Why is a AM signal likely to be more noisy than a FM signal upon transmission through a channel?
In case of AM, the instantaneous voltage of carrier waves is varied by the modulating wave voltage. So, during the transmission, nosie signals can also be added and receiver assumes noise a part of the modulating signal. In case of FM, the frequency of carrier waves is changed as the change in the instantaneous voltage of modulating waves. This can be done by mixing and not while the signal is transmitting in channel. So, noise does not affect FM signal.
Figure shows a communication system. What is the output power when input signal is of 1.01 mW ? [gain in $\left.\mathrm{dB}=10 \log _{10}\left(P_0 / P_i\right)\right]$
The distance travelled by the signal is 5 km
Loss suffered in path of transmission $=2 \mathrm{~dB} / \mathrm{km}$
So, total loss suffered in $5 \mathrm{~km}=-2 \times 5=-10 \mathrm{~dB}$
Total amplifier gain $=10 \mathrm{~dB}+20 \mathrm{~dB}=30 \mathrm{~dB}$
Overall gain in signal $=30-10=20 \mathrm{~dB}$
According to the question, gain in $\mathrm{dB}=10 \log _{10} \frac{P_0}{P_i}$
$\therefore \quad 20=10 \log _{10} \frac{P_0}{P_i}$
or $\quad \log _{10} \frac{P_0}{P_i}=2$
Here, $P_i=1.01 \mathrm{~mW}$ and $P_0$ is the output power.
$$\begin{array}{ll} \therefore & \frac{P_0}{P_i}=10^2=100 \\ \Rightarrow & P_0=P_i \times 100=1.01 \times 100 \\ \text { or } & P_0=101 \mathrm{~mW} \end{array}$$
Thus, the output power is 101 mW .
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
$$N_{\max }=\left(\frac{8 \times 10^6}{9}\right)^2=7.9 \times 10^{11} / \mathrm{m}^3$$