Mumbai University > Electronics Engineering > Sem4 > Fundamentals of Communication Engineering

Marks: 5M

Year: May 15

Power wastage in DSB-FC transmission

1. The carrier signal in the DSB-FC system does not convey any information. The information is obtained in the two sidebands only.
2. But the sidebands are images of each other and hence both of them contain the same information. Thus all the information can be conveyed by only one sideband

3. The total power transmitted by an AM wave is given by

   $Pt = PC + PUSB + PLSB \\ ∴ Pt = PC + \frac{m^2}4PC + \frac{m^2}4PC$

4. Out of the three terms in the Equation carrier compound does not contain any information and one side band is redundant

5. So out of the total power $P_t = [ 1 + \frac{m^2}4P_C]$

   Power wastage = $P_C + \frac{m^2}4$

   $Pc = [ 1 + \frac{m^2}4P_C]$

   This shows that we have to transmit much higher power than what is actually required. Hence DSB-FC system is a power inefficient system.