### 3.3HW 3

3.3.1  questions
3.3.2  Problem 2.7
3.3.3  Problem 2.8
3.3.4  Problem 2.9
3.3.5  Problem 2.15
3.3.6  Problem 2.17
3.3.7  Problem 2.18
3.3.8  Key solution
PDF (letter size)
PDF (legal size)

#### 3.3.2Problem 2.7

3.3.2.1  part(a)

Problem An average reading power meter is connected to output of transmitter. Transmitter output is fed into $$75\Omega$$ resistive load and the wattmeter read $$67W$$

(a) What is power in dBm units?

(b) What is power in dBk units?

(c) What is the value in dBmV units?

##### 3.3.2.1 part(a)

\begin{align*} P_{dbm} & =10\log _{10}P_{m}\\ & =10\log _{10}\left ( 67000\right ) \\ & =\fbox{$48.2607$}\ dbm \end{align*}

(b)\begin{align*} P_{dbk} & =10\log _{10}P_{k}\\ & =10\log _{10}\left ( 0.067\right ) \\ & =\fbox{$-11.7393$}\ dbk \end{align*}

(c)

$P=\frac{V^{2}}{R}$

Hence

$10\log _{10}P=20\log _{10}V-10\log _{10}R$

Hence

$20\log _{10}V=10\log _{10}P+10\log _{10}R$

so

\begin{align*} 20\log _{10}V & =10\log _{10}67000+10\log _{10}75000\\ & =\fbox{$97.0114$ dbmV} \end{align*}

#### 3.3.3Problem 2.8

Assume that a waveform with known rms value $$V_{rms}$$ is applied across a $$50\Omega$$ load. Derive a formula that can be used to computer the $$dbm$$ value from $$V_{rms}$$

$P\left ( watt\right ) =\frac{V_{rms}^{2}\left ( V\right ) }{R\left ( \Omega \right ) }$

Hence

\begin{align*} P_{dbm} & =10\log _{10}\left ( 10^{3}\times P_{watt}\right ) \\ & =10\log _{10}\frac{10^{3}\times V_{rms}^{2}\left ( V\right ) }{R\left ( \Omega \right ) }\\ & =10\left ( \log _{10}10^{3}V_{rms}^{2}-\log _{10}R\right ) \\ & =10\left ( \log _{10}10^{3}+\log _{10}V_{rms}^{2}-\log _{10}R\right ) \\ & =10\left ( 3+2\log _{10}V_{rms}-\log _{10}R\right ) \end{align*}

Hence

$\fbox{P_{dbm}=30+20\log _{10}V_{rms}-10\log _{10}R}$

When $$R=50\Omega$$, we obtain

\begin{align*} P_{dbm} & =30+20\log _{10}V_{rms}-10\log _{10}50\\ & =30+20\log _{10}V_{rms}-16.9897\\ & =13.0103+20\log _{10}V_{rms} \end{align*}

#### 3.3.4Problem 2.9

\begin{align*} Gain(db) & =10\log _{10}\frac{P_{L}}{P_{i}}\\ & =10\log _{10}\frac{\left ( \frac{V_{rms}^{2}}{R_{L}}\right ) }{I_{rms}^{2}R_{in}}\\ & =10\log _{10}\frac{\left ( \frac{10^{2}}{50}\right ) }{\left ( 0.5\times 10^{-3}\right ) ^{2}\times 2000}\\ & =10\log _{10}\frac{10^{5}}{25}\\ & =10\left ( \log _{10}10^{5}-\log _{10}25\right ) \\ & =10\left ( 5-1.39794\right ) \\ & =36.\,\allowbreak 021 \end{align*}

#### 3.3.5Problem 2.15

Using the convolution property ﬁnd the spectrum for $$w\left ( t\right ) =\sin 2\pi f_{1}t\ \cos 2\pi f_{2}t$$

Solution:

$$\digamma \left ( w\left ( t\right ) \right ) =\digamma \left ( \sin 2\pi f_{1}t\right ) \otimes \digamma \left ( \cos 2\pi f_{2}t\right ) \tag{1}$$ But \begin{align*} \digamma \left ( \sin 2\pi f_{1}t\right ) & =\frac{1}{2j}\left ( \delta \left ( f-f_{1}\right ) -\delta \left ( f+f_{1}\right ) \right ) \\ \digamma \left ( \cos 2\pi f_{2}t\right ) & =\frac{1}{2}\left ( \delta \left ( f-f_{2}\right ) +\delta \left ( f+f_{2}\right ) \right ) \end{align*}

