### 3.10HW 10

3.10.1  Problem 3.24
3.10.2  Part(b)
3.10.3  Problem 5.20
3.10.4  Problem 5.22
3.10.5  Problem 5.24
3.10.6  Problem 5.26
3.10.7  Key solution
PDF (letter size)
PDF (legal size)

#### 3.10.1Problem 3.24

3.10.1.1  Part(a)

$s_{1}\left ( t\right ) =A_{c}\cos \left ( \omega _{c}t+\phi \right )$

DSB-SC signal is

$s_{2}\left ( t\right ) =m\left ( t\right ) \cos \left ( \omega _{c}t\right )$

Hence by adding the above, we obtain

$s\left ( t\right ) =m\left ( t\right ) \cos \left ( \omega _{c}t\right ) +A_{c}\cos \left ( \omega _{c}t+\phi \right )$

The above signal is applied to an ideal envelope detector. The output of an envelope detector is given by

$a\left ( t\right ) =\sqrt{s_{I}^{2}\left ( t\right ) +s_{Q}^{2}\left ( t\right ) }$

Since $$s\left ( t\right )$$ is a bandpass signal, we need to ﬁrst write it in the canonical form $$s_{I}\left ( t\right ) \cos \left ( \omega _{c}t\right ) -s_{Q}\left ( t\right ) \sin \left ( \omega _{c}t\right )$$

Using $$\cos \left ( A+B\right ) =\cos A\cos B-\sin A\sin B$$, then we have

\begin{align*} s\left ( t\right ) & =m\left ( t\right ) \cos \left ( \omega _{c}t\right ) +A_{c}\left [ \cos \omega _{c}t\cos \phi -\sin \omega _{c}t\sin \phi \right ] \\ & =\left [ m\left ( t\right ) +A_{c}\cos \phi \right ] \cos \left ( \omega _{c}t\right ) -A_{c}\sin \omega _{c}t\sin \phi \end{align*}

Hence we see that \begin{align*} s_{I}\left ( t\right ) & =m\left ( t\right ) +A_{c}\cos \phi \\ s_{Q}\left ( t\right ) & =A_{c}\sin \phi \end{align*}

Now we can start answering parts (a) and (b)

##### 3.10.1.1 Part(a)

When $$\phi =0$$, then

\begin{align*} s_{I}\left ( t\right ) & =m\left ( t\right ) +A_{c}\\ s_{Q}\left ( t\right ) & =0 \end{align*}

Hence

\begin{align*} a\left ( t\right ) & =\sqrt{\left [ m\left ( t\right ) +A_{c}\right ] ^{2}+0^{2}}\\ & =m\left ( t\right ) +A_{c} \end{align*}

#### 3.10.2Part(b)

When $$\phi \neq 0$$ and $$\left \vert m\left ( t\right ) \right \vert <<\frac{A_{c}}{2}$$

\begin{align*} a\left ( t\right ) & =\sqrt{\left [ m\left ( t\right ) +A_{c}\right ] ^{2}+\left [ A_{c}\sin \phi \right ] ^{2}}\\ & =\sqrt{\left [ m^{2}\left ( t\right ) +A_{c}^{2}+2A_{c}m\left ( t\right ) \right ] +\left [ A_{c}^{2}\sin ^{2}\phi \right ] } \end{align*}

Since $$\left \vert m\left ( t\right ) \right \vert <<\frac{A_{c}}{2}$$, then $$m^{2}\left ( t\right ) +A_{c}^{2}+2A_{c}m\left ( t\right ) \simeq A_{c}^{2}$$ hence

\begin{align*} a\left ( t\right ) & \simeq \sqrt{A_{c}^{2}+A_{c}^{2}\sin ^{2}\phi }\\ & =A_{c}\sqrt{1+\sin ^{2}\phi } \end{align*}

#### 3.10.3Problem 5.20

3.10.3.1  Part(a)
3.10.3.2  Part(b)
3.10.3.3  Part(c)
3.10.3.4  part(d)

##### 3.10.3.1 Part(a)

An AM signal is $$s\left ( t\right ) =A_{c}\left [ 1+\mu \ m\left ( t\right ) \right ] \cos \left ( 2\pi f_{c}t+\theta \left ( t\right ) \right )$$.  Now compare this form with the one given above, which is $$s\left ( t\right ) =A_{c}\cos \left ( 2\pi f_{c}t+\theta \left ( t\right ) \right )$$. We see that $$\mu =0$$, i.e. no message source exist. Hence percentage of modulation is zero.

