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: 0.3 mW ? mW (Output) +44 dB This can be broken down as follows: 0.3 mW ? mW (Output) +50 dB -6 dB Remember that +50 dB is a multiplier of 105 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 X 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 C.2 dBm AND dBW networks into an equivalent single network regarding gain or loss. We are often faced with such a problem in the real world. Remember, we add the dB values in each network algebraically. +12 dB -28 dB +7 dB -11 dB Look what happens when we combine these four networks into one equivalent network. We just sum: +12 - 28 + 7 - 11 = -20, and -20 dB is a number we can readily handle. Thus the equivalent network looks like the following: -20 dB To see really how well you can handle dBs, the instructor might pose a difficult problem with several networks in series. The output power of the last network will be given and the instructor will ask the input power to the first network. Let us try one like that SO the instructor will not stump us. 40 mW (Output) +23 dB +15 dB -12 dB Input First sum the values to have an equivalent single network: +23 + 15 - 12 = +26 dB. Thus, 40 mW (Output) +26 dB Input We first must learn to ask ourselves: Is the input greater or smaller than the output? This network has gain, thus we know that the input must be smaller t