I have an input signal x (assumed to be iid Gaussian with $\mu=0$, $\sigma^2$) which is fed into two linear systems:

  • $y_1 = h_1 * x$
  • $y_2 = h_2 * x$

Now I would like to calculate $\mathbb{E}[y_1 y_2]$. What is the proper way to do this - preferably in the frequency domain?

Background: I want to calculate something like $\operatorname{var}(y_1 + y_2)$. This expands to $\operatorname{var}(y_1)+\operatorname{var}(y_2)+2\mathbb{E}[y_1 y_2]$. Now I am very familar with the first two components - the variance. I know that the variance of a stochastic signal is given by the autocorrelation sequence at position 0. With the Wiener-Kinchin theorem this translates to:

$$ \operatorname{var}(y_1) = r_{yy}(0) = \int_0^{\infty} \Phi_{yy}(f) df $$

and then:

$$ \cdots = \sigma_x^2 \int_0^{\infty} |H_1(f)|^2 df $$

Now for my problem - with $\mathbb{E}[y_1 y_2]$ I arrive at $\mathbb{E}[x^2 \cdots]$ (giving a variance) but the filter is not square magnitude.


You can rewrite the convolution operator as a matrix operation by building a Toeplitz matrix for the impulse response function. Your equations can be rewritten as $Y_1 = H_1 X$ and $Y_2 = H_2 X$, where $Y_1 = [y^1_1,\ldots,y^n_1]^T$, $Y_2 = [y^1_2,\ldots,y^n_2]^T$, $X = [x_1,\ldots,x_m]^T$, $m$ is the length of the input vector, and $n$ is the length of the convolution product which depends on length of the input vector and the length of the impulse response function.

Now, $Y_1 \sim \mathcal{N}(0, \sigma^2 H_1 H_1^T)$, $Y_2 \sim \mathcal{N}(0, \sigma^2 H_2 H_2^T)$, and $X \sim \mathcal{N}(0, \sigma^2 I)$.


\begin{align} \mathbb{E}(Y_1 Y_2^T) &= \mathbb{E}(H_1 X X^T H_2^T) \\ &= H_1 \mathbb{E}(X X^T) H_2^T \\ &= \sigma^2 H_1 H_2^T \end{align}

  • $\begingroup$ I was thinking about an approach such as that but couldn't figure out how to get $H$. Now I know. Thanks. $\endgroup$ – mark leeds Dec 19 '18 at 4:56

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.