3.1 3.1.1
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3.1.2 3.1.2.1 Let \(\digamma \left ( g\left ( t\right ) \right ) \) be the Fourier Transform of \(g\left ( t\right ) \) , i.e. \(\digamma \left ( g\left ( t\right ) \right ) =G\left ( f\right ) \) . First we use the given hint and note that \(g\left ( t\right ) \) can be written as
follows
\[ g\left ( t\right ) =A\cos \left ( \frac {\pi t}{T}\right ) \ rect\left ( \frac {t}{T}\right ) \]
Start by writing \(\frac {\pi t}{T}\) as \(2\pi f_{0}t\) , where \(f_{0}=\frac {1}{2T}\) . Now using the property that multiplication in time domain is
the same as convolution in frequency domain, we obtain
\begin{equation} G\left ( f\right ) =\digamma \left ( A\cos \left ( 2\pi f_{0}t\right ) \right ) \otimes \digamma \left ( rect\left ( \frac {t}{T}\right ) \right ) \tag {1}\end{equation}
\begin{align*} \digamma \left ( A\cos \left ( 2\pi f_{0}t\right ) \right ) & =A\ \digamma \left ( \cos \left ( 2\pi f_{0}t\right ) \right ) \\ & =A\ \digamma \left ( \frac {e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}}{2}\right ) \\ & =\frac {A}{2}\ \digamma \left ( e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}\right ) \\ & =\frac {A}{2}\left [ \ \digamma \left ( e^{j2\pi f_{0}t}\right ) +\ \digamma \left ( e^{-j2\pi f_{0}t}\right ) \right ] \end{align*}
But \(\digamma \left ( e^{j2\pi f_{0}t}\right ) =\delta \left ( f-f_{0}\right ) \) and \(\digamma \left ( e^{-j2\pi f_{0}t}\right ) =\delta \left ( f+f_{0}\right ) \) hence the above becomes
\begin{equation} \digamma \left ( A\cos \left ( 2\pi f_{0}t\right ) \right ) =\frac {A}{2}\left [ \ \delta \left ( f-f_{0}\right ) +\ \delta \left ( f+f_{0}\right ) \right ] \tag {2}\end{equation}
\[ G\left ( f\right ) =\frac {A}{2}\left [ \ \delta \left ( f-f_{0}\right ) +\ \delta \left ( f+f_{0}\right ) \right ] \otimes \digamma \left ( rect\left ( \frac {t}{T}\right ) \right ) \]
But \(\digamma \left ( rect\left ( \frac {t}{T}\right ) \right ) =T\operatorname {sinc}\left ( fT\right ) \) , hence the above
becomes
\[ \digamma \left ( g\left ( t\right ) \right ) =\frac {A}{2}\left [ \ \delta \left ( f-f_{0}\right ) +\ \delta \left ( f+f_{0}\right ) \right ] \otimes T\operatorname {sinc}\left ( fT\right ) \]
Now using the property of convolution with a delta, we obtain
\[ \fbox {$G\left ( f\right ) =\frac {AT}{2}\left [ \ \operatorname {sinc}\left ( \left ( f-f_{0}\right ) T\right ) +\ \operatorname {sinc}\left ( \left ( f+f_{0}\right ) T\right ) \right ] $}\]
note: by doing more
trigonometric manipulations, the above can be written as
\[ \fbox {$G\left ( f\right ) =\frac {2AT\cos \left ( \pi fT\right ) }{\pi \left ( 1-4f^{2}T^{2}\right ) }$}\]
3.1.2.2 Apply the time shifting property \(g\left ( t\right ) \Longleftrightarrow G\left ( f\right ) \) , hence \(g\left ( t-t_{0}\right ) \Longleftrightarrow e^{-j2\pi ft_{0}}G\left ( f\right ) \)
