In the question Estimating Filtered Noise Variance i've found that the variance of filtered noise can be found by

$$\min_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|^2\sigma_x^2\le \sigma_y^2\le\max_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|^2 \sigma_x^2\tag{1}$$

The derivation is based on spectral density functions and the relationship $$S_y(e^{j\theta})=S_x(e^{j\theta})\left|H(e^{j\theta})\right|^2\tag{2}$$ with $$\sigma_x^2=\frac{1}{2\pi}\int_{-\pi}^{\pi}S_x(e^{j\theta})d\theta= \frac{1}{\pi}\int_{\theta_1}^{\theta_2}S_x(e^{j\theta})d\theta\tag{3}$$

The important point for me is that the variances $\sigma_x^2$ and $\sigma_y^2$ are related by a quadratic relationship.

I'm interested in the relation of the standard deviations $\sigma_y$ and $\sigma_x$, not the variances. So the question is: How is the standard deviation $\sigma_y$ related to $\sigma_x$?

My naive idea (I'm yet not too experience with random variables in the frequency domain) is to take the square root and find $\sigma_y= |H(\omega)|\sigma_x$. So the variances should have a linear relationship.

Is this correct or am I missing some point?

  • $\begingroup$ Of course you can take the square root of inequality (1), but it still remains an inequality (i.e., lower and upper bounds); you will not get an expression as in your one but last paragraph, simply because $H$ is a function of frequency, not just a constant. $\endgroup$
    – Matt L.
    Commented Jun 8, 2016 at 13:01
  • $\begingroup$ @MattL.: Thanks. It should be $H(\theta)$. I'll correct it. $\endgroup$
    – snowflake
    Commented Jun 8, 2016 at 13:07
  • $\begingroup$ OK, but now that equation doesn't make any sense any more, because $\sigma_y$ can't be a function of frequency. $\endgroup$
    – Matt L.
    Commented Jun 8, 2016 at 13:24
  • $\begingroup$ Jup. And that's the point that confuses me. I sort of mixed up variance and spectral density... I now understand that $\sigma_x\ne\sqrt{S_x}$ $\endgroup$
    – snowflake
    Commented Jun 8, 2016 at 13:34

1 Answer 1


Thank's to MattL's comment I found the 'answer' myself... I was mixing up equations (1) and (2).

As MattL. pointed out, it makes sense to take the square root of equation (1), resulting in something like $$\min_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|\sigma_x\le \sigma_y\le\max_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right| \sigma_x\tag{1}$$ if all terms are positive.

If one takes equation (2), applies the integral and the same scaling as in (3) the result is $$\frac{1}{2\pi}\int_{-\pi}^{\pi}S_y(e^{j\theta})d\theta=\frac{1}{2\pi}\int_{-\pi}^{\pi}S_x(e^{j\theta}) \left|H(e^{j\theta})\right|^2 d\theta \tag{2}$$

The left hand side equals the variance of y, so taking the square root gives the standard deviation. However the right hand side is different, as $\left|H(e^{j\theta})\right|$ is a function of frequency.

Hence in general $\sigma_y\ne |H(e^{j\theta})|\sigma_x$.

  • $\begingroup$ @PeterK.: That could be the case, I have no idea. In that case I'd still advise chris_m to accept his own answer whenever it's possible. $\endgroup$
    – Matt L.
    Commented Jun 8, 2016 at 14:28

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