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A carrier wave of 1000 W is subjected to 100% modulation. Calculate (i) Power of modulated wave, (ii) Power in USB, (iii) Power in LSB. |
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Answer» Solution :(i) Total power of modulated WAVE `P_(tau)=P_(C)(1+(m^(2))/(2))=1000(1+(1^(2))/(2))=1500` watt (ii) Power in USB `=1/2P_("SB")` where power carried by side bands is given by Amplitude Modulation and Detection `P_("SB")=P_(C)((m^(2))/(2))=1000((1^(2))/(2))=500` watt `P_("USB")=1/2P_("SB")=1.2xx500=250` watt (iii) Since power in LSB = Power in USB `P_("LSB")=P_("USB")=250` watt |
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