Nyquist theorem
Nyquist theorem refers to representation of a function, continuous at least along the real axis, through the linear combination of equidistantly–displaced functions sinc:
$\displaystyle \mathrm{sinc}(z)=\frac{\sin(z)}{z} $
at zero sinc is defined to be unity; then sinc appears to be entire function.
For the Nyquist theorem it is convenient to scale the argument of the sinc function, working with function sink such that
$\displaystyle \mathrm{sink}(z)=\mathrm{sink}(\pi z)=\frac{\sin(\pi z)}{\pi z}$
The Nyquist theorem is important tool of the digital signal processing. Unfortunately, in the literature it is presented in a complicated way; it seems, the editors, who had edited the texts about Nyquist theorem, had neither proved it nor used it. For this reason, this theorem is loaded to TORI
Statement
Let function $f$ be defined at least for real values of the argument, and let it have restricted spectrum, id est, at least for real $x$, it can be represented as follows:
$\displaystyle f(x)=\int_{-\infty}^{\infty} \mathrm d k ~ F_k~ \theta(1\!-\!|2k|)~\exp(2 \pi \mathrm i k x)$
where $\theta$ is function, that is zero for negative argument and unity for positive argument.
Let
$\displaystyle u(x)=\sum_m ~f(m\!+\!\varphi) ~\mathrm{sink}(x\!-\!m\!-\!\varphi)$
be the approximation of function $f$ based on its values at the real equidistant points, diplaced with real constant $\varphi$ from the integer values. Summation here is performed on the set of integer values of $m$.
Then, at least for the real values of the argument, $u$ does not depend on $\varphi$ and, for real argument, $u=f$.
Proof
The proof is based on the inigueness of the Fourier representaion. It refers to the Fourier transforms of functions $f$ and $u$:
$\displaystyle F_k=\int_{-\infty}^{\infty} \exp(2\pi \mathrm i kx) ~f(x)~\mathrm d x $
$\displaystyle U_k=\int_{-\infty}^{\infty} \exp(2\pi \mathrm i kx) ~u(x)~\mathrm d x $
The last expression can be transformed in the following way:
$\displaystyle U_k=\int_{-\infty}^{\infty} \exp(2\pi \mathrm i kx) ~\sum_{m} ~f(m\!+\!\varphi)~\mathrm{sink}(x)~\mathrm d x$
Now I substitute representation of $f$ through its Fourier transform $F$, this gives
$ \displaystyle U_k=\int_{-\infty}^{\infty} \exp(2\pi \mathrm i kx) ~\sum_{m} ~ \int_{-\infty}^{\infty}~\mathrm d p F_p \theta(1-|2p|) \exp(-2\pi\mathrm i p (m\!+\!\varphi)) ~\mathrm{sink}(x)~\mathrm d x$
$=~ \sum_{m} ~f(m\!+\!\varphi) \int_{-\infty}^{\infty} \exp(2\pi \mathrm i kx) ~\mathrm{sink}(x\!-\!m\!-\!\varphi)~\mathrm d x $
In the last integral, replace $x$ to $~x\!+\!m\!+\!\varphi ~$; this gives
$\displaystyle U_k= \sum_{m} ~f(m\!+\!\varphi) ~\int_{-\infty}^{\infty} \exp(-2\pi \mathrm i k(x\!+\!m\!+\!\varphi)) ~\mathrm{sink}(x)~\mathrm d x$
$\displaystyle U_k= \sum_{m} ~f(m\!+\!\varphi) \exp(-2\pi \mathrm i k(\!m\!+\!\varphi)) ~\int_{-\infty}^{\infty} \exp(-2\pi \mathrm i k x) ~\mathrm{sink}(x)~\mathrm d x$
Due to the properties of function sink, the last integral gives $\theta(1-|2k|)$; so, the expression can be rewritten as
$\displaystyle U_k= \sum_{m} ~f(m\!+\!\varphi) ~ \exp(-2\pi \mathrm i k(\!m\!+\!\varphi)) ~ \theta(1-|2k|) $
Now, substitute in the expression the representaiton of $f$ through its courier-transform $F$; this gives
$\displaystyle U_k=\sum_{m} ~\int_{-\infty}^{\infty} \mathrm d p~F_p ~\theta(1-|2p|)~\exp(2\pi \mathrm i p(\!m\!+\!\varphi)) ~\exp(-2\pi \mathrm i k(\!m\!+\!\varphi)) ~ \theta(1-|2k|)$
$\displaystyle U_k=\int_{-\infty}^{\infty} \mathrm d p~F_p ~\theta(1-|2p|)~ \theta(1-|2k|)~\sum_{m} ~\exp(2\pi \mathrm i (p\!-\!k)(\!m\!+\!\varphi)) $
The last sum gives $\delta(p\!-\!k)$. The integration gives the identity
$U_k=F_k$
This leads to the equality of their inverse Fourier transforms, $u=f$.
(End of proof)
In such a way, the set of equidistant samples allows to recover the function, and the recovered function does not depend on the constant displacement $\mathrm \phi$ of the equidistant points of the evaluation.
Application
Function $f$ with restricted spectrum (that satisfies the conditions of the Nyquist theorem) can be prepared from any bounded function $g$, even for that defied as set of delta–function with displaced argument, by the appropriate filtration of the Fourier–transform:
$\displaystyle G_k=\int_{-\infty}^{\infty} \exp(2\pi \mathrm i kx) ~g(x)~\mathrm d x $
$ F_k=(1-\theta(2k)) G(k) $
Then we may reconstruct $f$ from $F$ by the inverse Fourier transform operation. No preliminary smoothing of function $g$ is necessary for such a filtration.
The coordinate $x$ in the formulas above may have sense of time. Practically, the oscilloscope cannot provide the unlimited set of data; some moment the registration is swithced on and some moment the registration is switched out. Often, the number of data does not exceed a million. However, one may assume, that the registration begins well before the interesting process begins, and ends well after the interesting process stops. The "missed" data may be set to zero; this allows the numerical approximations of the formulas above with finite number of operations. Finiteness of the number of operation is important quality of every computational algorithm.
References
http://redwood.berkeley.edu/bruno/npb261/aliasing.pdf Bruno A. Olshausen. Aliasing. PSC 129 Sensory processes. October 10, 2000.
http://en.wikipedia.org/wiki/Nyquist–Shannon_sampling_theorem
http://whatis.techtarget.com/definition/Nyquist-Theorem
