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Assuming I have samples of a colored noise. I know the empirical Variance of those samples. I also know the bandwidth of the filter which colorized the noise (Only the bandwidth, not the actual filter).

Now I want to estimate the variance of the noise if I filter those samples by a (Different) given filter.

Is there a way to estimate or bound from above and below the variance of the noise?

Thank You.

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I assume that you have a wide-sense stationary discrete-time input noise signal $x(n)$, a linear time-invariant filter with impulse response $h(n)$, and an output noise process $y(n)$. If we assume that $x(n)$ is zero mean (which is not necessary, but makes it easier) and if we model $x(n)$ as perfectly band-limited, we get for its variance

$$\sigma_x^2=\frac{1}{2\pi}\int_{-\pi}^{\pi}S_x(e^{j\theta})d\theta= \frac{1}{\pi}\int_{\theta_1}^{\theta_2}S_x(e^{j\theta})d\theta\tag{1}$$

where $S_x(e^{j\theta})$ is the power spectral density of $x(n)$, $\theta$ is the normalized frequency $\theta=2\pi f/f_s$, and $\theta_1$ and $\theta_2$ are the lower and upper band-edges, respectively. With the frequency response of the filter

$$H(e^{j\theta})=\sum_{n=-\infty}^{\infty}h(n)e^{-jn\theta}$$

we can write the power spectral density of the output noise as

$$S_y(e^{j\theta})=S_x(e^{j\theta})\left|H(e^{j\theta})\right|^2$$

and its variance is

$$\sigma_y^2=\frac{1}{2\pi}\int_{-\pi}^{\pi}S_y(e^{j\theta})d\theta= \frac{1}{2\pi}\int_{-\pi}^{\pi}S_x(e^{j\theta})\left|H(e^{j\theta})\right|^2d\theta= \frac{1}{\pi}\int_{\theta_1}^{\theta_2}S_x(e^{j\theta})\left|H(e^{j\theta})\right|^2d\theta$$

From this equation we can get a lower and upper bound for $\sigma_y^2$ as follows

$$\min_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|^2 \frac{1}{\pi}\int_{\theta_1}^{\theta_2}S_x(e^{j\theta})d\theta\le\sigma_y^2\le \max_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|^2 \frac{1}{\pi}\int_{\theta_1}^{\theta_2}S_x(e^{j\theta})d\theta$$

and from (1) we finally get

$$\min_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|^2\sigma_x^2\le \sigma_y^2\le\max_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|^2 \sigma_x^2\tag{2}$$

These bounds can be useful (i.e. relatively tight) if the noise bandwidth is small compared to the bandwidth of the filter $h(n)$. Unfortunately, they are almost useless if the filter bandwidth is small compared to the bandwidth of the input noise process $x(n)$.

In the latter case, i.e. if the filter bandwidth is much smaller than the noise bandwidth, you can come up with similar bounds as in (2). You just need to estimate the minimum and maximum of the power spectral density $S_x(e^{j\theta})$ within the filter bandwidth.

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  • $\begingroup$ Thank You. You haven't used any information about the bandwidth of the given filter (H). Could it mean a tighter boundaries could be achieved? $\endgroup$ – Royi Jun 21 '13 at 9:41
  • $\begingroup$ The band-edges $\theta_1$ and $\theta_2$ refer to the bandwidth of the colored noise, i.e. of the filter which shaped the noise spectrum. I guess this is the filter that you are talking about. So yes, I did make use of this information. $\endgroup$ – Matt L. Jun 21 '13 at 12:42
  • $\begingroup$ I saw you used the limited bandwidth property of the noise. I asked whether the bandwidth of the filter H could be used as well. Though it is there is some manner by asking for its "Max / Min". $\endgroup$ – Royi Jun 21 '13 at 13:22
  • $\begingroup$ Well, it is only the smaller bandwidth of the two which plays a significant role, assuming that the two bandwidths are not very similar. It is easy to give useful bounds for the two cases where one of the two bandwidths is significantly smaller than the other one. $\endgroup$ – Matt L. Jun 21 '13 at 13:25

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