From part(a) we found that \(\digamma \left ( g\left ( t\right ) \right ) =\frac {AT}{2}\left [ \ \operatorname {sinc}\left ( \left ( f-f_{0}\right ) T\right ) +\ \operatorname {sinc}\left ( \left ( f+f_{0}\right ) T\right ) \right ] \) , so in this part, the function in part(a) is shifted in time to the right
by amount \(\frac {T}{2}\) , let the new function be \(h\left ( t\right ) \,,\) hence we need to multiply \(G\left ( f\right ) \) by \(e^{-j2\pi f\frac {T}{2}}\) ,hence
\begin{align*} \digamma \left ( g\left ( t-\frac {T}{2}\right ) \right ) & =F\left ( h\left ( t\right ) \right ) \\ & =H\left ( f\right ) \\ & =e^{-j\pi fT}\left ( \frac {AT}{2}\left [ \ \operatorname {sinc}\left ( \left ( f-f_{0}\right ) T\right ) +\ \operatorname {sinc}\left ( \left ( f+f_{0}\right ) T\right ) \right ] \right ) \end{align*}
3.1.2.3 Using the time scaling property \(g\left ( t\right ) \Longleftrightarrow G\left ( f\right ) \) , hence \(g\left ( at\right ) \Longleftrightarrow \frac {1}{\left \vert a\right \vert }G\left ( \frac {f}{a}\right ) \) , and since we found in part(b) that \(H\left ( f\right ) =e^{-j\pi fT}\left ( \frac {AT}{2}\left [ \ \operatorname {sinc}\left ( \left ( f-f_{0}\right ) T\right ) +\ \operatorname {sinc}\left ( \left ( f+f_{0}\right ) T\right ) \right ] \right ) \) , hence
\[ \fbox {$\digamma \left \{ h\left ( at\right ) \right \} =\frac {1}{\left \vert a\right \vert }e^{-j\pi \frac {f}{a}T}\left ( \frac {AT}{2}\left [ \ \operatorname {sinc}\left ( \left ( \frac {f}{a}-f_{0}\right ) T\right ) +\ \operatorname {sinc}\left ( \left ( \frac {f}{a}+f_{0}\right ) T\right ) \right ] \right ) $}\]
3.1.2.4 Let \(f\left ( t\right ) \) be the function which is shown in figure 2.4c, we see that
\[ f\left ( t\right ) =-h\left ( -t\right ) \]
where \(h\left ( t\right ) \) is the function shown in
figure 2.4(b). We found in part(b) that
\[ H\left ( f\right ) =e^{-j\pi fT}\left ( \frac {AT}{2}\left [ \ \operatorname {sinc}\left ( \left ( f-f_{0}\right ) T\right ) +\ \operatorname {sinc}\left ( \left ( f+f_{0}\right ) T\right ) \right ] \right ) \]
Now using the property that \(h\left ( t\right ) \Longleftrightarrow H\left ( f\right ) \) then \(h\left ( -t\right ) \Longleftrightarrow \frac {1}{\left \vert -1\right \vert }H\left ( -f\right ) =H\left ( -f\right ) \) , hence
\[ \fbox {$\digamma \left \{ f\left ( t\right ) \right \} =-e^{j\pi fT}\left ( \frac {AT}{2}\left [ \ \operatorname {sinc}\left ( \left ( -f-f_{0}\right ) T\right ) +\ \operatorname {sinc}\left ( \left ( -f+f_{0}\right ) T\right ) \right ] \right ) $}\]
3.1.2.5 This function, call it \(g_{1}\left ( t\right ) ,\) is the sum of the functions shown in figure 2.4(b) and figure 2.4(c), then the
Fourier transform of \(g_{1}\left ( t\right ) \) is the sum of the Fourier transforms of the functions in these two figures
(using the linearity of the Fourier transforms). Hence
\begin{align*} \digamma \left ( g_{1}\left ( t\right ) \right ) & =e^{-j\pi fT}\left ( \frac {AT}{2}\left [ \ \operatorname {sinc}\left ( \left ( f-f_{0}\right ) T\right ) +\ \operatorname {sinc}\left ( \left ( f+f_{0}\right ) T\right ) \right ] \right ) \\ & -e^{j\pi fT}\left ( \frac {AT}{2}\left [ \ \operatorname {sinc}\left ( \left ( -f-f_{0}\right ) T\right ) +\ \operatorname {sinc}\left ( \left ( -f+f_{0}\right ) T\right ) \right ] \right ) \end{align*}
