Maximum voltage, V_max = \(\frac { 100 }{ 2 } \)= 50 V
Minimum voltage, V_min = \(\frac { 20 }{ 2 } \)= 10 V
(i) Percentage modulation,
\(\mu =\frac { { V }_{ max }-{ V }_{ min } }{ { V }_{ max }{ +V }_{ min } }\times100\)
\(=\frac { 50-10 }{ 50+10 } \times100=\frac { 40 }{ 60 } \times100=66.67%\)
(ii) Peak carrier voltage,
\({ V }_{ c }=\frac { { V }_{ max }+{ V }_{ min } }{ 2 } =\frac { 50+10 }{ 2 } =30\ V\) 
(iii) Peak value of information voltage,
\({ V }_{ m }=\mu { V }_{ c }=\frac { 66.67 }{ 100 } \times 30=20 \ V\)

Here, A_m = 0.1 V, A_c = 0.2 V,\(\mu \)=?
\(\mu \)= \(\frac { { A }_{ m } }{ { A }_{ c } } =\frac { 0.1 }{ 0.2 } =0.5\)

Examples of communication system which use space wave mode are television channel, UHF, VHF, etc.
Maximum distance upto which the signals can be transmitted, d = ?
\(\because \ d=\sqrt { 2hR } \)
\(where, \ R=radius \ of \ the \ earth\)
\(= 6400 \ km\)
\(=6.4\times { 10 }^{ 6 }m\)
\(\therefore \ d=\sqrt { 2\times 80\times 6.4\times { 10 }^{ 6 } } \)
\(=32\times { 10 }^{ 3 }m\)
\(=32 \ km \ \)

Let N₁ and N₂ be maximum electron densities of the ionosphere on first and second days corresponding to 10 MHz and 8 MHz, respectively.
\(\therefore \left( \frac { { N }_{ 1 } }{ { N }_{ 2 } } \right) ^{ 1/2 }=\frac { { V }_{ c1 } }{ { V }_{ c2 } } =\frac { 10 \ MHz }{ 8 \ MHz }\)
\(\left( \frac { { N }_{ 1 } }{ { N }_{ 2 } } \right) ^{ 1/2 }=\frac { 5 }{ 4 } =\frac { { N }_{ 1 } }{ { N }_{ 2 } } =\frac { 25 }{ 16 } \)
\({ N }_{ 1 }:{ N }_{ 2 }=25:16\)

The amplitude modulated radiowaves having frequency 530 kHz to 1710 kHz (or wavelength between 175 m to 566 m) which are travelling directly following the surface of earth are known as ground waves. The short wave communication over long distance is only possible via sky waves. It is not possible via ground waves because the ground waves can bend round the corners of the objects on earth and hence, their intensity falls with distance. Moreover the ground wave transmission becomes weaker as frequency increases.

Optical source frequency, \(v=\frac { c }{ \lambda } =\frac { 3\times { 10 }^{ 8 } }{ 800\times { 10 }^{ -9 } } =3.75\times { 10 }^{ 14 }Hz\)
Bandwidth of channel = 1% of source frequency = \(\frac { 1 }{ 100 } \times 3.75\times { 10 }^{ 14 }=3.75\times { 10 }^{ 12 }Hz\)
Number of channels for video T.V. signal = \(\frac { 3.75\times { 10 }^{ 12 } }{ 4.5\times { 10 }^{ 6 } } =8.3\times { 10 }^{ 5 }\)

Maximum distance covered by space wave communication .
 \(d=\sqrt { 2Rh } =\sqrt { 2\times6.4\times{ 10 }^{ 6 }\times400 } \)
\(=71.55\times{ 10 }^{ 3 }m=71.55 \ km\)
Since the distance between transmitter and receiver is 125 km, hence for the given frequency signals of 5 MHz and 100 MHz, the propagation is not possible via space wave communication.
For sky wave propagation,the critical frequency,
\({ v }_{ v }=9({ N }_{ max })^{ 1/2 }=9({ 10 }^{ 12 })^{ 1/2 }=9\times{ 10 }^{ 6 }Hz\)
\(=9 \ MHz\)
Since 5 MHz < 9 MHz, so the propagation of signal of frequency 5 MHz is possible via skywave. As 100 MHz > 9 MHz, hence the propagation of signal frequency 100 MHz is not possible via sky waves. For that satellite communication is used.

Here, \( (RC={ 10 }^{ 4 }\times{ 10 }^{ -11 }={ 10 }^{ -7 }s\ \frac { 1 }{ { v }_{ c } } =\frac { 1 }{ 50\times{ 10 }^{ 6 } } ={ 2 }\times10^{ -8 }s\), the circuit is good enough for the detection.

\({ v }_{ c }>>1\quad MHz\)
Here, \(RC={ 10 }^{ 6 }\times{ 10 }^{ -12 }={ 10 }^{ -6 }s\)
For demodulation \({ v }_{ c }>>\frac { 1 }{ { 10 }^{ -6 } } ={ 10 }^{ 6 }Hz=1\ MHz\)
\(\therefore \quad { V }_{ c }>>1\ MHz\)

Bandwidth required for each station
\(=2\times max. \ frq.\ of \ audio \ signal\)
\(=2\times \ 5 \ kHz=10 \ kHz\)
Max.number of stations, which can be broadcast within bandwidth of 5 MHz
\(=\frac { 5MHz }{ 10kHz } =\frac { 5\times{ 10 }^{ 6 }Hz }{ { 10 }^{ 4 }Hz } =500\)

Here,\({ A }_{ c }=2kV=2000V;\mu =?\)
Amplitude of each side band = \(\frac { \mu { A }_{ c } }{ 2 } \)
\(200=\frac { \mu *2000 }{ 2 } \mu =0.2\)

300 MHz
\(i=\frac { \lambda }{ 2 } =\frac { 1 }{ 2 } \left( \frac { c }{ v } \right)\)
\(\therefore v=\frac { c }{ 2l } =\frac { 3\times{ 10 }^{ 8 } }{ 2\times0.5 } =3\times{ 10 }^{ 8 }Hz=300\ MHz\)

Given, the equation of carrier wave,
 c(t) = 1 sin (8\(\pi \) t)        ..........(i)
(i) According to the figure,
Amplitude of modulating signal,
A_m = 1 V
Amplitude of carrier wave,
A_C =2
T_m = 1 s
From Eq.(i), we get
 \({ \omega }_{ m }=\frac { 2\pi }{ { T }_{ m } } =\frac { 2\pi }{ 1 } =2\pi \ rad/s\)    ....(ii)
 c(t) = 2 sin (8 \(\pi \) t)
So,  \({ \omega }_{ c }=4{ \omega }m_{ }\)
From Eq. (ii) , we get
So,   \({ \omega }_{ c }=4{ \omega }_{ m }\)
Amplitude of modulated wave,
A = A_m + A_c
 = 1 + 2= 3 V
The sketch of the amplitude modulated waveform is shown below:

For carrier signal, \({ \omega }\) = 8 \(\pi \) 
\(T=\frac { 2\pi }{ \omega } =\frac { 2\pi }{ 8\pi } =\frac { 1 }{ 4 } =0.25s\)
(ii) Modulation index, \(m=\frac{A_{m}}{A_{c}}=\frac{1}{2}=0.5\)