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)
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*}
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*}
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.
P_{av}=\frac{1}{2}A_{c}^{2}
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 )
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}
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*}
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*}
Modulation index \begin{align*} \beta & =\frac{\left ( \Delta f\right ) _{\max }}{f_{m}}\\ & =\frac{200}{1000}\\ & =0.2 \end{align*}
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
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
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
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
\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)