### 3.4HW 4

3.4.1  questions and hints
3.4.2  Problem 1
3.4.3  Problem 2
3.4.4  Problem 3
3.4.5  Problem 4
3.4.6  problem 5
3.4.7  Key solution
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#### 3.4.2Problem 1

Solution Using transfer function cascading, then the overall transfer function for the system can be written as$$H\left ( f\right ) =H_{1}\left ( f\right ) H_{1}\left ( f\right ) =\left [ H_{1}\left ( f\right ) \right ] ^{2} \tag{1}$$ Where $H_{1}\left ( f\right ) =\frac{Z\left ( f\right ) }{X\left ( f\right ) }$ Where \begin{align} Z\left ( f\right ) & =\digamma \left \{{\displaystyle \int \limits _{-\infty }^{t}} w\left ( \tau \right ) d\tau \right \} \nonumber \\ & =\frac{1}{j2\pi f}W\left ( f\right ) +\frac{W\left ( 0\right ) }{2}\delta \left ( f\right ) \tag{2} \end{align}

Where \begin{align} W\left ( f\right ) & =\digamma \left \{ x\left ( t\right ) -x\left ( t-T\right ) \right \} \nonumber \\ & =X\left ( f\right ) -X\left ( f\right ) e^{-j2\pi fT}\nonumber \\ & =X\left ( f\right ) \left [ 1-e^{-j2\pi fT}\right ] \tag{3} \end{align}

substitute (3) into (2) we obtain\begin{align} Z\left ( f\right ) & =\frac{1}{j2\pi f}X\left ( f\right ) \left [ 1-e^{-j2\pi fT}\right ] +\frac{X\left ( 0\right ) \overset{=0}{\overbrace{\left [ 1-e^{-j2\pi 0T}\right ] }}}{2}\delta \left ( f\right ) \nonumber \\ & =\frac{1}{j2\pi f}X\left ( f\right ) \left [ 1-e^{-j2\pi fT}\right ] \tag{4} \end{align}

Hence \begin{align*} H_{1}\left ( f\right ) & =\frac{Z\left ( f\right ) }{X\left ( f\right ) }\\ & =\frac{\frac{1}{j2\pi f}X\left ( f\right ) \left [ 1-e^{-j2\pi fT}\right ] }{X\left ( f\right ) } \end{align*}

Hence$\fbox{H_{1}\left ( f\right ) =\frac{1}{j2\pi f}\left [ 1-e^{-j2\pi fT}\right ] }$ Hence from (1) \begin{align*} H\left ( f\right ) & =\left ( \frac{1}{j2\pi f}\left [ 1-e^{-j2\pi fT}\right ] \right ) ^{2}\\ & =\frac{1}{-4\pi ^{2}f^{2}}\left [ 1-e^{-j2\pi fT}\right ] ^{2}\\ & =\frac{-1}{\left ( 2\pi f\right ) ^{2}}\left [ 1-2e^{-j2\pi fT}+e^{-j4\pi fT}\right ] \end{align*}

Hence$\fbox{H\left ( f\right ) =\frac{1}{\left ( 2\pi f\right ) ^{2}}\left [ 2e^{-j2\pi fT}-e^{-j4\pi fT}-1\right ] }$

#### 3.4.3Problem 2

3.4.3.1  Part(a)
3.4.3.2  Part (b)

##### 3.4.3.1 Part(a)

Transfer function for each stage is $$H_{i}\left ( f\right ) =\frac{Y_{i}\left ( f\right ) }{X_{i}\left ( f\right ) }=\frac{1}{1+j2\pi fRC}$$

