Filter Design'S examples...

Example 1:



For the given a Low-pass filter shown below, make frequency transformation to designa HP filter with cut-off frequency 0 =106 rad/sec and load resistance of 500 ohm

 

Given: L_n = 4/3 henry, C_n1 = 1/4 farad, C_n2 = 3/2 farad

Solution:

The normalized inductance is given by

L_h1 =1/w₀C_n1 =1/(10⁶x1/2) = 2 mh

The de-normalized value of L_h1  = 2 mh x R_L =1000 mh = 1 mh

Similarly,

L_h2 = R_L/w₀C_n2 = 500/(10⁶x3/2) = 0.334 mh

Normalized capacitance, C_h = 1/w₀L_n1 = 1/(10⁶x4/3) f

De-normalized C_h = 1/(10⁶x4/3)R_L = 1.5 nf

Example -2:

Transform the LP filter of example - 1 to a band-pass filter with BW=6x104 rad/sec and centre frequency 0 = 4x104 rad/sec


Solution:

Using the formulae for LP to BP transformation the series L,C elements are given by:

L2 = Lbs = Ln/BW = (4/3)/(6x104) =2/9 x10-4 h

C2 =Cbs =BW/02 Ln = (6x104)/(4x104 )2 (4/3) = 9/32 x10-4 f


Similarly, the parallel tank circuit elements are obtained by using following formulae:


Lbp = BW/(02 Cn)

Cbp = Cn/BW


L1 = L1bp = BW/(02 Cn1) = (6x104)/(4x104 )2 (1/2) =75 h

C1 =C1bp = Cn1/BW =(1/2)/(6x104) = 8.33 f



L3 =  L2bp = BW/(02 Cn2) = (6x104)/(4x104 )2 (3/2) =25 h

C3 =C2bp = Cn2/BW =(3/2)/(6x104) = 25 f

 







Formula for the Filter Transformation.

Low Pass Filter

High

 







Band Pass Filter