3.10 3.10.1
\[ s_{1}\left ( t\right ) =A_{c}\cos \left ( \omega _{c}t+\phi \right ) \]
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 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*}
3.10.2
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*}
3.10.3
3.10.3.1 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 \[ P_{av}=\frac {1}{2}A_{c}^{2}\]
But \(A_{c}=10\) , hence
\begin{align*} P_{av} & =\frac {100}{2}\\ & =50 \text {watt}\end{align*}
3.10.3.3 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 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
\[ \max \left ( \Delta f\right ) =10^{4}\ \text {Hz}\]
3.10.4
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 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 Modulation index
\begin{align*} \beta & =\frac {\left ( \Delta f\right ) _{\max }}{f_{m}}\\ & =\frac {200}{1000}\\ & =0.2 \end{align*}
3.10.5
\[ s\left ( t\right ) =A_{c}\cos \left ( 2\pi f_{c}t+2\pi k_{f}\int _{0}^{t}m\left ( x\right ) dx\right ) \]
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 \(9.375\times 10^{3}\pm 14980\)
3.10.5.1 To do
3.10.6
\[ s\left ( t\right ) =A_{c}\cos \left ( \omega _{c}t+20\cos \omega _{1}t\right ) \]
where \(A_{c}=500,f_{1}=1khz,f_{c}=100Mhz\)
3.10.6.1 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 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 \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)
3.10.7
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