Absolute convergence of periodic sinc interpolation

Question

An $N$-periodic complex discrete-time sequence $[x_0, \dots, x_{N-1}]$ can be resampled to an $M$-periodic sequence $[y_0, \dots, y_{M-1}]$ with $M>N$, using sinc interpolation:

$$\begin{align}y_m &= \sum_{n=-\infty}^\infty \operatorname{sinc}\left(\frac{Nm}{M} - n\right)x_{n\operatorname{mod}N} \\&= \sum_{k=-\infty}^{\infty}\sum_{n=0}^{N-1}\operatorname{sinc}\left(\frac{Nm}{M} - n - Nk\right)x_n\end{align}\tag{1}$$

where $\operatorname{mod}$ denotes the modulo operation and:

$$\operatorname{sinc}(x) = \begin{cases}1&\text{if }x=0,\\\frac{\sin(\pi x)}{\pi x}&\text{otherwise}.\tag{2}\end{cases}$$

Eq. 1 can be seen as resampling an $N$-periodic continuous-time signal from samples $x_n$ at times $n + Nk$ to samples $y_m$ at times $\frac{Nm}{M}$.

For example, a $2$-periodic complex discrete-time sequence $[x_0, x_1]$ can be resampled to a $4$-periodic sequence $[y_0, y_1, y_2, y_3]$:

$$\text{Eq. 1, }N=2,\,M=4$$ $$\Rightarrow\left\{\begin{align}y_0 &= x_0\\ y_1 &= \sum_{k=-\infty}^\infty\Bigg(\operatorname{sinc}\left(2k+\frac{3}{2}\right)x_0 + \operatorname{sinc}\left(2k+\frac{1}{2}\right)x_1\Bigg)\\ y_2 &= x_1\\ y_3 &= \sum_{k=-\infty}^\infty\Bigg(\operatorname{sinc}\left(2k+\frac{1}{2}\right)x_0 + \operatorname{sinc}\left(2k+\frac{3}{2}\right)x_1\Bigg) \end{align}\right.\tag{3}$$

The two series in Eq. 3 converge conditionally, with for example these possible rearrangements of the first series that give conflicting results if $x_0 \ne x_1$:

$$\begin{gather}\sum_{k=0}^\infty\bigg(f(-k) + f(k+1)\bigg)\\= \frac{x_0 + x_1}{2},\\ \sum_{k=0}^\infty\bigg(f(-k) + f(2k+1) + f(2k+2)\bigg)\\= \frac{(x_1-x_0)\ln(2)}{2\pi} + \frac{x_0 + x_1}{2},\\ \sum_{k=0}^\infty\bigg(f(-k) + f(3k+1) + f(3k+2) + f(3k+3)\bigg)\\= \frac{(x_1-x_0)\ln(3)}{2\pi} + \frac{x_0 + x_1}{2}, \end{gather}\tag{4}$$

with shorthand $f(k) = \operatorname{sinc}\left(2k+\frac{3}{2}\right)x_0 + \operatorname{sinc}\left(2k+\frac{1}{2}\right)x_1$.

Under which condition does the series given by Eq. 1 converge absolutely?

Answer 1

I'm top-editing this since it answers the question directly.

The sinc series is fundamentally a $C/x$, so you can extract as many absolutely convergent series out of it as you want, but what is left over is still only conditionally convergent. Also, you can rescale $x$ and it is still a $C/x$ series.

Saying you have a summation to or from infinity is an informality. Formally, you have a finite sum to some value, and take the limit as that value goes to infinity.

Therefore, your first and second series should have been done like this:

$$ \lim_{L \to \infty} \sum_{k=-L}^{L} f(k) = \lim_{L \to \infty} \left[ \sum_{k=0}^{L} f(-k) + f(k+1) \right] $$ $$ = \lim_{L \to \infty} \left[ \sum_{k=0}^{L} \left( f(-k) + f(2k+1) + f(2k+2) \right) + \sum_{k=0}^{L+1} f(-k-L-1) \right] $$

Likewise, your third should have added this:

$$ \sum_{k=0}^{L+1} \left( f(-k-L-1) + f(-k-2L-3) \right) $$

Sometimes it takes a while to get around to where you should have been in the first place. I'm deleting the rest. Whoever is curious can find it in the edit history.

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Proceeding informally....