Since $$RC=\tau _{0}$$, hence$H_{i}\left ( f\right ) =\frac{1}{1+j2\pi f\tau _{0}}$ Then, for $$N$$ stages, the overall transfer function is$H\left ( f\right ) =H_{1}\left ( f\right ) H_{2}\left ( f\right ) \cdots H_{N}\left ( f\right )$ Since they are identical stages, then the transfer function of each stage is the same, and the above becomes$H\left ( f\right ) =\left ( \frac{1}{1+j2\pi f\tau _{0}}\right ) ^{N}$ Hence the amplitude of the response is given by\begin{align*} \left \vert H\left ( f\right ) \right \vert & =\left ( \frac{1}{\left \vert 1+j2\pi f\tau _{0}\right \vert }\right ) ^{N}\\ & =\left ( \frac{1}{\sqrt{1^{2}+\left ( 2\pi f\tau _{0}\right ) ^{2}}}\right ) ^{N}\\ & =\left ( \frac{1}{\left ( 1+4\pi ^{2}f^{2}\tau _{0}^{2}\right ) ^{\frac{1}{2}}}\right ) ^{N}\\ & =\frac{1}{\left ( 1+4\pi ^{2}f^{2}\tau _{0}^{2}\right ) ^{\frac{N}{2}}} \end{align*}

Let $$\tau _{0}^{2}=\frac{\tau ^{2}}{4\pi ^{2}N}$$, the above becomes$$\left \vert H\left ( f\right ) \right \vert =\frac{1}{\left ( 1+\frac{f^{2}\tau ^{2}}{N}\right ) ^{\frac{N}{2}}} \tag{1}$$

##### 3.4.3.2 Part (b)

Let $$\alpha =f^{2}\tau ^{2},\beta =\frac{1}{2}$$, then (1) becomes$\left \vert H\left ( f\right ) \right \vert =\frac{1}{\left ( 1+\frac{\alpha }{N}\right ) ^{\beta N}}$ But $$\lim _{N\rightarrow \infty }\frac{1}{\left ( 1+\frac{\alpha }{N}\right ) ^{\beta N}}=e^{\alpha \beta },$$hence\begin{align*} \left \vert H\left ( f\right ) \right \vert & =\frac{1}{e^{\frac{f^{2}\tau ^{2}}{2}}}\\ & =e^{-\ \frac{f^{2}\tau ^{2}}{2}} \end{align*}

Which is what we are asked to show.

#### 3.4.4Problem 3

3.4.4.1  Part(a)
3.4.4.2  Part(b)

##### 3.4.4.1 Part(a)

(a) $$g\left ( t\right ) =\operatorname{sinc}\left ( t\right )$$

$$g_{+}\left ( t\right ) =g\left ( t\right ) +j\hat{g}\left ( t\right ) \tag{1}$$ Where $$\hat{g}\left ( t\right )$$ is Hilbert transform of $$g\left ( t\right )$$ deﬁned as $$\hat{g}\left ( t\right ) =g\left ( t\right ) \otimes \frac{1}{\pi t}$$\begin{align*} \hat{G}\left ( f\right ) & =-j\ sgn\left ( f\right ) \ G\left ( f\right ) \\ & =-j\ sgn\left ( f\right ) \ rect\left ( f\right ) \end{align*}

Now ﬁnd the inverse Fourier transform.

I derive the above to answer problem 4 part (b). The answer is the following (please see problem 4 part(b) for the derivation

$\hat{g}\left ( t\right ) =\frac{1}{\pi t}\left ( 1-\cos \pi t\right )$ In the above, I used $$\operatorname{sinc}\left ( t\right ) \equiv \frac{\sin \pi t}{\pi t}$$. If one uses $$\operatorname{sinc}\left ( t\right ) \equiv \frac{\sin t}{t}$$ then the answer becomes$$\hat{g}\left ( t\right ) =\frac{1}{t}\left ( 1-\cos t\right ) \tag{2}$$ The problem statement seems to want us to use the second deﬁnition of $$\operatorname{sinc}\left ( t\right )$$, so I will continue the rest of the solution using (1).

