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Home Electric Networks Filters HighPass Filter  


HighPass FilterSuch a network freely passes all signals above a certain critical or cutoff frequency and greatly attenuates signals below this frequency. Three T sections of a constantk highpass filter are shown in Fig. 32. In the highpass filter Z_{1} = 1/(jωC_{1}), since the condensers 2C_{1 }are in series, Z_{2} = jωL_{2}, and the ratio of these two impedances is Z_{1}/Z_{2} = 1/(ω^{2}L_{2}C_{1}). From the general filter theory, the frequency
band passed will lie between Z_{1}/Z_{2} = 0 and Z_{1}/Z_{2} = 4. When the first value is substituted, Z_{1}/Z_{2} = l/(ω^{2}C_{1}L_{2}) = 0, and f_{c}" will be infinite. When the second substitution is made, then l/ω^{2}L_{2}C_{1}) = 4, and
Thus, the highpass filter of Fig. 32 will freely pass all frequencies between f_{c}' and infinity and will greatly attenuate all frequencies below f_{c}'. The iterative impedance equations for the T section of Fig. 32 are found in the same general manner as for the lowpass filter. From equation 45,
In a similar manner the iterative impedance for a π section is found from equation 46 to be
As equations 79 and 80 indicate, Z_{K} varies with frequency in both the T and the π sections (see also Fig. 33). For both types of sections, Z_{K} = sqrt(L_{2}/C_{1}) at f = ∞. For the T section, Z_{KT} = 0 at the cutoff frequency; and for the π section, Z_{Kπ} = ∞ at the cutoff frequency.
The iterative impedance used for highpass filter design is at f = ∞; hence, for both the T and the π sections,
The design formulas are now easily found. Since Z_{K} = sqrt(^{/}L_{2}/C_{1}), l/(ω^{2}L_{2}C_{1}) = 4, and L_{2}/C_{1} = 4ω_{c}^{2}L_{2}^{2}, then Z_{K}^{2} = 4(ω_{c}^{2}L_{2}^{2}, and therefore
The value of C_{1} can now be readily found as follows
To consider a typical case, suppose that it is desired to design a highpass filter to operate between circuits of 600 ohms impedance, and to cut off at 1000 cycles. Then, L_{2} = 600/(4 x 3.1416 x 1000) = 0.0477 henry, and C_{1} = 1/(4 x 3.1416 x 1000 x 600) = 0.000000132 farad or 0.132 microfarad. If Z_{K} were desired, and L_{2} and C_{1} had been given, then Z_{K} = sqrt(L_{2}/C_{1}) = sqrt(0.0477/0.000000132) = 600 ohms, approximately. As shown in Fig. 32, two capacitors of 2C_{1} and an inductor L_{2} would be used for a T section. For a (π section, two inductors of 2L_{2} and one capacitor of C_{1} would be used. The attenuation of a constantk highpass filter can be calculated from equation 67. Thus,
when the value given by equation 82 is substituted for L_{2}, and the value given by equation 83 is substituted for C_{1}. At 500 cycles, for the highpass filter under consideration, α = 2 cosh^{1} (1000/500) = 2 cosh^{1} 2, and (α = 2 x 1.31 = 2.62 nepers, or 2.62 x 8.686 = 22.8 decibels. The dotted curve of Fig. 34 shows calculated values of attenuation for the filter under consideration (assumed to be composed of lossless elements), and the solid curve shows the attenuation measured on an actual filter, designed as explained in the preceding paragraph. The phase shift of a constantk highpass filter can be found from equation 63
when the same substitutions as in equation 84 are made. For the filter under consideration, and at a frequency of 2000 cycles, (β = 2 sin^{1}(1000/2000) = 2 sin^{1} 0.5 = 60°. The dotted curve of Fig. 35 shows calculated values, and the solid curve shows values measured with a cathoderay oscilloscope. As for the lowpass filter, an analysis of the characteristics of a highpass filter can be made from simple reactance sketches as in Fig. 36. In this figure Z_{1} = l/(2πfC_{1}) and Z_{2} = 2πfL_{2}, giving the curves positions different from Fig. 31. Transmission will occur for such frequencies that Z_{1} lies between 4Z_{2} and the frequency axis as indicated.


Home Electric Networks Filters HighPass Filter 