##### 3.10.3.2 Part(b)

$P_{av}=\frac{1}{2}A_{c}^{2}$ But $$A_{c}=10$$, hence \begin{align*} P_{av} & =\frac{100}{2}\\ & =50 \text{watt} \end{align*}

##### 3.10.3.3 Part(c)

From the general form for angle modulated signal

$s\left ( t\right ) =\cos \left ( \omega _{c}t+\theta \left ( t\right ) \right )$

Looking at $s\left ( t\right ) =A_{c}\cos \overset{Total\ Phase}{\overbrace{\left ( \overset{2\pi f_{c}}{\overbrace{\left ( 2\pi \times 10^{8}\right ) }}t+\overset{\theta \left ( t\right ) }{\overbrace{10\cos \left ( 2\pi \times 10^{3}t\right ) }}\right ) }}$

Phase deviation is $\theta \left ( t\right ) =10\cos \left ( 2\pi \times 10^{3}t\right )$ Which is maximum when $$\cos \left ( 2\pi \times 10^{3}t\right ) =1$$Hence maximum Phase deviation is $$10$$ radians.

##### 3.10.3.4 part(d)

Now, we know that the instantenouse frequency $$f_{i}$$ is given by

\begin{align*} f_{i}\left ( t\right ) & =\frac{1}{2\pi }\frac{d}{dt}\left ( \text{total phase}\right ) \\ & =\frac{1}{2\pi }\frac{d}{dt}\left [ \omega _{c}t+\theta \left ( t\right ) \right ] \\ & =\frac{1}{2\pi }\frac{d}{dt}\left [ 2\pi f_{c}t+10\cos \left ( 2\pi \times 10^{3}t\right ) \right ] \\ & =f_{c}-10\left ( 10^{3}\right ) \sin \left ( 2\pi \times 10^{3}t\right ) \end{align*}

The deviation of frequency is the diﬀerence between $$f_{i}$$ and the carrier frequency $$f_{c}$$. Hence from the above we see that the frequency deviation is \begin{align*} \Delta f & =f_{i}-f_{c}\\ & =-10\left ( 10^{3}\right ) \sin \left ( 2\pi \times 10^{3}t\right ) \end{align*}

So, maximum $$\Delta f$$ occures when $$\sin \left ( 2\pi \times 10^{3}t\right ) =-1$$, hence $\max \left ( \Delta f\right ) =10^{4}\ \text{Hz}$

#### 3.10.4Problem 5.22

3.10.4.1  Part(a)
3.10.4.2  Part(b)

The modulating waveform is $$m\left ( t\right )$$ Hence (I am assuming it is cos since it said sinusoidal)

\begin{align*} m\left ( t\right ) & =A_{m}\cos \left ( 2\pi f_{m}t\right ) \\ & =4\cos \left ( 2000\pi t\right ) \end{align*}

Since it is an FM signal, then

$s\left ( t\right ) =A_{c}\cos \left [ \overset{\theta \left ( t\right ) }{\overbrace{\omega _{c}t+2\pi k_{f}\int _{0}^{t}m\left ( x\right ) dx}}\right ]$

Where $$k_{f}$$ is the frequency deviation constant in cycle per volt-second. The gain here means the frequency gain, which is the frequency deviation (deviation from the $$f_{c}$$ frequency). Let $$\Delta f$$ be the frequency deviation in Hz, then \begin{align*} \Delta f & =f_{i}-f_{c}\\ & =\frac{1}{2\pi }\frac{d}{dt}\theta \left ( t\right ) \\ & =k_{f}m\left ( t\right ) \\ & =k_{f}\left [ 4\cos \left ( 2000\pi t\right ) \right ] \end{align*}

##### 3.10.4.1 Part(a)

max $$\Delta f$$ is

$\left ( \Delta f\right ) _{\max }=4k_{f}$

But $$k_{f}=50$$ hz/volt, hence

\begin{align*} \left ( \Delta f\right ) _{\max } & =4\times 50\\ & =200 \text{hz} \end{align*}

##### 3.10.4.2 Part(b)

Modulation index \begin{align*} \beta & =\frac{\left ( \Delta f\right ) _{\max }}{f_{m}}\\ & =\frac{200}{1000}\\ & =0.2 \end{align*}

#### 3.10.5Problem 5.24

3.10.5.1  Part (b)

$s\left ( t\right ) =A_{c}\cos \left ( 2\pi f_{c}t+2\pi k_{f}\int _{0}^{t}m\left ( x\right ) dx\right )$

We are told the carrier frequency has $$f_{c}=103.7$$ Mhz$$,$$ but there is a multiplier of 8$$,$$ and hence the center frequency of the bandpass ﬁlter must be $$\frac{1}{8}$$ of the carrier frequency. i.e.

center frequency of the bandpass ﬁlter is $$\frac{1}{8}103.7=\frac{103.7}{8}=12.963$$

Since peak deviation is $$75khz$$, which means the deviation from the central frequency has maximum of $$75khz$$, then $\frac{75}{8}=9.375 \text{ khz}$

Hence bandwidth from center of frequency of bandwidth ﬁlter is $$9.375$$ but we need to add frequency width of the audio which is $$15000-20=14980$$ Hz on both side, hence

