Nonuniform sampling

For a given function, it is possible to construct a polynomial of degree n which has the same value with the function at n + 1 points.[5]

Let the n + 1 points to be ${\displaystyle z_{0},z_{1},\ldots ,z_{n}}$, and the n + 1 values to be ${\displaystyle w_{0},w_{1},\ldots ,w_{n}}$.

In this way, there exists a unique polynomial ${\displaystyle p_{n}(z)}$ such that

${\displaystyle p_{n}(z_{i})=w_{i},{\text{ where }}i=0,1,\ldots ,n.}$[6]

Furthermore, it is possible to simplify the representation of ${\displaystyle p_{n}(z)}$ using the interpolating polynomials of Lagrange interpolation:

${\displaystyle I_{k}(z)={\frac {(z-z_{0})(z-z_{1})\cdots (z-z_{k-1})(z-z_{k+1})\cdots (z-z_{n})}{(z_{k}-z_{0})(z_{k}-z_{1})\cdots (z_{k}-z_{k-1})(z_{k}-z_{k+1})\cdots (z_{k}-z_{n})}}}$[7]

From the above equation:

${\displaystyle I_{k}(z_{j})=\delta _{k,j}={\begin{cases}0,&{\text{if }}k\neq j\\1,&{\text{if }}k=j\end{cases}}}$

As a result,

${\displaystyle p_{n}(z)=\sum _{k=0}^{n}w_{k}I_{k}(z)}$

${\displaystyle p_{n}(z_{j})=w_{j},j=0,1,\ldots ,n}$

To make the polynomial form more useful:

${\displaystyle G_{n}(z)=(z-z_{0})(z-z_{1})\cdots (z-z_{n})}$

In that way, the Lagrange Interpolation Formula appears:

${\displaystyle p_{n}(z)=\sum _{k=0}^{n}w_{k}{\frac {G_{n}(z)}{(z-z_{k})G'_{n}(z_{k})}}}$[8]

Note that if ${\displaystyle f(z_{j})=p_{n}(z_{j}),j=0,1,\ldots ,n,}$, then the above formula becomes:

${\displaystyle f(z)=\sum _{k=0}^{n}f(z_{k}){\frac {G_{n}(z)}{(z-z_{k})G'_{n}(z_{k})}}}$

Whittaker tried to extend the Lagrange Interpolation from polynomials to entire functions. He showed that it is possible to construct the entire function[9]

${\displaystyle C_{f}(z)=\sum _{n=-\infty }^{\infty }f(a+nW){\frac {\sin[\pi (z-a-nW)/W]}{[\pi (z-a-nW)/W]}}}$

which has the same value with ${\displaystyle f(z)}$ at the points ${\displaystyle z_{n}=a+nW}$

Moreover, ${\displaystyle C_{f}(z)}$ can be written in a similar form of the last equation in previous section:

${\displaystyle C_{f}(z)=\sum _{n=-\infty }^{\infty }f(z_{n}){\frac {G(z)}{G'(z_{n})(z-z_{n})}},{\text{ where }}G(z)=\sin[\pi (z-z_{n})/W]{\text{ and }}z_{n}=a+nW}$

When a = 0 and W = 1, then the above equation becomes almost the same as WSK theorem:[10]

If a function f can be represented in the form

${\displaystyle f(t)=\int _{-\sigma }^{\sigma }e^{jxt}g(x)\,dx\qquad (t\in \mathbb {R} ),\qquad \forall g\in L^{2}(-\sigma ,\sigma ),}$

then f can be reconstructed from its samples as following:

${\displaystyle f(t)=\sum _{k=-\infty }^{\infty }f\left({\frac {k\pi }{\sigma }}\right){\frac {\sin(\sigma t-k\pi )}{\sigma t-k\pi }}\qquad (t\in \mathbb {R} )}$

For a sequence ${\displaystyle \{t_{k}\}_{k\in \mathbb {Z} }}$ satisfying[11]

${\displaystyle D=\sup _{k\in \mathbb {Z} }|t_{k}-k|<{\frac {1}{4}},}$

then

${\displaystyle f(t)=\sum _{k=-\infty }^{\infty }f(t_{k}){\frac {G(t)}{G'(t_{k})(t-t_{k})}},\qquad \forall {}f\in B_{\pi }^{2},\qquad (t\in \mathbb {R} ),}$

where

• ${\displaystyle \textstyle G(t)=(t-t_{0})\prod _{k=1}^{\infty }\left(1-{\frac {t}{t_{k}}}\right)\left(1-{\frac {t}{t_{-k}}}\right),}$
• ${\displaystyle B_{\sigma }^{2}}$ is Bernstein space, and
• ${\displaystyle f(t)}$ is uniformly convergent on compact sets.[12]

The above is called the Paley–Wiener–Levinson theorem, which generalize WSK sampling theorem from uniform samples to non uniform samples. Both of them can reconstruct a band-limited signal from those samples, respectively.

• F. Marvasti, Nonuniform sampling: Theory and Practice. Plenum Publishers Co., 2001, pp. 123–140.