Page 330 - Physics Coursebook 2015 (A level)

Cambridge International A Level Physics
318
WORKED EXAMPLE
They are written here as lg and must not be confused with logarithms to base e, which are usually written as ln.
number of B = lg P2 P1
number of dB = 10 lg P2 P1
Overcoming attenuation
In long-distance cables, the attenuation is given as attenuation per unit length, with units such as dB km−1. The attenuation is found from the equation:
attenuation per unit length (dB km−1)
= attenuation (dB)
length of cable (km) When a signal travels along a cable, the level of the noise
is important. The signal must be distinguishable above the level of the noise. The signal-to-noise ratio, measured in decibels, is given by the expression:
signal-to-noise ratio = 10 lg signal power noise power
At regular intervals along a cable, repeaters amplify
the signal. If the signal is analogue then repeaters also amplify the noise. Multiplying both signal and noise
by the same amount keeps the signal-to-noise ratio the same. Regeneration of a digital signal at the same time as amplification removes most of the noise. This ensures that the signal-to-noise ratio remains high.
This shows that the total attenuation of the two cables is 50 dB, equal to the sum of the attenuations of the consecutive channels. Hence you can add attenuations to find the total attenuation (but be careful if a signal is being both amplified and attenuated).
Step3 WehaveP1=18mWandweneedtofindP3. Substituting gives:
50=10lg 18 P3
For example, suppose P2 is 1000 times greater than P1: number of dB = 10lg 1000 = 30
1
The number is positive because there is an increase in
power – the signal is amplified. Attenuation produces a negative number of decibels; for example, an attenuation of −30 dB means that the received signal is 1000 times smaller than the signal transmitted.
You may be much more familiar with logarithms to base e than with logarithms to base 10. All logarithms obey the same rules; some, which you should know, are:
log of a product log of a ratio log of a power
4 A signal of power 18.0 mW passes along one cable, where the attenuation is 20 dB. It then passes along another cable, where the attenuation is 30 dB. What is the power at the end of the two cables?
Step1 Applythedecibelequationtoeachcableinturn. Inthefirstcable,iftheinputisP1andtheoutputP2,
positive number since P1 > P2.
In the second cable, the input is P2, the output of the
log (ab) = log (a) + log (b) a
log b = log (a) − log (b) log(an) = nlog(a)
then: 20=10lg P2
P1
Hint: Notice that both sides of the equation produce a
so: lg
18 = 50 = 5 P3 10
first channel. If the output is P3, then: 30=10lg P2
Taking inverse logs, or pressing the inverse lg button on your calculator, gives:
18 5 P3 =10
P3=1.8×10−4mW
You could apply the decibel equation to each cable in
turn and use the output of the first cable as the input to the second cable. You should find that the result is the same.
P3
Step2 Addthetwoequations;thisgives:
P1 P2 50 = 10 lg P2 + lg P3
Applying the ‘log of a product rule’ gives:
50=10lg P1 ×P2 =10lg P1 P2 P3 P3