Substitute (2) into (1) we obtain\begin{align*} g_{+}\left ( t\right ) & =\operatorname{sinc}\left ( t\right ) +j\frac{1}{t}\left ( 1-\cos t\right ) \\ & =\frac{\sin \left ( t\right ) }{t}+j\frac{1}{t}\left ( 1-\frac{e^{jt}+e^{-jt}}{2}\right ) \\ & =\frac{1}{t}\frac{e^{jt}-e^{-jt}}{2j}+\frac{1}{t}\left ( j+\frac{e^{jt}+e^{-jt}}{2j}\right ) \\ & =\frac{1}{t}\frac{e^{jt}-e^{-jt}}{2j}+\frac{j}{t}+\frac{1}{t}\frac{e^{jt}+e^{-jt}}{2j}\\ & =\frac{1}{t}\frac{e^{jt}}{2j}+\frac{j}{t}+\frac{1}{t}\frac{e^{jt}}{2j} \end{align*}

Hence$\fbox{g_{+}\left ( t\right ) =\frac{1}{t}\left ( j+e^{jt}\right ) }$

##### 3.4.4.2 Part(b)

$g\left ( t\right ) =\left [ 1+k\cos 2\pi f_{m}t\right ] \cos \left ( 2\pi f_{c}t\right )$

$g_{+}\left ( t\right ) =g\left ( t\right ) +j\hat{g}\left ( t\right )$ Where $$\hat{g}\left ( t\right )$$ is Hilbert transform of $$g\left ( t\right )$$ deﬁned as $$\hat{g}\left ( t\right ) =g\left ( t\right ) \otimes \frac{1}{\pi t}.$$$G_{+}\left ( f\right ) =\left \{ \begin{array} [c]{ccc}2G\left ( f\right ) & & f>0\\ G\left ( 0\right ) & & f=0\\ 0 & & f<0 \end{array} \right .$ But $$G\left ( f\right ) =\digamma \left [ 1+k\cos 2\pi f_{m}t\right ] \otimes \digamma \left [ \cos \left ( 2\pi f_{c}t\right ) \right ] \tag{1}$$ But $\digamma \left [ \cos \left ( 2\pi f_{c}t\right ) \right ] =\frac{1}{2}\left [ \delta \left ( f-f_{c}\right ) +\delta \left ( f+f_{c}\right ) \right ]$ and $\digamma \left [ 1+k\cos 2\pi f_{m}t\right ] =\delta \left ( f\right ) +\frac{k}{2}\left [ \delta \left ( f-f_{m}\right ) +\delta \left ( f+f_{m}\right ) \right ]$ Hence (1) becomes\begin{align*} G\left ( f\right ) & =\left \{ \delta \left ( f\right ) +\frac{k}{2}\left [ \delta \left ( f-f_{m}\right ) +\delta \left ( f+f_{m}\right ) \right ] \right \} \otimes \frac{1}{2}\left [ \delta \left ( f-f_{c}\right ) +\delta \left ( f+f_{c}\right ) \right ] \\ & =\delta \left ( f\right ) \otimes \frac{1}{2}\left [ \delta \left ( f-f_{c}\right ) +\delta \left ( f+f_{c}\right ) \right ] \\ & +\frac{k}{2}\left [ \delta \left ( f-f_{m}\right ) +\delta \left ( f+f_{m}\right ) \right ] \otimes \frac{1}{2}\left [ \delta \left ( f-f_{c}\right ) +\delta \left ( f+f_{c}\right ) \right ] \\ & \\ & =\frac{1}{2}\delta \left ( f\right ) \otimes \delta \left ( f-f_{c}\right ) +\\ & \frac{1}{2}\delta \left ( f\right ) \otimes \delta \left ( f+f_{c}\right ) +\\ & \frac{k}{4}\delta \left ( f-f_{m}\right ) \otimes \delta \left ( f-f_{c}\right ) +\\ & \frac{k}{4}\delta \left ( f-f_{m}\right ) \otimes \delta \left ( f+f_{c}\right ) +\\ & \frac{k}{4}\delta \left ( f+f_{m}\right ) \otimes \delta \left ( f-f_{c}\right ) +\\ & \frac{k}{4}\delta \left ( f+f_{m}\right ) \otimes \delta \left ( f+f_{c}\right ) \end{align*}

