1
\$\begingroup\$

For this circuit:

enter image description here

We get this transfer function (this is the solution of another student, but I did it by hand and I get the same thing):

enter image description here

Where the first damping ratio is negative. I read on another topic (https://math.stackexchange.com/questions/23314/negative-damping-ratio-for-second-order-system) that "Practically this is nonsense since such an oscillator will saturate the amplifier almost instantly and be of no practical use."

Is it plausible we get something like this for our circuit? Is it another explanation? Or did we just both make a mistake calculating this transfer function?

Also, if this is possible that it isn't a mistake, how should we considerate a negative damping ratio when drawing the bode diagram? (Possibility of resonance for some values? How does it work?)

EDIT: as requested, this is the complete calculation I get for the transfer function: enter image description here

We can apply a nodal analysis to node B, given that the current i− = 0. enter image description here

We can apply a nodal analysis to node Vx: enter image description here

We can apply a nodal analysis to node A, given that the current i is present: enter image description here

The transfer function can be rewritten in a general form as follows: enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ Thank you. I've just edited it with the complete calculation. \$\endgroup\$
    – c.leblanc
    Jun 8, 2023 at 13:34
  • \$\begingroup\$ This is a 2nd-order allpass circuit. The resistors appearing in the numerator must be chosen so that the damping ratio will be negative. Hence : 2R1<RR2/R3. \$\endgroup\$
    – LvW
    Jun 8, 2023 at 13:56
  • 1
    \$\begingroup\$ @c.leblanc I get:$$\begin{align*}\omega_{_0}&=\frac1{C\,\sqrt{R_1\,R_2}}\\\\\zeta&=\sqrt{\frac{R_1}{R_2}}\\\\H\left(s\right)&=\left[\vphantom{\frac{R_2\,R}{R_1\,R_3}}-1\right]\cdot&\frac{\omega_{_0}^2}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^2}\\\\&+\left[\frac12\frac{R_2\,R}{R_1\,R_3}-1\right]\cdot&\frac{2\zeta\,\omega_{_0}s}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^2}\\\\&+\left[\vphantom{\frac{R_2\,R}{R_1\,R_3}}-1\right]\cdot&\frac{s^2}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^2}\end{align*}$$Plugging in your result I get the same except for the sign of the gains. \$\endgroup\$ Jun 8, 2023 at 21:18

1 Answer 1

1
\$\begingroup\$

Is it plausible we get something like this for our circuit? Is it another explanation?

Your transfer function is correct.

Also, if this is possible that it isn't a mistake, how should we considerate a negative damping ratio when drawing the bode diagram?

Draw the Bode diagram as you would any other Bode diagram. The magnitude of \$H(\omega)\$ is the ratio of the magnitudes of the numerator and the denominator complex polynomials. The phase \$\angle H(\omega)\$ is the difference of the angles of the numerator and the denominator. Use Microcap, LTspice, Excel. Any of these will allow you to explore this interesting circuit.

\$R_3\$ can be used to adjust the numerator damping ratio. It is very interesting to watch the the response go from a band stop to a band pass just by adjusting \$R_3\$.

If the denominator damping ratio is unity and

  1. the numerator damping ratio is unity. the magnotude is 0dB across the band, but the phase goes from \$0^0\$ to \$-360^0\$
  2. the numerator damping ratio is -1, the magnitude and phase are flat at 0dB and 0 degrees respectively.

For the damping ratios shown in the OP and a centre frequency of 1, the Bode diagram should look as shown below.

enter image description here

I read on another topic...(link)

The transfer function considered in the link provided is not the same as the OP. A negative damping ratio in the characteristic polynomial (denominator) is bad news. The system will be unstable.

In the numerator the negative damping ratio can be helpful. Correctly choosing the numerator damping ratio can make an almost perfect allpass filter within the limitations of the op-amp.

Addition:By adjusting just \$R_3\$, the circuit can be made to perform as a notch filter, bandpass filter, constant gain amplifier, and an all=pass filter.

\$\endgroup\$
4
  • \$\begingroup\$ Should I consider negative damping ratio in the numerator as highly underdamped system with high resonance ? \$\endgroup\$
    – c.leblanc
    Jun 8, 2023 at 21:59
  • \$\begingroup\$ also didn't you mean notch filter instead of all pass ? \$\endgroup\$
    – c.leblanc
    Jun 8, 2023 at 22:26
  • \$\begingroup\$ The damping ratio is local to the quadratic poles of the characteristic polynomial. Be careful trying to generalize to the entire transfer function. It is better to use \$2\zeta=\frac{BW}{\omega_0. Then you will find a better understanding to the damping question. Really, plot the TF on a simulator. \$\endgroup\$
    – RussellH
    Jun 8, 2023 at 22:59
  • 1
    \$\begingroup\$ No, I meant allpass. The TF in the OP is a notch. See my update \$\endgroup\$
    – RussellH
    Jun 8, 2023 at 23:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.