I've been trying to study the Multiple Feedback Bandpass Filter.

enter image description here

I managed to find the transfer function and the natural frequency of oscillation.


$$\frac{V_\text{out}}{V_\text{in}}(s) = \frac{2\zeta(2\pi f_0)K_s}{s^2 + 2\zeta(2\pi f_0)s + (2\pi f_0)^2}$$

$$Q = \frac{1}{2\zeta}$$

$$K = \frac{R_3}{R_1}\frac{-C_2}{C_1 + C_2}$$

How do I write the transfer function in the following format?

$$H(j\omega) = \frac{-H_0}{1 = jQ\left(\frac{\omega}{\omega_0}-\frac{\omega_0}{\omega}\right)}$$

And assuming that I know the values of the circuit components, I wish to calculate the natural frequency for the max gain.

  • \$\begingroup\$ I'm guessing (because I don't see all the necessary information laid out) that you "managed to find" this TF by searching books or the web. Not deriving it, yourself, I'm betting. The equation given for Q is "by definition" and is true for all 2nd order. The value for K is specific to this circuit and what you have is correctly stated. That said, you didn't find (it seems) anything about how to get \$\zeta\$ or Q or \$\omega_{_0}\$. And so, it's no wonder you are lost. Given you are studying this circuit, is it because of homework or is it because you really want to analyze it on your own? \$\endgroup\$ Mar 12 at 17:21
  • \$\begingroup\$ @periblepsis I'm sorry my question appeared a little bit abstract. I actually derived the TF (after many headaches) in the standard for any 2n order band-pass filter. here's the derivation, not that it matters Again, my point of confusion is if there is another analytical solution that reaches a TF with the form I specified before, what is it? And on what basis do we calculate the natural frequency to achieve max gain? \$\endgroup\$
    – sumato
    Mar 12 at 19:00
  • 1
    \$\begingroup\$ It matters a lot to me that you performed a derivation. I see you are avoiding the use of 's' and just replacing it with 'j w'. That's your first problem, as it then conflates some values that you'll need to then later tease apart, again. (Not a good thing to look forward to.) It's much easier to parse these things if you keep 's' and avoid prematurely converting it to 0+jw. Since you've shown me your derivation I'll show you mine. \$\endgroup\$ Mar 12 at 19:05

2 Answers 2


Here, in sympy:

eqv1 = Eq(v1/r1+v1/r2+s*v1*c1+s*v1*c2,vi/r1+s*v2*c2+s*vo*c1)
eqv2 = Eq(s*v2*c2+v2/r3,s*v1*c2+vo/r3)
eqvo = Eq(vo/r3+s*vo*c1,io+s*v1*c1+v2/r3)
eqv2z = Eq(v2,0)
mfb = solve([eqv1,eqv2,eqvo,eqv2z],[vo,io,v1,v2])
{vo: -c2*r2*r3*s*vi/(c1*c2*r1*r2*r3*s**2 + c1*r1*r2*s + c2*r1*r2*s + r1 + r2),
 io: (-c1*c2*r2*r3*s**2*vi - c1*r2*s*vi - c2*r2*s*vi)/(c1*c2*r1*r2*r3*s**2 + c1*r1*r2*s + c2*r1*r2*s + r1 + r2),
 v1: r2*vi/(c1*c2*r1*r2*r3*s**2 + c1*r1*r2*s + c2*r1*r2*s + r1 + r2),
 v2: 0}
-c2*r2*r3*s/(c1*c2*r1*r2*r3*s**2 + c1*r1*r2*s + c2*r1*r2*s + r1 + r2)

Now, that's nicer (to me.)

I know that the general form should be \$K\frac{2\zeta\omega_{_0}s}{s^2+2\zeta\omega_{_0}s+\omega_{_0}^2}\$ or else \$K\frac{\frac1{Q}\frac{s}{\omega_{_0}}}{\left(\frac{s}{\omega_{_0}}\right)^2+\frac1{Q}\left(\frac{s}{\omega_{_0}}\right)+1}\$.

The first thing I look at is the denominator. That's because if I treat it as \$b_2s^2+b_1s+b_0\$ then it follows that \$\omega_{_0}=\sqrt{\frac{b_0}{b_2}}\$. In this case, \$\omega_{_0}=\sqrt{\frac{R_1+R_2}{R_1\,R_2\,R_3\,C_1\,C_2}}=\frac1{\sqrt{\left(R_1\,\mid\mid\,R_2\right)\,R_3\,C_1\,C_2}}\$. See how easy it is when you keep \$s\$ isolated?

The next thing is that I know \$Q=\frac{\sqrt{b_2 \,b_0}}{b_1}\$. I'll let you plug things in.

The final thing is that I know \$K=\frac{a_1}{b_1}\$ (given that the numerator -- did we forget about it? -- is \$a_2s^2+a_1s+a_0\$.) Again, you can plug things in here, too. (I think you will find that it agrees with your own example.)

(I can prove the above without difficulty. But it would take up space.)

Assuming all that's been done then you can make your substitutions:

$$\begin{align*} K\cdot\frac{\frac1{Q}\frac{s}{\omega_{_0}}}{\left(\frac{s}{\omega_{_0}}\right)^2+\frac1{Q}\left(\frac{s}{\omega_{_0}}\right)+1} \\\\ K\cdot\frac{\frac1{Q}\frac{j\,\omega}{\omega_{_0}}}{\left(\frac{j\,\omega}{\omega_{_0}}\right)^2+\frac1{Q}\left(\frac{j\,\omega}{\omega_{_0}}\right)+1} \\\\ K\cdot\frac{\frac1{Q}\frac{\omega}{\omega_{_0}}j}{\frac1{Q}\left(\frac{\omega}{\omega_{_0}}\right)j+1-\left(\frac{\omega}{\omega_{_0}}\right)^2} \end{align*}$$

You can work the algebra from there. But perhaps multiplying by \$\frac{-Q\frac{\omega_{_0}}{\omega}j}{-Q\frac{\omega_{_0}}{\omega}j}\$ might come to mind.

By the way, the \$H_{_0}\$ shown in the picture isn't quite the same as \$K\$. Instead, I believe \$H_{_0}=-K\$.


Your derivation is in the sinsuoidal domain, but the OP jumps to the Laplace Domain, then provides an answer in the sinusoidal domain.

You have what you need in your derivation. Just identify the damping ratio and the natural frequency. You can do this by starting with $$H(s)=\frac{2\zeta\omega_{0}Ks}{s^{2}+2\zeta\omega_{0}s+\omega_{0}^{2}}$$ as you provided.

Substitute \$s=j\omega\$ then solve to match the result in your derivation.. Then it is just a matter of algebraic manipulation to match:$$H(j\omega)=\frac{H_0}{1+jQ\left(\frac{\omega_{0}}{\omega}-\frac{\omega}{\omega_{0}}\right)}$$

  • \$\begingroup\$ Thanks. That's exactly what I did to calculate the coefficients. Also, do you happen to have a clue on how to derive the TF in this format \$\endgroup\$
    – sumato
    Mar 12 at 20:04
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    \$\begingroup\$ It is just algebraic manipulation to make the numerator match the format. \$\endgroup\$
    – RussellH
    Mar 12 at 20:16
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    \$\begingroup\$ Notice that \$H_0\$ is a real constant \$\endgroup\$
    – RussellH
    Mar 12 at 20:19

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