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Fraction of total power carried by side bands is given by-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))` |
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Answer» Correct Answer - D `("side band power")/("total power") =(P_(s))/(P_(T))=(m^(2))/(2+m^(2))` |
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