### 3.2HW 2

3.2.1  Questions
3.2.2  Problem 2.30
3.2.3  Problem 2.32
3.2.4  Problem 2.33
3.2.5  Problem 2.35
3.2.6  extra Problem
3.2.7  Key solution
PDF (letter size)
PDF (legal size)

#### 3.2.2Problem 2.30

3.2.2.1  part(b)
3.2.2.2  part(c)

Problem

Determine and sketch the autocorrelation function of the following

(b) $$g\left ( t\right ) =e^{-a\left \vert t\right \vert }$$

(c) $$g\left ( t\right ) =e^{-at}u\left ( t\right ) -e^{at}u\left ( -t\right )$$

##### 3.2.2.1 part(b)

$g\left ( t\right ) =\left \{ \begin{array} [c]{ccc}e^{-at} & & t>0\\ 1 & & t=0\\ e^{at} & & t<0 \end{array} \right .$

Assume $$a>0$$ for the integral to be deﬁned. From deﬁnition, autocorrelation of a function $$g\left ( t\right )$$ is

$R\left ( \tau \right ) =\int _{-\infty }^{\infty }g\left ( t\right ) g^{\ast }\left ( t-\tau \right ) dt$

Since $$g\left ( t\right )$$ in this case is real, then $$g^{\ast }\left ( t-\tau \right ) =g\left ( t-\tau \right )$$, hence

$R\left ( \tau \right ) =\int _{-\infty }^{\infty }g\left ( t\right ) g\left ( t-\tau \right ) dt$

Consider the 3 cases, $$\tau <0$$ and $$\tau >0$$ and when $$\tau =0$$

case $$\tau >0$$

Break the integral over the 3 regions, $$\left \{ -\infty ,0\right \} ,\left \{ 0,\tau \right \} ,\left \{ \tau ,\infty \right \}$$

$R\left ( \tau \right ) =\int _{-\infty }^{0}e^{at}e^{a(t-\tau )}dt+\int _{0}^{\tau }e^{-at}e^{a(t-\tau )}dt+\int _{\tau }^{\infty }e^{-at}e^{-a(t-\tau )}dt$

But $$\int _{-\infty }^{0}e^{at}e^{a(t-\tau )}dt=e^{-a\tau }\int _{-\infty }^{0}e^{2at}dt=e^{-a\tau }\frac{\left [ e^{2at}\right ] _{-\infty }^{0}}{2a}=e^{-a\tau }\frac{\left [ 1-0\right ] }{2a}=\frac{e^{-a\tau }}{2a}$$

and $$\int _{0}^{\tau }e^{-at}e^{a(t-\tau )}dt=e^{-a\tau }\int _{0}^{\tau }1dt=\tau e^{-a\tau }$$

and $$\int _{\tau }^{\infty }e^{-at}e^{-a(t-\tau )}dt=e^{a\tau }\int _{\tau }^{\infty }e^{-2at}dt=e^{a\tau }\frac{\left [ e^{-2at}\right ] _{\tau }^{\infty }}{-2a}=e^{a\tau }\frac{\left [ 0-e^{-2a\tau }\right ] }{-2a}=\frac{e^{-a\tau }}{2a}$$

Hence for $$\tau >0$$ we obtain

\begin{align*} R\left ( \tau \right ) & =\frac{e^{-a\tau }}{2a}+\tau e^{-a\tau }+\frac{e^{-a\tau }}{2a}\\ & =\frac{e^{-a\tau }}{a}+\tau e^{-a\tau }\\ & =\fbox{$e^{-a\tau }\left ( \frac{1}{a}+\tau \right )$} \end{align*}

case $$\tau <0$$

Break the integral over the 3 regions, $$\left \{ -\infty ,\tau \right \} ,\left \{ \tau ,0\right \} ,\left \{ 0,\infty \right \}$$

$R\left ( \tau \right ) =\int _{-\infty }^{\tau }e^{at}e^{a(t-\tau )}dt+\int _{\tau }^{0}e^{-at}e^{a(t-\tau )}dt+\int _{0}^{\infty }e^{-at}e^{-a(t-\tau )}dt$

