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Home Electric Networks Filters BandPass Filter  


BandPass FilterThis filter is designed to pass a given band of frequencies, and to attenuate all other frequencies. A bandpass filter of the constantk type is shown in Fig. 37. The arm Z_{1} is resonant for some frequency f_{c}, and Z_{2} is antiresonant at this same value. At the resonant frequency f_{r}, Z_{1} offers zero impedance and Z_{2} offers an infinite impedance. Although more generalized structures are possible,^{4,5} only the simple "confluent" bandpass filter passing only one band will be considered. In this structure, L_{1}C_{1} = L_{2}C_{2}.
For the filter of Fig. 37,
and
The ratio Z_{1}/Z_{2} (when L_{1}C_{1} = L_{2}C_{2}, and L_{2} = L_{1}C_{1}/C_{2} are substituted) is
A network of this general type will pass frequencies between Z_{1}/Z_{2} = 4 and Z_{1}/Z_{2} = 0. Accordingly, equation 86 becomes
As equation 87 indicates, an upper and a lower cutoff frequency are obtained; that is, and
These cutoff points were found by Z_{1}/Z_{2} = 4. When the ratio is equated to zero, Then,
since L_{1}C_{1} = L_{2}C_{2}. This is not a true cutoff frequency,^{4} but lies at the middle of the band passed. It is the frequency for which the series arm is resonant and the shunt arm antiresonant as previously considered. In a generalized filter there are two bands passed; but with L_{1}C_{1} = L_{2}C_{2} these two bands "flow together" and the filter is of the simple confluent type. If the two expressions of equation 88 are subtracted,
since L_{1}C_{1} = L_{2}C_{2}. Then,
where Z_{K} = sqrt(L_{2}/C_{1}) = sqrt(L_{1}/C_{2}) and is the iterative impedance taken at the resonant frequency. By making similar substitutions,
As an illustration of the design of a simple confluent constantk bandpass filter, assume that it is desired to construct a filter such that Z_{K} = 600 ohms, f_{c}' = 900 cycles, and f_{c}" = 1100 cycles. Then,
The values here computed are for the entire series and parallel arms of Fig. 37. If a T section is desired, it should be constructed as indicated in Fig. 38. Also, the values for a π section may be readily determined.
The attenuation curve for a bandpass filter of this type would be approximately as shown in Fig. 39. The pass bands can also be computed by reactance sketches^{4,5} as in Fig. 40. This sketch is not for the simple confluent type just considered. It will therefore be noted that 4Z_{2} values determined by the parallel or antiresonant circuit do not pass through infinity at the same point that the Z_{1} curve determined by the series elements passes through zero. The two transmission bands T and T' are accordingly not confluent and, as the diagram indicates, exist separately. This figure represents a double bandpass filter.


Home Electric Networks Filters BandPass Filter 