\[ s\left ( t\right ) =\overset{\text{in-phase component}}{\overbrace{A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) }}\cos \omega _{c}t \] Assume \(k_{a}=1\) in this problem. \(m\left ( t\right ) =A_{1}\left ( \cos \omega _{1}t+\cos 2\omega _{1}t\right ) \), then \(s\left ( t\right ) \) becomes\begin{equation} s\left ( t\right ) =\overset{\text{in-phase component}}{\overbrace{A_{c}\left ( 1+A_{1}\left ( \cos \omega _{1}t+\cos 2\omega _{1}t\right ) \right ) }}\cos \omega _{c}t \tag{1} \end{equation} But \(s\left ( t\right ) \) can be written as \begin{equation} s\left ( t\right ) =s_{I}\left ( t\right ) \cos \omega _{c}t-s_{Q}\left ( t\right ) \sin \omega _{c}t \tag{2} \end{equation} Where \(s_{I}\left ( t\right ) \) is the inphase component and \(s_{Q}\left ( t\right ) \) is the quadrature component of \(s\left ( t\right ) \). Compare (1) to (2), we see that \begin{align*} s_{I}\left ( t\right ) & =A_{c}\left [ 1+A_{1}\left ( \cos \omega _{1}t+\cos 2\omega _{1}t\right ) \right ] \\ s_{Q}\left ( t\right ) & =0 \end{align*}
Now, the complex envelope \(\tilde{s}\left ( t\right ) \)of \(s\left ( t\right ) \) is given by \[ \tilde{s}\left ( t\right ) =s_{I}\left ( t\right ) +js_{Q}\left ( t\right ) \] Hence replacing the value found for \(s_{I}\left ( t\right ) \) and \(s_{Q}\left ( t\right ) \) we obtain \begin{equation} \tilde{s}\left ( t\right ) =A_{c}\left [ 1+A_{1}\left ( \cos \omega _{1}t+\cos 2\omega _{1}t\right ) \right ] \tag{3} \end{equation} Now, we can find \(A_{c}\) since the average power in the carrier signal is given as \(50000\) watt as follows\[ P_{\text{av\_carrier}}=\frac{A_{c}^{2}}{2\left ( 50\right ) }=50000 \] Hence \[ A_{c}=\sqrt{100\times 50000}=2236.1 \text{volt}\] Then (3) becomes \begin{equation} \tilde{s}\left ( t\right ) =2236.1 \left [ 1+A_{1}\left ( \cos \omega _{1}t+\cos 2\omega _{1}t\right ) \right ] \tag{4} \end{equation} The above is the complex envelope in terms of \(A_{1}\) and \(\omega _{1}\) only as required to show.
\begin{equation} \mu =\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }} \tag{5} \end{equation} Need to find angle at which \(\cos \omega _{1}t+\cos 2\omega _{1}t\) is Max and at which it is min. then Let \(\Delta =\cos \omega _{1}t+\cos 2\omega _{1}t\)
We see that when \(\omega _{1}t=2\pi \), then \(\Delta =1+1=2\), hence \[ A_{\max }=A_{c}\left ( 1+2A_{1}\right ) \] Need to find \(A_{\min }\) hence we need to find \(\Delta _{\min \text{. }}\)For this case we must use calculus as it is not obvious where this is minimum \begin{align*} \frac{\partial \Delta }{\partial t} & =0=-\omega _{1}\sin \omega _{1}t-2\omega _{1}\sin 2\omega _{1}t\\ 0 & =-\omega _{1}\sin \omega _{1}t-2\omega _{1}\left ( 2\sin \left ( \omega _{1}t\right ) \cos \left ( \omega _{1}t\right ) \right ) \\ & =-\omega _{1}\sin \omega _{1}t-4\omega _{1}\sin \left ( \omega _{1}t\right ) \cos \left ( \omega _{1}t\right ) \\ \frac{-1}{4} & =\cos \left ( \omega _{1}t\right ) \end{align*}