Hence (1) becomes\begin{align} \digamma \left ( w\left ( t\right ) \right ) & =\left \{ \frac{1}{2j}\left ( \delta \left ( f-f_{1}\right ) -\delta \left ( f+f_{1}\right ) \right ) \right \} \otimes \left \{ \frac{1}{2}\left ( \delta \left ( f-f_{2}\right ) +\delta \left ( f+f_{2}\right ) \right ) \right \} \nonumber \\ & =\frac{1}{4j}\left \{ \delta \left ( f-f_{1}\right ) -\delta \left ( f+f_{1}\right ) \right \} \otimes \left \{ \delta \left ( f-f_{2}\right ) +\delta \left ( f+f_{2}\right ) \right \} \tag{2} \end{align}

Applying the distributed property of convolution, i.e. $$a\otimes \left ( b+c\right ) =a\otimes b+a\otimes c$$ on equation (2) we obtain$$4j\ \digamma \left ( w\left ( t\right ) \right ) =\delta \left ( f-f_{1}\right ) \otimes \delta \left ( f-f_{2}\right ) +\delta \left ( f-f_{1}\right ) \otimes \delta \left ( f+f_{2}\right ) -\delta \left ( f+f_{1}\right ) \otimes \delta \left ( f-f_{2}\right ) -\delta \left ( f+f_{1}\right ) \otimes \delta \left ( f+f_{2}\right ) \tag{3}$$ Now \begin{align} \delta \left ( f-f_{1}\right ) \otimes \delta \left ( f-f_{2}\right ) & ={\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( \lambda -f_{1}\right ) \delta \left ( f-\left ( \lambda -f_{2}\right ) \right ) d\lambda \nonumber \\ & =\delta \left ( f+f_{2}-f_{1}\right ){\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( f-\left ( \lambda -f_{2}\right ) \right ) d\lambda \nonumber \\ & =\delta \left ( f+f_{2}-f_{1}\right ) \tag{4} \end{align}

And\begin{align} \delta \left ( f-f_{1}\right ) \otimes \delta \left ( f+f_{2}\right ) & ={\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( \lambda -f_{1}\right ) \delta \left ( f-\left ( \lambda +f_{2}\right ) \right ) d\lambda \nonumber \\ & =\delta \left ( f-f_{2}-f_{1}\right ){\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( f-\left ( \lambda -f_{2}\right ) \right ) d\lambda \nonumber \\ & =\delta \left ( f-f_{2}-f_{1}\right ) \tag{5} \end{align}

And\begin{align} \delta \left ( f+f_{1}\right ) \otimes \delta \left ( f-f_{2}\right ) & ={\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( \lambda +f_{1}\right ) \delta \left ( f-\left ( \lambda -f_{2}\right ) \right ) d\lambda \nonumber \\ & =\delta \left ( f+f_{2}+f_{1}\right ){\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( f-\left ( \lambda -f_{2}\right ) \right ) d\lambda \nonumber \\ & =\delta \left ( f+f_{2}+f_{1}\right ) \tag{6} \end{align}

And\begin{align} \delta \left ( f+f_{1}\right ) \otimes \delta \left ( f+f_{2}\right ) & ={\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( \lambda +f_{1}\right ) \delta \left ( f-\left ( \lambda +f_{2}\right ) \right ) d\lambda \nonumber \\ & =\delta \left ( f-f_{2}+f_{1}\right ){\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( f-\left ( \lambda -f_{2}\right ) \right ) d\lambda \nonumber \\ & =\delta \left ( f-f_{2}+f_{1}\right ) \tag{7} \end{align}

Substitute (4,5,6,7) into (3) we obtain

$\fbox{\ \digamma \left ( w\left ( t\right ) \right ) =\frac{1}{4j}\left [ \delta \left ( f+f_{2}-f_{1}\right ) +\delta \left ( f-f_{2}-f_{1}\right ) -\delta \left ( f+f_{2}+f_{1}\right ) -\delta \left ( f-f_{2}+f_{1}\right ) \right ] }$

or$$\fbox{\ \digamma \left ( w\left ( t\right ) \right ) =\frac{1}{4j}\left [ \delta \left ( f+\left ( f_{2}-f_{1}\right ) \right ) +\delta \left ( f-\left ( f_{2}+f_{1}\right ) \right ) -\delta \left ( f+\left ( f_{2}+f_{1}\right ) \right ) -\delta \left ( f-\left ( f_{2}-f_{1}\right ) \right ) \right ] } \tag{8}$$ This problem can also be solved as follows

