Assuming stationary process, \[ R_{x}\left ( \tau \right ) \Leftrightarrow S_{x}\left ( f\right ) \]
But \(S_{x}\left ( f\right ) =\delta \left ( f\right ) +tri\left ( \frac{f}{2f_{0}}\right ) \), hence
\begin{align*} R_{x}\left ( \tau \right ) & =\digamma ^{-1}\left ( \delta \left ( f\right ) +tri\left ( \frac{f}{2f_{0}}\right ) \right ) \\ & ={\displaystyle \int \limits _{-\infty }^{\infty }} \left [ \delta \left ( f\right ) +tri\left ( \frac{f}{2f_{0}}\right ) \right ] e^{j2\pi f\tau }df \end{align*}
But \(\digamma ^{-1}\left ( tri\left ( \frac{f}{2f_{0}}\right ) \right ) =f_{0}\frac{\sin ^{2}\left ( f_{0}\pi \tau \right ) }{f_{0}^{2}\pi ^{2}\tau ^{2}}\), and \(\digamma ^{-1}\left ( \delta \left ( f\right ) \right ) =1\), hence the above becomes
Hence
\[ R_{x}\left ( \tau \right ) =\overset{\text{dc part}}{\overbrace{1}}+\overset{\text{AC part}}{\overbrace{f_{0}\operatorname{sinc}^{2}\left ( f_{0}\tau \right ) }} \]
\[ P_{x}\left ( 0\right ) =1+f_{0}\] Hence DC power in \(X\left ( t\right ) \) is given \(1\) watt.
The AC power is \(f_{0}\) watt.
Since \(R_{x}\left ( \tau \right ) =1+f_{0}\operatorname{sinc}^{2}\left ( f_{0}\tau \right ) \), we need to make this zero. But this has no real root as solution (assuming \(f_{0}\geq 0\))
To obtain a solution, I will only consider the AC part.
Hence we need to solve for \(\tau \) in \[ R_{x}\left ( \tau \right ) =f_{0}\operatorname{sinc}^{2}\left ( f_{0}\tau \right ) =0 \]
i.e. the AC part only.
This is zero when \(\operatorname{sinc}^{2}\left ( f_{0}\tau \right ) =0\) or when \(\sin \left ( \pi f_{0}\tau \right ) =0\) or when
\(\pi f_{0}\tau =k\pi \), \(k=\pm 1,\pm 2,\cdots \).
Hence when \[ \tau =\pm \frac{1}{f_{0}},\pm \frac{2}{f_{0}},\cdots \]
(see graded HW for solution)
A random telegraph signal \(X(t)\) charaterized by the autocorrelation function \[ R_X(\tau ) - e^{-2\nu |tau|} \]
Let \(S_{y}\left ( f\right ) \) be the psd of the output, then
\[ S_{y}\left ( f\right ) =S_{x}\left ( f\right ) \left \vert H\left ( f\right ) \right \vert ^{2}\]
But \begin{align*} S_{x}\left ( f\right ) & =\digamma \left ( R_{x}\left ( \tau \right ) \right ) \\ & ={\displaystyle \int \limits _{-\infty }^{0}} e^{2v\tau }e^{-j2\pi f\tau }d\tau +{\displaystyle \int \limits _{0}^{\infty }} e^{-2v\tau }e^{-j2\pi f\tau }d\tau \\ & ={\displaystyle \int \limits _{-\infty }^{0}} e^{\tau \left ( 2v-j2\pi f\right ) }d\tau +{\displaystyle \int \limits _{0}^{\infty }} e^{\tau \left ( -2v-j2\pi f\right ) }d\tau \\ & =\frac{\left [ e^{\tau \left ( 2v-j2\pi f\right ) }\right ] _{-\infty }^{0}}{2v-j2\pi f}+\frac{\left [ e^{\tau \left ( -2v-j2\pi f\right ) }\right ] _{0}^{\infty }}{-2v-j2\pi f}\\ & =\frac{1}{2v-j2\pi f}+\frac{-1}{-2v-j2\pi f}\\ & =\frac{1}{2v-j2\pi f}+\frac{1}{2v+j2\pi f}\\ & =\frac{4v}{4v^{2}+4\pi ^{2}f^{2}} \end{align*}
Now we need to find \(H\left ( f\right ) \). Using voltage divider \(H\left ( f\right ) =\frac{Y\left ( f\right ) }{X\left ( f\right ) }=\frac{\frac{1}{j2\pi fC}}{R+\frac{1}{j2\pi fC}}\)
hence \[ H\left ( f\right ) =\frac{1}{j2\pi fRC+1}\]
Hence
\[ \left \vert H\left ( f\right ) \right \vert =\frac{1}{\sqrt{1+\left ( 2\pi fRC\right ) ^{2}}}\]
Then
\begin{align*} S_{y}\left ( f\right ) & =S_{x}\left ( f\right ) \left \vert H\left ( f\right ) \right \vert ^{2}\\ & =\left ( \frac{4v}{4v^{2}+4\pi ^{2}f^{2}}\right ) \left ( \frac{1}{1+\left ( 2\pi fRC\right ) ^{2}}\right ) \\ & =\frac{4v}{\left ( 4v^{2}+4\pi ^{2}f^{2}\right ) \left ( 1+4\pi ^{2}f^{2}R^{2}C^{2}\right ) }\\ & =\frac{4v}{4v^{2}+4v^{2}\left ( 2\pi fRC\right ) ^{2}+4\pi ^{2}f^{2}+4\pi ^{2}f^{2}\left ( 2\pi fRC\right ) ^{2}}\\ & =\frac{4v}{4v^{2}+16v^{2}\pi ^{2}f^{2}R^{2}C^{2}+4\pi ^{2}f^{2}+16\pi ^{2}f^{2}\pi ^{2}f^{2}R^{2}C^{2}}\\ & =\frac{v}{v^{2}+4v^{2}\pi ^{2}f^{2}R^{2}C^{2}+\pi ^{2}f^{2}+4\pi ^{4}f^{4}R^{2}C^{2}} \end{align*}
Now, \(R_{y}\left ( \tau \right ) \) is the inverse Fourier transform of the above.
(see graded HW for solution)