Low-Pass Filter

In Fig. 27 the series impedances are Z₁ = jωL₁ and the shunt impedances are Z₂ = l/(jωC₂). Then, Z₁/Z₂ = (jωL₁)(jωC₂) = -ω²L₁C₂. When Z₁/Z₂ = 0, the lower cutoff frequency is found; thus, since ω = 2πf, Z₁/Z₂ = -ω_c²L₁C₂ = -(2πfc')²L₁C₂ = 0, and f_c' = 0. When Z₁/Z₂ = -ω_C²L₁C₂ = -(2πfc")²L₁C₂ = -4, then the upper cutoff frequency is

Figure 27. Three constant-k low-pass filter sections.

The low-pass filter will accordingly pass currents of all frequencies from direct current (f = 0) up to the value given by equation 70.

The variations of the iterative impedances of T and π low-pass filter sections will now be investigated. From equation 45,

when -ω_c²L₁C₂ = -4 and ω_c" = 2πf_c" are substituted. This equation gives the iterative impedance of the low-pass T section of Fig. 27 at any given frequency ratio f/f_c". By a similar process it can be shown that, for the π section, equation 46 becomes

The equations for these two impedance curves are shown plotted in Fig. 28. They show that, for f = 0,

for both the T and the π sections. They further show that the impedance values then diverge and that at the value f = f_c" (that is, at the cutoff frequency) Z_KT = 0 and Z _Kπ = ∞. It is common practice to design a low-pass filter so that the iterative impedance Z_K computed at f = 0 and given by equation 73 equals the impedance of the line in which it is to operate.

It is now possible to derive the design equations for low-pass filters. The upper cutoff frequency was found from the relation -ω_c²L₁C₂ ⁼ -4. If 2πf_c" is substituted for ω_c, this relation can be arranged L₁/C₂ = (2πf_c"L₁)²/4. But Z_K = sqrt(L₁/C₂), and thus

Therefore,

Using these two equations, simple low-pass constant-k filters can be designed as will be illustrated.

Suppose that it is desired to construct a low-pass filter having an iterative impedance of 600 ohms and a cutoff frequency of 1000 cycles.

From equation 74, L₁ = 600/(3.1416 x 1000) = 0.191 henry, and from equation 75, C₂ = 1/(3.1416 x 1000 x 600) = 0.000000530 farad or 0.53 microfarad. That is, the series element Z₁ of Fig. 27 must be 0.191 henry inductance and the shunt element Z₂ of Fig. 27 must be 0.53 microfarad capacitance.

Figure 28. Theoretical curves are shown by broken lines for lossless low-pass constant-k filters. These are plots of equations 71 and 72 for 600-ohm filters having the values calculated on this page. The solid curves are the resistance and reactance components as measured on an actual π section having the values calculated on this page. The measurements were made with an impedance bridge, and the impedance found from the relation ZK = sqrt(ZocZsc).

The filter section can be connected either as a T or as a π section. If connected as a T section, two inductors each having an inductance of 0.191/2 or 0.0955 henry, and one capacitor having a capacitance of 0.53 microfarad, will be required. If connected as a π section, one 0.191-henry inductor and two 0.53/2 = 0.265-microfarad capacitors will be needed. These relations will be evident from Figs. 4, 5, and 6. It should also be noted that

Of particular interest in filter studies are the iterative impedance that has been considered, and the attenuation and phase shift that will now be treated. If the filter elements are pure reactances, the attenuation will be zero for frequencies within the band passed and will be as given by equation 67 for frequencies beyond the band passed. For the low-pass filter of Fig. 27,

when the value given by equation 74 is substituted for L₁, and the value given by equation 75 is substituted for C₂. Thus, at 2000 cycles the attenuation of the low-pass filter under consideration will be α = 2 cosh-¹ (2000/1000) = 2 cosh^-1 2, and α = 2 x 1.31 = 2.62 nepers, the value 1.31 being determined from a table of hyperbolic functions. The corresponding loss in decibels is db = 2.62 x 8.686 = 22.8 decibels. The dotted curve of Fig. 29 was computed in this way. The solid curve of Fig. 29 was measured on an actual filter section, the elements of which had some loss. For methods of calculating the attenuation in which the losses of the filter elements are considered, reference 4 should be consulted.

If the elements of a low-pass filter are pure reactances, there will be a varying phase shift over the band passed, and a constant phase shift of 180° beyond the band passed. From equation 63, and for the low-pass filter,

Figure 29. Attenuation curves as calculated using equation 76 and as measured. These curves are for one π section of the constant-k low-pass filter calculated on page 170. Small variations in the measured values below cutoff have been neglected.

when the substitutions as in equation 76 are made. Thus, for the low-pass filter under consideration, and at 500 cycles, β = 2 sin^-1 (500/1000) = 2 sin^-1 0.5 = 60°. The dotted curve of Fig. 30 was

Figure 30. Phase-shift curves for one π section of a constant-k low-pass filter as calculated by equation 77, and as measured with a cathode-ray oscilloscope. The input voltage was impressed on one set of deflecting plates, the output was impressed on the other set, and the phase angle was determined from the resulting figure. (Page 310.)

computed in this manner, and the solid curve was measured on an actual filter section, the elements of which had some loss. The phase measurements were made with a cathode-ray oscilloscope.

Figure 31. Reactance curves for a low-pass filter.

The transmission and attenuation bands of low-pass filters can be determined from simple reactance sketches as in Fig. 31. Thus, if curves for Z₁ (equals 2πL₁) and -4Z₂ (where Z₂ = l/(2πf/C₂)) are plotted as indicated, the transmission band T will occur where Z₁ lies between the frequency axis and the curve -4Z₂, and the band A where Z₁ lies without curve -4Z₂ will be attenuated.