Embibe Experts Solutions for Chapter: Communication Systems, Exercise 3: Exercise-3

A radar has a power of $1 \mathrm{kW}$ and is operating at a frequency of $10 \mathrm{GHz}$. It is located on a mountain top of height $500 \mathrm{m}$. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth$=6.4\times {10}^6 \mathrm{m}$) is:

A signal of $5 \mathrm{kHZ}$ frequency is amplitude modulated on a carrier wave of frequency $2 \mathrm{MHz}$. The frequencies of the resultant signal is/are:

In amplitude modulation, the sinusoidal carrier frequency used is denoted by $\omega_{\rm{c}}$ and the signal frequency is denoted by ${\omega }_{\mathrm{m}}$. The bandwidth $\left(\Delta {\omega }_{\mathrm{m}}\right)$ of the signal is such that $\Delta {\omega }_{\mathrm{m}}\ll {\omega }_{\mathrm{c}}$. Which of the following frequencies is not contained in the modulated wave?

A telephonic communication service is working at carrier frequency of $10 \mathrm{GHz}$. Only $10\%$ of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of $5 \mathrm{kHz}$?

A TV transmission tower has a height of $140 \mathrm{m}$ and the height of the receiving antenna is $40 \mathrm{m}$. What is the maximum distance up to which signals can be broadcasted from this tower is LOS (Line of Sight) mode? (Given: radius of earth$=6.4\times {10}^6 \mathrm{m}$).

A $100 V$ carrier wave is made to vary between $160 V$ and $40 V$ by a modulating signal. What is the modulation index?

This question has Statement$-1$ and statement$-2$. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement$-1$: Skywave signals are used for long-distance radio communication. These signals are in general, less stable than ground wave signals.

Statement$-2$: The state of the ionosphere varies from hour to hour, day to day and season to season.

An amplitude modulated signal is given by $V\left(t\right)=10\left[1+0.3 \cos \left(2.2\times {10}^{4}t\right)\right]\sin \left(5.5\times {10}^{5}t\right)$. Here $t$ is in seconds. The side-band frequencies (in $\mathrm{kHz}$) are: (Given $\mathrm{\pi }=22/7$).