### 3.5HW 5

3.5.1  Problem 1
3.5.2  Problem 2
3.5.3  Problem 3
3.5.4  Problem 4
3.5.5  Key solution
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#### 3.5.1Problem 1

3.5.1.1  Part(a)
3.5.1.2  Part(b)
3.5.1.3  Part(c)
3.5.1.4  Part(d)

##### 3.5.1.1 Part(a)

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

$R_{x}\left ( \tau \right ) =\overset{\text{dc part}}{\overbrace{1}}+\overset{\text{AC part}}{\overbrace{f_{0}\operatorname{sinc}^{2}\left ( f_{0}\tau \right ) }}$

##### 3.5.1.2 Part(b)

$P_{x}\left ( 0\right ) =1+f_{0}$ Hence DC power in $$X\left ( t\right )$$ is given $$1$$ watt.

##### 3.5.1.3 Part(c)

The AC power is $$f_{0}$$ watt.

##### 3.5.1.4 Part(d)

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 $\tau =\pm \frac{1}{f_{0}},\pm \frac{2}{f_{0}},\cdots$

#### 3.5.2Problem 2

(see graded HW for solution)

#### 3.5.3Problem 3

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.

#### 3.5.4Problem 4

(see graded HW for solution)