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# The Band-Elimination Filter

A filter of this type is designed to pass freely currents of all frequencies except those within a definite band. Three sections of a constant-k band-elimination filter are shown in Fig. 41. By comparison with the band-pass filter,

 Figure 39. Attenuation curve for three constant-k band-pass filter sections such as Fig. 38. Terminated in 600 ohms resistance.

and

since, for the constant-k filter, L1C1 = L2C2- A filter of this type will pass all frequencies such that Z1/Z2 will lie between 0 and -4. Thus,

the solution of which is

 Figure 40. Reactance curves for a non-confluent double band-pass filter, passing the two bands T and T'. This figure is not for the filter of Figs. 38 and 39.

 Figure 41. A band-elimination filter of the constant-k type. Two cutoff frequencies are obtained,

and

To find the upper and lower frequency limits, the impedance ratio is equated to zero. Thus, Z1/Z2 = -ωc2C2L1/(1 - ωc2L1C1) = 0, and from this, two values ωc = 0 and ωc = ∞ are obtained. From these two relations and from the preceding paragraph it follows that the band-elimination filter will pass all frequencies from zero to fc' cycles per second, will attenuate all frequencies from fc' to fc", and will pass all currents having frequencies between fc" and infinity. This filter then eliminates the band fc' to fc".

The method of finding the design equations is similar to that used for the band-pass filter. If the two expressions of equation 93 are subtracted, it will be found that (fc" - fc') = 1/(4πsqrt(L2C1)). Since, for the constant-k filter, L1C1 = L2C2 and ZK = sqrt(L2/C1) = sqrt(L1/C2). this equation can be solved for C1, and

Infinite and zero values occur at

Using

these relations and the equations just obtained, it can easily be shown that

Calculations for the various inductors and capacitors for a simple band-elimination filter are made from equations 94 and 95.

Last Update: 2011-05-18