Transfer Function of Multiple Feedback Bandpass Filter in a specific format

Question

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

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.

Answer 1

Here, in sympy:

vi,vo,v1,v2,r1,r2,c1,c2,r3,io=symbols('vi,vo,v1,v2,r1,r2,c1,c2,r3,io',real=True)
s=symbols('s')
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])
mfb
{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}
mfb[vo]/vi
-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\$.

Answer 2

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:$$H(j\omega)=\frac{H_0}{1+jQ\left(\frac{\omega_{0}}{\omega}-\frac{\omega}{\omega_{0}}\right)}$$