Hence\begin{align*} G\left ( f\right ) & =\frac{1}{2}\delta \left ( f+f_{c}\right ) +\\ & \frac{1}{2}\delta \left ( f-f_{c}\right ) +\\ & \frac{k}{4}\delta \left ( f-f_{m}+f_{c}\right ) +\\ & \frac{k}{4}\delta \left ( f-f_{m}-f_{c}\right ) +\\ & \frac{k}{4}\delta \left ( f+f_{m}+f_{c}\right ) +\\ & \frac{k}{4}\delta \left ( f+f_{m}-f_{c}\right ) \end{align*}

Hence for $$f>0$$ ,$$G_{+}\left ( f\right ) =2G\left ( f\right )$$ and we obtain$G_{+}\left ( f\right ) =\delta \left ( f-f_{c}\right ) +\frac{k}{2}\left [ \delta \left ( f-f_{m}+f_{c}\right ) +\delta \left ( f-f_{m}-f_{c}\right ) +\delta \left ( f+f_{m}+f_{c}\right ) +\delta \left ( f+f_{m}-f_{c}\right ) \right ]$ Then (since carrier frequency $$f_{c}>f_{m}$$), we could simplify the above, by keeping positive frequencies $$f$$

$G_{+}\left ( f\right ) =\delta \left ( f-f_{c}\right ) +\frac{k}{2}\left [ \delta \left ( f-f_{m}-f_{c}\right ) +\delta \left ( f+f_{m}-f_{c}\right ) \right ]$

or

$G_{+}\left ( f\right ) =\delta \left ( f-f_{c}\right ) +\frac{k}{2}\left [ \delta \left ( f-\left ( f_{m}+f_{c}\right ) \right ) +\delta \left ( f-\left ( f_{c}-f_{m}\right ) \right ) \right ]$

Hence

\begin{align*} g_{+}\left ( t\right ) & =e^{j2\pi f_{c}t}+\frac{k}{2}\left ( e^{j2\pi \left ( f_{m}+f_{c}\right ) t}+e^{j2\pi \left ( f_{c}-f_{m}\right ) t}\right ) \\ & =e^{j2\pi f_{c}t}+\frac{k}{2}\left ( e^{j2\pi f_{m}t}e^{j2\pi f_{c}t}+e^{j2\pi f_{c}t}e^{-j2\pi f_{m}t}\right ) \\ & =e^{j2\pi f_{c}t}\left [ 1+\frac{k}{2}\left ( e^{j2\pi f_{m}t}+e^{-j2\pi f_{m}t}\right ) \right ] \\ & =e^{j2\pi f_{c}t}\left [ 1+\frac{k}{2}\left ( 2\cos \left ( 2\pi f_{m}t\right ) \right ) \right ] \\ & =e^{j2\pi f_{c}t}\left [ 1+k\cos \left ( 2\pi f_{m}t\right ) \right ] \end{align*}

#### 3.4.5Problem 4

3.4.5.1  Part(a)
3.4.5.2  Part(b)

##### 3.4.5.1 Part(a)

$$g\left ( t\right ) =\delta \left ( t\right )$$

\begin{align*} \hat{g}\left ( t\right ) & =g\left ( t\right ) \otimes \frac{1}{\pi t}\\ & =\frac{1}{\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( \tau \right ) \frac{1}{t-\tau }d\tau \\ & =\frac{1}{\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( \tau \right ) \frac{1}{t}d\tau \\ & =\frac{1}{\pi t}{\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( \tau \right ) d\tau \\ & =\frac{1}{\pi t} \end{align*}

##### 3.4.5.2 Part(b)

And Since $$sgn\left ( f\right ) =-1$$ for $$f<0$$ and $$sgn\left ( f\right ) =1$$ for $$f>0$$ then