$w\left ( t\right ) =\sin 2\pi f_{1}t\ \cos 2\pi f_{2}t$

Using $$\sin \alpha \cos \beta =\frac{1}{2}\left ( \sin \left ( \alpha -\beta \right ) +\sin \left ( \alpha +\beta \right ) \right )$$, hence\begin{align} w\left ( t\right ) & =\frac{1}{2}\left ( \sin \left ( 2\pi f_{1}t-2\pi f_{2}t\right ) +\sin \left ( 2\pi f_{1}t+2\pi f_{2}t\right ) \right ) \nonumber \\ & =\frac{1}{2}\left ( \sin \left ( 2\pi \left ( f_{1}-f_{2}\right ) t\right ) +\sin \left ( 2\pi \left ( f_{1}+f_{2}\right ) t\right ) \right ) \nonumber \\ & =\frac{1}{2}\left ( \frac{1}{2j}\left ( \delta \left ( f-\left ( f_{1}-f_{2}\right ) \right ) -\delta \left ( f+\left ( f_{1}-f_{2}\right ) \right ) \right ) +\frac{1}{2j}\left ( \delta \left ( f-\left ( f_{1}+f_{2}\right ) \right ) -\delta \left ( f+\left ( f_{1}+f_{2}\right ) \right ) \right ) \right ) \nonumber \\ & =\frac{1}{4j}\left \{ \delta \left ( f-\left ( f_{1}-f_{2}\right ) \right ) -\delta \left ( f+\left ( f_{1}-f_{2}\right ) \right ) +\delta \left ( f-\left ( f_{1}+f_{2}\right ) \right ) -\delta \left ( f+\left ( f_{1}+f_{2}\right ) \right ) \right \} \nonumber \\ & =\frac{1}{4j}\left \{ \delta \left ( f+\left ( f_{2}-f_{1}\right ) \right ) +\delta \left ( f-\left ( f_{2}+f_{1}\right ) \right ) -\delta \left ( f+\left ( f_{2}+f_{1}\right ) \right ) -\delta \left ( f-\left ( f_{2}-f_{1}\right ) \right ) \right \} \tag{9} \end{align}

Compare (8) and (9) we see they are the same.

#### 3.3.6Problem 2.17

$w\left ( t\right ) =4\ rect\left ( \frac{t}{4}\right ) -2\ rect\left ( \frac{t}{2}\right )$

By linearity of Fourier Transform$$\digamma \left ( w\left ( t\right ) \right ) =4\times \digamma \left ( rect\left ( \frac{t}{4}\right ) \right ) -2\times \digamma \left ( rect\left ( \frac{t}{2}\right ) \right ) \tag{1}$$ Since

$\digamma \left ( rect\left ( \frac{t}{4}\right ) \right ) =4\operatorname{sinc}\left ( 4f\right )$

and

$\digamma \left ( rect\left ( \frac{t}{2}\right ) \right ) =2\operatorname{sinc}\left ( 2f\right )$

Then (1) becomes\begin{align*} \digamma \left ( w\left ( t\right ) \right ) & =4\times 4\operatorname{sinc}\left ( 4f\right ) -2\times 2\operatorname{sinc}\left ( 2f\right ) \\ & =\fbox{$16\operatorname{sinc}\left ( 4f\right ) -4\operatorname{sinc}\left ( 2f\right )$} \end{align*}

Or in terms of just the $$\sin$$ function, the above becomes\begin{align*} \digamma \left ( w\left ( t\right ) \right ) & =16\frac{\sin \left ( 4\pi f\right ) }{4\pi f}-4\frac{\sin \left ( 2\pi f\right ) }{2\pi f}\\ & =4\frac{\sin \left ( 4\pi f\right ) }{\pi f}-2\frac{\sin \left ( 2\pi f\right ) }{\pi f}\\ & =\fbox{$\frac{4\sin \left ( 4\pi f\right ) -2\sin \left ( 2\pi f\right ) }{\pi f}$} \end{align*}

#### 3.3.7Problem 2.18

3.3.7.1  Part(a)
3.3.7.2  Part(b)
3.3.7.3  Part(c)

If $$w\left ( t\right )$$ has the Fourier Transform $$W\left ( f\right ) =\frac{j2\pi f}{1+j2\pi f}$$ ﬁnd $$X\left ( f\right )$$ for the following waveforms

