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Double T rejector filter

1. Circuit definition

This filter contains 3 resistors and 3 capacitors, in the ratios shown in the figure. The diagram also has an rc load resistor.

Double T rejector filter

We use the Mathematica module:

Get["../simulin/simulineaire.m"];

The following function defines the circuit, with arguments R, C and the charge conductance.

doubleT[r_,c_,gc_]:=Module[{n,A,B},
    n=5;
    A=Table[0,{n},{n}];
    B=Table[0,{n}];
    A=ajouterResistance[A,1,2,r];
    A=ajouterResistance[A,2,4,r];
    A=ajouterResistance[A,3,5,r/2];
    A=ajouterCapacite[A,1,3,c];
    A=ajouterCapacite[A,3,4,c];
    A=ajouterCapacite[A,2,5,2*c];
    A=ajouterDipole[A,4,5,gc];
    {A,B}=ajouterMasse[A,B,5];
    {A,B}=definirEntree[A,B,1];
    Return[{A,B}];
]
            

2. Transfer function

The transfer function is obtained with zero charge conductance (open circuit at output):

{A,B}=doubleT[R,C,0];
H=transfert[A,B,4]/.{s->I*omega}
            

1-C2 ω2 R2-C2 ω2 R2+4 i C ω R+1

We recognize a rejector filter, presenting an infinite attenuation for the pulsation: ωc = 1RC

Let’s define values to obtain a 50 Hz rejection:

fc=50;
r=10^2;
c=1/(2*N[Pi]*r*fc)
            
0.00003183098861837907

Plot of Bode’s diagram:

 
{A,B}=doubleT[r,c,0];
h=transfert[A,B,4];
            
bodeGain[h,0,4,-60,0]
plot1.png
CURVE
bodePhase[h,0,4]
plot2.png
CURVE

The output impedance can be obtained by calculating the current in a very low load resistance (very large conductance):

{A,B}=doubleT[r,c,10^9];
v=LinearSolve[A,-B];
icc=1/r*(v[[2]]-v[[4]])+c*s*(v[[3]]-v[[4]]);
icc=Abs[icc/.s->I*2*N[Pi]*100]
        
0.006708203932412162

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