The intensity of a light pulse travelling along a communication channel decreases exponentially with distance $x$ according to the relation $I=I_0 e^{-\alpha x}$, where $I_0$ is the intensity at $x=0$ and $\alpha$ is the attenuation constant.
(a) Show that the intensity reduces by $75 \%$ after a distance of $\left(\frac{\ln 4}{\alpha}\right)$.
(b) Attenuation of a signal can be expressed in decibel ( dB ) according to the relation $\mathrm{dB}=10 \log _{10}\left(\frac{I}{I_0}\right)$. What is the attenuation in $\mathrm{dB} / \mathrm{km}$ for an optical fibre in which the intensity falls by $50 \%$ over a distance of 50 km ?
(a) Given, the intensity of a light pulse $I=I_0 \mathrm{e}^{-\alpha x}$
where, $I_0$ is the intensity at $x=0$ and $\alpha$ is constant.
According to the question, $I=25 \%$ of $I_0=\frac{25}{100} \cdot I_0=\frac{I_0}{4}$
Using the formula mentioned in the question,
$$\begin{aligned} I & =I_0 \mathrm{e}^{-\alpha x} \\ \frac{I_0}{4} & =I_0 \mathrm{e}^{-\alpha x} \end{aligned}$$
or $$\frac{1}{4}=e^{-\alpha x}$$
Taking log on both sides, we get
$$\begin{aligned} & \ln 1-\ln 4=-\alpha x \ln e \quad(\because \ln e=1) \\ & -\ln 4=-\alpha x \\ & x=\frac{\ln 4}{\alpha} \end{aligned}$$
Therefore, at distance $x=\frac{\ln 4}{\alpha}$, the intensity is reduced to $75 \%$ of initial intensity.
(b) Let $\alpha$ be the attenuation in $\mathrm{dB} / \mathrm{km}$. If $x$ is the distance travelled by signal, then
$$10 \log _{10}\left(\frac{I}{I_0}\right)=-\alpha x\quad\text{.... (i)}$$
where, $I_0$ is the intensity initially.
According to the question, $I=50 \%$ of $I_0=\frac{I_0}{2}$ and $x=50 \mathrm{~km}$ Putting the value of $x$ in Eq. (i), we get
$$\begin{aligned} 10 \log _{10} \frac{I_0}{2 I_0} & =-\alpha \times 50 \\ 10[\log 1-\log 2] & =-50 \alpha \\ \frac{10 \times 0.3010}{50} & =\alpha \end{aligned}$$
$\therefore$ The attenuation for an optical fibre
$$\alpha=0.0602 \mathrm{~dB} / \mathrm{km}$$
A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above Earth's surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should size and placement of receiving and transmitting antenna be?
Let the receiver is at point A and source is at B.
$$\begin{aligned} & \text { Velocity of waves }=3 \times 10^8 \mathrm{~m} / \mathrm{s} \\ & \text { Time to reach a receiver }=4.04 \mathrm{~ms}=4.04 \times 10^{-3} \mathrm{~s} \\ & \text { Let the height of satellite is } \quad h_s=600 \mathrm{~km} \\ & \text { Radius of earth }=6400 \mathrm{~km} \\ & \text { Size of transmitting antenna }=h_T \\ & \text { We know that } \frac{\text { Distance travelled by wave }}{\text { Time }}=\text { Velocity of waves } \\ & \frac{2 x}{4.04 \times 10^{-3}}=3 \times 10^8 \end{aligned}$$
$$\begin{aligned} &\begin{aligned} \text { or }\quad x & =\frac{3 \times 10^8 \times 4.04 \times 10^{-3}}{2} \\ & =6.06 \times 10^5=606 \mathrm{~km} \end{aligned} \end{aligned}$$
Using Phythagoras theorem,
$$\begin{aligned} d^2=x^2-h_s^2 & =(606)^2-(600)^2=7236 \\ \text{or}\quad d & =85.06 \mathrm{~km} \end{aligned}$$
So, the distance between source and receiver $=2 d$
$$=2 \times 85.06=170 \mathrm{~km}$$
The maximum distance covered on ground from the transmitter by emitted EM waves
$$d=\sqrt{2 R h_T}$$
or $$\frac{d^2}{2 R}=h_T$$
$$\begin{aligned} \text {or size of antenna } h_T & =\frac{7236}{2 \times 6400} \\ & =0.565 \mathrm{~km}=565 \mathrm{~m} \end{aligned}$$
An amplitude modulated wave is as shown in figure. Calculate
(i) the percentage modulation,
(ii) peak carrier voltage and
(iii) peak value of information voltage
From the diagram,
Maximum voltage $V_{\max }=\frac{100}{2}=50 \mathrm{~V}$
Minimum voltage $V_{\min }=\frac{20}{2}=10 \mathrm{~V}$
(i) Percentage modulation, $\propto=\frac{V_{\max }-V_{\min }}{V_{\max }+V_{\min }} \times 100=\frac{50-10}{50+10} \times 100$
$$=\frac{40}{60} \times 100=66.67 \%$$
(ii) Peak carrier voltage, $\quad V_c=\frac{V_{\max }+V_{\min }}{2}=\frac{50+10}{2}=30 \mathrm{~V}$
(iii) Peak value of information voltage,
$$V_m=\propto V_c=\frac{66.67}{100} \times 30=20 \mathrm{~V}$$
(i) Draw the plot of amplitude versus $\omega$ for an amplitude modulated were whose carrier wave $\left(\omega_c\right)$ is carrying two modulating signals, $\omega_1$ and $\omega_2 \left(\omega_2>\omega_1\right)$.