(a) $$x\left ( t\right ) =w\left ( 2t+2\right )$$

(b)$$\ x\left ( t\right ) =w\left ( t-1\right ) e^{-jt}$$

(c) $$x\left ( t\right ) =w\left ( 1-t\right )$$

##### 3.3.7.1 Part(a)

$w\left ( t\right ) \Leftrightarrow \frac{j2\pi f}{1+j2\pi f}$ Then \begin{align*} w\left ( 2t\right ) & \Leftrightarrow \frac{1}{2}X\left ( \frac{f}{2}\right ) \\ w\left ( 2t+2\right ) & \Leftrightarrow \frac{1}{2}X\left ( \frac{f}{2}\right ) e^{j2\pi \frac{f}{2}\left ( 2\right ) } \end{align*}

Hence\begin{align*} w\left ( 2t+2\right ) & \Leftrightarrow \frac{1}{2}\left ( \frac{j2\pi \frac{f}{2}}{1+j2\pi \frac{f}{2}}\right ) e^{j2\pi f}\\ & \Leftrightarrow \frac{1}{2}\left ( \frac{j\pi f}{1+j\pi f}\right ) e^{j2\pi f} \end{align*}

This can be simpliﬁed to$\fbox{w\left ( 2t+2\right ) \Leftrightarrow \frac{\pi f\ }{2\left ( \pi f-j\right ) }e^{j2\pi f}}$

##### 3.3.7.2 Part(b)

$w\left ( t\right ) \Leftrightarrow \frac{j2\pi f}{1+j2\pi f}$\begin{align*} w\left ( t-1\right ) & \Leftrightarrow X\left ( f\right ) e^{-j2\pi f\left ( -1\right ) }\\ w\left ( t-1\right ) & \Leftrightarrow X\left ( f\right ) e^{j2\pi f} \end{align*}

Now Let $$e^{-jt}=e^{-j2\pi f_{0}t}$$, hence $$2\pi f_{0}=1$$ or $$f_{0}=\frac{1}{2\pi }$$, then $w\left ( t-1\right ) e^{-j2\pi f_{0}t}\Leftrightarrow X\left ( f+f_{0}\right ) e^{j2\pi \left ( f+f_{0}\right ) }$ Hence\begin{align*} w\left ( t-1\right ) e^{-jt} & \Leftrightarrow \frac{j2\pi \left ( f+f_{0}\right ) }{1+j2\pi \left ( f+f_{0}\right ) }e^{j2\pi \left ( f+f_{0}\right ) }\\ w\left ( t-1\right ) e^{-jt} & \Leftrightarrow \frac{j2\pi \left ( f+\frac{1}{2\pi }\right ) }{1+j2\pi \left ( f+\frac{1}{2\pi }\right ) }e^{j2\pi \left ( f+\frac{1}{2\pi }\right ) }\\ w\left ( t-1\right ) e^{-jt} & \Leftrightarrow \frac{j2\pi \left ( 2\pi f+1\right ) }{2\pi +j2\pi \left ( 2\pi f+1\right ) }e^{j\left ( 2\pi f+1\right ) }\\ w\left ( t-1\right ) e^{-jt} & \Leftrightarrow \frac{j4\pi ^{2}f+j2\pi }{2\pi +j4\pi ^{2}f+j2\pi }e^{j2\pi f}e^{j}\\ w\left ( t-1\right ) e^{-jt} & \Leftrightarrow \frac{2\pi f+1}{-j+2\pi f+1}e^{j2\pi f}e^{j} \end{align*}

Hence$\fbox{w\left ( t-1\right ) e^{-jt}\Leftrightarrow \frac{2\pi f+1}{1-j+2\pi f}e^{j\left ( 2\pi f+1\right ) }}$

##### 3.3.7.3 Part(c)

$w\left ( t\right ) \Leftrightarrow \frac{j2\pi f}{1+j2\pi f}$ $w\left ( -t\right ) \Leftrightarrow X\left ( -f\right )$ Then \begin{align*} w\left ( -t+1\right ) & \Leftrightarrow X\left ( -f\right ) e^{j2\pi f\left ( 1\right ) }\\ w\left ( 1-t\right ) & \Leftrightarrow \frac{-j2\pi f}{1-j2\pi f}e^{j2\pi f} \end{align*}