$\hat{G}\left ( f\right ) =-j\left [ -rect\left ( \frac{f+\frac{1}{4}}{\frac{1}{2}}\right ) +rect\left ( \frac{f-\frac{1}{4}}{\frac{1}{2}}\right ) \right ]$ Hence$$\hat{g}\left ( t\right ) =j\digamma ^{-1}\left [ rect\left ( \frac{f+\frac{1}{4}}{\frac{1}{2}}\right ) -rect\left ( \frac{f-\frac{1}{4}}{\frac{1}{2}}\right ) \right ] \tag{1}$$ But $$\digamma ^{-1}\left ( rect\left ( \frac{f+\frac{1}{4}}{\frac{1}{2}}\right ) \right ) =\frac{1}{2}\operatorname{sinc}\left ( \frac{1}{2}t\right ) e^{-j2\pi \frac{1}{4}t}$$ and $$\digamma ^{-1}\left ( rect\left ( \frac{f-\frac{1}{4}}{\frac{1}{2}}\right ) \right ) =\frac{1}{2}\operatorname{sinc}\left ( \frac{1}{2}t\right ) e^{+j2\pi \frac{1}{4}t}$$, hence (1) becomes

\begin{align*} \hat{g}\left ( t\right ) & =j\left [ \frac{1}{2}\operatorname{sinc}\left ( \frac{1}{2}t\right ) e^{-j2\pi \frac{1}{4}t}-\frac{1}{2}\operatorname{sinc}\left ( \frac{1}{2}t\right ) e^{+j2\pi \frac{1}{4}t}\right ] \\ & =\frac{1}{2}\operatorname{sinc}\left ( \frac{1}{2}t\right ) \left [ j\left ( e^{-j2\pi \frac{1}{4}t}-e^{j2\pi \frac{1}{4}t}\right ) \right ] \\ & =\frac{1}{2}\operatorname{sinc}\left ( \frac{1}{2}t\right ) \left [ \frac{e^{-j\frac{\pi }{2}t}-e^{j\frac{\pi }{2}t}}{-j}\right ] \\ & =\operatorname{sinc}\left ( \frac{1}{2}t\right ) \left [ \frac{e^{j\frac{\pi }{2}t}-e^{-j\frac{\pi }{2}t}}{2j}\right ] \\ & =\operatorname{sinc}\left ( \frac{1}{2}t\right ) \left [ \sin \frac{\pi }{2}t\right ] \end{align*}

But $$\operatorname{sinc}\left ( \frac{1}{2}t\right ) =\frac{\sin \frac{\pi t}{2}}{\frac{\pi t}{2}}$$ hence

\begin{align*} \hat{g}\left ( t\right ) & =\frac{\sin \frac{\pi t}{2}}{\frac{\pi t}{2}}\sin \frac{\pi }{2}t\\ & =\frac{2}{\pi t}\sin ^{2}\frac{\pi }{2}t\\ & =\frac{2}{\pi t}\left ( \frac{1}{2}-\frac{1}{2}\cos \pi t\right ) \\ & =\frac{1}{\pi t}\left ( 1-\cos \pi t\right ) \end{align*}

#### 3.4.6problem 5

$S_{g}\left ( f\right ) =rect\left ( \frac{f}{4}\right ) +rect\left ( \frac{f}{2}\right )$

$R_{g}\left ( \tau \right ) =\digamma ^{-1}\left ( S_{g}\left ( f\right ) \right )$ Hence\begin{align*} R_{g}\left ( \tau \right ) & =\digamma ^{-1}\left ( rect\left ( \frac{f}{4}\right ) +rect\left ( \frac{f}{2}\right ) \right ) \\ & =\digamma ^{-1}\left [ rect\left ( \frac{f}{4}\right ) \right ] +\digamma ^{-1}\left [ rect\left ( \frac{f}{2}\right ) \right ] \\ & =4\operatorname{sinc}\left ( 4t\right ) +2\operatorname{sinc}\left ( 2t\right ) \end{align*}