Now $$\int _{-\infty }^{\tau }e^{at}e^{a(t-\tau )}dt=e^{-a\tau }\int _{-\infty }^{\tau }e^{2at}dt=e^{-a\tau }\frac{\left [ e^{2at}\right ] _{-\infty }^{\tau }}{2a}=e^{-a\tau }\frac{\left [ e^{2a\tau }-0\right ] }{2a}=\frac{e^{a\tau }}{2a}$$

and $$\int _{\tau }^{0}e^{-at}e^{a(t-\tau )}dt=e^{-a\tau }\int _{\tau }^{0}1dt=-\tau e^{-a\tau }$$

and $$\int _{0}^{\infty }e^{-at}e^{-a(t-\tau )}dt=e^{a\tau }\frac{\left [ e^{-2at}\right ] _{0}^{\infty }}{-2a}=\frac{e^{a\tau }}{-2a}\left ( 0-1\right ) =\frac{e^{a\tau }}{2a}$$

Hence \begin{align*} R\left ( \tau \right ) & =\frac{e^{a\tau }}{2a}-\tau e^{-a\tau }+\frac{e^{a\tau }}{2a}\\ & =\fbox{$e^{a\tau }\left ( \frac{1}{a}-\tau \right )$} \end{align*}

When $$\tau =0$$

$$R\left ( 0\right )$$ gives the the maximum power in the signal $$g\left ( t\right )$$. Now evaluate this

\begin{align*} R\left ( \tau \right ) & =\int _{-\infty }^{0}e^{at}e^{at}dt+\int _{0}^{\infty }e^{-at}e^{-at}dt\\ & =\frac{\left [ e^{2at}\right ] _{-\infty }^{0}}{2a}+\frac{\left [ e^{-2at}\right ] _{0}^{\infty }}{-2a}\\ & =\frac{1}{a} \end{align*}

Hence

$R\left ( \tau \right ) =\left \{ \begin{array} [c]{ccc}e^{-a\tau }\left ( \frac{1}{a}+\tau \right ) & & \tau >0\\ \frac{1}{a} & & \tau =0\\ e^{a\tau }\left ( \frac{1}{a}-\tau \right ) & & \tau <0 \end{array} \right .$

Or we could write

$\fbox{R\left ( \tau \right ) =e^{-\left \vert \tau \right \vert a}\left ( \frac{1}{a}-(-\left \vert \tau \right \vert \right ) }$

This is a plot of $$R\left ( \tau \right )$$, ﬁrst plot is for $$a=1$$ and the second for $$a=4$$

##### 3.2.2.2 part(c)

$g\left ( t\right ) =e^{-at}u\left ( t\right ) -e^{at}u\left ( -t\right )$

Assume $$a>0.$$

Consider the 3 cases, $$\tau <0$$ and $$\tau >0$$ and when $$\tau =0$$

case $$\tau >0$$

Break the integral into 3 parts, $$\left \{ -\infty ,0\right \} ,\left \{ 0,\tau \right \} ,\left \{ \tau ,\infty \right \}$$

\begin{align*} R\left ( \tau \right ) & =\int _{-\infty }^{0}g\left ( t\right ) g\left ( t-\tau \right ) dt+\int _{0}^{\tau }g\left ( t\right ) g\left ( t-\tau \right ) dt+\int _{\tau }^{\infty }g\left ( t\right ) g\left ( t-\tau \right ) dt\\ & =\int _{-\infty }^{0}-e^{at}\left ( -e^{a\left ( t-\tau \right ) }\right ) dt+\int _{0}^{\tau }e^{-at}\left ( -e^{a\left ( t-\tau \right ) }\right ) dt+\int _{\tau }^{\infty }e^{-at}\left ( e^{-a\left ( t-\tau \right ) }\right ) dt\\ & =e^{-a\tau }\int _{-\infty }^{0}e^{2at}dt-e^{-a\tau }\int _{0}^{\tau }1dt+e^{a\tau }\int _{\tau }^{\infty }e^{-2at}dt\\ & =e^{-a\tau }\frac{\left [ e^{2at}\right ] _{-\infty }^{0}}{2a}-\tau e^{-a\tau }+e^{a\tau }\frac{\left [ e^{-2at}\right ] _{\tau }^{\infty }}{-2a}\\ & =e^{-a\tau }\frac{\left [ 1-0\right ] }{2a}-\tau e^{-a\tau }+e^{a\tau }\frac{\left [ 0-e^{-2a\tau }\right ] }{-2a}\\ & =\frac{e^{-a\tau }}{2a}-\tau e^{-a\tau }+\frac{e^{-a\tau }}{2a}\\ & =e^{-a\tau }\left ( \frac{1}{2a}-\tau +\frac{1}{2a}\right ) \\ & =e^{-a\tau }\left ( \frac{1}{a}-\tau \right ) \end{align*}

