Electronics and Communication Engineering - Communication Systems - Discussion

When m = 0.5 total power is given by Pt(total power) = pc(power of carrier) * (1 + m^2/2).

= Pc(1 + (0.5)^2/2) = 1.125Pc.

One sideband power:

Psb = Pc.m^2/4.

= Pc(0.5)^2/4 = 0.0625Pc.

Savings In power:

(1.125 - 0.0625)/1.125.

= 94.4%

Can anyone have a simple solution for this?

Assume,

Pt(AM)=100W,
FOR AM, Pt(AM)=Pc(1+m^2/2),
Put the value of m and get Pc,
FOR SSBC,Pt(ssbc) =Pc(m^2/4),
Pt(AM) - Pt(ssbc) = 94.4%.

How it is possible?

Explain in details.

pt= pc(1+m2/2)= pc(1+(.5*.5)/2)= 1.125 pc.

psb =pc m2/4 = pc (.5*.5)/4 = 0.0625 pc.

% power saving = (1.125-0.0625)/1.125 = 94.44%.

But Psb = pc(m2/2) not psb = pc m2/4 = pc (.5*.5)/4 = 0.0625 pc.

This signal is ssb-sc. Because a carrier & one side band is absent.

Formula for power saving: p= 4+m^2/4+2m^2*100

Where m=modulation index.

You can use this formula.

P@carrier = Pc.
P@uppersideband = (Pc*m^2)/4.
P@lowersideband = (Pc*m^2)/4.

Then assign values and do the relationship and then compare it to a Double Sideband Full Carrier.

Like this:
Let: Pc = 10.
m = 0.5.

**We are gonna solve for the Power of DSBFC**
P@carrier = Pc = 10.
P@uppersideband = (Pc*m^2)/4 = .625.
P@lowersideband = (Pc*m^2)/4 = .625.
Total Power DSBFC = 10+.625+.625= 11.25.

**Now, let's solve the AM Wave given by the problem: Single Sideband Suppressed Carrier.

P@carrier = 0, because it is suppressed.
P@uppersideband = (Pc*m^2)/4 = .625.
P@lowersideband = 0, because it is suppressed.
Total Power SSBSC = .625.

Now to solve the total power % that was saved.

Total Power SSBSC/Total Power DSBFC.
.625/11.25 = .0555555555555556.
1-.0555555555555556 = .94444444444444.

.94444444444444x100 = 94.44%.

@Bharath.

Where have you subtracted carrier power?