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Home Electric Networks Filters Transmitted Bands from Propagation Constant
 
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Transmitted Bands from Propagation Constant

As shown by equation 58, the propagation constant for a T section such as Fig. 24 is γ = 2 sinh-1 (0.5*sqrt(Z1/Z2)). Since the elements of a filter are assumed to be without resistance and to cause no energy losses, Z1 and Z2 must be pure reactances. Accordingly, the ratio Z1/Z2 must have an angle that is either 0 or 180°; that is, the ratio must be a real positive quantity, or a real negative quantity, but cannot be complex. For instance, if Z1 and Z2 are both inductors with negligible resistance, the ratio Z1/Z2 will have a zero angle; if Z1 is inductive and Z2 capacitive, then Z1/Z2 will have an angle of 180°. The effects of these different ratios of Z1/Z2 on the propagation constant will be considered now.5

Case I. When Z1/Z2 is positive. The propagation constant y is composed of a real and an imaginary term; that is, γ = α + jβ. When Z1/Z2 of equation 58 gives a zero angle, y is a real number since the jβ part becomes zero. There will accordingly be undesired attenuation in the filter for all positive values of Z1/Z2.

Case II. When Z1/Z2 is negative and less than -4. If the ratio Z1/Z2 has an angle of 180° it is negative in sign; and, therefore, 0.5*sqrt(Z1/Z2) of equation 58 has an angle of 90° and both 0.5*sqrt(Z1/Z2) and γ are pure imaginaries. Equation 58 can be written

From hyperbolic trigonometry, sinh (x + jy) = sinh x cos y + j cosh x sin y. Thus,

Since it was previously shown that 0.5*sqrt(Z1/Z2) and γ are pure imaginaries for the case being considered, then the real part of equation 62, that is, sinh α/2 cosβ/2, must be zero. For this to be true, either sinh α/2 or cosβ/2 must equal zero. If sinh α/2 = 0, then cosh α/2 = 1, because of the numerical relations between these two hyperbolic functions. When these relations exist, equation 62 becomes

therefore,

If Z1/Z2 is greater than -4, then 0.5*(sqrt(-Z1/Z2)) would exceed 1, and equation 63 cannot hold because the sine of an angle cannot exceed unity. Thus, from equation 64 a filter will transmit without attenuation when Z1/Z2 is negative and less than -4 in magnitude.

Case III. When Z1/Z2 is more negative than -4. For values of Z1/Z2 more negative than -4, the real part sinh α/2 cosβ/2 of equation 62 can be made zero by cos β/2 = 0. At the values cos β/2 = 0, sin β/2 = ±1. Thus the angle β/2 must be some odd multiple of 90° to give these zero and unity values. Expressed in terms of radians (90° = π/2 radians),

where K is any whole number. This relation must hold to give cosβ/2 = 0 values.

When cos β/2 = 0 and sinβ/2 = ±1 are placed in equation 62, this becomes

From this relation

Then, from equations 65 and 67, the propagation constant becomes

This relation shows that, for negative values of Z1/Z2 greater than -4, the filter will offer attenuation.

From the foregoing, it appears that a non-dissipative recurrent structure of the type shown in Fig. 4 having series impedances Z1 and shunt impedances Z2 will pass readily only those signals of frequencies such that the ratio Z1/Z2 will lie between zero and -4. The values of Z1 and Z2 depend on the frequency because in filters Z1 and Z2 are inductors and capacitors.


Home Electric Networks Filters Transmitted Bands from Propagation Constant

Last Update: 2011-05-18