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I'm trying to design second order multiple feedback low-pass filter with cut-off frequency at 1500 Hz and Chebyshev approximation.

enter image description here

Transfer function for this scheme is defined as:

enter image description here

Where:

enter image description here

And condition for real values of resistors:

enter image description here

However using Matlab and cheby1 function I get extremely high values of coefficients and as a result high values of resistors and capacitors.

hertz = 1500; % passband in hertz
rad_s = hertz*2*pi; % convert to rad/s
order = 2;
ripple = 3; % dB

[b, a] = cheby1(order, ripple, rad_s, 's');

% Extract coefficients for computing sheme values
a1 = a(2);
b1 = a(3);
H0 = -b(3);
fprintf('Coefficients:\n\ta1=%d\n\tb1=%d\n\tH0=%d\r\n', a1, b1, H0);

% Condition for capacitors
KCmin = (4*b1*(1-H0))/(a1^2);
fprintf('Capacitors condition:\n\tC2 / C1 >= %d\r\n', KCmin);

% Choose satisfying capasitors
C2 = 10^(-3); % F
C1 = 10^(-12); % F
fprintf('Chosen capacitors:\n\tC1=%d\n\tC2=%d\n\tCk=%d\r\n', C1, C2, C2/C1);

R2 = (a1*C2 - sqrt((a1^2)*(C2^2) - 4*C1*C2*b1*(1-H0))) / (2*hertz*C1*C2);
R1 = -R2 / H0;
R3 = b1 / ((hertz^2)*C1*C2*R2);

fprintf('Resistors:\n\t%d\n\t%d\n\t%d\n', R1, R2, R3);

And output is:

Coefficients:
    a1=6.078036e+03
    b1=6.288448e+07
    H0=-4.451880e+07

Capacitors condition:
    C2 / C1 >= 3.031241e+08

Chosen capacitors:
    C1=1.000000e-12
    C2=1.000000e-03
    Ck=1000000000

Resistors:
    7.518520e+03
    3.347155e+11
    8.349974e+04

Also I noticed that whatever ripple I chose condition C2/C1 was always of 10^8 order.

What am I doing wrong?

UPDATE

I believe I know what the problem is: given equations are in normalized form and I need coefficients for normalized form too. However MATLAB gives coefficients which are totally ready to be used. Therefore as I understand I have to remove w from the equations but it has not helped me in some reason.

Also I found normalized coefficients in my textbook: enter image description here

The last column is normalized coefficients for 2d order 3dB Chebyshev filter. But I can't understand where from I have to take H0.

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  • \$\begingroup\$ For sure you are having some troubles in generating coefficients. Ho=-4E7 is totally nonsense, it's a DC gain over 150dB! From this on anything is just meaningless: 1mF, 1pF capacitors, 335Gohm resistor... \$\endgroup\$ – carloc Dec 18 '16 at 17:11
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I am afraid that your transfer function is not correct. Cancel the wp² and the wp in the denominator - and the results will be OK. Your error can be verified very easily: All three expressions in the denominator must have the unit "1" - and this is not the case with your function.

The correct equations are like this:

1/wp²=R2R3C1C2;

1/(wp*Qp)=(R2+R3+R2R3/R1)C1.

Update:

The above two equations result from a comparison between the (corrected !) transfer function of YOUR specific circuit and the general second-order transfer function:

H(s)=Ao/[1+s/(wpQp)+s²/wp²]

For Chebyshev (3 dB ripple) the pole data are: Qp=1.30656 and the normalized pole frequency is wp/wo=0.8409 (wo=cut-off).

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  • \$\begingroup\$ I don't believe it's wrong because I took it from my textbook by Ulrich Titse and Christoph Shank (I can't find out how to write them right). But I came up with the same solution but it did not help me. See my update please. \$\endgroup\$ – Long Smith Dec 18 '16 at 19:09
  • \$\begingroup\$ My recommendation: Do NOT use the mentioned book (Tietze/Schenk). It is not wrong - but the given filter coefficients are valid for the 3dB cutoff only (even for Chebyshev responses, which is VERY unusual). Instead, use the pole data wp and Qp (together with both equations I gave you) which are available for different Chebyshev ripple values. See also my update. \$\endgroup\$ – LvW Dec 18 '16 at 19:23
  • \$\begingroup\$ And what are conditions on capacitors this way? Regardless of chosen transfer function, current conditions with general coefficients give wrong result. \$\endgroup\$ – Long Smith Dec 18 '16 at 19:25
  • \$\begingroup\$ Recommended selection (option): Ao=1 with R1=R2 and C2/C1=9Qp². \$\endgroup\$ – LvW Dec 18 '16 at 19:31
  • \$\begingroup\$ I forgot: For the above selection: wp=1/3R1C1Qp. \$\endgroup\$ – LvW Dec 18 '16 at 20:14

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