A TV transmission tower antenna is at a height of 20 m . How much service area can it cover if the receiving antenna is (i) at ground level, (ii) at a height of 25 m ? Calculate the percentage increase in area covered in case (ii) relative to case (i).
Given, height of antenna $h=20 \mathrm{~m}$
Radius of earth $=6.4 \times 10^6 \mathrm{~m}$
At the ground level,
$$\begin{aligned} & \text {(i) Range }=\sqrt{2 h R}=\sqrt{2 \times 20 \times 6.4 \times 10^6} \\ & =16000 \mathrm{~m}=16 \mathrm{~km} \\ & \text { Area covered } A=\pi(\text { range })^2 \\ & =3.14 \times 16 \times 16=803.84 \mathrm{~km}^2 \end{aligned}$$
(ii) At a height of $H=25 \mathrm{~m}$ from ground level
$$\begin{aligned} \text { Range } & =\sqrt{2 h R}+\sqrt{2 H R} \\ & =\sqrt{2 \times 20 \times 6.4 \times 10^6}+\sqrt{2 \times 25 \times 6.4 \times 10^6} \\ & =16 \times 10^3+17.9 \times 10^3 \\ & =33.9 \times 10^3 \mathrm{~m} \\ & =33.9 \mathrm{~km} \end{aligned}$$
$$\begin{aligned} &\begin{aligned} \text { Area covered } & =\pi(\text { Range })^2 \\ & =3.14 \times 33.9 \times 33.9 \\ & =3608.52 \mathrm{~km}^2 \\ \text { Percentage increase in area } & =\frac{\text { Difference in area }}{\text { Initial area }} \times 100 \\ & =\frac{(3608.52-803.84)}{803.84} \times 100 \\ & =348.9 \% \end{aligned}\\ &\text { Thus, the percentage increase in area covered is 348.9\% } \end{aligned}$$
If the whole earth is to be connected by LOS communication using space waves (no restriction of antenna size or tower height), what is the minimum number of antennas required? Calculate the tower height of these antennas in terms of earth's radius.
Consider the figure given below to solve this question.
Suppose the height of transmitting antenna or receiving antenna in order to cover the entire surface of earth through communication is $h_t$ and radius of earth is $R$
Then, maximum distance
$$\begin{array}{l} d_m^2 =\left(R+h_t\right)^2+\left(R+h_t\right)^2 \\ =2\left(R+h_t\right)^2 \\ \therefore \quad d_m =\sqrt{2 h_t R}+\sqrt{2 h_t R}=2 \sqrt{2 h_t R} \\ \Rightarrow \quad 8 h_t R =2\left(R+h_t\right)^2 \\ \Rightarrow \quad 4 h_t R =R^2+2 R h_t+h_t^2 \\ \Rightarrow \quad R^2-2 h_t R+h_t^2 =0 \\ \Rightarrow \quad \left(R-h_t\right)^2 =0 \\ \Rightarrow \quad R =h_t. \end{array}$$
Since, space wave frequency is used so $\lambda<< h_t$, hence only tower height is to be taken into consideration. In three dimensions of earth, 6 antenna towers of each of height $h_t=R$ would be used to cover the entire surface of earth with communication programme.
The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be $f_{\max }=9\left(N_{\max }\right)^{1 / 2}$, where $N_{\max }$ is the maximum electron density at that layer of the ionosphere. On a certain day it is observed that signals of frequencies higher than 5 MHz are not received by reflection from the $F_1$ layer of the ionosphere while signals of frequencies higher than 8 MHz are not received by reflection from the $F_2$ layer of the ionosphere. Estimate the maximum electron densities of the $F_1$ and $F_2$ layers on that day.
The maximum frequency for reflection of sky waves
$$f_{\max }=9\left(N_{\max }\right)^{1 / 2}$$
where, $N_{\max }$ is a maximum electron density.
For $\mathrm{F}_1$ layer,
$$\begin{aligned} f_{\max } & =5 \mathrm{MHz} \\ \text{So,}\quad 5 \times 10^6 & =9\left(N_{\max }\right)^{1 / 2} \end{aligned}$$
Maximum electron density
$$N_{\max }=\left(\frac{5}{9} \times 10^6\right)^2=3.086 \times 10^{11} / \mathrm{m}^3$$
For $F_2$ layer, $$f_{\max }=8 \mathrm{MHz}$$
So, $$8 \times 10^6=9\left(N_{\max }\right)^{1 / 2}$$
Maximum electron density
$$N_{\max }=\left(\frac{8 \times 10^6}{9}\right)^2=7.9 \times 10^{11} / \mathrm{m}^3$$
On radiating (sending out) and AM modulated signal, the total radiated power is due to energy carried by $\omega_c, \omega_c-\omega_m$ and $\omega_c+\omega_m$. Suggest ways to minimise cost of radiation without compromising on information.
In amplitude modulated signal, only side band frequencies contain information. Thus only $\left(\omega_c+\omega_m\right)$ and $\left(\omega_c-\omega_m\right)$ contain information.
Now, according to question, the total radiated power is due to energy carried by
$$\omega_c,\left(\omega_C-\omega_m\right) \text { and }\left(\omega_c+\omega_m\right) .$$
Thus to minimise the cost of radiation without compromising on information $\omega_c$ can be left and transmitting. $\left(\omega_c+\omega_m\right),\left(\omega_c-\omega_m\right)$ or both $\left(\omega_c+\omega_m\right)$ and $\left(\omega_c-\omega_m\right)$.
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
$$\alpha=0.0602 \mathrm{~dB} / \mathrm{km}$$