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An AM wave has `1800` watt of total power content, For `100%` modulation the carrier should have power content equal toA. `1000` wattB. `1200` wattC. `1500` wattD. `1600` watt |
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Answer» Correct Answer - b `P_(t)=P_(c)(1+(m_(a)^(2))/(2))`, Here `m_(a)=1` `implies 1800=P_(c)(1+((1)^(2))/(2))impliesP_(c)=1200W` |
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