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The specific chapter about this is here - Sampling Theory
Unlike what I read anywhere else, it defines Shah as: $$Ш_T(x)=T\sum\nolimits_i{\delta{(x-Ti)}}$$ And the T is still present in the reconstructed function: $$f\tilde(x)=T\sum\limits_{i=-\infty}^\infty{f(iT)r(x-iT)}$$ where r(x) is a reconstruction filter.
Everywhere else I can find gives: $$s_T{(x)}=\sum\nolimits_i{\delta{(x-Ti)}}$$ Can someone work out the maths for me? Why is the difference?

I haven't been doing maths for years, and even back in my school days, I don't think I was a good student, so please be gentle.

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  • $\begingroup$ Why is the difference? - they're just using a different definition with a scaling factor, it shouldn't really matter. $\endgroup$
    – lightxbulb
    Oct 26 '20 at 17:10
  • $\begingroup$ @lightxbulb Thanks for your reply! I thought so at first, but the triangle filter r(x) = max(0, 1 - |x|) that they give as an example in 7.1.2, doesn't seem to unscale it (while the box function given later does). Maybe it's simply too trivial that they overlooked? Still what good does it do to have the T factor and then 1/T it? $\endgroup$
    – Eugene
    Oct 27 '20 at 1:42
  • $\begingroup$ It doesn't change anything if you have $T$ and then unscale it. It could be that they wanted a specific relationship to the Fourier transform. Similar to how one would choose different scaling factors for the DFT, e.g. to make it unitary. $\endgroup$
    – lightxbulb
    Oct 27 '20 at 7:17
  • $\begingroup$ @lightxbulb Thanks. I'm convinced. $\endgroup$
    – Eugene
    Oct 28 '20 at 11:34

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