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title: Notes on noise
tags: [Notes on Noise]

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tags: Notes on Noise
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# Notes on noise

Let $f(t)= e^{i\omega_0 t}f(t)$ the function we like to sample with a step T. After sampling we obtain $f_n = f(nT)$;  $n\in  \mathbb{N}$, with
$$
f_n = e^{i\omega_0nT}f(nT).
$$
The poisson reconstruction formula is
$$
f(t) = \sum_n f_n \text{sinc}\left(\frac{t-nT}{T}\right) = \sum_n e^{i\omega_0nT}f(nT)\text{sinc}\left(\frac{t-nT}{T}\right)
$$

Now we introduce somme uncertainty on the step $T$, $T=T+\epsilon$ where $\epsilon$ is  distributed  according to a probability distribution $P(\epsilon)$, centered on $\epsilon=0$ and with a width $\sigma$.

With this noise, the experimental sampled function $f^e_n$ would be:

$$
f^e_n(\epsilon) = e^{i\omega_0n(T+\epsilon)}f(n(T+\epsilon))
$$
and the reconstruction formula becomes
$$
f^e(t) = \sum_n e^{i\omega_0n(T+\epsilon)}f(n(T+\epsilon))\text{sinc}\left(\frac{t-nT}{T}\right)
$$
where the $\epsilon$ is not included in the argument of the $\text{sinc}$ function because the sinc function is added after the sampling which is believed to occur at points $t=nT$.
After averaging over several realisations of the data acquisition, we obtain
$$
\moy{f^e}(t) = 
\int d\epsilon \sum_n \text{sinc}\left(\frac{t-nT}{T}\right)e^{i\omega_0n(T+\epsilon)}f(n(T+\epsilon)) P(\epsilon)
$$
which can be written as

$$
\moy{f^e}(t) = \sum_n \text{sinc}\left(\frac{t-nT}{T}\right)e^{i\omega_0nT}
\int d\epsilon e^{i\omega_0n\epsilon} f(n(T+\epsilon)) P(\epsilon)
$$
Now, if we suppose that the width $\sigma$ of $P(\epsilon)$ is small enough, such that $\sigma \ll T$ and thus in the integral we have $\epsilon \ll T$ and therefore  $f((n(T+\epsilon)) \simeq f(nT)=f_n$, then we can write
$$
\moy{f^e}(t) \simeq \sum_n f_n \text{sinc}\left(\frac{t-nT}{T}\right)e^{i\omega_0nT}
\int d\epsilon e^{i\omega_0n\epsilon}  P(\epsilon)
$$
where we have negelcted the variation of function $f((n(T+\epsilon))$ on the range of integration.
The resulting integral  is the Fourier transform of $P(\epsilon)$ calculated at $n\omega_0$:
$$
\int d\epsilon e^{i\omega_0n\epsilon}  P(\epsilon)= \hat{P}(n\omega_0)
$$
The Fourier transform  $\hat{P}(\omega)$ is centered on $\omega=0$ and its width is of the order $1/\sigma$. 
We conclude that we must have $1/\sigma > n\omega_0$. If not, the integral would be zero.
Thus the error $\sigma$ must be small not only compared to $T$ but also compared to $1/(n\omega_0)$.

Here I have considered that we have the same error $\epsilon$ for all the points $nT$, and we average over different sequences of acquisition.
The probabilistic model must be generalized such that to consider error fluctuation for each $n$, but I think that it would give the same result that $\sigma$  must be small compared to $1/\omega_0$. 






