 each network making up the whole is the same as that of the original network. We have a good example with the preceding network displaying a gain of +9 dB. Obviously 3 X 3 = 9. We break down the +9-dB network into three networks in series, each with a gain of +3 dB. This is shown in the following diagram: ?mW (Output) +3 dB +3 dB +3 dB We should be able to do this now by inspection. Remember that +3 dB is double the power; the power at the output of a network with +3-dB gain has 2x the power level at the input. Obviously, the output of the first network is 12 mW (point A above). The 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 X 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. 15 mW ? mW (Output) +7 dB We can break this down into two networks using dB values that are familiar to us: 15 mW ? mW (Output) +10 dB -3 dB 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 X 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 X 