Moving from deterministic signals to stochastic signals in s-domain (Power Spectral Density)

Question

Assume we have the following system (coming from control systems theory, hence in s-domain)

$ Y(s) = H_A (s) \cdot A(s) - H_B (s) \cdot B(s) $

I now wish to consider $a(t)$ and $b(t)$ as white noise of unit variance, and I'm interested in the Power spectral density of $y(t)$ (rather the RMS of y(t) derived via the integral of the PSD of $y(t)$, but regardless).

Intuition tells me, that I should get something along the lines of

$ |Y(j\omega)|^2 = |H_A (j\omega)|^2 \cdot 1 + |H_B (j\omega)|^2 \cdot 1 $

But I cannot show how. Especially the switch from subtraction to addition stumps me.

Answer 1

You have to look at the autocorrelation function of $y(t)$:

$$R_y(\tau)=E\{y(t)y(t+\tau)\}\tag{1}$$

with

$$y(t)=(h_A\star a)(t) - (h_B\star b)(t)\tag{2}$$

where $\star$ denotes convolution. If you write out $(2)$ with integrals and plug it into $(1)$ then, with the given assumptions on $a(t)$ and $b(t)$, you'll see that the mixed terms with the negative sign are zero, and the two remaining terms with a positive sign become

$$R_y(\tau)=\sigma_a^2r_{h_A}(\tau)+\sigma_b^2r_{h_B}(\tau)\tag{3}$$

where $\sigma_a^2$ and $\sigma_b^2$ are the variances of the processes $a(t)$ and $b(t)$, respectively, and $r_{h_A}(\tau)$ and $r_{h_B}(\tau)$ are the deterministic autocorrelation functions of $h_A(t)$ and $h_B(t)$, respectively. Finally, taking the Fourier transform of $(3)$ gives

$$S_y(\omega)=\sigma_a^2|H_A(j\omega)|^2+\sigma_b^2|H_B(j\omega)|^2\tag{4}$$

for the power spectrum of $y(t)$.

Answer 2

Assuming LTI properties you can break down the output $Y(s)$ into the linear combination of the output of two subsytems $Y_A(s)$ and $Y_B(s)$ as

$$Y(s) = Y_A(s) + Y_B(s)$$

Where

$$Y_A(s) = A(s)H_A(s)$$

$$Y_B(s) = -B(s)H_B(s)$$

If the processes are wide-sense stationary (WSS) then you have the input-output relationship of the PSD's given the frequency responses of the input $X(f)$ and system $H(f)$ as

$$S_Y(f) = S_X(f)|H(f)|^2$$

Applying this result to $Y(s) = Y_A(s) + Y_B(s)$

$$S_Y(f) = S_{Y_A}(f) + S_{Y_B}(f)$$ $$= S_A(f)|H_A(f)|^2 + S_B(f)|H_B(f)|^2$$ $$= |H_A(f)|^2 + |H_B(f)|^2$$

Since $S_A(f)$ and $S_B(f)$ are unity for all values of $f$.