Modulation is a process in which one of the characteristics (amplitude, frequency, phase) of a high frequency carrier wave is made to change in accordance with a give low frequency message signal.
Modulated wave 1 : Frequency Modulation
Modulated wave 2 : Amplitude Modulation
Two advantages of FM over AM :
(i) Lower noise, better power efficiency
(ii) Higher operating range
(iii) Higher fidelity reception.
[Alternatively, Two advantages of AMover FM]
(i) Simple circuits are required.
(ii) Lower frequency space for transmission

Diode Bis reverse biased.
When diode is reverse based, the barrier height increases as the direction of applied voltage (V) and the direction of barrier potential (V₀) is same. The effective barrier height under reverse biased is (V₀ + V).
Diode C is forward biased.
As the direction of applied voltage is opposite to the barrier potential, therefore, the effective barrier height is reduced to (V₀ - V).

The equivalent gates in the two cases are the OR gate and the AND gate respectively.
(i)A combination of three NAND gates, connected in the manner shown, would be equivalent to an OR gate.

(ii) A combination of three NOR gates connected in the manner shown would be equivalent to an AND gate.

(i) Standard AM broadcast 540-1600kHz
(ii) Television                     54-890MHz
(iii) Satellite communication 5.925-6.425GHz uplink 3.7-4.2GHz Downlink

First draw the truth table of C _1.

Therefore, the output signal C₁ is shown as:

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.

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 = \(\frac { 3.75\times { 10 }^{ 12 } }{ 4.5\times { 10 }^{ 6 } } =8.3\times { 10 }^{ 5 }\)

Three modes of propagation of electromagnetic waves are
(i) Ground wave propagation
(ii) Space wave propagation
(ii) Sky wave propagation 
The mode of propagation used in the frequency range from a few 40 MHz is sky wave propagation.

3.75 m
\(l=\frac { \lambda }{ 4 } =\frac { 1 }{ 4 } \left( \frac { c }{ v } \right) =\frac { 1 }{ 4 } \times \left( \frac { 3\times{ 10 }^{ 8 } }{ 20\times{ 10 }^{ 6 } } \right) =3.75\ m\)