I am studying shot noise characteristics from this source:

Here the author writes that autocorrelation function is given by: $$R_I(\tau)=\bar{h}*h*R_Z(\tau)$$ where $R_Z(\tau)=q^2(\lambda^2+\lambda \delta(\tau))$ and $\bar{h}(t)=h(-t)$ now applying the value of $R_Z(\tau)$ in the integral we have $$=q^2\lambda^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\text{d}s\text{d}t+q^2\lambda\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\delta(t)\text{d}s\text{d}t$$ $$=q^2\lambda^2(\int_{-\infty}^{\infty}{h}(t)\text{d}t)^2+q^2\lambda\bar{h}*h(\tau)$$ the second term I am able to derive as I know that $$q^2\lambda\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\delta(t)\text{d}s\text{d}t$$ $$=q^2\lambda\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s)\text{d}s$$ however, I can not get that how: $$q^2\lambda^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\text{d}s\text{d}t$$ turns out to be $$=q^2\lambda^2(\int_{-\infty}^{\infty}{h}(t)\text{d}t)^2$$ can somebody please help me out with this?

PS: As per the answer of Matt L., as inner integral is independent of $s$ so I can do certain manipulations to get requisite output, however, I have a doubt here that I want to clear: $$q^2\lambda^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\text{d}s\text{d}t$$ is precisely $$q^2\lambda^2\int_{t=-\infty}^{\infty}\int_{s=-\infty}^{\infty}\bar{h}(\tau-s)h(s-t)\text{d}s\text{d}t$$ now repeating the steps as suggested in the answer I have: $$q^2\lambda^2\int_{s=-\infty}^{\infty}\bar{h}(\tau-s)\left[\int_{t=-\infty}^{\infty}h(s-t)\text{d}t\right]\text{d}s$$ I believe that the inner integral should be some function of $s$ let us say $f(s)$ and then we should have $$q^2\lambda^2\int_{s=-\infty}^{\infty}\bar{h}(\tau-s)f(s)\text{d}s$$ The reason I believe so is because $s$ may be constant for inner integral but outer integral needs $s$, how I can I simply replace $s$ without changing the outer integral limits and $\text{d}s$ in outer integral.

Also taking the answer's view: $$q^2\lambda^2\int_{s=-\infty}^{\infty}\bar{h}(\tau-s)\left[\int_{t=-\infty}^{\infty}h(s-t)\text{d}t\right]\text{d}s$$ we replace $s-t$ with some dummy variable $t'$, but if we do so, shouldn't the outer integral will change accordingly as it also contains $s$?


1 Answer 1


Note that


The second equality is true because the inner integral is independent of $s$. Similarly, the integral of $h(s-\tau)$ in the third line is independent of $\tau$.

  • $\begingroup$ I have a doubt when you talk about second equality then you basically assume that $s$ to be constant and I assume that you replace $s-t$ with some dummy variable (let us say $t'$). However, I believe it will also affect the outer integral because that involves $s$ and also the limits of outer integral are from $s$ ranging from $-\infty$ to $\infty$. I assume that you may have missed several steps which precisely I am trying to figure out. $\endgroup$ Jul 28, 2021 at 18:14
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    $\begingroup$ @UserHuffmann: Since the inner integral doesn't depend on $s$ (because you integrate over the real line), it is constant and can be taken out of the integral over $s$. $\endgroup$
    – Matt L.
    Jul 28, 2021 at 19:07
  • $\begingroup$ can you please check my edited question wherein I have tried to be more precise about my doubt. I will be grateful. Thanks $\endgroup$ Jul 29, 2021 at 6:07
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    $\begingroup$ @UserHuffmann: You can indeed see $$\int_{-\infty}^{\infty}h(s-t)dt$$ as a function of $s$. However, it should be easy to see that this function is actually constant. Just think about it as integrating over a function $h(-t)$ that is shifted by some parameter $s$. Regardless of the shift $s$, the result is always the integral over that same function, because shifting doesn't change the value of the integral (the area), as long as we integrate from $-\infty$ to $+\infty$. Now, since we see that what you call $f(s)$ is actually a constant, we can take it out of the integral. $\endgroup$
    – Matt L.
    Jul 29, 2021 at 11:11
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    $\begingroup$ Now, I got your point. The limit in this expression from $-\infty$ to $\infty$ is what is leading to this peculiar solution. Thanks $\endgroup$ Jul 30, 2021 at 8:11

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