First rearrange it:

$$ \begin{aligned} y_m &= \sum_{n=0}^{N-1} x[n] \sum_{k=-\infty}^{\infty} \operatorname{sinc} \left( \frac{Nm}{M} - n - Nk \right) \\ &= \sum_{n=0}^{N-1} x[n] W_m[n] \end{aligned} $$

One way to look at that is a resampled value is a linear combination (weighted average) of the sample points.

Another way is that you now have $N$ separate infinite series, all of the form:

$$ \begin{aligned} W_m[n] &= \sum_{k=-\infty}^{\infty} \operatorname{sinc} \left( \frac{Nm}{M} - n - Nk \right) \\ &= \sum_{k=-\infty}^{\infty} \frac{ \sin \left( ( Nm/M - n - Nk ) \pi\right) }{ (Nm/M - n - Nk) \pi } \\ \end{aligned} $$

Even $N$ Case:

$$ W_m[n] = \sin \left( ( Nm/M - n ) \pi\right) \sum_{k=-\infty}^{\infty} \frac{ 1 }{ (Nm/M - n - Nk) \pi } $$

Odd $N$ Case:

$$ W_m[n] = \sin \left( ( Nm/M - n ) \pi\right) \sum_{k=-\infty}^{\infty} \frac{ (-1)^k }{ (Nm/M - n - Nk) \pi } $$

Clearly, both are cases of $C/x$ series and not absolutely convergent. If $Nm/M$ is an integer all the terms are zero except perhaps the zeroth terms.

As for the second comment, if I remember correctly (and I've already proven I didn't remember well), doing it formally does away with all the rearrangement tricks. And yes, If I remember correctly, absolutely convergent series are immune to rearrangement tricks.

This too:

A series converges if and only if the sequence of partial sums converges.

A sequence converges if and only if for any given $\epsilon$ there exist a $\delta$ so for every $k > \delta$ the absolute value of the difference of the limit and the sequence value is less than $\epsilon$.

Stamp it on your forehead for formal occasions.

Disclaimer: Been a long time ...

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As clearly as I think I can say it:

The only conditions for which the series in Olli's Eq (1) will converge absolutely is when all the terms heading towards infinity are zero, since then their absolute values are zero. This happens when all the $x_n$ are zero (the trivial solution) or $Nm/M$ is an integer. Both the even and odd cases under any different conditions can be rearranged to be summations of alternating monotonically decreasing sequences, therefore they converge conditionally since they diverge absolutely.

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Epilogue:

There is no need to do the infinite summation at all. Direct closed form expressions exist for the odd and then even case based on the interpolation functions found when considering an inverse DFT as a continuous function. The derivation of the functions can be found in the epilogue of my answer here:

How to get Fourier coefficients to draw any shape using DFT?

The derivation is based on the definitions of the DFT, the inverse DFT, and a finite geometric summation.

Resampling the continuous function at $M$ evenly spaced (in the cycle domain) points can be done by a simple variable substitution.

$$ t = \frac{m}{M} 2\pi $$

The direct sample set to sample set equations are then as follows.

Odd case:

$$ y_m = \sum_{n=0}^{N-1} x[n] \left[ \frac{ \sin \left( N \left( \frac{m}{M} - \frac{n}{N} \right) \pi \right) } { N \sin \left( \left( \frac{m}{M} - \frac{n}{N} \right) \pi \right) } \right] $$

Even case, evenly split Nyquist bin:

$$ y_m = \sum_{n=0}^{N-1} x[n] \left[ \frac{ \sin \left( N \left( \frac{m}{M} - \frac{n}{N} \right) \pi \right) } { N \sin \left( \left( \frac{m}{M} - \frac{n}{N} \right) \pi \right) } \right] \cos \left( \left( \frac{m}{M} - \frac{n}{N} \right) \pi \right) $$

These are mathematically equivalent to taking the DFT of size $N$, zero padding it at the Nyquist frequency to size $M$ (splitting the Nyquist bin in the even case), then taking the inverse DFT to recover a $M$ point upsampled sequence. All the upsampled points lie on the underlying continuous interpolation function no matter what the point count.