case $$\tau <0$$

Break the integral into 3 parts, $$\left \{ -\infty ,\tau \right \} ,\left \{ \tau ,0\right \} ,\left \{ 0,\infty \right \}$$

\begin{align*} R\left ( \tau \right ) & =\int _{-\infty }^{\tau }g\left ( t\right ) g\left ( t-\tau \right ) dt+\int _{\tau }^{0}g\left ( t\right ) g\left ( t-\tau \right ) dt+\int _{0}^{\infty }g\left ( t\right ) g\left ( t-\tau \right ) dt\\ & =\int _{-\infty }^{\tau }-e^{at}\left ( -e^{a\left ( t-\tau \right ) }\right ) dt+\int _{\tau }^{0}-e^{at}e^{-a\left ( t-\tau \right ) }dt+\int _{0}^{\infty }e^{-at}e^{-a\left ( t-\tau \right ) }dt\\ & =e^{-a\tau }\int _{-\infty }^{\tau }e^{2at}dt-e^{a\tau }\int _{\tau }^{0}1dt+e^{a\tau }\int _{0}^{\infty }e^{-2at}dt\\ & =e^{-a\tau }\frac{\left [ e^{2at}\right ] _{-\infty }^{\tau }}{2a}+\tau e^{a\tau }+e^{a\tau }\frac{\left [ e^{-2at}\right ] _{0}^{\infty }}{-2a}\\ & =e^{-a\tau }\frac{\left [ e^{2a\tau }-0\right ] }{2a}+\tau e^{a\tau }+e^{a\tau }\frac{\left [ 0-1\right ] }{-2a}\\ & =\frac{e^{a\tau }}{2a}+\tau e^{a\tau }+\frac{e^{a\tau }}{2a}\\ & =e^{a\tau }\left ( \frac{1}{a}+\tau \right ) \end{align*}

At $$\tau =0$$, we see that $$R\left ( 0\right ) =\frac{1}{a}$$, hence the ﬁnal answer is

$R\left ( \tau \right ) =\left \{ \begin{array} [c]{ccc}e^{-a\tau }\left ( \frac{1}{a}-\tau \right ) & & \tau >0\\ \frac{1}{a} & & \tau =0\\ e^{a\tau }\left ( \frac{1}{a}+\tau \right ) & & \tau <0 \end{array} \right .$

Or we could write

$\fbox{R\left ( \tau \right ) =e^{-\left \vert \tau \right \vert a}\left ( \frac{1}{a}-\left \vert \tau \right \vert \right ) }$

This is a plot of $$R\left ( \tau \right )$$, ﬁrst plot is for $$a=1$$ and the second for $$a=4$$

#### 3.2.3Problem 2.32

problem: Determine the autocorrelation function of $$g\left ( t\right ) =A\operatorname{sinc}\left ( 2Wt\right )$$ and sketch it

solution:

$R\left ( \tau \right ) ={\displaystyle \int \limits _{-\infty }^{\infty }} g\left ( t\right ) g^{\ast }\left ( t-\tau \right ) dt$

The above is diﬃcult to do directly, hence we use the second method.

Since the function $$g\left ( t\right )$$ is an energy function, hence $$R\left ( \tau \right )$$ and the energy spectrum density $$\Psi _{g}\left ( f\right )$$ of $$g\left ( t\right )$$make a Fourier transform pairs.

$R\left ( \tau \right ) \Leftrightarrow \Psi _{g}\left ( f\right )$

Therefore, to ﬁnd $$R\left ( \tau \right )$$, we ﬁrst ﬁnd $$\Psi _{g}\left ( f\right )$$, then ﬁnd the Inverse Fourier Transform of $$\Psi _{g}\left ( f\right )$$, i.e.

$$R\left ( \tau \right ) =\digamma ^{-1}\left ( \Psi _{g}\left ( f\right ) \right ) \tag{1}$$