Hence \(\omega _{1}t=\cos ^{-1}\left ( \frac{-1}{4}\right ) \rightarrow \omega _{1}t=104.477^{0}\) (using calculator). hence \begin{align*} \Delta _{\min } & =\cos \left ( 104.477^{0}\right ) +\cos \left ( 2\times 104.477^{0}\right ) \\ & =-0.2499-0.875\\ & =-1.\,\allowbreak 124\,9 \end{align*}
Then \(A_{\min }=A_{c}\left ( 1-1.\,\allowbreak 124\,9A_{1}\right ) \), so from (5) above\begin{align*} \mu & =\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }}\\ 0.9 & =\frac{A_{c}\left ( 1+2A_{1}\right ) -A_{c}\left ( 1-1.\,\allowbreak 124\,9A_{1}\right ) }{A_{c}\left ( 1+2A_{1}\right ) +A_{c}\left ( 1-1.\,\allowbreak 124\,9A_{1}\right ) }\\ & =\frac{\left ( 1+2A_{1}\right ) -\left ( 1-1.\,\allowbreak 124\,9A_{1}\right ) }{\left ( 1+2A_{1}\right ) +\left ( 1-1.\,\allowbreak 124\,9A_{1}\right ) }\\ & =\frac{1+2A_{1}-1+1.\,\allowbreak 124\,9A_{1}}{1+2A_{1}+1-1.\,\allowbreak 124\,9A_{1}}\\ & =\frac{3.\,\allowbreak 124\,9A_{1}}{2+0.875\,1A_{1}} \end{align*}
Hence \begin{align*} 1.8+0.9\left ( 0.8751 A_{1}\right ) -3.9 A_{1} & =0\\ 1.8-2.3 A_{1} & =0 \end{align*}
Then\[ A_{1}=0.770 \]
Since \begin{align*} A_{\max } & =A_{c}\left ( 1+2A_{1}\right ) \\ & =2236.\,\allowbreak 1\left ( 1+2\times 0.770\,12\right ) \\ & =\allowbreak 5680.\,\allowbreak 2~\text{volts} \end{align*}
Then from Ohm’s law, \(V=RI\), \begin{align*} I_{\max } & =\frac{V_{\max }}{R}\\ & =\frac{5680.\,\allowbreak 2}{50}\\ & =113.\,\allowbreak 6\text{ amps} \end{align*}
Since mean voltage is zero, then average current is zero.
answer For normal modulation, let\[ s_{am}\left ( t\right ) =A_{c}\left ( 1+m\left ( t\right ) \right ) \cos \omega _{c}t \] Maximum envelop is \(2A_{c}\) (i.e. when \(m_{\max }\left ( t\right ) =1\))\(,\) this means that \(A_{p}=2A_{c}\)
But \[ s_{am}\left ( t\right ) =\overset{\text{carrier}}{\overbrace{A_{c}\cos \omega _{c}t}}+\overset{\text{side band}}{\overbrace{A_{c}m\left ( t\right ) \cos \omega _{c}t}}\] So max of sideband is \(A_{c}\) or \(\frac{A_{p}}{2}\). Hence maximum power of sideband is \(\frac{1}{2}\left ( \frac{A_{p}}{2}\right ) ^{2}= \frac{A_{p}^{2}}{8}\) and for DSB-SC, where now use \(A_{p}\) in place of what we normally use \(A_{c}\) then we obtain \[ s\left ( t\right ) =A_{p}m\left ( t\right ) \cos \omega _{c}t \] Hence maximum for sideband is \(\frac{1}{2}A_{p}^{2}\)
Hence we see that power of sideband of DSB-SC to the power of sideband of AM is\[ \frac{\frac{1}{2}A_{p}^{2}}{\frac{A_{p}^{2}}{8}}=4 \]
\[ m\left ( t\right ) =5\cos \omega _{1}t \] \(\hat{m}\left ( t\right ) \) is Hilbert transform of \(m\left ( t\right ) \) defined as \(\hat{m}\left ( t\right ) ={\displaystyle \int \limits _{-\infty }^{\infty }} m\left ( \tau \right ) \frac{1}{t-\tau }d\tau \). Or we can use the frequency approach where \(\hat{m}\left ( t\right ) =\digamma ^{-1}\left [ -j\ sign\left ( f\right ) \ M\left ( f\right ) \right ] \) where \(M\left ( f\right ) \) is the Fourier transform of \(m\left ( t\right ) \). We can carry out this easily, but since this is a phase 90 change, and \(m\left ( t\right ) \) is a cosine function, then \[ \hat{m}\left ( t\right ) =5\sin \omega _{1}t \]
