Inverse function of the Nyquist-Shannon sampling theorem

[d2x]

I'm currently carrying out an analysis on waveforms produced by a particular particle detector. The Nyquist-Shannon sampling theorem has been very useful for making an interpolation over the original sample points obtained from the oscilloscope. The theorem (for a finite set of samples) is given by:

$$x(t) = \sum_{n=1}^{N} x(nT)sinc\bigg(\frac{t-nT}{T}\bigg)$$

It seems that it would also be very useful to have a function that can do the opposite operation (i.e. you give it a value of x and get t):

$$t(x)$$

I'm wondering whether or not this can be easily achieved. I'm not sure where to start due to the sum on the RHS. Any help or tips on how to start would be greatly appreciated.

 

[jim mcnamara]

Hmm. I'm not sure about the applicability of "reversing interpolation" and definitely have no immediate solution to the math problem. But. There may be some help available from @StoneTemplePython or @Stephen Tashi

 

[marcusl]

Any signal of interest is likely to be multi-valued as a function of time (think about a sine wave, for example, which takes on the same value at many different times). Since these functions are not one to one across the two domains, your expression cannot be inverted uniquely. A waveform that is not multivalued in time, on the other hand, must be constant or monotonically increasing or decreasing, and is therefore of little interest to real detection problems.

As an aside, note that the sinc function used for interpolation has an infinite extent. (I think that Shannon considered the problem of exactly reconstructing an infinite waveform.) As a result, its use to reconstruct any finite length sampled series is only approximate. The approximation is worst for short sequences (a small number of samples), and at the beginning and end of any finite sequence.

BTW, Nyquist did examine the band-limited frequency content of finite sequences, but did not work on sampling or reconstruction. The sampling theorem is therefore Shannon’s alone. (If other names must be included, they would be Whittaker and Kotelnikov.)