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Let $X(t)$ and $Y(t)$ be two orthogonal processes with power spectral densities $$S_{xx}(f) = S_{yy}(f)=\begin{cases} 1-\lvert f\rvert, & \lvert f\rvert<1 \\[1ex] 0,& \text{otherwise} \end{cases}$$

Define a new process $Z(t) = Y(t)-X(t-1)$. Determine and sketch the power spectral density $S_{zz}(f)$.

The solution is the following: The autocorrelation function of the process is $$R_{zz}= E \bigg\{\big[Y(t-\tau)-X(t+\tau-1)\big]\big[Y(t)-X(t-1)\big] \bigg\} $$ And now comes the part which I do not understand at all. \begin{align} R_{zz}&=R_{yy}(\tau)-R_{yx}(\tau+1)-R_{xy}(\tau-1)+R_{xx}(\tau)\\ &=R_{yy}(\tau) + R_{xx}(\tau)\ \quad\text{since}\quad R_{yx} = R_{xy} = 0 \quad\text{from orthogonality.} \end{align}

Therefore, $S_{zz}(f)=2S_{yy}(f)=2S_{xx}(f)$ as shown below. enter image description here

Can anyone explain it like I am five?

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HINT:

Expand the terms inside the expectation $E\{\ldots\}$, and then use the fact that the expectation of a sum equals the sum of the individual expectations (due to linearity).

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  • $\begingroup$ So $Y(t+\tau)Y(t)-Y(t+\tau)X(t-1)-X(t+\tau-1)Y(t)+X(t+\tau-1)X(t-1)$ and then i can see in my notes that $R_{xx}(\tau) = E[X(t+\tau)X(t)]$ so it gives me the first line i dont understand. But how is $Ryx=Rxy=0$? $\endgroup$ – Xraycat922 Dec 29 '17 at 20:57
  • $\begingroup$ @Xraycat922: As you said in your question: "from orthogonality". $\endgroup$ – Matt L. Dec 29 '17 at 21:13
  • $\begingroup$ ah. Okay, last question why is the peak amplitude 2? (why $2 S_{yy}(f)$) $\endgroup$ – Xraycat922 Dec 29 '17 at 21:35
  • $\begingroup$ @Xraycat922: You get $S_{zz}=S_{xx}+S_{yy}$, and since $S_{xx}=S_{yy}$ you have $S_{zz}=2S_{xx}=2S_{yy}$. That was easy, wasn't it? $\endgroup$ – Matt L. Dec 29 '17 at 22:56
  • $\begingroup$ Hi Matt: I think the confusion ( atleast for me. maybe not the OP ) was that the relations between the correlations then can be used in the relations for the respective spectral densities. I imagine this is true because, for any stochastic process, the spectral density at $\omega$ is t the moment generating function of the autocorrelations evaluated at $\omega$. I should say that I think this is the reason but I'm not positive. thanks for clarification. $\endgroup$ – mark leeds Dec 30 '17 at 8:02

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