\[ \,s_{SSB}\left ( t\right ) =A_{c}\left [ m\left ( t\right ) \cos \omega _{c}t\mp \hat{m}\left ( t\right ) \sin \omega _{c}t\right ] \] Where the negative sign for upper sided band, and positive sign for the lower sided band, hence\begin{align*} \,s_{LSSB}\left ( t\right ) & =A_{c}\left [ m\left ( t\right ) \cos \omega _{c}t+\hat{m}\left ( t\right ) \sin \omega _{c}t\right ] \\ & =5A_{c}\left [ \cos \omega _{1}t\cos \omega _{c}t+\sin \omega _{1}t\sin \omega _{c}t\right ] \\ & =5A_{c}\left [ \cos \left ( \omega _{c}-\omega _{1}\right ) t\right ] \end{align*}
We can plug in numerical values given\[ s_{LSSB}\left ( t\right ) =5\left [ \cos \left ( \omega _{c}-\omega _{1}\right ) t\right ] \]
To find the RMS value of the SSB, pick the above lower side band. First find \(P_{av}\).\[ s_{LSSB}\left ( t\right ) =5\left [ \cos \left ( \omega _{1}-\omega _{c}\right ) t\right ] \] Hence\begin{align*} RMS\ \text{value of signal} & =\frac{5}{\sqrt{2}}\\ & =3.535\,5\text{ volt} \end{align*}
Then maximum of \(5\left [ \cos \left ( \omega _{1}-\omega _{c}\right ) t\right ] \) is when \(\cos \left ( \omega _{1}-\omega _{c}\right ) t=1\), hence \[ s_{LSSB_{\max }}\left ( t\right ) =5 \text{volt} \]
\begin{align*} P_{av} & =\frac{1}{2}A_{c}^{2}\\ & =\frac{1}{2}\times 25\\ & =12.5 \text{watt} \end{align*}
\begin{align*} PEP & =\frac{1}{2}s_{LSSB_{\max }}^{2}\left ( t\right ) \\ & =\frac{5^{2}}{2}\\ & =12.5\text{ watt} \end{align*}
This is a detector for USSB (Upper side band). i.e. \[ s\left ( t\right ) =A_{c}\left ( m\left ( t\right ) \cos \omega _{c}t-\hat{m}\left ( t\right ) \sin \omega _{c}t\right ) \] Note, I wrote \(A_{c}\) and not \(\frac{A_{c}}{2}\) in the above. As long this is a constant, it gives the same analysis.
The reason is because at point \(H\) the signal is \(-\frac{1}{2}m\left ( t\right ) \) and at the \(C\) point the signal is \(+\frac{1}{2}m\left ( t\right ) \) , hence due to subtraction at the audio output end we obtain \(m\left ( t\right ) \). To receive LSSB, we should change the sign to positive at the audio output end.
\[ s\left ( t\right ) =A_{c}\left ( m\left ( t\right ) \cos \omega _{c}t-\hat{m}\left ( t\right ) \sin \omega _{c}t\right ) \] at point B\begin{align*} s_{B}\left ( t\right ) & =s\left ( t\right ) \ast \overset{\text{local oscillator}}{\overbrace{A_{c}^{^{\prime }}\cos \omega _{c}t}}\\ & =A_{c}^{^{\prime }}A_{c}\left ( m\left ( t\right ) \cos \omega _{c}t-\hat{m}\left ( t\right ) \sin \omega _{c}t\right ) \cos \omega _{c}t\\ & =A_{c}^{^{\prime }}A_{c}\left ( m\left ( t\right ) \cos ^{2}\omega _{c}t-\hat{m}\left ( t\right ) \sin \omega _{c}t\cos \omega _{c}t\right ) \\ & =A_{c}^{^{\prime }}A_{c}\left ( m\left ( t\right ) \left ( \frac{1}{2}+\frac{1}{2}\cos 2\omega _{c}t\right ) -\frac{1}{2}\hat{m}\left ( t\right ) \sin 2\omega _{c}t\right ) \\ & =\overset{\text{low pass}}{\overbrace{\frac{A_{c}^{^{\prime }}A_{c}}{2}m\left ( t\right ) }}+\overset{\text{high pass}}{\overbrace{\frac{A_{c}^{^{\prime }}A_{c}}{2}m\left ( t\right ) \cos 2\omega _{c}t}}-\overset{\text{high pass}}{\overbrace{\frac{A_{c}^{^{\prime }}A_{c}}{2}\hat{m}\left ( t\right ) \sin 2\omega _{c}t}} \end{align*}
