Global interpolation or sinc interpolation is an ideal filter since its frequency response is a rect function. The impulse response of this filter is the sinc function (same as the coefficients of the interpolator).

Local interpolation methods are constructed by applying some window to the sinc function. For example, according to https://ccrma.stanford.edu/~jos/Interpolation/Relation_Lagrange_Interpolation_Windowed.html, for uniformly spaced samples and finite $N$ , Lagrange interpolaton is equivalent to windowed sinc interpolation using a binomial window. The impulse response of this filter are the coefficients of the interpolator.

I am interested to know what is the relationship between the impulse response, accuracy of the interpolation and frequency response of the filter/local interpolator, especifically for Lagrange interpolation. I cannot see this clearly.


any interpolation polynomial that goes through the two adjacent samples to the left and right of the area being interpolated will result in an effective impulse response that is a piecewise-polynomial function going through zero at all integers except zero (like the sinc function does).

e.g. linear interpolation:

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third-order lagrange:

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third-order hermite:

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note all of these effective impulse responses go through the $(x,y)$ points of $(0,1)$ and $(n,0)$ for any other integer $n \ne 0$, just like the sinc function does.

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so you can represent that interpolation function as something times a sinc function and the "something" can be obtained by dividing that interpolation function, point-by-point, with the sinc function. the only division by zero is when the numerator is also zero and that can be, what we like to call in the hand-waving department, a "removeable singularity". (maybe we can find a decent limit at the $\tfrac00$ divisions.)

any interpolation that forces the smooth interpolated function to go through the discrete points can always be described as a windowed-sinc function. all you have to do is divide to get the effective window.

  • $\begingroup$ If I compare the frequency response of any local interpolator, something times a sinc function, with the frequency response of the sinc function, can i say that the more that the frequency response of my interpolator looks like the ideal filter, the more accurate is the interpolator is? Does this also means that if I want to have an accurate interpolator (up to an acceptable error) with a narrow impulse response, I need to find the appropriate window that makes the lobes of sinc decay exponentially such that the frequency response of interpolator look as close as the rect function? $\endgroup$ May 4 '18 at 10:47
  • $\begingroup$ it's not any interpolators. like an interpolator you might get from the Parks-McClellan algorithm would not likely be representable as a windowed sinc, because the effective impulse response would not go through zero at those integer values of $x$. $\endgroup$ May 4 '18 at 10:49
  • $\begingroup$ it all depends on how you wanna measure error. sometimes a windowed sinc function is better, sometimes what you get from firpm() or firls() is better. $\endgroup$ May 4 '18 at 10:51
  • $\begingroup$ Thanks for pointing this out. I am new in this topic and i was not aware of such algorithms. What i understand is that in filter design we want to have an efficient approximation of the ideal filter. From the point of view of interpolation, this to me looks that we are trying to get the optimal interpolation. Thus, if I want to have an accurate interpolator (with the error computed pointwise), what I need to do is to find a filter which frequency response is as close as possible to the ideal low pass filter, no matter which method is used (windowed sinc, firpm of firls). $\endgroup$ May 4 '18 at 12:25

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