What is the meaning of the double left-right arrow in the following context? This is from a signal processing paper, Alias-Free Digital Synthesis of Classic Analog Waveforms.

For a general discrete-time signal $x(nT_s)$ which has been uniformly sampled at twice its highest frequency, exact bandlimited interpolation, which we call sinc interpolation, is carried out as $$x(t) = \sum_{n=-\infty}^\infty x(nT_s)h_s(t\Leftrightarrow nT_s)$$ $$=\frac{sin(\pi F_st)}{\pi F_s} \sum_{n=-\infty}^\infty x(nT_s)\frac{(\Leftrightarrow 1)^n}{t\Leftrightarrow nT_s}$$ where $$h_s(t) \triangleq sinc(F_st) \triangleq \frac{sin(\pi F_st)}{\pi F_st}.$$

I have read in Wikipedia's list of mathematical symjbols, and in other questions that it means if and only if. However, this does not make sense to me here.

There is also an answer suggesting "the $p\Leftrightarrow q$ symbol is just a strange font substitution for $p−q$", which I would indeed find very strange. Most of the answer is specific to a different context which I don't understand well enough to verify it.

I'm especially confused about its use in the second line: $(\Leftrightarrow 1)^n$ where there isn't even a left-hand term.

This must be a specialised usage. What is it?


If it is being used as iff, could someone walk me through how to interpret that? eg. Could I unpack the first line of the equation as

$$\sum_{n=-\infty}^\infty x(nT_s) \begin{cases} h_s(t), & t = nT_s \\ [...], & otherwise \end{cases} $$

and what would go in place of $[...]$? How to then unpack the expression $(\Leftrightarrow 1)^n$?

  • $\begingroup$ If that is the case, how do I evaluate it when the condition is not satisfied? I've added a follow-up question $\endgroup$ – Igid Dec 2 '18 at 11:24
  • $\begingroup$ It's a typo for a '$-$' sign. The expression is a convolution. $\endgroup$ – Andy Walls Dec 2 '18 at 11:45
  • $\begingroup$ @AndyWalls As in minus?? $\endgroup$ – Igid Dec 2 '18 at 11:48

It's a '$-$' sign.

$\sin(\pi F_s [t - nT_s]) =\sin(\pi F_s t)\cos(\pi F_s nT_s) - \cos(\pi F_s t)\sin(\pi F_s nT_s) = \sin(\pi F_s t)(-1)^n -\cos(\pi F_s t)(0)$

since $F_sT_s =1$ by definition.

The overall expression is a convolution.

  • $\begingroup$ Thank you, that's a very helpful expansion. But why on earth would someone print that for a minus sign? That's just pure obfuscation. $\endgroup$ – Igid Dec 2 '18 at 12:21
  • $\begingroup$ Blame computers and software and editors. That's not something a signal processing expert did. $\endgroup$ – Andy Walls Dec 2 '18 at 12:56
  • $\begingroup$ Fair enough I suppose. Now I’m really curious to know how that confusion could arise. Surely if you’re marking up in LaTeX there’s no proximity between the two. $\endgroup$ – Igid Dec 2 '18 at 17:11

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