The above can be simplified to
\begin{align*} \digamma \left ( g_{1}\left ( t\right ) \right ) & =\frac {AT}{2}\left ( \operatorname {sinc}\left ( \left ( f+f_{0}\right ) T\right ) \left [ e^{j\pi fT}+\ e^{-j\pi fT}\right ] +\operatorname {sinc}\left ( \left ( f-f_{0}\right ) T\right ) \left [ e^{j\pi fT}+e^{-j\pi fT}\right ] \right ) \\ & =\frac {AT}{2}\left ( \operatorname {sinc}\left ( \left ( f+f_{0}\right ) T\right ) \left [ 2\cos \left ( \pi fT\right ) \right ] +\operatorname {sinc}\left ( \left ( f-f_{0}\right ) T\right ) \left [ 2\cos \left ( \pi fT\right ) \right ] \right ) \end{align*}
Hence
\[ \fbox {$\digamma \left ( g_{1}\left ( t\right ) \right ) =AT\cos \left ( \pi fT\right ) \left [ \operatorname {sinc}\left ( \left ( f+f_{0}\right ) T\right ) +\operatorname {sinc}\left ( \left ( f-f_{0}\right ) T\right ) \right ] $}\]
3.1.3
Given \(g\left ( t\right ) =e^{-t}\sin \left ( 2\pi f_{c}t\right ) u\left ( t\right ) \) find \(\digamma \left ( g\left ( t\right ) \right ) \) Answer:
\begin{equation} \digamma \left ( g\left ( t\right ) \right ) =\digamma \left ( e^{-t}u\left ( t\right ) \right ) \otimes \digamma \left ( \sin \left ( 2\pi f_{c}t\right ) \right ) \tag {1}\end{equation}
\begin{equation} \digamma \left ( \sin \left ( 2\pi f_{0}t\right ) \right ) =\frac {1}{2j}\left [ \delta \left ( f-f_{c}\right ) -\delta \left ( f+f_{c}\right ) \right ] \tag {2}\end{equation}
\begin{align} \digamma \left ( e^{-t}u\left ( t\right ) \right ) & ={\displaystyle \int \limits _{0}^{\infty }} e^{-t}e^{-j2\pi ft}dt={\displaystyle \int \limits _{0}^{\infty }} e^{-t\left ( 1+j2\pi f\right ) }dt\nonumber \\ & =\frac {\left [ e^{-t\left ( 1+j2\pi f\right ) }\right ] _{0}^{\infty }}{-\left ( 1+j2\pi f\right ) }=\frac {0-1}{-\left ( 1+j2\pi f\right ) }\nonumber \\ & =\frac {1}{1+j2\pi f} \tag {3}\end{align}
Substitute (2) and (3) into (1) we obtain
\begin{align*} \digamma \left ( g\left ( t\right ) \right ) & =\frac {1}{2j}\left [ \delta \left ( f-f_{c}\right ) -\delta \left ( f+f_{c}\right ) \right ] \otimes \frac {1}{1+j2\pi f}\\ & =\frac {1}{2j}\left [ \frac {1}{1+j2\pi \left ( f-f_{c}\right ) }-\frac {1}{1+j2\pi \left ( f+f_{c}\right ) }\right ] \end{align*}
3.1.4 3.1.4.1 \begin{align*} g\left ( t\right ) & =A\ rect\left ( \frac {t}{T}-\frac {1}{2}\right ) \\ & =A\ rect\left ( \frac {t-\frac {T}{2}}{T}\right ) \end{align*}
hence it is a rect function with duration \(T\) and centered at \(\frac {T}{2}\) and it has height \(A\)
\begin{align} g_{e} & =\frac {g\left ( t\right ) +g\left ( -t\right ) }{2}\tag {1}\\ g_{o} & =\frac {g\left ( t\right ) -g\left ( -t\right ) }{2}\nonumber \end{align}
Hence \(g_{e}=\frac {1}{2}\left [ A\ rect\left ( \frac {t}{T}-\frac {1}{2}\right ) +A\ rect\left ( \frac {-t}{T}-\frac {1}{2}\right ) \right ] \) which is a rectangular pulse of duration \(2T\) and centered at zero and height \(A\)
