ZL equals Zo, we can easily express insertion loss as a voltage ratio: Insertion lossdB = 20 log(E0/E2). C.9 RETURN LOSS If the network were one that furnished gain, such as an amplifier, we would invert the ratio and write: Insertion gaindB=10 log(p2/P0) or, for the case of voltage, Insertion gaindB=20 log(E2/E0). This may seem to the reader somewhat redundant to our introductory explanation of dBs. The purpose of this section is to instill the concepts of insertion loss and insertion gain. If we say that waveguide section had an insertion loss of 3.4 dB, we know that the power would drop 3.4 dB from the input to the output of that waveguide section. If we said that the LNA (low noise amplifier) had an insertion gain of 30 dB, we would expect the output to have a power 30 dB greater than the input. C.9 RETURN LOSS Return loss is an important concept that sometimes confuses the student, particularly when dealing with the telephone network. We must remember that we achieve a maximum power transfer in an electronic circuit when the output impedance of a device (network) is exactly equal to the impedance of the device or transmission line connected to the output port. Return loss tells us how well these impedances match; how close they are to being equal in value (ohms) to each other. Consider the following network's output port and its termination. The characteristic impedance (Z) of the output of the network is 600 S2. Network We have terminated this network in its characteristic impedance (Zo). Let us assume for this example that it is 600 S. How well does the network's output port match its characteristic impedance? Return loss tells us this. Using the notation in the preceding example, return loss is expressed by the following formula: Return = First let us suppose that Zn is exactly 600 S. If we substitute that in the equation, what do we get? We have then in the denominator 0. Anything divided by zero is infinity. Here we have the ideal case, an infinite return loss; a perfect match. Suppose Zn were 700 S2. W