Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 3 (Q.No. 33)
33.
In an AM wave the carrier and one of the side bands is suppressed. If m = 1, the percentage saving in power is
Answer: Option
Explanation:
.
Discussion:
3 comments Page 1 of 1.
Povi said:
8 years ago
Anyone explain this.
Amod Kumar jha said:
8 years ago
% power save=(power save)/(total power) * 100.
Now,
AM power = Pc(1+u^2/2)=Pc+Pc*u^2/2
Here Pc = career power and 2nd term is a power of both side band since in AM both side band and career is present.
So,
Power of one of the side band = Pc * u^2/4.
% save power = (Pc + Pc * u^2/4) ÷ (Pc + Pc * u^2/2),
= (1+u^2/4) ÷ (1+u^2/2) = (4+u^2)÷(4+2u^2),
Here given u=1, put it into last equation,
Ans = 83.3%.
Now,
AM power = Pc(1+u^2/2)=Pc+Pc*u^2/2
Here Pc = career power and 2nd term is a power of both side band since in AM both side band and career is present.
So,
Power of one of the side band = Pc * u^2/4.
% save power = (Pc + Pc * u^2/4) ÷ (Pc + Pc * u^2/2),
= (1+u^2/4) ÷ (1+u^2/2) = (4+u^2)÷(4+2u^2),
Here given u=1, put it into last equation,
Ans = 83.3%.
R Das said:
8 years ago
As carrier and one of the sideband is suppressed,
%power saving= [Power in Carrier+Power in one sidebands]/Total Power
=[P(C){1+((m^2)/4}]/[P(C){1+((m^2)/2}]={1+((m^2)/4}/{1+((m^2)/2} = {1+(1/4)}/{1+(1/2)}= 1.25/1.5 = 0.8333.
=83.33%.
m= modulation index. P(C)= Carrier power.
%power saving= [Power in Carrier+Power in one sidebands]/Total Power
=[P(C){1+((m^2)/4}]/[P(C){1+((m^2)/2}]={1+((m^2)/4}/{1+((m^2)/2} = {1+(1/4)}/{1+(1/2)}= 1.25/1.5 = 0.8333.
=83.33%.
m= modulation index. P(C)= Carrier power.
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