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?



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).

| improve this answer | |
  • $\begingroup$ 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 $Ryx=Rxy=0$? $\endgroup$ – XRaycat 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 $2 S_{yy}(f)$) $\endgroup$ – XRaycat 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.