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i was studying SSB demodulation and learned that a local 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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When m=.5 total power is given by Pt(total power)=pc(power of carrier)*(1+m²/2)

=Pc(1+(0.5)²/2) =1.125Pc

One sideband power

Psb=Pc.m²/4

=Pc.(0.5)²/4 =0.0625Pc

Savings In power

(1.125-0.0625)/1.125

=94.4%

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  • \$\begingroup\$ Thanks! I calculated power efficiency instead of power saving. That cleared it up :) \$\endgroup\$ – Anjana Verma Nov 14 '14 at 9:27
  • \$\begingroup\$ If the carrier is suppressed only to side bands power remains and which is equal to =Pc.m²/4 about 66% savings will be done.Here Pc is Carrier power .And if One of the sideband is also suppressed the Remaining power is Pc.m²/4 a further saving of 50% power will be achieved \$\endgroup\$ – Siddhartha B Nov 14 '14 at 9:28
  • \$\begingroup\$ Your question is of Power saving or of power efficiency? \$\endgroup\$ – Siddhartha B Nov 14 '14 at 9:30
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The main advantage would be almost doubling spectral power density and almost halving bandwidth with the main disadvantage receivers not ready for the bandwidth and requiring linearity in the transmitter. Calculation is not very interesting in this cases to decide.

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