Processing math: 100%

3.5  HW 5

PDF (letter size)

PDF (legal size)

3.5.1  Problem 1

Figure 3.13:the Problem statement

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

3.5.2  Problem 2

Figure 3.14:the Problem statement

(see graded HW for solution)

3.5.3  Problem 3

A random telegraph signal $X(t)$ charaterized by the autocorrelation function $$R_X(\tau ) - e^{-2\nu |tau|}$$

Figure 3.15:the Problem statement

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.

3.5.4  Problem 4

Figure 3.16:the Problem statement

(see graded HW for solution)

3.5.5  Key solution

PDF