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Consider the following system:

Simple Wien-Hammerstein System

What is the variance of $y$, $\mathbb{E}(y^2)$ ?

(EDIT: I know input signal has infinite power but will be made bandlimited by $H$. Both $H$ and $G$ are simple rational filters with lowpass characteristics. Feel free to use discrete-time versions if it's easier)

If $H(s)=G(s)=1$, it is trivially $\mathbb{E}((x^3)^2)=15\sigma^6$.

If only $G(s)=1$, I can pretend that $z$ is still white with a variance decreased by $H(s)$ and apply the formla from before. This is mathematically not sound but matches at least with numerical experiments.

Similarly, I can "hack" an expression which matches numerics when only $H(s)=1$. However, this is even less sound and when I combine them ($H(s)\neq 1, G(s)\neq 1$) it falls apart.

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  • $\begingroup$ 1) Where does the 15 come from in $\mathbb E((x^3)^2)$? 2) Have you tried applying Wiener-Khinchine to find the variance of $z$ as a function of $H(s)$? You could then apply the same idea to $G(s)$. 3) Why do you mention a "filtered polynomial" in your question? It seems like you're filtering Gaussian noise, not a polynomial (whatever that means)? $\endgroup$ – MBaz Nov 26 '16 at 14:47
  • $\begingroup$ 1.) $\mathbb{E}(x^p) = (p-1)!! \sigma^p$. 2.) Yes of course, but the problem is the nonlinearity 3.) Because the system is not linear. The output of the first filter is cubed and then fed into another filter. Technically, the distribution of $z^3$ is even indeterminite. $\endgroup$ – divB Nov 27 '16 at 8:42

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