For the $N=2$, $M=4$ case:

$$ \begin{aligned} y_0 &= x_0 ( 1 ) + x_1 ( 0 ) = x_0 \\ y_1 &= x_0 \left( \frac{ \sin( \pi / 2 ) }{ 2 \sin( \pi / 4 ) } \cos( \pi / 4 ) \right) + x_1 \left( \frac{ \sin( -\pi / 2 ) }{ 2 \sin( -\pi / 4 ) } \cos( -\pi / 4 ) \right) \\ &= \frac{1}{2} ( x_0 + x_1 ) \\ y_2 &= x_0 ( 0 ) + x_1 ( 1 ) = x_1 \\ y_3 &= x_0 \left( \frac{ \sin( 3 \pi / 2 ) }{ 2 \sin( 3 \pi / 4 ) } \cos( 3 \pi / 4 ) \right) + x_1 \left( \frac{ \sin( \pi / 2 ) }{ 2 \sin( \pi / 4 ) } \cos( \pi / 4 ) \right) \\ &= \frac{1}{2} ( x_0 + x_1 ) \end{aligned} $$ Which should be the results you are expecting.

An infinite number of sinc functions can now take the day off.

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Suppose that instead of doing halfsies on the Nyquist bin we apportioned them as $(1/2+g)$ and $(1/2-g)$, this would alter the continuous interpolation function as follows.

$$ \begin{aligned} D(t_n) &= \left( \frac{1}{2} + g \right) e^{i(N/2) t_n } + \left( \frac{1}{2} - g \right) e^{i(-N/2) t_n } + \sum_{l=0}^{N-2} e^{i ( l - N/2 + 1 ) t_n } \\ &= \cos \left( \frac{N}{2} t_n \right) + i 2 g\sin \left( \frac{N}{2} t_n \right) + \frac{ \sin( t_n N /2 ) } { \sin( t_n / 2 ) } \cos( t_n / 2 ) - \cos( t_n N /2 ) \\ &= \frac{ \sin( N t_n/2 ) }{ \sin( t_n / 2 ) } \cos( t_n / 2 ) + i 2g\sin \left( \frac{N}{2} t_n \right) \end{aligned} $$

The extra term introduced is purely imaginary. That can be folded in, but I prefer to leave it separate when put back into the function definition.

$$ \begin{aligned} z(t) &= \sum_{n=0}^{N-1} x[n] \left[ \frac{ \sin( N (t - \frac{n}{N}2\pi) / 2 ) } { N \tan( (t - \frac{n}{N}2\pi) / 2 ) } + i \frac{2g}{N}\sin \left( N (t - \frac{n}{N}2\pi) / 2 \right) \right] \end{aligned} $$

If is obvious that any non-zero value of $g$ will add "energy" to the signal, thus the $g=0$ solution, corresponding to halfsies on the Nyquist bin, is the most natural solution, or lowest energy, out of a whole family of solutions of periodic bandlimited at $N/2$ functions.

The more significant convincer for me is that it also introduces imaginary values into what is other wise fully real set of weighting values.

Whether R B-J's series converges uniquely to this "natural" solution, or the "natural solution" is a unique solution (it is not) are two totally separate issues.

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Olli, I hope this makes you smile.

Start with the discrete resampling formula for the odd $N$ case.

$$ y_m = \sum_{n=0}^{N-1} x[n] \left[ \frac{ \sin \left( N \left( \frac{m}{M} - \frac{n}{N} \right) \pi \right) } { N \sin \left( \left( \frac{m}{M} - \frac{n}{N} \right) \pi \right) } \right] $$

Since the sequence of $N$ points is periodic $( x[n] = x[n+N] )$ and all the points are covered, we can shift the summation range to be zero centered.

$$ L = (N-1) / 2 $$

Also, the $m$th point can be located on the $n$ scale.

$$ w = m \frac{N}{M} = \frac{m}{M} N $$

Since the $M$ resampled points are evenly spaced along the cycle, they too can be arbitrarily shifted to be zero centered, though strictly that is not necessary.

Since "$t$" has already been used above, the scale of the domain of the continuous interpolation function, both will get new names. "$z(t)$" and "$Y(\omega)$" describe the same function. Plug all the defined values in.

$$ \begin{aligned} y_m = Y(w) &= \sum_{n=-L}^{L} x[n] \left[ \frac{ \sin \left( \left( w - n \right) \pi \right) } { N \sin \left( \frac{1}{N} \left( w - n \right) \pi \right) } \right] \\ &= \sum_{n=-L}^{L} x[n] \left[ \frac{\frac{\sin \left( \left( w - n \right) \pi \right)}{ \left( \omega - n \right) \pi }} {\frac{\sin \left( \frac{1}{N} \left( w - n \right) \pi \right)}{\frac{1}{N} \left( w - n \right) \pi }} \right] \\ &= \sum_{n=-L}^{L} x[n] \left[ \frac{\operatorname{sinc} \left( w - n \right) } {\operatorname{sinc} \left( \frac{1}{N} \left( w - n \right) \right)} \right] \\ \end{aligned} $$

