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i was studying SSB demodulation and learned that a carrier is added to the SSB signal to make sure that envelope detection is possible. I came across a problem to find the percentage of power saved in this type of modulation if the modulation index is .5.

I tried to derive the expression, % saving = (Useful power ) / Total power

However i am not able to reach the final expression given in the answer. Please help !

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I think there is quite a bit of confusion in your question, which could have led to not being able to solve it. The Power Saving you obtain is certainly not in %. For that there should be a factor 100 around.

learned that a local carrier

This is done in the receiver ('local'), so I'm not sure if you are really talking about a 'local' carrier.

SSB is (or was?) transmitted at times with a carrier, to allow detection even in AM receivers, but then the carrier is not local.

The formula you mention, is that the correct solution or the one you arrived at?

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    $\begingroup$ Sorry, local carrier part was a mistake. It is actually the carrier added at modulator so that envelope detection will be possible at receiver. The answer is the one they have given . Actually i have reached at the answer. Please correct me if i am wrong. Total power, Pt = Pc (1 + 0.5* m^2) P(side band) = (Pc * m^2 )/4 Power saved = (Pt-Pssb)/Pt I was able to reach that expression with this procedure. $\endgroup$ – Anjana Verma Nov 14 '14 at 16:07
  • $\begingroup$ Yes, that is correct. Congrats for getting there yourself... Now, I still am not quite sure if it makes sense changing the modulation index of the modulation you are comparing to. $\endgroup$ – jcoppens Nov 14 '14 at 21:33
  • $\begingroup$ I've taken the liberty of editing out the 'local'. Maybe someone else finds it clearer this way... $\endgroup$ – jcoppens Nov 14 '14 at 21:37

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