High-Pass Filter

In the high-pass filter Z₁ = 1/(jωC₁), since the condensers 2C₁ are in series, Z₂ = jωL₂, and the ratio of these two impedances is Z₁/Z₂ = -1/(ω²L₂C₁). From the general filter theory, the frequency

Figure 32. Three high-pass filter sections.

band passed will lie between Z₁/Z₂ = 0 and Z₁/Z₂ = -4. When the first value is substituted, Z₁/Z₂ = -l/(ω²C₁L₂) = 0, and f_c" will be infinite. When the second substitution is made, then -l/ω²L₂C₁) = -4, and

Thus, the high-pass 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 low-pass 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(L₂/C₁) at f = ∞. For the T section, Z_KT = 0 at the cutoff frequency; and for the π section, Z_Kπ = ∞ at the cutoff frequency.

Figure 33. Theoretical curves are shown by broken lines for lossless high-pass constant-k filters. These are plots of equations 79 and 80 for 600-ohm filters having the values calculated on page 175. The solid curves are the resistance and reactance components as measured on an actual T section having the values calculated on page 175. (See explanation accompanying Fig. 28 for method of making measurements.)

The iterative impedance used for high-pass 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(^/L₂/C₁), l/(ω²L₂C₁) = 4, and L₂/C₁ = 4ω_c²L₂², then Z_K² = 4(ω_c²L₂², and therefore

The value of C₁ can now be readily found as follows

To consider a typical case, suppose that it is desired to design a high-pass filter to operate between circuits of 600 ohms impedance, and to cut off at 1000 cycles. Then, L₂ = 600/(4 x 3.1416 x 1000) = 0.0477 henry, and C₁ = 1/(4 x 3.1416 x 1000 x 600) = 0.000000132 farad or 0.132 microfarad. If Z_K were desired, and L₂ and C₁ had been given, then Z_K = sqrt(L₂/C₁) = sqrt(0.0477/0.000000132) = 600 ohms, approximately. As shown in Fig. 32, two capacitors of 2C₁ and an inductor L₂ would be used for a T section. For a (π section, two inductors of 2L₂ and one capacitor of C₁ would be used.

The attenuation of a constant-k high-pass filter can be calculated from equation 67. Thus,

when the value given by equation 82 is substituted for L₂, and the value given by equation 83 is substituted for C₁. At 500 cycles, for the high-pass 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 constant-k high-pass 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 cathode-ray oscilloscope.

As for the low-pass filter, an analysis of the characteristics of a high-pass filter can be made from simple reactance sketches as in Fig. 36. In this figure Z₁ = l/(2πfC₁) and Z₂ = 2πfL₂, giving the curves positions different from Fig. 31. Transmission will occur for such frequencies that Z₁ lies between -4Z₂ and the frequency axis as indicated.