o derive dB values for power ratios that lie in between 1 and 10, 10 and 100, 0.1 and 0.01, and SO on. C.1 LEARNING DECIBEL BASICS One excellent recourse is the scientific calculator. Here we apply a formula (C1.1). For example, let us deal with the following situation: Network Because the output of this network is greater than the input, the network has a gain. Keep in mind we are in the power domain; we are dealing with mW. Thus: dB value = 10 log 4/2 = 10 log 2 = 10 X 0.3010 = +3.01 dB. We usually roundoff this dB value to +3 dB. If we were to do this on our scientific calculator, we enter 2 and press the log button. The value 0.3010-appears on the display. We then multiply (x) this value by 10, arriving at the +3.010 dB value. This relationship should be memorized. The amplifying network has a 3-dB gain because the output power was double the input power (i.e., the output is twice as great as the input). For the immediately following discussion, we are going to show that under many situations a scientific calculator is not needed and one can carry out these calculations in his or her head. We learned the 3-dB rule. We learned the +10, +20, +30 dB; -10, -20, -30 (etc.) rules. One should be aware that with the 3-dB rule, there is a small error that occurs two places to the right of the decimal point. It is SO small that it is hard to measure. With the 3-dB rule, multiples of 3 are easy. If we have power ratios of 2, 4, and 8, we know that the equivalent (approximate) dB values are +3 dB, +6 dB, and +9 dB, respectively. Let us take the +9 dB as an example problem. A network has an input of 6 mW and a gain of +9 dB. What power level in mW would we expect to measure at the output port? mW (Output) +9 dB One thing that is convenient about dBs is that when we have networks in series, each with a loss or gain given in dB, we can simply sum the values algebraically. Likewise, we can do the converse: We can break down a network into hypothetical networks in series, SO long as the algebraic sum in dB of the gain/loss of