Now it's time to take the big step, that is, big stroll out to infinity. The cycle of $N$ points grows until one cycle spans negative to positive infinity. As it gets bigger, the circular nature gets more remote.

$$ \begin{aligned} \lim_{N \to \infty} y_m &= \lim_{N \to \infty} Y(w) \\ &= \lim_{N \to \infty} \sum_{n=-L}^{L} x[n] \left[ \frac{\operatorname{sinc} \left( w - n \right) } {\operatorname{sinc} \left( \frac{1}{N} \left( w - n \right) \right)} \right] \\ &= \sum_{n=-\infty}^{\infty} x[n] \left[ \frac{\operatorname{sinc} \left( w - n \right) } {1} \right] \\ &= \sum_{n=-\infty}^{\infty} x[n] \operatorname{sinc} \left( w - n \right) \\ &= \sum_{n=-\infty}^{\infty} x[n] \operatorname{sinc} \left( \frac{Nm}{M} - n \right) \end{aligned} $$

Now look at that. The Whittaker–Shannon interpolation formula has been derived from scratch and we are right at your starting point.

The even case can be done similarly and ends up with the same formula.

1. Definition of DFT of $N$ samples
2. Inverse DFT used as Fourier Series Coefficients for interpolation function
3. Dirichlet Kernel form of interpolation function
4. Interpolation function used for $M$ samples
5. Even and Odd Discrete Weighted Average Resampling Formulas
6. N goes to infinity
7. Whittaker–Shannon emerges
8. Whittaker–Shannon applied to a repeating sequence of $N$
9. Convergence questioned

I hope realizing using step 7 to achieve what step 2 has already answered will put a smile on R B-J as well. Your proof lies there.

For $ N = 2 $

$$ \begin{aligned} y_m &= \sum_{n=0}^{1} x[n] \left[ \frac{ \sin \left( 2 \left( \frac{m}{M} - \frac{n}{2} \right) \pi \right) } { 2 \sin \left( \left( \frac{m}{M} - \frac{n}{2} \right) \pi \right) } \right] \cos \left( \left( \frac{m}{M} - \frac{n}{2} \right) \pi \right) \\ &= \sum_{n=0}^{1} x[n] \cos^2 \left( \left( \frac{m}{M} - \frac{n}{2} \right) \pi \right) \\ &= x_0 \cos^2 \left( \frac{m}{M} \pi \right) + x_1 \sin^2 \left( \frac{m}{M} \pi \right) \end{aligned} $$

For $ x_0 = 1 $ and $ x_1 = -1 $

$$ \begin{aligned} y_m &= \cos^2 \left( \frac{m}{M} \pi \right) - \sin^2 \left( \frac{m}{M} \pi \right) \\ &= \cos \left( \frac{m}{M} 2 \pi \right) \end{aligned} $$

I'm going to have to be done with this for a while. Neat stuff.

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Olli, thanks for the bounty points.

This little exercise has deepened my understanding of W-S considerably. I hope that is true for you and Robert (and others) too.

It is still a precarious foundation though. I wanted to convince myself that it would work for a sinusoid of any frequency. To wit:

$$ x[n] = M \cos( \alpha n + \phi ) $$

$$ \begin{aligned} x(t) &= \sum_{n=-\infty}^{\infty} x[n] \operatorname{sinc}(t-n) \\ &= \sum_{n=-\infty}^{\infty} M \cos( \alpha n + \phi ) \operatorname{sinc}(t-n) \\ &= \sum_{n=-\infty}^{\infty} M \cos( \alpha t + \phi - \alpha( t - n ) ) \operatorname{sinc}(t-n) \\ &= \sum_{n=-\infty}^{\infty} M \left[ \cos( \alpha t + \phi ) \cos( \alpha( t - n ) ) + \sin( \alpha t + \phi ) \sin( \alpha( t - n ) ) \right] \operatorname{sinc}(t-n) \\ &= M \cos( \alpha t + \phi ) \sum_{n=-\infty}^{\infty} \cos( \alpha( t - n ) ) \operatorname{sinc}(t-n) \\ & \qquad \qquad + M \sin( \alpha t + \phi ) \sum_{n=-\infty}^{\infty} \sin( \alpha( t - n ) ) \operatorname{sinc}(t-n) \\ &= M \cos( \alpha t + \phi ) \cos( \alpha( t - t ) ) + M \sin( \alpha t + \phi ) \sin( \alpha( t - t ) ) \\ &= M \cos( \alpha t + \phi ) \cdot 1 + M \sin( \alpha t + \phi ) \cdot 0 \\ &= M \cos( \alpha t + \phi ) \end{aligned} $$

