I saw such an example in a book. Assume that the $w(t)$ is a white Gaussian noise with zero-mean and power spectrum density $N_0/2$. Now, consider the sample function:

$$n(t)=\sqrt{2\over T}\int_{0}^{T}w(t)\cos(2\pi ft)dt, 0<t<T$$

I found that the $\sqrt {\frac{2}{T}}\cos(2\pi ft),0<t<T$ is often used as one of the orthogonal basis functions. So, the $n(t)$ also can be seen as the coefficient of the basis function. It is said that the $n(t)$ is iid zero-mean Gaussian random variables with variance $\frac{N_0}{2}$. My question is about the variance. The variance of $n(t)$ is given by:

$$var[n(t)]=E[\frac{2}{T} \int_{0}^{T}\int_{0}^{T} w(t_1)\cos(2\pi ft_1)w(t_2)\cos(2\pi ft_2) dt_1dt_2]$$ $$=\frac{2}{T} \int_{0}^{T}\int_{0}^{T} E[w(t_1)w(t_2)]\cos(2\pi ft_1)\cos(2\pi ft_2) dt_1dt_2$$ $$=\frac{2}{T} \int_{0}^{T}\int_{0}^{T} R_w(t_1,t_2)\cos(2\pi ft_1)\cos(2\pi ft_2) dt_1dt_2$$

where $R_w(t_1,t_2)$ is the autocorrelation function of the white noise $w(t)$. The autocorrelation function is $\frac{N_0}{2}g(t_1-t_2)$, where $g(t)$ is the dirac delta function.

Then, the variance becomes

$$=\frac{N_0}{2} \frac{2}{T} \int_{0}^{T}\int_{0}^{T} g(t_1-t_2)\cos(2\pi ft_1)\cos(2\pi ft_2) dt_1dt_2$$ $$=\frac{N_0}{2} \frac{2}{T} \int_{0}^{T} \cos^2(2\pi ft_2)dt_2$$ $$=\frac{N_0}{2}$$

Doesn't it mean that the $n(t)$ is not independent at different times $t_1$ and $t_2$? And why it is wrong to calculate the variance by letting $t_1 = t_2$? If $t_1 = t_2$, the variance of the $n(t)$ is infinite, which is the same as the variance of the white Gaussian noise. Thanks.

  • $\begingroup$ To answer your first question: indeed, sampling $n(t)$ will produce samples that are not independent. However, for certain basis functions $n(t_0)$ and $n(t_0+T)$ will be uncorrelated for some value of $T$. $\endgroup$ – MBaz Feb 27 '19 at 16:54

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