Search Search Any Topic from Any Website Search
1. Start with normal AM For AM: P T = P C ( 1 + μ 2 2 ) P_T=P_C\left(1+\frac{\mu^2}{2}\right) The two sidebands together have power: P S B = μ 2 2 P C P_{SB}=\frac{\mu^2}{2}P_C And because the two sidebands are equal: P U S B = P L S B = μ 2 4 P C P_{USB}=P_{LSB}=\frac{\mu^2}{4}P_C So for μ = 1 \mu=1 : P U S B = P L S B = 1 4 P C P_{USB}=P_{LSB}=\frac14P_C 2. What does VSB do? VSB = Vestigial Sideband. It keeps: one complete sideband a small portion of the other sideband For example, suppose we transmit: complete USB → 1 / 4 P C 1/4P_C 50% of LSB → 1 / 8 P C 1/8P_C Then total sideband power is: P S B , V S B = 1 4 P C + 1 8 P C P_{SB,VSB}=\frac14P_C+\frac18P_C = 3 8 P C =\frac38P_C Therefore total VSB power is: P T = P C + 3 8 P C P_T=P_C+\frac38P_C P T = 11 8 P C \boxed{P_T=\frac{11}{8}P_C} for μ = 1 \mu=1 and a 50% vestige . 3. General idea If the vestigial sideband contains a fraction k k of one sideband, then: P S B , V S B = μ 2 4 P C + k μ 2 4 P C P_{SB,VSB} = \frac{\m...