# Negative damping ratio for second order transfer function

For this circuit:

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

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:

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

We can apply a nodal analysis to node Vx:

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

The transfer function can be rewritten in a general form as follows:

• Thank you. I've just edited it with the complete calculation. Jun 8, 2023 at 13:34
• 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.
– LvW
Jun 8, 2023 at 13:56
• @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. Jun 8, 2023 at 21:18

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

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.

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.