at point C, after LPF we obtain\[ s_{c}\left ( t\right ) =A_{c}^{^{\prime }}A_{c}\frac{m\left ( t\right ) }{2}\] at point F we have\begin{align*} s_{f}\left ( t\right ) & =s\left ( t\right ) A_{c}^{^{\prime }}\sin \omega _{c}t\\ & =A_{c}^{^{\prime }}A_{c}\left ( m\left ( t\right ) \cos \omega _{c}t-\hat{m}\left ( t\right ) \sin \omega _{c}t\right ) \sin \omega _{c}t\\ & =A_{c}^{^{\prime }}A_{c}\left ( m\left ( t\right ) \cos \left ( \omega _{c}t\right ) \sin \left ( \omega _{c}t\right ) -\hat{m}\left ( t\right ) \sin ^{2}\omega _{c}t\right ) \\ & =A_{c}^{^{\prime }}A_{c}\left ( m\left ( t\right ) \frac{1}{2}\sin \left ( 2\omega _{c}t\right ) -\hat{m}\left ( t\right ) \left ( \frac{1}{2}-\frac{1}{2}\cos 2\omega _{c}t\right ) \right ) \\ & =\frac{A_{c}^{^{\prime }}A_{c}}{2}\left ( m\left ( t\right ) \sin \left ( 2\omega _{c}t\right ) -\hat{m}\left ( t\right ) \left ( 1-\cos 2\omega _{c}t\right ) \right ) \end{align*}
at point G after LPF\[ s_{g}\left ( t\right ) =-\frac{A_{c}^{^{\prime }}A_{c}}{2}\hat{m}\left ( t\right ) \] at point H after \(-90^{0}\) phase shift\[ s_{h}\left ( t\right ) =+\frac{A_{c}^{^{\prime }}A_{c}}{2}m\left ( t\right ) \] at point \(I\), we sum \(s_{h}\left ( t\right ) \) and \(s_{c}\left ( t\right ) \), hence \(s_{i}\left ( t\right ) =A_{c}^{^{\prime }}A_{c}\frac{m\left ( t\right ) }{2}+\frac{A_{c}^{^{\prime }}A_{c}}{2}m\left ( t\right ) =A_{c}^{^{\prime }}A_{c}m\left ( t\right ) \)
\[ s\left ( t\right ) =A_{c}\left ( m\left ( t\right ) \cos \omega _{c}t+\hat{m}\left ( t\right ) \sin \omega _{c}t\right ) \] This the same as part (b), except now since there is a sign difference, this carries all the way to point \(I\), and then we obtain \[ s_{i}\left ( t\right ) =A_{c}^{^{\prime }}A_{c}\frac{m\left ( t\right ) }{2}-\frac{A_{c}^{^{\prime }}A_{c}}{2}m\left ( t\right ) =0 \] This if this circuit is used as is to demodulate an LSSB AM signal, then the signal will be lost. So, instead of adding at point \(I\) we should now subtract to counter the effect of the negative sign.
Since SSB has bandwidth of \(3kHz\) then this means the width of upper (or lower) band is \(3khz\). This means the signal has \(3khz\) bandwidth. This diagram shows the LPF requirement
Hence LPF is centered at zero frequency and have bandwidth of \(3khz\) (may be make it a little over \(3khz\) band width?)
The IF filter is centered at \(455+\left ( \frac{3}{2}\right ) \) for the upper band of the positive band, and centered at \(-455-\left ( \frac{3}{2}\right ) \) for the upper band of the negative band. (i.e. for the \(USSB\)).
For \(LSSB\), IF should be centered at \(455-\left ( \frac{3}{2}\right ) \) for the lower band of the positive band, and centered at \(-455+\left ( \frac{3}{2}\right ) \) for the lower band of the negative band. (This works if there is a guard band around \(455\), small one, to make the design of IF possible).