# 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:

first method:

attenuation=190x10^-3x14=2.66dB

ratio/dB=-10lg(P2/P1) I can't understand why there is negative sign

2.66=-10lg(Pout/Pin)

Pout/Pin=0.54

frictional loss=1-(Pout/Pin)=1-0.54-0.46 and I don understand this step either.

second method:

2.66=10lg(Pin/Pou) why positive sign for Pin/Pout?

Pin/Pout =1.85

frictional loss =(Pin-Pout)/Pin=(1.85-1)/1.85=0.46 also don understan...

can anyone help me?

### 1 Answer

- Steve4PhysicsLv 71 month ago
It requires some careful thought. If you work through this, it should help.

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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)

E.g.

lg(50) = 1.699

lg(1/50) =lg(0.02) = -1.699

So in your question that means:

lg(Pout/Pin) = -lg(Pin/Pout)

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Suppose x halves.

The change, in dB, is

10lg(final value/initial value)

= 10log(x/(2x))

= 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.

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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

In symbols:

Fractional loss = (Pin – Pout)/Pin

. . . . . . . . . . . . = 1 – Pout/Pin