How to find poles of transfer function by looking at the step response?

How to find poles of transfer function by looking at the step response?

Given a step response graph like such:

How would I find the sketch for its poles on the complex plane? The only thing I can gather are the rise time, settling time, overshoot, etc. But how do I derive the poles from that?

• Just by looking? Or are you able to Laplace transform it? Sep 27, 2012 at 2:51
• @ColinK just by looking, the actual plot has numbers on the axis Sep 27, 2012 at 2:52
• Interesting question. An answer would provide a better intuition for the Laplace/Z transform than I currently have. Sep 27, 2012 at 2:57
• This might help: facstaff.bucknell.edu/mastascu/econtrolhtml/Ident/…
– Alfred Centauri
Sep 27, 2012 at 3:01

There are two things to determine: the frequency $f_n$ (or angular frequency $\omega_n$) and the $Q$. Approximately (for a high $Q$ system like the one pictured):

1) period $T$ is the time between successive peaks. $f_n=1/T$, and $\omega_n=2 \pi f_n$.

2) ratio of successive positive (or negative) peaks (measured relative to steady-state) $1- \delta = e^{-2 \pi /(2Q)}$, or, for high $Q$, $Q \approx \frac{\pi}{\delta}$. [Example: If $\delta=0.1$ (10% amplitude decay per cycle), $Q \approx 30$.]

The poles are at $\omega_n \left[-\frac{1}{2Q} \pm i \sqrt{1-\frac{1}{(2Q)^2}} \right]$, or in radial coordinates: $(\omega_n, \pi \pm \tan^{-1}{\sqrt{4Q^2-1}})$.

[There's also a pole at zero for the step response. Per bobuhito's comment, this pole is not part of the transfer function.]

• the output has a pole at zero, but the transfer function does not
– bobuhito
Sep 27, 2012 at 14:21
• @bobuhito: Yes, good point. I'll edit appropriately.
– Art Brown
Sep 27, 2012 at 15:17

The output here for t>0 looks like a constant plus a decaying sine wave, so mathematically it's the real part of $A+Be^{st}$ for three complex constants (A,B,s). Since the input for t>0 is a constant, the non-constant output is from a pole (that's the definition of pole: zero input for some s can have finite output for that s), so you just need to estimate s from the plot (you technically have two poles which are complex conjugates, as needed to create a "real part").