3.9HW 9

3.9.1  Problem 5-5
3.9.2  Problem 5-8
3.9.3  Problem 5-13
3.9.4  Problem 5-18
3.9.5  Key solution
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3.9.1Problem 5-5

3.9.1.1  part(a)
3.9.1.2  part(b)
3.9.1.3  part(c)

3.9.1.1 part(a)

$s\left ( t\right ) =\overset{\text{in-phase component}}{\overbrace{A_{c}\left ( 1+k_{a}m\left ( t\right ) \right ) }}\cos \omega _{c}t$ 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$$s\left ( t\right ) =\overset{\text{in-phase component}}{\overbrace{A_{c}\left ( 1+A_{1}\left ( \cos \omega _{1}t+\cos 2\omega _{1}t\right ) \right ) }}\cos \omega _{c}t \tag{1}$$ But $$s\left ( t\right )$$ can be written as $$s\left ( t\right ) =s_{I}\left ( t\right ) \cos \omega _{c}t-s_{Q}\left ( t\right ) \sin \omega _{c}t \tag{2}$$ 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 $$\tilde{s}\left ( t\right ) =A_{c}\left [ 1+A_{1}\left ( \cos \omega _{1}t+\cos 2\omega _{1}t\right ) \right ] \tag{3}$$ 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 $$\tilde{s}\left ( t\right ) =2236.1 \left [ 1+A_{1}\left ( \cos \omega _{1}t+\cos 2\omega _{1}t\right ) \right ] \tag{4}$$ The above is the complex envelope in terms of $$A_{1}$$ and $$\omega _{1}$$ only as required to show.

3.9.1.2 part(b)

$$\mu =\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }} \tag{5}$$ 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$A_{1}=0.770$

3.9.1.3 part(c)

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.

3.9.2Problem 5-8

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$\frac{\frac{1}{2}A_{p}^{2}}{\frac{A_{p}^{2}}{8}}=4$

3.9.3Problem 5-13

3.9.3.1  part(a)
3.9.3.2  part(b)
3.9.3.3  Part(c)
3.9.3.4  part(d)
3.9.3.5  part(e)
3.9.3.6  Part(f)

3.9.3.1 part(a)

$m\left ( t\right ) =5\cos \omega _{1}t$ $$\hat{m}\left ( t\right )$$ 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 $\hat{m}\left ( t\right ) =5\sin \omega _{1}t$

3.9.3.2 part(b)

$\,s_{SSB}\left ( t\right ) =A_{c}\left [ m\left ( t\right ) \cos \omega _{c}t\mp \hat{m}\left ( t\right ) \sin \omega _{c}t\right ]$ 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$s_{LSSB}\left ( t\right ) =5\left [ \cos \left ( \omega _{c}-\omega _{1}\right ) t\right ]$

3.9.3.3 Part(c)

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*}

3.9.3.4 part(d)

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 $s_{LSSB_{\max }}\left ( t\right ) =5 \text{volt}$

3.9.3.5 part(e)

\begin{align*} P_{av} & =\frac{1}{2}A_{c}^{2}\\ & =\frac{1}{2}\times 25\\ & =12.5 \text{watt} \end{align*}

3.9.3.6 Part(f)

\begin{align*} PEP & =\frac{1}{2}s_{LSSB_{\max }}^{2}\left ( t\right ) \\ & =\frac{5^{2}}{2}\\ & =12.5\text{ watt} \end{align*}

3.9.4Problem 5-18

3.9.4.1  part(a)
3.9.4.2  part(b)
3.9.4.3  Part(c)
3.9.4.4  part(d)

3.9.4.1 part(a)

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.

3.9.4.2 part(b)

$s\left ( t\right ) =A_{c}\left ( m\left ( t\right ) \cos \omega _{c}t-\hat{m}\left ( t\right ) \sin \omega _{c}t\right )$ 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 $$s_{i}\left ( t\right ) =A_{c}^{^{\prime }}A_{c}\frac{m\left ( t\right ) }{2}+\frac{A_{c}^{^{\prime }}A_{c}}{2}m\left ( t\right ) =A_{c}^{^{\prime }}A_{c}m\left ( t\right )$$

3.9.4.3 Part(c)

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

3.9.4.4 part(d)

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).