Question

When 80% AM is transmitted, total power is 10KW. If carrier and one side band are suppressed, then power saved is :

• A. 8.28 kW
• B. 8.08 kW
• C. 7.88 kW
• D. 8.78 kW

Answer 1

The correct option is D 8.78 kW

Given: $m_a=80$% $=0.8$

$P_t=10kW$

Now by:

$P_t=P_c(1+\frac{m_a^2}{2})$

$10=P_c(1+\frac{{0.8}^2}{2})$

$P_c=\frac{10\times 200}{264}=7.57kW$

Sideband power, $P_{SB}=P_t-P_c$

$P_{SB}=10-7.57=2.43kW$

Power of one sideband, $=P_{SB}/2=2.43/2=1.21kW$

As one sideband and carrier are supressed therefore power saved:

$P=7.57+1.21=8.78kW$