I seem to have accomplished my goal. However, there is nothing in this proof that prohibits $\alpha \ge \pi$, though that is a condition for the validity of the theorem. So, knowing that, you are okay. If you didn't know that, the formula itself does not reveal it. To me, that's troubling.

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Reply to R B-J:

First off, no where is it stipulated that $x[n]$ must be real. Even for a real valued function, you don't have to split the Nyquist bin halfsies to get a real interpolation function. Just pick $g$ to be a multiple of $i$ above.

Suppose you have the function:

$$ z(\tau) = \sum_{k=-L}^{L} c_k e^{ik\tau} $$

Its band limit is $L$ or less. Every k term, except 0, can be paired up with it's conjugate bin and the sum can be decomposed into a cosine and sine term.

let $ A = \frac{c_k + c_{-k}}{2} $ and $ B = \frac{c_k - c_{-k}}{2} $

$$ \begin{aligned} c_k e^{ik\tau} + c_{-k} e^{-ik\tau} &= (A+B) e^{ik\tau} + (A-B) e^{-ik\tau} \\ &= 2A \cos(\tau) - i 2B \sin(\tau) \end{aligned} $$

For a regular bin, we can only say $X[k] = c_k$ if $k+N>L$, otherwise I have more than one k in the bin and cannot separate them. At the Nyquist bin $X[k] = c_k + c_{-k}$

Think in terms of degrees of freedom. For a complex signal, $c_k + c_{-k}$ has four and the Nyquist bin two. Therefore there are two free. Just enough to put a complex parameter on the Sine function at the Nyquist frequency. With a real signal, $c_k + c_{-k}$ has two degrees of freedom and the Nyquist bin value restricts one of those leaving one left over. Just enough for a real valued parameter times the Sine function to remain a real valued signal.

I showed earlier the translation between not doing halfsies and the consequence on the interpolation function. Nothing prohibits that and it doesn't increase the bandwidth of the solution one iota.

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R B-J asks:

// //"But we do know A will be zero in the halfsies and W-S reconstructions."// how do you know that? //

The halfsies is easy. Without loss of generality, consider the $N=2$ case.

$$ x[n] = [1,-1] $$

$$ \frac{1}{N} X[k] = [0,1] $$

Halfsies on the Nyquist of 1. Doing an unfurled inverse DFT with split Nyquist:

$$ x[n] = \frac{1}{2} e^{i\pi n} + \frac{1}{2} e^{-i\pi n} = \cos(\pi n) $$

Now allow $n$ to be real, call it $t$ to indicate the change. This defines an interpolation function (still called $x$).

$$ x(t) = \cos(\pi t) $$

For every other even N, the unnormalized DFT will be (0,0,0,....,N), so the result remains the same.

For the W-S summation, look at the section where "omega" temporarily lived, the "Sinc is the limit of the Dirichlet Kernel" section. The left side $y_m=Y(w)$ is known to be $ \cos( \pi w ) $. I even did the specific $N=2$ case after the dependency list. Just set "M=2" which makes $w = m$. The limit reached at the end of the second chunk gives your summation. Just reverse the order of the equation and you get:

$$ \sum_{n=-\infty}^{\infty} (-1)^n \operatorname{sinc}( w - n ) = \cos( \pi w ) $$

The fact that your summation is the limit of something is why proving it differently has been hard.

I think your time reversal argument is good, too. The sampled points are time reversible on the discrete $n$ scale, but that does not mean the source x(t) is, but it does mean Y(w) is.

P.S. From now on, when a fresh context can be established, I'm going to use $\tau$ for a $ 0 \to 2\pi $ cycle scale, $t$ to be on the sampling scale ($=n$).

