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Home Electric Networks Filters Transmitted Band from Iterative Impedances  


Transmitted Band from Iterative ImpedancesThe frequencies transmitted by a filter can be determined from the iterative impedances of the filter sections by the following method. Assume that a resistance load is connected to a source through a filter composed of inductors and capacitors having negligible losses. Then, the filter itself cannot absorb power. Now suppose that a frequency is chosen such that the iterative impedance of the filter is purely resistive. At this frequency, the source will send power to the filter, and, since the filter cannot absorb power, it must reach the load. Next assume that at a different frequency the iterative impedance of the filter is purely reactive. This prevents the source from sending any power into the connected circuit, and thus no power reaches the load at this frequency. From these facts it may be concluded that a filter will transmit without attenuation those frequencies for which the iterative impedance Z_{K }is resistive (Fig. 25). The iterative impedance given by equation 45 is
For unattenuated transmission this must be a positive real number to be resistive. This equation consists of two parts: X_{a} = Z_{1} and X_{b} = (Z_{2}+Z_{1}/4). Thus, Z_{KT} = sqrt(X_{a}X_{b}). Pure reactances only are being considered; and for Z_{KT} to be positive, if X_{a} represents inductive reactance and is +jN_{a}, then X_{b} must be capacitive reactance or jX_{b}, and vice versa, within the band of frequencies transmitted. For these relations Z_{KT} = sqrt((+jX_{a})(jX_{b})), and Z_{KT} will be positive. To satisfy these conditions, Z_{2} must be, first, the opposite type of reactance to Z_{1}, and second, greater in numerical value than Z_{1}/4. Or, in other words, the ratio Z_{1}/4Z_{2} from equation 45 must lie between 0 and 1 for the frequencies to be transmitted, as shown in Fig. 26.


Home Electric Networks Filters Transmitted Band from Iterative Impedances 