= log I1/I2. We see, in this case, that we multiply the log by the factor 20 rather than the factor 10 as we did in the power domain (i.e., 20 log VS. 10 log) because we really are dealing with power. Power is the function of the square of the signal voltage (E2/R or signal current (I2R). We use traditional notation for voltage and current. Voltage is measured in volts (E); current, in amperes (I). We must impress on the reader two important points: (1) Equations as written are only valid when R1 = R2, and (2) validity holds only for terminations in pure resistance (there are no reactive components). Consider these network examples: Current (I): R1 Network Voltage (E): E1 Network (R,=R2) E1 and E2 are signal voltage drops across R1 and R2, respectively. The incisive reader will tell us that signals at the input are really terminated in an impedance (Z), which should equal the characteristic impedance, Z (specified impedance). Such an impedance could be 600 S2, for example. An impedance usually has a reactive component. Our argument is only valid if, somehow, we can eliminate the reactive component. The validity only holds true for a pure resistance. About the closest thing we can find to a "pure" resistance is a carbon resistor. Turning back to our discussion, the input in the two cases cited may not be under our control, and there may be some reactive component. The output can be under our control. We can terminate the output port with a pure resistor, whose ohmic value equals the characteristic impedance. Our purpose for this discussion is to warn of possible small errors when reading input voltage or current. Let's discuss the calculation of decibels dealing with a gain or loss by an example. A certain network with equal impedances at its input and output ports displays a signal C.5 CALCULATING A NUMERIC VALUE GIVEN A dB VALUE voltage of 10 V at the input and 100 V at the output. The impedances are entirely resistive. What is the gain of the network? 100 V Network GaindB = 20 log 100/10 = 20 dB. A similar netw