Answer 2

Some remarks. The series in Eq. 1 of the question:

$$y_m = \sum_{k=-\infty}^{\infty}\sum_{n=0}^{N-1}\operatorname{sinc}\left(\frac{Nm}{M} - n - Nk\right)x_n$$

explicitly means this (see this answer to the Mathematics Stack Exchange question: Notation of double-sided infinite sum):

$$\begin{align}y_m &= \lim_{K_2\to\infty}\lim_{K_1\to\infty}\sum_{k=-K_1}^{K_2}\sum_{n=0}^{N-1}\operatorname{sinc}\left(\frac{Nm}{M} - n - Nk\right)x_n\\ &= \lim_{K_1\to\infty}\lim_{K_2\to\infty}\sum_{k=-K_1}^{K_2}\sum_{n=0}^{N-1}\operatorname{sinc}\left(\frac{Nm}{M} - n - Nk\right)x_n,\end{align}\tag{1}$$

which is only a valid statement if all of those limits exist and the two definitions (with the limits in different order) are equal.

Answer 3

FYI: This was the question I put to the math guys, but here I changed the notation from what might be most conventional to the math guys to one that is more conventional to EEs. (I am using that post as a starting point to sorta exhaustively deal with Olli's question, but in mathematical terms that are easier for me to grok, so i am not exactly following Olli's math. This ain't done yet.)

This has to do with the Nyquist-Shannon sampling and reconstruction theorem and the so-called Whittaker–Shannon interpolation formula. I had previously asked an ancillary question about this here but this is about a specific nagging issue that seems to "periodically" crop up.

Let's begin with a periodic infinite sequence of real numbers, $x[n] \in\mathbb{R}$, having period $N>0\in\mathbb{Z}$. That is:

$$ x[n+N]=x[n] \qquad \forall \ n\in\mathbb{Z}. $$

So there are only $N$ unique values of $x[n]$.

Imagine these discrete (but ordered) samples as equally spaced on the real number line (with a sampling period of 1) and being interpolated (between integer $n$) as

$$x(t) = \sum_{n=-\infty}^{\infty} x[n] \, \operatorname{sinc}(t-n),$$

where

$$ \operatorname{sinc}(u) \triangleq \begin{cases} \dfrac{\sin(\pi u)}{\pi u} & \text{if } u \ne 0, \\ \ 1 & \text{if } u = 0. \end{cases} $$

Clearly $x(t)$ is periodic with the same period $N$:

$$ x(t+N) = x(t) \qquad \forall \ t \in \mathbb{R}. $$

All terms are bandlimited to a maximum frequency of $\frac{1}{2}$, so the summation is bandlimited to the same bandlimit. And, in any case, we have

$$ x(t) \Big|_{t = n} = x[n], $$

so the reconstruction works out exactly at the sampling instances.

$$\begin{align} x(t) &= \sum_{n=-\infty}^{\infty} x[n] \, \operatorname{sinc}(t-n) \\ &= \sum_{m=-\infty}^{\infty} \sum_{n=0}^{N-1} x[n+mN] \, \operatorname{sinc}\big(t - (n+mN)\big) \\ &= \sum_{m=-\infty}^{\infty} \sum_{n=0}^{N-1} x[n] \, \operatorname{sinc}\big(t - (n+mN)\big) \\ &= \sum_{n=0}^{N-1} x[n] \, \sum_{m=-\infty}^{\infty} \operatorname{sinc}\big(t - (n+mN)\big). \\ \end{align}$$

Substituting $u \triangleq t-n$ gives

$$ x(t) = \sum_{n=0}^{N-1} x[n] \, g(t-n), $$

where

$$ g(u) = \sum_{m=-\infty}^{\infty} \operatorname{sinc}(u-mN). $$

Clearly the continuous (and real) $g(u)$ is periodic with period $N$:

$$ g(u+N) = g(u) \qquad \forall u \in \mathbb{R}. $$

What is the closed-form expression for $g(u)$ in terms of $u$ and $N$?

