1
$\begingroup$

In an other post 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 devation $\sigma_y$ 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?

Improve this question
$\endgroup$
4
  • $\begingroup$ 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. Jun 8 '16 at 13:01
  • $\begingroup$ @MattL.: Thanks. It should be $H(\theta)$. I'll correct it. $\endgroup$ – snowflake Jun 8 '16 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. Jun 8 '16 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 Jun 8 '16 at 13:34
2
$\begingroup$

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

Improve this answer
$\endgroup$
1
  • $\begingroup$ @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. Jun 8 '16 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.