# 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.