Fraction of total power carried by side bands is given by-

Options

A. `(P_(s))/(P_(T))=m^(2)`
B. `(P_(s))/(P_(T))=1/(m^(2))`
C. `(P_(s))/(P_(T))=(2+m^(2))/(m^(2))`
D. `(P_(s))/(P_(T))=(m^(2))/(2+m^(2))`

Answer

D

Explanation

`("side band power")/("total power") =(P_(s))/(P_(T))=(m^(2))/(2+m^(2))`