In the analysis of a passive 2nd order RC high pass filter, I find that the damping ratio zeta = 3/2 and the Quality Factor Q = 1/3. These values do not depend on R and C. I was under the impression that if the damping ratio is greater than one, then the system step response does not overshoot its steady state value. However, when performing step response in MATLAB to my system, I find that the circuit does overshoot slightly. I am analyzing the circuit below for equal R and C (10k and 100nF respectively).
MATLAB Code:
MATLAB Output:
I believe the calculations for zeta and Q to be correct, something must be wrong in my understanding of their role in the step response. Why does the circuit overshoot if it is "overdamped"?
I have repeated this procedure with a passive 2nd order RC LPF (same R and C values) and I do not see any overshoot. This agrees with my intuition. Why is the HPF different?
Thanks for any help that can be provided.
Best Answer
The double zero at \$ s\small =0\$ gives rise to the undershoot.
To make the analysis easier, it's best to normalise the TF to \$\omega_n=1\$, thus divide \$\omega_n\$ by \$\small 1000\$ to give: $$G(s)=\dfrac{s^2}{s^2+3s+1}$$
Multiply by \$\dfrac{1}{s}\$ to obtain the step response: \$R(s)=s\left(\dfrac{1}{s^2+3s+1}\right)\$
Now find the inverse Laplace transform of the bracketed term and then differentiate (multiply by \$s\$ = differentiation) to determine \$\small r(t)\$, thus:
$$\dfrac{1}{s^2+3s+1}=\dfrac{0.45}{s+0.38}-\dfrac{0.45}{s+2.62}\rightarrow 0.45e^{-0.38t}-0.45e^{-2.62t}$$ differentiate to give the step response:
$$r(t)=\dfrac{d}{dt}\left( 0.45e^{-0.38t}-0.45e^{-2.62t}\right)=-0.17e^{-0.38t}+1.17e^{-2.62t}$$
This starts at \$\small r(0)=1.0\$, then falls to a minimum (undershoot), \$\small r(1.72)=-0.0755\$, and settles at \$\small r(\infty)=0\$
Finally, scale the time axis by the normalising factor, \$\small 1/1000\$, giving the undershoot of \$\small -0.0755\$ at \$\small t=1.72ms\$