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$\DeclareMathOperator{\sinc}{sinc}\DeclareMathOperator{\asinc}{asinc}$I'm trying to follow the proof of the Fourier Transform for an Impulse train given in Julius Smith's textbook. I come across the following:

\begin{align} (2M+1)\asinc_{2M+1}(2\pi f) &= \frac{\sin [ \pi f(2M+ 1) ]}{\sin(\pi f)} \\ &= \sum_{k=-\infty}^{\infty} \sinc(2Mf - k) \tag{1} \\ &\rightarrow \sum_{k=-\infty}^{\infty} \delta (f -k) \tag{2} \end{align} where I suppose $\rightarrow$ in (2) denotes the limit as $M \rightarrow \infty$, and $\asinc$ denotes the aliased sinc. I'm not able to understand both of (1) and (2). In particular, it seems to me that $$ \lim_{M \rightarrow \infty} \sinc (2Mf - k) = \lim_{M \rightarrow \infty} \delta(f - k/2M) = \delta(f). $$

Can someone help me?

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    $\begingroup$ Beautified your math a bit, it was already pretty nice. Could you check whether you intended to add the $a$ in front of the first sinc? $\endgroup$ Nov 2, 2022 at 8:11
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    $\begingroup$ @MarcusMüller: The $a$ is supposed to mean "aliased" for "aliased sinc". Smith's notation ... $\endgroup$
    – Matt L.
    Nov 2, 2022 at 8:52
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    $\begingroup$ @MattL. thanks! Because I was surprised, I added that definition to the text. Hope it helps more than it clutters. $\endgroup$ Nov 2, 2022 at 8:55
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    $\begingroup$ @MarcusMüller: No, it's a good thing because it's definitely non-standard notation. $\endgroup$
    – Matt L.
    Nov 2, 2022 at 9:18

1 Answer 1

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In the given proof, first a finite length symmetrical impulse train of length $2M+1$ is transformed, and then the limit $M\to\infty$ is computed. The Fourier transform of the length $2M+1$ impulse train is computed as

$$\frac{\sin\big[\pi f(2M+1)\big]}{\sin(\pi f)}\tag{1}$$

which is a scaled Dirichlet kernel, also known as periodic sinc function, or - at least in the book you cited - aliased sinc. Note that a sampling interval $T_s=1$ is implied here.

A finite length impulse train can be interpreted as a sampled rectangular pulse of width $(2M+1)T_s$, where we use $T_s=1$. The Fourier transform of such a rectangular pulse is a sinc function:

$$(2M+1)\textrm{sinc}\big[(2M+1)f\big]\tag{2}$$

Sampling that rectangular pulse results in aliasing, i.e., a periodic repetition of the spectrum $(2)$ (with period $1$ because $T_s=1$):

$$(2M+1)\sum_{k=-\infty}^{\infty}\textrm{sinc}\big[(2M+1)\left(f-k\right)\big]\tag{3}$$

Both, $(1)$ and $(3)$ represent the Fourier transform of a symmetric finite length impulse train with length $2M+1$. Consequently, we have

$$\frac{\sin\big[\pi f(2M+1)\big]}{\sin(\pi f)}=(2M+1)\sum_{k=-\infty}^{\infty}\textrm{sinc}\big[(2M+1)\left(f-k\right)\big]\tag{4}$$

This is why aliased sinc is an appropriate name for the function given in $(1)$.

The last thing we need to know in order to arrive at the desired result is that the limit of a (scaled!) sinc function is a Dirac delta impulse:

$$\lim_{\tau\to\infty}\tau\,\textrm{sinc}(\tau x)=\delta(x)\tag{5}$$

Using $(5)$, the limit of the right-hand side of Equation $(4)$ is

$$\lim_{M\to\infty}(2M+1)\sum_{k=-\infty}^{\infty}\textrm{sinc}\big[(2M+1)\left(f-k\right)\big]=\sum_{k=-\infty}^{\infty}\delta(f-k)\tag{6}$$

which is the desired result, namely the Fourier transform of an impulse train of infinite length.

If you compare my derivation to the one of Smith, you'll see that there are a few constants that are different. Especially, the limit of the sinc function doesn't seem to be correct in Smith's book, because you need that constant factor $2M+1$ in order for $(4)$ to converge to a sum of Dirac impulses.

For an alternative derivation of this result, see this question and its answer.

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