Question 20.WE.4: A signal of power 18.0 mW passes along one cable, where the ...
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?
Step 1 Apply the decibel equation to each cable in turn. In the first cable, if the input is P_{1} and the output P_{2} ,
then:
20 = 10 lg \left\lgroup \frac{P_{1}}{P_{2}}\right\rgroupHint: Notice that both sides of the equation produce a positive number since P_{1} > P_{2}.
In the second cable, the input is P_{2} , the output of the first channel. If the output is P_{3} , then:
Step 2 Add the two equations; this gives:
50 = 10 \left[ lg \left\lgroup \frac{P_{1}}{P_{2}}\right\rgroup + lg \left\lgroup \frac{P_{2}}{P_{3}}\right\rgroup \right ]Applying the ‘log of a product rule’ gives:
50 = 10 lg \left\lgroup \frac{P_{1}}{P_{2}} \times \frac{P_{2}}{P_{3}}\right\rgroup = 10 lg \left\lgroup \frac{P_{1}}{P_{3}} \right\rgroupThis 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).
Step 3 We have P_{1} = 18 mW and we need to find P_{3} . Substituting gives:
so:
lg \left\lgroup\frac{18}{P_{3}}\right\rgroup = \frac{5}{10} = 5Taking inverse logs, or pressing the inverse lg button on your calculator, gives:
\left\lgroup\frac{18}{P_{3}}\right\rgroup = 10^{5}P_{3} = 1.8 \times 10^{-4} mW
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