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} But \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} Substitute (2) into (1) we obtain 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 ) }$}
Apply the time shifting property g\left ( t\right ) \Longleftrightarrow G\left ( f\right ) , hence g\left ( t-t_{0}\right ) \Longleftrightarrow e^{-j2\pi ft_{0}}G\left ( f\right )
From part(a) we found that \digamma \left ( g\left ( t\right ) \right ) =\frac{AT}{2}\left [ \ \operatorname{sinc}\left ( \left ( f-f_{0}\right ) T\right ) +\ \operatorname{sinc}\left ( \left ( f+f_{0}\right ) T\right ) \right ] , so in this part, the function in part(a) is shifted in time to the right by amount \frac{T}{2}, let the new function be h\left ( t\right ) \,,hence we need to multiply G\left ( f\right ) by e^{-j2\pi f\frac{T}{2}} ,hence\begin{align*} \digamma \left ( g\left ( t-\frac{T}{2}\right ) \right ) & =F\left ( h\left ( t\right ) \right ) \\ & =H\left ( f\right ) \\ & =e^{-j\pi fT}\left ( \frac{AT}{2}\left [ \ \operatorname{sinc}\left ( \left ( f-f_{0}\right ) T\right ) +\ \operatorname{sinc}\left ( \left ( f+f_{0}\right ) T\right ) \right ] \right ) \end{align*}
Using the time scaling property g\left ( t\right ) \Longleftrightarrow G\left ( f\right ) , hence g\left ( at\right ) \Longleftrightarrow \frac{1}{\left \vert a\right \vert }G\left ( \frac{f}{a}\right ) , and since we found in part(b) that H\left ( f\right ) =e^{-j\pi fT}\left ( \frac{AT}{2}\left [ \ \operatorname{sinc}\left ( \left ( f-f_{0}\right ) T\right ) +\ \operatorname{sinc}\left ( \left ( f+f_{0}\right ) T\right ) \right ] \right ) , hence \fbox{$\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 ) $}
Let f\left ( t\right ) be the function which is shown in figure 2.4c, we see that f\left ( t\right ) =-h\left ( -t\right ) where h\left ( t\right ) is the function shown in figure 2.4(b). We found in part(b) that H\left ( f\right ) =e^{-j\pi fT}\left ( \frac{AT}{2}\left [ \ \operatorname{sinc}\left ( \left ( f-f_{0}\right ) T\right ) +\ \operatorname{sinc}\left ( \left ( f+f_{0}\right ) T\right ) \right ] \right ) Now using the property that h\left ( t\right ) \Longleftrightarrow H\left ( f\right ) then h\left ( -t\right ) \Longleftrightarrow \frac{1}{\left \vert -1\right \vert }H\left ( -f\right ) =H\left ( -f\right ) , hence \fbox{$\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 ) $}
This function, call it g_{1}\left ( t\right ) , is the sum of the functions shown in figure 2.4(b) and figure 2.4(c), then the Fourier transform of g_{1}\left ( t\right ) is the sum of the Fourier transforms of the functions in these two figures (using the linearity of the Fourier transforms). Hence \begin{align*} \digamma \left ( g_{1}\left ( t\right ) \right ) & =e^{-j\pi fT}\left ( \frac{AT}{2}\left [ \ \operatorname{sinc}\left ( \left ( f-f_{0}\right ) T\right ) +\ \operatorname{sinc}\left ( \left ( f+f_{0}\right ) T\right ) \right ] \right ) \\ & -e^{j\pi fT}\left ( \frac{AT}{2}\left [ \ \operatorname{sinc}\left ( \left ( -f-f_{0}\right ) T\right ) +\ \operatorname{sinc}\left ( \left ( -f+f_{0}\right ) T\right ) \right ] \right ) \end{align*}
The above can be simplified to\begin{align*} \digamma \left ( g_{1}\left ( t\right ) \right ) & =\frac{AT}{2}\left ( \operatorname{sinc}\left ( \left ( f+f_{0}\right ) T\right ) \left [ e^{j\pi fT}+\ e^{-j\pi fT}\right ] +\operatorname{sinc}\left ( \left ( f-f_{0}\right ) T\right ) \left [ e^{j\pi fT}+e^{-j\pi fT}\right ] \right ) \\ & =\frac{AT}{2}\left ( \operatorname{sinc}\left ( \left ( f+f_{0}\right ) T\right ) \left [ 2\cos \left ( \pi fT\right ) \right ] +\operatorname{sinc}\left ( \left ( f-f_{0}\right ) T\right ) \left [ 2\cos \left ( \pi fT\right ) \right ] \right ) \end{align*}
Hence \fbox{$\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 ] $}
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} But \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} and\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*}
\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
\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 ) $}
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 ) $}