But $$\Psi _{g}\left ( f\right ) =\left \vert G\left ( f\right ) \right \vert ^{2}\tag{2}$$

and we know that $A\operatorname{sinc}\left ( 2Wt\right ) \Leftrightarrow \frac{A}{2W}rect\left ( \frac{f}{2W}\right )$

Hence $G\left ( f\right ) =\frac{A}{2W}rect\left ( \frac{f}{2W}\right )$

The (2) becomes

\begin{align*} \Psi _{g}\left ( f\right ) & =\left \vert \frac{A}{2W}rect\left ( \frac{f}{2W}\right ) \right \vert ^{2}\\ & =\left ( \frac{A}{2W}\right ) ^{2}\left \vert rect\left ( \frac{f}{2W}\right ) \right \vert ^{2} \end{align*}

But $$\left \vert rect\left ( \frac{f}{2W}\right ) \right \vert ^{2}=rect\left ( \frac{f}{2W}\right )$$, since it has height of 1, so

$\fbox{\Psi _{g}\left ( f\right ) =\left ( \frac{A}{2W}\right ) ^{2}rect\left ( \frac{f}{2W}\right ) }$

Hence from (1)

\begin{align*} R\left ( \tau \right ) & =\digamma ^{-1}\left ( \left ( \frac{A}{2W}\right ) ^{2}rect\left ( \frac{f}{2W}\right ) \right ) \\ & =\left ( \frac{A}{2W}\right ) ^{2}\digamma ^{-1}\left [ rect\left ( \frac{f}{2W}\right ) \right ] \end{align*}

Hence

$\fbox{R\left ( \tau \right ) =\left ( \frac{A}{2W}\right ) ^{2}\operatorname{sinc}\left ( 2W\tau \right ) }$

This is a plot of the above function, for $$W=4$$, and $$A=1$$

#### 3.2.4Problem 2.33

The Fourier transform of a signal is deﬁned by $$\left \vert \operatorname{sinc}\left ( f\right ) \right \vert$$. Show that $$R\left ( \tau \right )$$ of the signal is triangular in form.

Since $R\left ( \tau \right ) \Leftrightarrow \left \vert G\left ( f\right ) \right \vert ^{2}$

Then

\begin{align*} R\left ( \tau \right ) & \Leftrightarrow \left \vert \operatorname{sinc}\left ( f\right ) \right \vert ^{2}\\ & \Leftrightarrow \operatorname{sinc}^{2}\left ( f\right ) \end{align*}

Hence to ﬁnd $$R\left ( \tau \right )$$ we need to ﬁnd the inverse Fourier transform of $$\operatorname{sinc}^{2}\left ( f\right )$$

But \begin{align*} \digamma ^{-1}\left ( \operatorname{sinc}^{2}\left ( f\right ) \right ) & =\digamma ^{-1}\left ( \operatorname{sinc}\left ( f\right ) \times \operatorname{sinc}\left ( f\right ) \right ) \\ & =\digamma ^{-1}\left \{ \operatorname{sinc}\left ( f\right ) \right \} \otimes \digamma ^{-1}\left \{ \operatorname{sinc}\left ( f\right ) \right \} \end{align*}

But $$\digamma ^{-1}\left \{ \operatorname{sinc}\left ( f\right ) \right \} =rect\left ( t\right )$$, hence

\begin{align*} \digamma ^{-1}\left ( \operatorname{sinc}^{2}\left ( f\right ) \right ) & =rect\left ( t\right ) \otimes rect\left ( t\right ) \\ & ={\displaystyle \int \limits _{-\infty }^{\infty }} rect\left ( \tau \right ) rect\left ( t-\tau \right ) d\tau \end{align*}

This integral has the value of $$tri\left ( t\right )$$ (we also did this in class) Hence

$tri\left ( \tau \right ) \Leftrightarrow \operatorname{sinc}^{2}\left ( f\right )$

Hence $R\left ( \tau \right ) =tri\left ( \tau \right )$

Where $$tri\left ( \tau \right )$$ is the triangle function, deﬁned as

$tri\left ( t\right ) =\left \{ \begin{array} [c]{ccc}1-\left \vert t\right \vert & & \left \vert t\right \vert <0\\ 0 & & otherwise \end{array} \right .$

#### 3.2.5Problem 2.35

Consider the signal $$g\left ( t\right )$$ deﬁned by $g\left ( t\right ) =A_{0}+A_{1}\cos \left ( 2\pi f_{1}t+\theta \right ) +A_{2}\cos \left ( 2\pi f_{2}t+\theta \right )$

