(i) Skywave and space wave propagation

1. Sky wave propagation is due to ionospheric reflection of radio waves back to the earth.
2. Space wave propagation is by line of sight propagation, directly between transmitter to receiver / or by satellite.
3. Frequency range of sky wave - few MHz to 40 MHz
4. Frequency range of space wave - above 40 MHz

Long distance communication between two points on the earth is achieved through reflection of electromagnetic waves by ionosphere. Such waves are called sky waves. Sky wave propagation takes place up to frequency of about 40 MHz. A space wave travels in a straight line from transmitting antenna to the receiving antenna. It is used for line of sight (LOS) communication as well as satellite communication.
(a) The ionospheric layer acts as reflector for a certain range of frequencies (3 to 30 MHz). Electro-magnetic waves of frequencies higher than 30 MHz (upto 40 MHz) penetrate the ionosphere and escape.
(b) In television broadcast, microwave links and satellite communication, space wave mode of propagation is used.

Give ,carrier wave frequency
\({ f }_{ c }=20MHz=20\times { 10 }^{ 6 }Hz\)
Bandwidth required for modulation is
\({ 2f }_{ m }=3kHz=3\times { 10 }^{ 3 }Hz\)
\(\Rightarrow \ \ { f }_{ m }=\frac { 3\times { 10 }^{ 3 } }{ 2 } =1.5\times { 10 }^{ 3 }Hz\) s
Demodulation by a diode is possible,if the condition 
Thus,\(\frac { 1 }{ { f }_{ c } } =\frac { 1 }{ 20\times { 10 }^{ 6 }Hz } =0.5\times { 10 }^{ -7 }s\) .....(i)
and \(\frac { 1 }{ { f }_{ m } } =\frac { 1 }{ 1.5\times { 10 }^{ 3 }Hz } =0.7\times { 10 }^{ -3 }s\)    ......(ii)
Now,gain through all the option of R and C one by one,we get
(i)  \(RC=1k\Omega \times 0.01\mu F\)
\(\\ ={ 10 }^{ 3 }\Omega \times ({ 0.01 }\times { 10 }^{ -6 }F)={ 10 }^{ -5 }s\)
Here, condition \(\frac { 1 }{ { f }_{ c } } <\) is satisfied  
Hence, it can be demodulated.
(ii) \((RC=1k\Omega \times 0.01\mu F={ 10 }^{ 4 }\Omega \times { 10 }^{ -8 }F={ 10 }^{ -4 }s)\)
Here, condition\(\frac { 1 }{ { f }_{ c } }\) < is satisfied
 Hence, it  can be demodulated
(iii) \(RC=10k\Omega \times 1\mu F={ 10 }^{ 4 }\Omega \times { 10 }^{ -6 }F={ 10 }^{ -2 }s\)
Here, condition\(\frac { 1 }{ { f }_{ c } } >RC,\) so this cannot be demodulated.

Given \(\lambda =800nm,c=3\times { 10 }^{ 8 }\quad m/s\)
Source frequency, \(v=\frac { c }{ \lambda } =\frac { 3\times { 10 }^{ 8 } }{ 8\times { 10 }^{ -7 } } =3.75\times { 10 }^{ 14\quad }Hz\)
Now, total bandwidth of a channel
=1% of optical source frequency
=\(\frac { 1 }{ 100 } \times 3.75\times { 10 }^{ 14 }=3.75\times { 10 }^{ 12 }Hz\)
Number of channels
=\(\frac { Total \ bandwidth \ of \ channel }{ Bandwidth \ needed \ per \ channel }\)
(i) Number of channels for bandwidth of 8kHz
 \(=\frac { 3.75\times { 10 }^{ 12 } }{ 8\times { 10 }^{ 3 } } =4.7\times { 10 }^{ 8 }\)
(ii) Number of channels for video bandwidth of 4.5 MHz
 \(=\frac { 3.75\times { 10 }^{ 12 } }{ 4.5\times { 10 }^{ 6 } } =8.3\times { 10 }^{ 5 }\)

The distance travelled by the signal is 5km. Loss suffered in path of transmission = 2dB/km
So, total loss suffered in 5 km = \(-2\times 5=-10dB\)
Total amplifier gain = 10dB + 20dB = 30dB
Overall gain in signal = 30 - 10 = 30dB
According to the question, gain in dB = \(10{ log }_{ 10 }\frac { { P }_{ 0 } }{ { P }_{ 1 } } \)
\(\because \ 20=10{ log }_{ 10 }\frac { { P }_{ 0 } }{ P_{ i } } or\quad { log }_{ 10 }\frac { { P }_{ 0 } }{ P_{ i } } =2\)
Where\({ P }_{ i }=1.01mW\ and\ { P }_{ 0 }\) is the output power.  
\(\because \ \frac { { P }_{ 0 } }{ { P }_{ i } } ={ 10 }^{ 2 }=100\)
\(\\ \Rightarrow { P }_{ 0 }={ p }_{ i }\times 100=1.01\times 100\ or\ { p }_{ 0 }=101mW\)
Thus the output is 101 mW.