The maximum amplitude of an AM wave is found to be 15 V while its minimum amplitude is found to be 3 V . What is the modulation index?
Let $A_c$ and $A_m$ be the amplitudes of carrier wave and modulating wave respectively. So,
Maximum amplitude $\longrightarrow A_{\text {max }}=A_c+A_m=15 \mathrm{~V}\quad\text{.... (i)}$
Minimum amplitude $\longrightarrow A_{\min }=A_c-A_m=3 \mathrm{~V}\quad\text{.... (ii)}$
Adding Eqs. (i) and (ii), we get
$$2 A_C=18$$
$$\begin{aligned} \text{or}\quad & A_c=9 \mathrm{~V} \\ \text{and}\quad & A_m=15-9=6 \mathrm{~V} \end{aligned}$$
Modulating index of wave $\propto=\frac{A_m}{A_c}=\frac{6}{9}=\frac{2}{3}$
Compute the $L C$ product of a tuned amplifier circuit required to generate a carrier wave of 1 MHz for amplitude modulation.
Given, the frequency of carrier wave is 1 MHz .
Formula for the frequency of tuned amplifier,
$$\begin{aligned} \frac{1}{2 \pi \sqrt{L C}} & =1 \mathrm{MHz} \\ \sqrt{L C} & =\frac{1}{2 \pi \times 10^6} \\ L C & =\frac{1}{\left(2 \pi \times 10^6\right)^2}=2.54 \times 10^{-14} \mathrm{~s} \end{aligned}$$
Thus, the product of $L C$ is $2.54 \times 10^{-14} \mathrm{~s}$.
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
$$\begin{aligned} \text { Range } & =\sqrt{2 h R}+\sqrt{2 H R} \\ & =\sqrt{2 \times 20 \times 6.4 \times 10^6}+\sqrt{2 \times 25 \times 6.4 \times 10^6} \\ & =16 \times 10^3+17.9 \times 10^3 \\ & =33.9 \times 10^3 \mathrm{~m} \\ & =33.9 \mathrm{~km} \end{aligned}$$
$$\begin{aligned} &\begin{aligned} \text { Area covered } & =\pi(\text { Range })^2 \\ & =3.14 \times 33.9 \times 33.9 \\ & =3608.52 \mathrm{~km}^2 \\ \text { Percentage increase in area } & =\frac{\text { Difference in area }}{\text { Initial area }} \times 100 \\ & =\frac{(3608.52-803.84)}{803.84} \times 100 \\ & =348.9 \% \end{aligned}\\ &\text { Thus, the percentage increase in area covered is 348.9\% } \end{aligned}$$