(a) determine $$R\left ( \tau \right )$$

(b) what is $$R\left ( 0\right )$$

(c) has any information been lose in obtaining $$R\left ( \tau \right ) ?$$

(a)

Take the Fourier transform of $$g\left ( t\right )$$ we obtain

$G\left ( f\right ) =A_{0}\delta \left ( f\right ) +\frac{A_{1}}{2}\left [ e^{j\theta }\delta \left ( f-f_{1}\right ) +e^{-j\theta }\delta \left ( f+f_{1}\right ) \right ] +\frac{A_{2}}{2}\left [ e^{j\theta }\delta \left ( f-f_{2}\right ) +e^{-j\theta }\delta \left ( f+f_{2}\right ) \right ]$

Hence $$\left \vert G\left ( f\right ) \right \vert ^{2}=G\left ( f\right ) G^{\ast }\left ( f\right )$$, so we need to ﬁnd $$G^{\ast }\left ( f\right )$$

$G^{\ast }\left ( f\right ) =A_{0}\delta \left ( f\right ) +\frac{A_{1}}{2}\left [ e^{-j\theta }\delta \left ( f-f_{1}\right ) +e^{j\theta }\delta \left ( f+f_{1}\right ) \right ] +\frac{A_{2}}{2}\left [ e^{-j\theta }\delta \left ( f-f_{2}\right ) +e^{j\theta }\delta \left ( f+f_{2}\right ) \right ]$

So $G\left ( f\right ) G^{\ast }\left ( f\right ) =A_{0}^{2}\delta \left ( f\right ) +\frac{A_{1}^{2}}{4}\left [ \delta \left ( f-f_{1}\right ) +\delta \left ( f+f_{1}\right ) \right ] +\frac{A_{2}^{2}}{4}\left [ \delta \left ( f-f_{2}\right ) +\delta \left ( f+f_{2}\right ) \right ]$

So $S_{g}\left ( f\right ) =A_{0}^{2}\delta \left ( f\right ) +\frac{A_{1}^{2}}{4}\left [ \delta \left ( f-f_{1}\right ) +\delta \left ( f+f_{1}\right ) \right ] +\frac{A_{2}^{2}}{4}\left [ \delta \left ( f-f_{2}\right ) +\delta \left ( f+f_{2}\right ) \right ]$

So \begin{align*} R\left ( \tau \right ) & =\digamma ^{-1}\left ( S_{g}\left ( f\right ) \right ) \\ & =\digamma ^{-1}\left ( A_{0}^{2}\delta \left ( f\right ) \right ) +\frac{A_{1}^{2}}{4}\digamma ^{-1}\left [ \delta \left ( f-f_{1}\right ) +\delta \left ( f+f_{1}\right ) \right ] +\frac{A_{2}^{2}}{4}\digamma ^{-1}\left [ \delta \left ( f-f_{2}\right ) +\delta \left ( f+f_{2}\right ) \right ] \end{align*}

Hence

$$\fbox{R\left ( \tau \right ) =A_{0}^{2}+\frac{A_{1}^{2}}{2}\cos 2\pi f_{1}\tau +\frac{A_{2}^{2}}{2}\cos 2\pi f_{2}\tau }\tag{1}$$

Part (b)

\begin{align*} R\left ( 0\right ) & =A_{0}^{2}+\frac{A_{1}^{2}}{2}+\frac{A_{2}^{2}}{2}\\ & =\frac{1}{2}\left ( 2A_{0}^{2}+A_{1}^{2}+A_{2}^{2}\right ) \end{align*}

part(c)

In obtaining $$R\left ( \tau \right )$$ we have lost the phase information in the original signal as can be seen from (1) above

#### 3.2.6extra Problem

(a) ﬁnd $$\xi \left ( t\right ) \otimes \xi \left ( t\right )$$ where $$\xi \left ( t\right )$$ is unit step function

(b)Find $$t\xi \left ( t\right ) \otimes e^{at}\xi \left ( t\right )$$ where $$a>0$$

(c)ﬁnd $$u\left ( t\right ) \otimes h\left ( t\right )$$ where $$h\left ( t\right ) =e^{-3t}u\left ( t\right )$$ and $$u\left ( t\right )$$ is as shown

To DO