Explanation relevant to the filter response ($\mathrm{sinc}$ interpolation) - Equation $(2)$ in my question

Question

At present, I am learning the theory of operation of resampling for bandlimited periodic discrete signals using $\mathrm{sinc}$ interpolation. I am developing a design flow and having difficulty in understanding a particular expression/equation, hence I am stuck! Here I go:

Suppose we have a samples $x(nT_s)$ of a continuous time signal $x(t)$, where $t$ is time in seconds and $n$ is sample points that ranges over integers and ofcourse $T_s$ is the sampling period. $x(t)$ is bandlimited to $\pm F_s/2$, where $F_s = 1/T_s$.

Now, we know that the original signal $x(t)$ can be reconstructed or synthesized by the samples of $x(nT_s)$ using the equation:

$$ x'(t) = x(nT_s)+H_s(t-nT_s) = x(t), \quad \text{for}\quad n =\ldots,-3,-2,-1,0,1,2,3,\ldots\tag1 $$

and

$$ H_s(t) = \mathrm{sinc}(\pi F_s t) $$

To resample $x(t)$ for the new rate i.e. $F'_s = 1/T'_s$, we have to evaluate equation $(1)$ at integer multiples of $T'_s$ i.e. $nT'_s$.

We know that, when the new sampling rate $F'_s$ is less than the original sampling rate or frequency $F_s$, the lowpass cutoff must be placed below half the new lower sampling rate.

Doubt - Based on the highlighted statement made above, the ideal low pass is given by:

$$ H_s(t) = \min\left\{1,F'_s/F_s\right\}\mathrm{sinc}\left(\pi\min\left\{F'_s,F_s\right\}t\right) \tag2 $$

where the scale factor maintains unity gain in the passband.

What does the equation $(2)$ represent? I understand that, if $F'_s$ is lower than $F_s$, then the filters cut off frequency should be half the $F'_s$, however, I am having trouble understanding this equation $(2)$ and how to benefit from it in the design process. Any explanation regarding this would be appreciated.

Source/Reference:

$\scriptstyle{\textrm{Julius O.Smith & Phil Gosett "A Flexible Sampling Rate Conversion Method" IEEE, 1984}}$.

Answer 1

A Sinc function is both a reconstruction formula for a band-limited waveform and the time domain representation (impulse response) of an ideal "brick-wall" low-pass filter. Since it does both, if you "fatten" the Sinc and then use it as an interpolation kernel, you get the reconstruction of the result of low-pass filtering the original waveform.

Answer 2

Eq$(1)$ doesn't look right. It should be:

$$x'(t) = \sum_{n=-\infty}^{+\infty} x(nT_s) \mathrm{sinc}(t-nT_s)$$

For your doubt, consider subsampling (lowering $F_s$) as equivalent to reconstructing $x(t)$ and then sampling it at a lower frequency.

Eq$(2)$ comes from the combination of $2$ lowpass filters:

1. The ideal reconstruction filter (from eq.$1$)
2. a new anti-alias filter previos to sampling at the lower $F_s$

These $2$ filters combined should result in eq$(2)$ (several lowpass filters in cascade is equivalent to one filter with the minimum bandwidth, and the product of the gains).