### 3.1HW 1

3.1.1  Questions
3.1.2  Problem 2.1
3.1.3  Problem 2.2
3.1.4  Problem 2.3
3.1.5  Problem 2.4
3.1.6  Key solution
PDF (letter size)
PDF (legal size)

#### 3.1.2Problem 2.1

3.1.2.1  part(a)
3.1.2.2  part(b)
3.1.2.3  part(c)
3.1.2.4  part(d)
3.1.2.5  part(e)

##### 3.1.2.1 part(a)

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 $$G\left ( f\right ) =\digamma \left ( A\cos \left ( 2\pi f_{0}t\right ) \right ) \otimes \digamma \left ( rect\left ( \frac{t}{T}\right ) \right ) \tag{1}$$ 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$$\digamma \left ( A\cos \left ( 2\pi f_{0}t\right ) \right ) =\frac{A}{2}\left [ \ \delta \left ( f-f_{0}\right ) +\ \delta \left ( f+f_{0}\right ) \right ] \tag{2}$$ 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$\fbox{G\left ( f\right ) =\frac{2AT\cos \left ( \pi fT\right ) }{\pi \left ( 1-4f^{2}T^{2}\right ) }}$

##### 3.1.2.2 part(b)

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 part(c)

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 part(d)

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 part(e)

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 $\fbox{\digamma \left ( g_{1}\left ( t\right ) \right ) =AT\cos \left ( \pi fT\right ) \left [ \operatorname{sinc}\left ( \left ( f+f_{0}\right ) T\right ) +\operatorname{sinc}\left ( \left ( f-f_{0}\right ) T\right ) \right ] }$

#### 3.1.3Problem 2.2

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:$$\digamma \left ( g\left ( t\right ) \right ) =\digamma \left ( e^{-t}u\left ( t\right ) \right ) \otimes \digamma \left ( \sin \left ( 2\pi f_{c}t\right ) \right ) \tag{1}$$ But $$\digamma \left ( \sin \left ( 2\pi f_{0}t\right ) \right ) =\frac{1}{2j}\left [ \delta \left ( f-f_{c}\right ) -\delta \left ( f+f_{c}\right ) \right ] \tag{2}$$ 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*}

#### 3.1.4Problem 2.3

3.1.4.1  part(a)
3.1.4.2  part(b)

##### 3.1.4.1 part(a)

\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 part(b)

\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$\fbox{\digamma \left ( g_{o}\left ( t\right ) \right ) =-jAT\operatorname{sinc}\left ( fT\right ) \sin \left ( \pi fT\right ) }$

#### 3.1.5Problem 2.4

$G\left ( f\right ) =\left \vert G\left ( f\right ) \right \vert e^{j\arg \left ( G\left ( f\right ) \right ) }$ 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$\fbox{g\left ( t\right ) =We^{-\frac{j\pi ^{2}W}{2}}\left ( \operatorname{sinc}\left ( \left ( t+\frac{\pi }{2}\right ) W\right ) e^{-j\pi Wt}-\operatorname{sinc}\left ( \left ( t-\frac{\pi }{2}\right ) W\right ) e^{j\pi Wt}\right ) }$