Inter-point interpolation using FIR filter Suppose I have discrete noisy signal $X = (0.096, -0.0632, 0.351, 0.531, 0.360, 0.006, -0.320)$ sampled at discrete time points $T = (1, 2, 3, 4, 5, 6, 7)$. Filtering (zero-padded) $X$ with symmetrical FIR $G = (\frac{1}{4}, \frac{1}{2}, \frac{1}{4})$ gives me a good approximation of underlying $f(t)$ but only at discrete time points $T$. Is it possible to use filter coefficients to approximate the values of $f(t)$ between discrete values, e.g. estimate $f(\sqrt 2)$?

In Interpolation with an FIR filter user Hilmar mentions Whittaker–Shannon interpolation formula, but it is sensitive to noise. Is it possible to construct smooth but more noise-robust interpolation by using FIR coefficients somehow?

Or, separate interpolation step (e.g. spline or polynomial interpolation) after FIR application is the only practical way to go?
From Hilmar's answer to the linked question, the Whittaker–Shannon interpolation formula is:


  $$x(t)=\sum_{n=-\infty }^{\infty }x[n]\cdot \operatorname{sinc}\left(\frac{t-nT}{T}\right).$$


If the signal is low-pass and the noise is high-pass, to filter high-pass noise, we can make a low-pass version of the interpolation formula:

$$x(t)=\sum_{n=-\infty }^{\infty }x[n]\cdot 2f_c/f_s\cdot \operatorname{sinc}\left(\frac{t-2f_c/f_s\cdot nT}{T}\right),$$

where $f_c$ is the cutoff frequency and $f_s$ is the sampling frequency. The highest signal-to-noise ratio is obtained by setting the cutoff frequency to a frequency for which the signal and the noise have equal power. Actually the Whittaker–Shannon interpolation formula is low-pass as well, with $f_c=f_s/2$.