\(g_{o}=\frac {1}{2}\left [ A\ rect\left ( \frac {t}{T}-\frac {1}{2}\right ) -A\ rect\left ( \frac {-t}{T}-\frac {1}{2}\right ) \right ] \) which is shown in the figure below
3.1.4.2 \begin{align} \digamma \left ( g\left ( t\right ) \right ) & =\digamma \left ( A\ rect\left ( \frac {t-\frac {T}{2}}{T}\right ) \right ) \nonumber \\ & =AT\ \operatorname {sinc}\left ( fT\right ) \ e^{-j2\pi f\frac {T}{2}}\nonumber \\ & =AT\ \operatorname {sinc}\left ( fT\right ) \ e^{-j\pi fT} \tag {2}\end{align}
Now using the property that \(g\left ( t\right ) \Leftrightarrow G\left ( f\right ) \) , then \(g\left ( -t\right ) \Leftrightarrow G\left ( -f\right ) \) , then we write
\begin{align} \digamma \left ( g\left ( -t\right ) \right ) & =G\left ( -f\right ) \nonumber \\ & =AT\ \operatorname {sinc}\left ( -fT\right ) \ e^{j\pi fT} \tag {3}\end{align}
Now, using linearity of Fourier transform, then from (1) we obtain
\begin{align*} \digamma \left ( g_{e}\left ( t\right ) \right ) & =\digamma \left ( \frac {g\left ( t\right ) +g\left ( -t\right ) }{2}\right ) \\ & =\frac {1}{2}\left [ \digamma \left ( g\left ( t\right ) \right ) +\digamma \left ( g\left ( -t\right ) \right ) \right ] \\ & =\frac {1}{2}\left [ AT\ \operatorname {sinc}\left ( fT\right ) \ e^{-j\pi fT}+AT\ \operatorname {sinc}\left ( -fT\right ) \ e^{j\pi fT}\right ] \\ & =\frac {AT}{2}\left [ \operatorname {sinc}\left ( fT\right ) \ e^{-j\pi fT}+\operatorname {sinc}\left ( -fT\right ) \ e^{j\pi fT}\right ] \end{align*}
now \(\operatorname {sinc}\left ( -fT\right ) =\frac {\sin \left ( -\pi fT\right ) }{-\pi fT}=\frac {-\sin \left ( \pi fT\right ) }{-\pi fT}=\operatorname {sinc}\left ( fT\right ) \) , hence the above becomes
\begin{align*} \digamma \left ( g_{e}\left ( t\right ) \right ) & =\frac {AT\operatorname {sinc}\left ( fT\right ) }{2}\left [ \ e^{-j\pi fT}+\ e^{j\pi fT}\right ] \\ & =\frac {AT\operatorname {sinc}\left ( fT\right ) }{2}\left [ \ 2\cos \left ( \pi fT\right ) \right ] \end{align*}
Hence
\[ \fbox {$\digamma \left ( g_{e}\left ( t\right ) \right ) =AT\operatorname {sinc}\left ( fT\right ) \cos \left ( \pi fT\right ) $}\]
Now to find the Fourier transform of the odd part
\[ g_{o}=\frac {g\left ( t\right ) -g\left ( -t\right ) }{2}\]
Hence
\begin{align*} \digamma \left ( g_{o}\left ( t\right ) \right ) & =\digamma \left ( \frac {g\left ( t\right ) -g\left ( -t\right ) }{2}\right ) \\ & =\frac {1}{2}\left [ \digamma \left ( g\left ( t\right ) \right ) -\digamma \left ( g\left ( -t\right ) \right ) \right ] \\ & =\frac {1}{2}\left [ AT\ \operatorname {sinc}\left ( fT\right ) \ e^{-j\pi fT}-AT\ \operatorname {sinc}\left ( -fT\right ) \ e^{j\pi fT}\right ] \\ & =\frac {AT}{2}\left [ \operatorname {sinc}\left ( fT\right ) \ e^{-j\pi fT}-\operatorname {sinc}\left ( fT\right ) \ e^{j\pi fT}\right ] \\ & =\frac {AT\operatorname {sinc}\left ( fT\right ) }{2}\left [ \ e^{-j\pi fT}-\ e^{j\pi fT}\right ] \\ & =\frac {-AT\operatorname {sinc}\left ( fT\right ) }{2}\left [ \ e^{j\pi fT}-e^{-j\pi fT}\right ] \\ & =\frac {-AT\operatorname {sinc}\left ( fT\right ) }{2}\left [ \ 2j\sin \left ( \pi fT\right ) \right ] \end{align*}
