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It is given to us some white noise as $A z(t)$ and the autocorrelation of $A z(t)$ is given as

$\phi(t)= A^2 \delta(t)$ where $\delta(t)$ is the Dirac delta function

Now one signal with $y(t)= B \cos(\omega t)$ with autocorrelation $\phi (t) =[B^2 \cos(\omega t)]/2$ get mixed with the above white noise to form a new impure signal named $g(t)$

And I got $g(t)=A z(t) + B \cos(wt)$ Now how can I find the autocorrelation function for this new impure signal ?

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  • $\begingroup$ How is this a physics question? $\endgroup$ – Jannik Pitt Jun 28 at 18:43
  • $\begingroup$ Dirac delta function is very well known phenomenon in quantum physics and we use autocorrelation for various disciplines of physics in research while handling the lag of data from expected outcome and currently I am working on waves and their behaviour so I guess it's question of physics $\endgroup$ – Abhiraj Jun 29 at 7:18
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Note that $y$ is a deterministic signal and thus independent of $Az(t)$. The answer should just be the sum of the two autocorrelations. More precisely, you want to compute the following expectation:

$$\mathbb{E}_{\tau}[(Az(\tau)+B\cos(\omega \tau))(Az(\tau+t)+B\cos(\omega(\tau +t)))]$$

By linearity of the expectation you get the two autocorrelations that you already have plus the mixed term:

$$\mathbb{E}_{\tau}[ABz(\tau)\cos(\omega(\tau +t))+AB\cos(\omega\tau)z(\tau+t)]$$

Now the two signals factor out in the expectation (by independence) and they both have 0 expectation i.e.:

$$\mathbb{E}_{\tau}[ABz(\tau)\cos(\omega(\tau+t))]=AB\mathbb{E}_{\tau}z(\tau)\times\mathbb{E}_{\tau}[\cos(\omega(\tau+t))]=0$$

And likewise for the other mixed term $\mathbb{E}_{\tau}[AB\cos(\omega\tau)z(\tau+t)]$.

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  • $\begingroup$ Wow...thanks!!!!! I got it now $\endgroup$ – Abhiraj Jun 29 at 7:12
  • $\begingroup$ I also got the same answer , I have done it graphically by comparing graph of autocorrelation of added signals with the graph of sum of autocorrelations of two signals $\endgroup$ – Abhiraj Jun 29 at 7:14

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