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I have been looking about how to calculate poles if I have experimental result of a step response in Matlab that looks like an overdamped system. So far I only found information (through internet or in textbooks) on how to calculate poles for underdamped systems but not for overdamped system.

For example, the results from recent experiment looks like something like this:

enter image description here

I am trying to compare the pole locations between those two. I was able to find the poles for the underdamped system, but not for the overdamped system.

I know for the overdamped system the poles should be two real distinct poles and can be calculated if I know the damping ratio and natural frequency, which is:

$$p_{1,2}=(\zeta\pm\sqrt{\zeta^2-1})\omega_n$$

If this is case then how do I figure out the damping ratio and the natural frequency?

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    \$\begingroup\$ Why are you assuming it to be over damped second-order system from the outset? Will a first-order system suffice to capture the dynamics? Only if not, would I look at approximating it as a critically or over damped system. \$\endgroup\$ – Suba Thomas Nov 16 '15 at 23:18
  • \$\begingroup\$ I agree with Suba that your red graph doesn't look like a 2nd order system. To see one of those look at the 2nd graph in electronics.stackexchange.com/questions/199955/… According to Chu's calculation (which checks out modolo a sign) you'd need to have a real maximum (or minimum), but I don't really see that in your red graph. \$\endgroup\$ – Fizz Nov 17 '15 at 0:09
  • \$\begingroup\$ See also dcontrolsystems.com/system-performance/2nd-order-systems/… \$\endgroup\$ – Fizz Nov 17 '15 at 0:19
  • \$\begingroup\$ Are you sure its overdamped? Since both graph tend towards to approximately the same value, wouldn't this be a critically-damped system instead? \$\endgroup\$ – K. Rmth Nov 17 '15 at 8:27
  • \$\begingroup\$ @K. Rmth, Provided the system is stable, damping and final value are not related. Damping relates to the transient; final value is the steady-state. \$\endgroup\$ – Chu Nov 17 '15 at 13:30
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Write the TF as \$ \small G(s)=\large\frac{ab}{(s+a)(s+b)}\$, where \$a\$ and \$b\$ are real and distinct.

This factorises to: \$\small G(s)=\large\frac{ab}{(b-a)} \large(\frac{1}{s+a}-\frac{1}{s+b}\large)\$

From which the inverse LT, i.e. the unit impulse response, is:

\$g(t)= \large\frac{ab}{(b-a)} \normalsize (e^{-at}-e^{-bt})\$

Differentiating and setting to zero determines the time to peak value, \$t_P\$:

\$t_P =\large\frac{ln(\frac{b}{a})}{(b-a)}\$, with corresponding peak value: \$g(t_P)=be^{-b t_P}\$

Hence, measuring \$t_P\$ and \$g(t_P)\$ will yield the value of \$b\$; and substituting back into the expression for \$t_P\$ will yield \$a\$

Note that the unit impulse response can be obtained by differentiating the unit step response, but the differentiator must be carefully designed to avoid introducing excessive noise.

Typical overdamped step and impulse responses are shown below.

enter image description here

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  • \$\begingroup\$ It looks to me like you have a sign wrong there. \$\endgroup\$ – Fizz Nov 16 '15 at 23:43
  • \$\begingroup\$ @Respawned Fluff, yes, I think it's fixed now. Didn't affect the result. Thank you \$\endgroup\$ – Chu Nov 17 '15 at 0:25
  • \$\begingroup\$ It might not help him much though because his red graph looks very overdamped (if it's even a 2nd order system), so finding the maximum in that noise seems rather difficult. \$\endgroup\$ – Fizz Nov 17 '15 at 0:38
  • \$\begingroup\$ @Respawned Fluff, looks 1st order; initial slope is not zero, so there may not be a maximum. Analysis may be useful in the future though. \$\endgroup\$ – Chu Nov 17 '15 at 0:44
  • \$\begingroup\$ Curious, how do you determine peak time in overdamped system? \$\endgroup\$ – vxs8122 Nov 17 '15 at 14:04

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