of $\frac{1}{T}$

$$\begin{align} X(f) & = \operatorname{sinc}(f) \cdot \frac1T\sum_{k=-\infty}^{\infty}\delta(f-\tfrac{k}{T}) \\ & = \frac1T\sum_{k=-\infty}^{\infty}\operatorname{sinc}(\tfrac{k}{T}) \cdot \delta(f-\tfrac{k}{T}) \\ \end{align} $$

Now when $T=1$, only one of those dirac functions in the spectrum $X(f)$ remains, which is the $\delta(f)$ at DC.

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