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.

  • $\begingroup$ Hint: What does the PSD look like for white noise? $\endgroup$
    – Envidia
    Oct 18 '20 at 3:42
  • $\begingroup$ @Envidia it's constant, in my case even one. I already hinted myself at this fact with the 1s in the last equation, but I'm still stuck with the sign. $\endgroup$
    – user53750
    Oct 18 '20 at 14:20

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



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

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


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


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

  • $\begingroup$ Thanks! I knew the solution would have to go via the time domain, but I was curious if it was possible to show this without going back to the convolution. Nevertheless, I'm happy :) $\endgroup$
    – user53750
    Oct 19 '20 at 17:06
  • $\begingroup$ Shouldn't it say $jw$ in (4) for $H_a$ and $H_b$ though? $\endgroup$
    – user53750
    Oct 19 '20 at 17:07
  • $\begingroup$ @XaserIII: In order to be consistent with your question, yes, otherwise it doesn't matter. Both conventions are used. $\endgroup$
    – Matt L.
    Oct 19 '20 at 19:39
  • $\begingroup$ @DilipSarwate $a(t)$ and $b(t)$ are stated as white-noise of unit variance. And MattL concludes that they are uncorrelated (cross-corr being null), which is the conclusion I would also make, as they are white noise. Yet I wonder if you have an objection to this too? For example, doesn't the identity $b(t) = c \cdot a(t)$ make them correlated (at lag zero), eventhough $b(t)$ is no more unit-variance? $\endgroup$
    – Fat32
    Oct 22 '20 at 22:32

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

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


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

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

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 $f$.

  • $\begingroup$ I like the simplicity of your answer, but it's essentially saying "just pull the minus into the modulus", which is not clear to me why this is allowed, or rather what stops me from keeping it outside? $\endgroup$
    – user53750
    Oct 19 '20 at 17:05
  • $\begingroup$ @XaserIII I see your concern, and maybe I should edit my answer to show the underlying math as to why this negative sign goes away. What I tried to use was an equivalent control system. It's valid to move the negative sign to the transfer function and add at the junction just as the original setup where you subtract at the junction. The intent is what might confuse us because simply looking at the whole system we don't know if we really are meaning to subtract at the junction, or the path really does have a negative gain. Regardless, Matt's answer yields the result you're looking for :) $\endgroup$
    – Envidia
    Oct 19 '20 at 17:13

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