618
LEARNING DECIBELS AND THEIR APPLICATIONS
input to the second network is now 12 mW and this network again doubles the power.
The power level at point B, the output of the second network, is 24 mW. The third
network--double the power still again. The power level at point C is 48 mW.
Thus we see that a network with an input of 6 mW and a 9-dB gain, will have an
output of 48 mW. It multiplied the input by 8 times (8
× 6 = 48). That is what a 9-dB
gain does. Let us remember:
+3 dB is a two-times multiplier; +6 dB is a four-times
multiplier, and
+9 dB is an eight-times multiplier.
Let us carry this thinking one step still further. We now know how to handle 3 dB,
whether
+ or -, and 10 dB (+ or -), and all the multiples of 10 such as 100,000 and
0.000001. Here is a simple network. Let us see what we can do with it.
We can break this down into two networks using dB values that are familiar to us:
If we algebraically sum the
+10 dB and the -3 dB of the two networks in series
shown above, the result is
+7 dB, which is the gain of the network in question. We have
just restated it another way. Let us see what we have here. The first network multiplies
its input by 10 times (
+10 dB). The result is 15 × 10 or 150 mW. This is the value of
the level at A. The second network has a 3-dB loss, which drops its input level in half.
The input is 150 mW and the output of the second network is 150
× 0.5, or 75 mW.
This thinking can be applied to nearly all dB values except those ending with a 2, 5,
or 8. Even these values can be computed without a calculator, but with some increase
in error. We encourage the use of a scientific calculator, which can provide much more
accurate results, from 5 to 8 decimal places.
Consider the following problem:
This can be broken down as follows:
Remember that
+50 dB is a multiplier of 10
5
and
-6 dB is a loss that drops the power
to one quarter of the input to that second network. Now the input to the first network is
0.3 mW and so the output of the first network (A) is 0.3 mW
× 100,000 or 30,000 mW
(30 W). The output of the second network (B) is one-quarter of that value (i.e.,
-6 dB),
or 7500 mW.
Now we will do a practice problem for a number of networks in series, each with its
own gain or loss given in dB. The idea is to show how we can combine these several