(ii) Is the plot symmetrical about $\omega_c$ ? Comment especially about plot in region $\omega<\omega_c$.
(iii) Extrapolate and predict the problems one can expect if more waves are to be modulated.
(iv) Suggest solutions to the above problem. In the process can one understand another advantage of modulation in terms of bandwidth?
(i) The plot of amplitude versus $\omega$ can be shown in the figure below
(ii) From figure, we note that frequency spectrum is not symmetrical about $\omega_{\mathrm{c}}$. Crowding of spectrum is present for $\omega<\omega_c$.
(iii) If more waves are to be modulated then there will be more crowding in the modulating signal in the region $\omega<\omega_c$. That will result more chances of mixing of signals.
(iv) To accommodate more signals, we should increase bandwidth and frequency carrier waves $\omega_c$. This shows that large carrier frequency enables to carry more information (i.e., more $\omega_m$ ) and the same will in turn increase bandwidth.
An audio signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3 kHz . Could this wave be demodulated by a diode detector which has the values of $R$ and $C$ as
(i) $R=1 \mathrm{k} \Omega, C=0.01 \times \mathrm{F}$.
(ii) $R=10 \mathrm{k} \Omega, C=0.01 \propto \mathrm{~F}$.
(iii) $R=10 \mathrm{k} \Omega, C=0.1 \times \mathrm{F}$.
$$\begin{aligned} \text { Given, carrier wave frequency } f_C & =20 \mathrm{MHz} \\ & =20 \times 10^6 \mathrm{~Hz} \end{aligned}$$
Bandwidth required for modulation is
$$\begin{aligned} 2 f_m & =3 \mathrm{kHz}=3 \times 10^3 \mathrm{~Hz} \\ f_m & =\frac{3 \times 10^3}{2}=1.5 \times 10^3 \mathrm{~Hz} \end{aligned}$$
Demodulation by a diode is possible if the condition $\frac{1}{f_c}<< R C<\frac{1}{f_m}$ is satisfied
$$\begin{aligned} \text{Thus,}\quad & \frac{1}{t_c}=\frac{1}{20 \times 10^6}=0.5 \times 10^{-7} \quad\text{.... (i)}\\ \text{and}\quad & \frac{1}{f_m}=\frac{1}{1.5 \times 10^3} \mathrm{~Hz}=0.7 \times 10^{-3} \mathrm{~s}\quad\text{.... (ii)} \end{aligned}$$
Now, gain through all the options of $R$ and $C$ one by one, we get
(i) $R C=1 \mathrm{k} \Omega \times 0.01 \propto \mathrm{~F}=10^3 \Omega \times\left(0.01 \times 10^{-6} \mathrm{~F}\right)=10^{-5} \mathrm{~S}$
Here, condition $\frac{1}{f_c} \ll R C<\frac{1}{f_m}$ is satisfied.
Hence it can be demodulated.
(ii) $R C=10 \mathrm{k} \Omega \times 0.01 \propto \mathrm{~F}=10^4 \Omega \times 10^{-8} \mathrm{~F}=10^{-4} \mathrm{~s}$
Here condition $\frac{1}{f_c} \ll R C<\frac{1}{f_m}$ is satisfied.
Hence, it can be demodulated.
(iii) $R C=10 \mathrm{k} \Omega \times 1 \propto \propto \mathrm{~F}=10^4 \Omega \times 10^{-12} \mathrm{~F}=10^{-8} \mathrm{~S}$
Here, condition $\frac{1}{f_c}>R C$, so this cannot be demodulated.