Filter Design'S

• Frequency denormalization:



Let, sn = the normalized frequency variable
s = denormalized frequency variable,
0 = normalizing constant


For a Low pass filter 0 is the cut-off frequency and sn=s/0
The impedance of an element will remain same under frequency normalization.

So for an inductance snLn = sL=0. snL


Where Ln = normalized inductance
L= denormalized inductance

So that L = Ln/0


For a capacitor,

Xc = 1/snCn =1/sC =1/(s0C)

Thus, C = Cn/0






• Impedance Denormalization:


Let, Zn = Rn = the normalized impedance or resistance
R= Denormalized resistance
Rn = Normalizing constant

Then, Rn = R/R0

R=R0Rn

Thus for an inductance: sL = (sLn) R0

L = LnR0

For a capacitor, 1/sC = R0/sC C=Cn/R0




• Frequency Transformation: LP to HP


For a HP Filter, sn=w0/s





 And    |sn| = w0/w        0 < |sn| <1



s=w0/sn

         

s+jw = w0/(sn+ jwn) = w0{(sn- jwn)} /(sn2+ jwn2)



Let sn =0,        then jw = w0{(- jwn)} /( jwn2)



Or,       w= - w0/wn



Or,       wn= - w0/w



Thus, wn=±1 corresponds to  w0=±w



wn=1 corresponds to  w0=w

wn<1 corresponds to  w>w0

 s-plane for Low-pass

s-plane for High-pass

For the following Low-pass Filter the transformed HP filter elements can be found as:

For a capacitor:

1/snCn = 1/Cn(0/s) = s/( Cn0) = sLh

where subscript h refers to the high pass filter element.
i.e. a capacitor changes to an inductance  such that

Lh = 1/ Cn0=normalized value of the inductance for HPF


Similarly,

For an inductor, snLn = (0/s)Ln = 1/sCh

So,  Ch = 1/(0Ln) = normalized value of the capacitor for the HPF
The transformed high-pass filter with normalized elements is shown below:

•    LP to Band-pass Transformation :

Since a band-pass filter is combination of a Low-pass and High pass, the normalized value of s for bandpass signal is given by



Where the bandwidth of the BP filter is BW = C2 - C1 and 0 = (C1C2)

 So the normalized low-pass filter elements are to be modified such that their normalized impedance remains same after frequency transformation:



Where subscript b denotes bandpass element, so the series inductance Lnof the LPF transforms to a series tank circuit of Lbs and Cbs given by

Lbs = Ln/BW

Cbs =BW/02 Ln


s denotes series branch.



For the shunt capacitance of the Low-pass filter, applying the above transformation we get,



Thus the shunt capacitance C¬n of the Low-pass filter transforms to a parallel tank circuit with inductance Lbp and capacitance Cbp such that

Lbp = BW/(02 Cn)

Cbp = Cn/BW

p denotes parallel branch.

If the series branches are denotes by L1, C1 and parallel branches by L2, C2 and L3, C3 , then the circuit for the Band-pass filter is given  as:

 





The component values obtained above are normalized values which correspond to a BP Filter with bandwidth BW and centre frequency 0 and for a load resistance of normalized value of 1 ohm.
To get the de-normalized values of the components for a given load resistance RL (say, 100ohm), impedance de-normalization is to be carried out as such that

L = Ln . RL

C=Cn / RL

where L and C denote the de-normalized values