Anonymous asked in Science & MathematicsPhysics · 1 month ago

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:


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.

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

  • 1 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(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.


    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

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