I can extend the Discrete Fourier Transform (DFT) a little and relate it to the continuous Fourier series:

$$ X[k] \triangleq \sum_{n=0}^{N-1} x[n] e^{-j 2 \pi n k/N} $$

and

$$ x[n] = \frac{1}{N} \sum_{k=0}^{N-1} X[k] e^{+j 2 \pi n k/N} $$

We know that both infinite sequences $x[n]$ and $X[k]$ are periodic with period $N$. This means that the samples of $x[n]$ or $X[k]$ can be any adjacent $N$ samples:

$$ X[k] \triangleq \sum_{n=n_0}^{n_0+N-1} x[n] e^{-j 2 \pi n k/N} \qquad \forall n_0 \in \mathbb{Z}$$

and

$$ x[n] = \frac{1}{N} \sum_{k=k_0}^{k_0+N-1} X[k] e^{+j 2 \pi n k/N} \qquad \forall k_0 \in \mathbb{Z} $$

Now, the continuous Fourier series for $x(t)$ is

$$ x(t) = \sum\limits_{k=-\infty}^{\infty} c_k \, e^{+j 2 \pi (k/N) t}, $$

and, because $x(t) \in \mathbb{R}$, we know we have conjugate symmetry

$$ c_{-k} = (c_k)^* \qquad \forall \ k \in \mathbb{Z}. $$

Being "bandlimited" means that

$$ c_k = 0 \qquad \forall \ |k| > \tfrac{N}{2}. $$

From this we know that

$$\begin{align} x(t) &= \sum\limits_{k=-\infty}^{\infty} c_k \, e^{+j 2 \pi (k/N) t} \\ \\ &= \sum\limits_{k=-\lfloor N/2 \rfloor}^{\lfloor N/2 \rfloor} c_k \, e^{+j 2 \pi (k/N) t} \\ \end{align}$$

where $\lfloor \cdot \rfloor$ is the floor() operator that essentially rounds down to the nearest integer. If $N$ is even $\lfloor \frac{N}{2} \rfloor = \frac{N}{2}$. If $N$ is odd $\lfloor \frac{N}{2} \rfloor = \frac{N-1}{2}$.

We could combine the $-k$ and $+k$ terms in the summation but we may have to subtract an extra $c_0$ because that term gets added twice with both summations.

For $N$ odd,

$$\begin{align} x(t)\bigg|_{t=n} &= \sum\limits_{k=-\infty}^{\infty} c_k \, e^{+j 2 \pi (k/N) n} \\ \\ &= \sum\limits_{k=-(N-1)/2}^{(N-1)/2} c_k \, e^{+j 2 \pi (k/N) n} \\ \\ &= \sum\limits_{k=-(N-1)/2}^{(N-1)/2} \tfrac{1}{N} X[k] \, e^{+j 2 \pi (k/N) n} \\ \\ &= \sum\limits_{k=-(N-1)/2}^{(N-1)/2} \tfrac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-j 2 \pi n k/N} \, e^{+j 2 \pi (k/N) n} \\ \\ &= \tfrac{1}{N} \sum_{n=0}^{N-1} x[n] \sum\limits_{k=-(N-1)/2}^{(N-1)/2} e^{-j 2 \pi n k/N} \, e^{+j 2 \pi (k/N) n} \\ \\ \end{align}$$

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For $N$ odd, we get the Dirichlet kernel:

$$ g(u) = \frac{\sin(\pi u)}{N \sin(\pi u/N)}. $$

But when $N$ is even, what should $g(u)$ be? Now there is potentially a non-zero component to the DFT value at what we EEs call the "Nyquist frequency"; namely $X[\tfrac{N}{2}]$ exists and might not be zero.

The expression for $g(u)$ I get when $N$ is even is

$$ g(u) = \frac{\sin(\pi u)}{N \tan(\pi u/N)}. $$

But the question is: can it be, in the case that $N$ is even, that

$$ x(t) = \sum_{n=0}^{N-1} a_n \, g(t-n) + A \sin(\pi t),$$

where $A$ can be any real and finite number?

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So my most concise question is: for $N$ even and $x[n] \in\mathbb{R}$ having period $N>0\in\mathbb{Z}$, namely

$$ x[n+N]=x[n] \qquad \forall \ n\in\mathbb{Z}, $$

is it true that

$$\sum_{n=-\infty}^{\infty} x[n] \, \operatorname{sinc}(t-n) = \sum_{n=0}^{N-1} x[n] \frac{\sin\big(\pi (t-n)\big)}{N \tan\big(\pi (t-n)/N\big)} $$

??

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Another way of looking at the question is this special case. Can anyone prove that

$$\sum_{n=-\infty}^{\infty} (-1)^n \, \frac{\sin\big(\pi(t-n) \big)}{\pi(t-n)} = \cos(\pi t) $$

??