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Published on: 23/05/2021
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Questions + Answers key
Take MCQ Physics Test1.
What is space wave communication? Write the range of frequenices suitable for space wave communication.
2.
On radiating (sending out) an AM modulated signal, the total radiated power is due to the energy carried by \({ \omega }_{ c },{ \omega }_{ c }-{ \omega }_{ m }\) and Suggest ways to minimize the cost of radiation without compromising on information.
3.
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 F1 layer of the ionosphere while signals of frequencies higher than 8 MHz are not received by reflection from the F2 layer of the ionosphere. Estimate the maximum electron densities of the F1 and F2 layers on that day.
4.
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 (i) relative to case.
1.
Space wave propagation.
Range of frequencies for space wave propagation is used to transmit
VHF band 30MHz to 300 MHz
UHF band 300MHz to 3 GHz
Microwaves 100 GHz
e.g. TV signals 80 to 200 MHz and RADAR beans (microwaves).
2.
In amplitude modulated signal, out of \(\left( { \omega }_{ c }-{ \omega }_{ m } \right) \), \({ \omega }_{ c }\) and \(\left( { \omega }_{ c }+{ \omega }_{ m } \right) \) only \(\left( { \omega }_{ c }+{ \omega }_{ m } \right) \)and \(\left( { \omega }_{ c }-{ \omega }_{ m } \right) \) contain informations. Hence cost can be reduced by transmitting \(\left( { \omega }_{ c }+{ \omega }_{ m } \right) \), \(\left( { \omega }_{ c }-{ \omega }_{ m } \right) \) or both \(\left( { \omega }_{ c }+{ \omega }_{ m } \right) \) and \(\left( { \omega }_{ c }-{ \omega }_{ m } \right) \).
3.
For F1 layer : 5 x 106 = 9 (Nmax)1/2 or \({ N }_{ max }=\left( \frac { 5 }{ 9 } \times { 10 }^{ 6 } \right) ^{ 2 }=3.086\times { 10 }^{ 11 }{ m }^{ -3 }\)
For F2 layer : 8 x 106 = 9(Nmax)1/2 or \({ N }_{ max }=\left( \frac { 8 }{ 9 } \times { 10 }^{ 6 } \right) ^{ 2 }=7.9\times { 10 }^{ 11 }{ m }^{ -3 }\)
4.
Here, h1 = 20 m, h2 = 25 m
(i) \(d=\sqrt { 2{ h }_{ 1 }R } =\sqrt { 2\times 20\times \left( 6.4\times { 10 }^{ 6 } \right) } =16\times { 10 }^{ 3 }m=16km\)
Area covered, \(A=\pi { d }^{ 2 }=\frac { 22 }{ 7 } \times { \left( 16 \right) }^{ 2 }\simeq 804.6{ km }^{ 2 }\)
(ii) Range, \({ d }_{ 1 }=\sqrt { 2{ h }_{ 1 }R } +\sqrt { 2{ h }_{ 2 }R } =\sqrt { 2\times 20\times \left( 6.4\times { 10 }^{ 6 } \right) } +\sqrt { 2\times 25\times 6.4\times { 10 }^{ 6 } } \)
\(=16km+17.9km=33.9km\)
Area covered, \({ A }_{ 1 }=\pi { d }_{ 1 }^{ 2 }=\frac { 22 }{ 7 } \times { \left( 33.9 \right) }^{ 2 }=3611.8{ km }^{ 2 }\)
% increase in area = \(\frac { { A }_{ 1 }-A }{ A } \times 100=\left( \frac { 3611.8-804.6 }{ 804.6 } \right) \times 100=\) 348.9%
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