Problem
Determine and sketch the autocorrelation function of the following
(b) \(g\left ( t\right ) =e^{-a\left \vert t\right \vert }\)
(c) \(g\left ( t\right ) =e^{-at}u\left ( t\right ) -e^{at}u\left ( -t\right ) \)
\[ g\left ( t\right ) =\left \{ \begin{array} [c]{ccc}e^{-at} & & t>0\\ 1 & & t=0\\ e^{at} & & t<0 \end{array} \right . \]
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 \(\tau =0\)
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*}
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*}
\(R\left ( 0\right ) \) 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 \(a=4\)
\[ g\left ( t\right ) =e^{-at}u\left ( t\right ) -e^{at}u\left ( -t\right ) \]
Assume \(a>0.\)
Consider the 3 cases, \(\tau <0\) and \(\tau >0\) and when \(\tau =0\)
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*}
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 \(a=4\)
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.
\begin{equation} R\left ( \tau \right ) =\digamma ^{-1}\left ( \Psi _{g}\left ( f\right ) \right ) \tag{1} \end{equation}
But \begin{equation} \Psi _{g}\left ( f\right ) =\left \vert G\left ( f\right ) \right \vert ^{2}\tag{2} \end{equation}
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 \(A=1\)
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.
Answer:
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
\[ tri\left ( t\right ) =\left \{ \begin{array} [c]{ccc}1-\left \vert t\right \vert & & \left \vert t\right \vert <0\\ 0 & & otherwise \end{array} \right . \]
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 \(R\left ( \tau \right ) ?\)
Answer:
(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
\begin{equation} \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} \end{equation}
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
(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