Hence
\[ \fbox {$\digamma \left ( g_{o}\left ( t\right ) \right ) =-jAT\operatorname {sinc}\left ( fT\right ) \sin \left ( \pi fT\right ) $}\]
3.1.5
\[ G\left ( f\right ) =\left \vert G\left ( f\right ) \right \vert e^{j\arg \left ( G\left ( f\right ) \right ) }\]
Hence from the diagram given, we write
\[ G\left ( f\right ) =\left \{ \begin {array} [c]{ccc}1\times e^{j\frac {\pi }{2}} & & -W\leq f<0\\ 1\times e^{-j\frac {\pi }{2}} & & 0\leq f\leq W \end {array} \right . \]
Therefore, we can use a rect function now to express \(G\left ( f\right ) \) over
the whole \(f\) range as follows
\[ G\left ( f\right ) =e^{j\frac {\pi }{2}}\ rect\left ( \frac {f+\frac {W}{2}}{W}\right ) -e^{-j\frac {\pi }{2}}rect\left ( \frac {f-\frac {W}{2}}{W}\right ) \]
Now, noting that \(\delta \left ( t-t_{0}\right ) \Leftrightarrow e^{-j2\pi t_{0}}\) and \(\delta \left ( t+t_{0}\right ) \Leftrightarrow e^{j2\pi t_{0}}\) and \(W\operatorname {sinc}\left ( tW\right ) \Leftrightarrow rect\left ( \frac {f}{W}\right ) \) and noting that shift in frequency by
\(\frac {W}{2}\) becomes multiplication by \(e^{-j2\pi t\frac {W}{2}}\) , then now we write
\begin{align*} g\left ( t\right ) & =\digamma ^{-1}\left ( e^{j\frac {\pi }{2}}\ rect\left ( \frac {f+\frac {W}{2}}{W}\right ) \right ) -\digamma ^{-1}\left ( e^{-j\frac {\pi }{2}}rect\left ( \frac {f-\frac {W}{2}}{W}\right ) \right ) \\ & =\digamma ^{-1}\left ( e^{j\frac {\pi }{2}}\right ) \ \otimes \digamma ^{-1}\left ( rect\left ( \frac {f+\frac {W}{2}}{W}\right ) \right ) -\digamma ^{-1}\left ( e^{-j\frac {\pi }{2}}\right ) \ \otimes \digamma ^{-1}\left ( rect\left ( \frac {f-\frac {W}{2}}{W}\right ) \right ) \end{align*}
Hence
\begin{align*} g\left ( t\right ) & =\left [ \delta \left ( t+\frac {\pi }{2}\right ) \otimes W\operatorname {sinc}\left ( tW\right ) e^{-j2\pi t\frac {W}{2}}\right ] -\left [ \delta \left ( t-\frac {\pi }{2}\right ) \otimes W\operatorname {sinc}\left ( tW\right ) e^{j2\pi t\frac {W}{2}}\right ] \\ & =W\operatorname {sinc}\left ( \left ( t+\frac {\pi }{2}\right ) W\right ) e^{-j2\pi \left ( t+\frac {\pi }{2}\right ) \frac {W}{2}}-W\operatorname {sinc}\left ( \left ( t-\frac {\pi }{2}\right ) W\right ) e^{j2\pi \left ( t-\frac {\pi }{2}\right ) \frac {W}{2}}\\ & =W\operatorname {sinc}\left ( \left ( t+\frac {\pi }{2}\right ) W\right ) e^{-j\pi Wt-j\pi W\frac {\pi }{2}}-W\operatorname {sinc}\left ( \left ( t-\frac {\pi }{2}\right ) W\right ) e^{j\pi Wt-j\pi W\frac {\pi }{2}}\end{align*}
Hence
\[ \fbox {$g\left ( t\right ) =We^{-\frac {j\pi ^{2}W}{2}}\left ( \operatorname {sinc}\left ( \left ( t+\frac {\pi }{2}\right ) W\right ) e^{-j\pi Wt}-\operatorname {sinc}\left ( \left ( t-\frac {\pi }{2}\right ) W\right ) e^{j\pi Wt}\right ) $}\]
3.1.6
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