alevel physics ?
the coaxial cable connecting an aerial to a receiver has length 14m.the cable has an attenuation per unit length of 190dB/km
calculate the fractional loss in signal power during transmission of the signal along the cable.
on the mark scheme, there are 2 methods to solve it:
ratio/dB=-10lg(P2/P1) I can't understand why there is negative sign
frictional loss=1-(Pout/Pin)=1-0.54-0.46 and I don understand this step either.
2.66=10lg(Pin/Pou) why positive sign for Pin/Pout?
frictional loss =(Pin-Pout)/Pin=(1.85-1)/1.85=0.46 also don understan...
can anyone help me?
- Steve4PhysicsLv 71 month ago
It requires some careful thought. If you work through this, it should help.
First note that lg(A/B) = -lg(B/A) (equation 1)
lg(x) is positive is x is greater than 1.
lg(x) is negative is x is less than 1 (but x must be greater than 0)
lg(50) = 1.699
lg(1/50) =lg(0.02) = -1.699
So in your question that means:
lg(Pout/Pin) = -lg(Pin/Pout)
Suppose x halves.
The change, in dB, is
10lg(final value/initial value)
= 10log(1/2) ≈ -3dB
The *change* is -3dB. The negative sign tells us the value of x has reduced.
BUT we could say the *reduction* is 3dB (because the word ‘reduction’ takes the minus sign into account).
Your question gives the *attenuation* (reduction), so this takes
the minus sign into accourt.
Suppose you start with £100 (Pin) and lose some. You are left with £30 (Pout).
Your loss is £100 - £30 = £70
Your fractional loss is £70/£100 = 0.7
Fractional loss = (Pin – Pout)/Pin
. . . . . . . . . . . . = 1 – Pout/Pin