Absolute convergence of periodic sinc interpolation

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?

• That's an interesting question and one which has no straightforward answer indeed. Given its mathematical nature, I wonder if it wouldn't be a better fit to Math.Stackexchange? But let's see if some of the more mathematically inclined users here can help out. – Florian Jul 4 at 8:16
• isn't this about the same sorta topic as this or this? i call this "bandlimited reconstruction of periodic discrete-time sampled functions" and the formula is slightly different for even $N$ vs. odd $N$. – robert bristow-johnson Jul 5 at 19:06
• @robertbristow-johnson yes (thanks for the links), and this. I was not fully convinced about the justification of Nyquist bin halving so started to look at the same thing in time domain, but so far things don't seem any more conclusive. – Olli Niemitalo Jul 5 at 20:25
• however you divide up the Nyquist bin, it must add to the original $X[\frac{N}2]$. if $\tilde{x}[n]$ is real, then Hermitian symmetry applies which means that $X[\frac{N}2]$ must be purely real (as well as $X$). – robert bristow-johnson Jul 5 at 20:42
• @robertbristow-johnson This is what I have so far: You can add the positive and negative nyquist bins in any linear combination that adds up to one and still be just as valid. The only justification for splitting them at 1/2 is that the resulting series will produce real only results for real only sequences. I'm not feeling that strong about it. See my answer and comments here as well dsp.stackexchange.com/questions/59265/…. – Cedron Dawg Jul 5 at 20:45

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.

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 $$Nm/N$$ 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 ...

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.

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.

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 signifinant 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.

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.

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.

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.

// //"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 ($$=n$$).

• about $$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$ i remember this first proved to me with a geometric argument. remember $x$ is radians and radians is arc length. – robert bristow-johnson Jul 6 at 2:17
• @robertbristow-johnson Yeah, it's my favorite example to use to teach limits, especially to those who "aren't getting it". You can start with a drawing, then ask "Does this sentence make sense to you?" Then say: "The limit as the number of sides approaches infinity of the perimeter of the inscribed polygon is the circumference of the circle." Then write: $$\lim_{n \to \infty } P_n = C$$ Then you take it from there and end up at the $sin(\theta)/ \theta$ limit – Cedron Dawg Jul 6 at 2:36
• Formally, you should also do the circumscribed polygon formula and sandwich the circumference between them. Then the guy from Real Analysis comes along and says, "Hey, you only proved it on the reciprocals of the integers." – Cedron Dawg Jul 6 at 2:40
• I don't think i ever said "non-bandlimited", Ced. i have always meant "bandlimited" (as well as real and periodic). if it were mathematically elegant, i would find it convenient that "bandlimited" mean "absolutely no energy at Nyquist or higher", but in general, the real and discrete-time periodic sequence $x[n]$ where $$x[n+N] = x[n] \qquad \forall n\in\mathbb{Z}$$ will not guarantee that there is no energy in the Nyquist bin (or in the DC bin). but if $x[n]$ is real, there is nothing imaginary in either of those DFT bins. – robert bristow-johnson Jul 7 at 3:21
• @robertbristow-johnson I have reversed your "omega" for "w" edits. "w" is a stand in for "t" on a different scale. "w" goes from 0 to N, and "t" (used earlier in the answer) goes from 0 to \$\2pi" for each cycle. I didn't want to reuse "t" so I choose "w" (rather than "s") because it is an upside down "m" which it is based on. "omega" is usually used as a frequency variable and that make it more confusing. – Cedron Dawg Jul 15 at 12:33

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 have to subtract $$c_0$$ 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}

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?

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

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

??

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

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

??

• Even N case: $$g( t - n )= \frac{ \sin \left( \left( t - n \right) \pi \right) } { N \sin \left( \frac{1}{N} \left( t - n \right) \pi \right) }$$ Odd N case: $$g( t - n )= \frac{ \sin \left( \left( t - n \right) \pi \right) } { N \tan \left( \frac{1}{N} \left( t - n \right) \pi \right) }$$ See my answer for the derivations and more fun observations. – Cedron Dawg Jul 11 at 5:33
• But this is a more informative arrangement in the even case: $$g( t - n )= \frac{ \sin \left( \left( t - n \right) \pi \right) } { N \sin \left( \frac{1}{N} \left( t - n \right) \pi \right) } \cos \left( \frac{1}{N} \left( t - n \right) \pi \right)$$ Makes it look like a "window function" of the odd case. – Cedron Dawg Jul 11 at 5:37
• Oop, reverse the even and odd in the first comment. Sigh. – Cedron Dawg Jul 11 at 5:47
• we agree. that thing about factoring out the cosine is, of course true. i was aware of that here. dunno exactly how it's useful. well, anyway, i haven't finished this answer (and it will be rearranged, i just wanted to copy what i wrote in the math.se and change it to EE nomenclature. more to come. – robert bristow-johnson Jul 11 at 20:44
• Did you know that when two tones that are close in frequency are added together there is a beat phenomenon? The sum appears to be a tone at the average of the two source frequencies attenuated by a cosine function envelope. The proof is easy, comes straight from the angle addition formulas. – Cedron Dawg Jul 12 at 3:20

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.

• So, I wasn't quite formal enough. In your example, for the odd case, it doesn't matter, as the condition of an alternating monotonically decreasing series holds independently on "both arms". For the even case, it's a little trickier as the series has to be rearranged by "folding it" to achieve the alternating monotonically decreasing series. In either case, I think the point is moot because the series is known to converge to a closed form expression as the result of taking the limit of the discrete case. Therefore you know it converges, and you know it doesn't converge absolutely. – Cedron Dawg Jul 13 at 16:33