Signal reconstruction and interpolation using cubic B-splines I am new to reconstruction and interpolation. From what I understand, interpolation using cubic B-splines can be viewed in two ways: 1) through construction of a linear system of equations whose solution uniquely determines the interpolating cubic spine, and 2) in a filter/convolution type framework. My question is of the latter.
When viewing the problem as filter/convolution, the reconstruction for a uniformly sampled data set is given by something like:
\begin{align}
\sum_i s(i)*h(i) \beta_3 (x - i)
\end{align}
Where $s(i)$ is the sampled signal, h(i) is an interpolation filter, and $\beta_3$ is a third order B-spline (or cubic spine obtained by 3 convolutions of the box function). The reason this filter is needed is to remove the high frequency "copies" of the signal in the frequency domain by: S($\omega)$H($\omega$). My question is the following:
Is this filter H($\omega$) uniquely determined based on the fact that I am using cubic B-splines? Or can I just take it to be the box function?
\begin{align}
H(\omega) = 
\left\{
\begin{array}{c}
1, |\omega| < \omega_{max}  \\
0, |\omega| \geq \omega_{max}\\
\end{array}
\right.
\end{align}
I came across one possible explanation here:
http://shulgadim.blogspot.ca/2014/01/1-d-b-spline-interpolation-via-digital.html
but am unsure if it is exactly related to my problem since in their sum, they only have $c(l)$, where as I have $s(i)*h(i)$.
 A: A simpler explanation of the filter $h$ is that it is required to get an interpolation of the discrete signal $s$. (Note that the support of $\beta_3$ is the interval $[-2, 2]$ so the supports overlap when shifted over integers.)
Another way to read this interpolation formula is $$\sum_{k\in \mathbb{Z}} s(k) \cdot (h \ast \beta_3)(x-k)$$ where $h \ast \beta_3$ is a piecewise cubic function with infinite support that vanishes at all integers except at $0$ where it has value $1$. In this form it resembles the Shannon interpolation formula $$\sum_{k\in \mathbb{Z}} s(k) \cdot \operatorname{sinc}(x-k)$$ where $$\operatorname{sinc}(x) = \frac{\sin(\pi x)}{\pi x}.$$ The box filter is the Fourier transform of $\operatorname{sinc}$. However, the Fourier transform of $h \ast \beta_3$ is not a box filter (since $h \ast \beta_3 \neq \operatorname{sinc}$) but it is in fact $$ \frac{3 \operatorname{sinc}^4(x)}{\cos(2\pi x) + 2}.$$ Plotting this function reveals that it approximates a box function already pretty closely, which is why cubic splines are so popular for interpolation. See the excellent paper by Michael Unser for more background.
