Mumbai University > Electronics Engineering > Sem4 > Fundamentals of Communication Engineering

Marks: 6M

Year: May 15

PC =100 kw

M = 0.75

The total power transmitted by an AM wave is given by

$Pt = P_C + P_{USB} + P_{LSB} \\ P_{USB} = P_{LSB} = \frac{m^2}4 \times PC$

Sideband Powers =$P_{LSB}$ and $P_{USB}$

Sideband Powers $= 100KW \times \frac{0.75^2}4 = 14.0625 K$

Total Side Band Powers = 2x 14.0625 = 28.125 KW

$P_t = P_C + P_{USB} + P_{LSB}$ = 100 + 28.125 = 128.125 KW

Total Power = 128.125 KW

Side Band Power = 28.125 KW