Bandwidth of BPF is $$9.375\times 10^{3}\pm 14980$$

To do

#### 3.10.6Problem 5.26

3.10.6.1  Part(a)
3.10.6.2  Part(b)
3.10.6.3  Part(c)

$s\left ( t\right ) =A_{c}\cos \left ( \omega _{c}t+20\cos \omega _{1}t\right )$

where $$A_{c}=500,f_{1}=1khz,f_{c}=100Mhz$$

##### 3.10.6.1 Part(a)

The general form of the above PM signal is

$s\left ( t\right ) =A_{c}\cos \left ( \omega _{c}t+\overset{\text{phase deviation}}{\overbrace{k_{p}m\left ( t\right ) }}\right )$

Where $$k_{p}m\left ( t\right )$$ is the phase deviation, and $$k_{p}$$ is the phase deviation constant in radians per volt.  Hence we write

$k_{p}m\left ( t\right ) =20\cos \omega _{1}t$

Then

$m\left ( t\right ) =\frac{20\cos \omega _{1}t}{k_{p}}$

But we are given that $$k_{p}=100$$ rad/voltage and $$f_{1}=1000hz$$, then the above becomes

\begin{align*} m\left ( t\right ) & =\frac{20\cos \left ( 2000\pi t\right ) }{100}\\ & =0.2\cos \left ( 2000\pi t\right ) \end{align*}

its frequency is $$1$$ khz and its peak value is $$0.2$$ volts

##### 3.10.6.2 Part(b)

The general form of the above FM signal is

$s\left ( t\right ) =A_{c}\cos \left ( \omega _{c}t+k_{f}\int _{0}^{t}m\left ( x\right ) dx\right )$

Where $$k_{f}$$ is the frequency deviation constant in radians per volt-second

Hence

$k_{f}\int _{0}^{t}m\left ( x\right ) dx=20\cos \omega _{1}t$

Solve for $$m\left ( t\right )$$ in the above, given that $$k_{f}=10^{6}$$radians per volt-second, hence

\begin{align*} k_{f}\int _{0}^{t}m\left ( x\right ) dx & =20\cos \omega _{1}t\\ \int _{0}^{t}m\left ( x\right ) dx & =\frac{20\cos \left ( 2000\pi t\right ) }{10^{6}} \end{align*}

Take derivative of both sides, we obtain

\begin{align*} m\left ( t\right ) & =\frac{20}{10^{6}}\left [ -\sin \left ( 2000\pi t\right ) \times 2000\pi \right ] \\ & =-\frac{20\times 2000\pi }{10^{6}}\sin \left ( 2000\pi t\right ) \\ & =-0.126\sin \left ( 2000\pi t\right ) \end{align*}

Hence its peak value is $$0.126$$ and its frequency is $$1$$ khz

##### 3.10.6.3 Part(c)

\begin{align*} P_{av} & =\frac{\left \langle s^{2}\left ( t\right ) \right \rangle }{50}\\ & =\frac{\frac{1}{2}A_{c}^{2}}{50}\\ & =\frac{500^{2}}{100}\\ & =2500 \text{watt} \end{align*}

PEP is average power obtained if the complex envelope is held constant at its maximum values. i.e. (the normalized PEP) is

$PEP=\frac{1}{2}\left [ \max \left ( \left \vert \tilde{s}\left ( t\right ) \right \vert \right ) \right ] ^{2}$

Since \begin{align*} s\left ( t\right ) & =A_{c}\cos \left ( \omega _{c}t+20\cos \omega _{1}t\right ) \\ & =A_{c}\left [ \cos \omega _{c}t\cos \left ( 20\cos \omega _{1}t\right ) -\sin \omega _{c}t\sin \left ( 20\cos \omega _{1}t\right ) \right ] \\ & =\overset{s_{I}\left ( t\right ) }{\overbrace{A_{c}\cos \left ( 20\cos \omega _{1}t\right ) }}\cos \omega _{c}t-\overset{s_{Q}\left ( t\right ) }{\overbrace{A_{c}\sin \left ( 20\cos \omega _{1}t\right ) }}\sin \omega _{c}t \end{align*}

Hence \begin{align*} \tilde{s}\left ( t\right ) & =s_{I}\left ( t\right ) +js_{Q}\left ( t\right ) \\ & =A_{c}\cos \left ( 20\cos \omega _{1}t\right ) +jA_{c}\sin \left ( 20\cos \omega _{1}t\right ) \end{align*}

Then\begin{align*} \left \vert \tilde{s}\left ( t\right ) \right \vert & =\sqrt{\left [ A_{c}\cos \left ( 20\cos \omega _{1}t\right ) \right ] ^{2}+\left [ A_{c}\sin \left ( 20\cos \omega _{1}t\right ) \right ] ^{2}}\\ & =A_{c}\sqrt{\cos ^{2}\left ( 20\cos \omega _{1}t\right ) +\sin ^{2}\left ( 20\cos \omega _{1}t\right ) }\\ & =A_{c} \end{align*}

Hence the non-normalized PEP is

\begin{align*} PEP & =\frac{\frac{1}{2}\left [ A_{c}\right ] ^{2}}{50}\\ & =\frac{500^{2}}{100}\\ & =2500 \text{watt} \end{align*}

ps. is there an easier or more direct way to ﬁnd PEP than what I did? (assuming it is correct)