How do I find the second order transfer function from this step response diagram?

Question

I have the following diagram of a system's step response:

I'm having trouble understanding how to calculate the system's transfer function, given this diagram. Specifically, I don't understand how exactly I can calculate the natural frequency and damping ratio.

Nothing I've read on this has helped me get a clear picture of what I should do. Can anyone help me understand step-by-step how to think about this problem?

Answer 1

From the step response plot, the peak overshoot, defined as

$$M_p = \frac{y_{\text{peak}}-y_{\text{steady-state}}}{y_{\text{steady-state}}}\approx\frac{1.25-0.92}{0.92}=0.3587$$ Also, the relationship between \$M_p\$ and damping ratio \$\zeta\$ (\$0\leq\zeta<1\$) is given by:

$$M_p=e^\frac{-\pi\zeta}{\sqrt{1-\zeta^2}}$$

Or, in terms of \$\zeta\$:

$$\zeta=\sqrt{\frac{\ln^2M_p}{\ln^2M_p+\pi^2}}$$ So, replacing that estimated \$M_p\$: $$\zeta\approx0.31$$ Also, from the step response plot, the damped natural frequency is aoprox. 0.5 Hz or \$\pi\$ rad/s. The relationship with the undamped natural frequency is: $$\omega_n=\frac{\omega_d}{\sqrt{1-\zeta^2}}\approx3.3\text{ rad/s}$$ Finally, the gain \$G_{DC}=y_{\text{steady-state}}\approx0.92\$

A standard second order transfer function has the form: $$H(s)=G_{DC}\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$$ Putting the obtained values: $$H(s)\approx\frac{10}{s^2+2s+11}$$ Compare the step response below with that supplied by you:

Answer 2

Calculating the natural frequency and the damping ratio is actually pretty simple.

If you look at that diagram you see that the output oscillates around some constant value finally settling on it: the frequency of these oscillations is the damped frequency. To measure it from the diagram you should measure the distance between the points where the output crosses the settling value, that's half the period that is the inverse of the damped frequency. Looking at your graph I'd say that that distance is about one second, so the damped frequency should be approximately 0.5Hz, i.e. \$f_\mathrm{d}=0.5\$Hz. Let's just keep this in mind for now.

Now for the damping ratio. That's a bit trickier, the damping ratio measures how fast the oscillations decay, i.e. how fast the output settles. If the damping ratio is 0 they don't settle, if it's above unity you don't have oscillations but just a nice exponential. What you need to do is fit a curve on the output maxima, the curve being of the family \$Ae^{-\frac{t}{\tau}}+C\$. You need three points, you have three points, what you need is that tiny \$\tau\$. I don't really want to fit the curve so I'll make a wild guess and say that \$\tau=3s\$. Now let's recap: $$ f_\mathrm{d}=0.5\text{ Hz}\\ \tau=3\text{ s} $$ Since that's an under damped system the following holds: $$ \omega_\mathrm{d} = \omega_0 \sqrt{1 - \zeta^2 }\\ \tau = \frac{1}{\omega_0\zeta} $$ And you guessed it:

• \$\omega_\mathrm{d}=2\pi f_\mathrm{d}\$
• \$\omega_0\$ is the natural pulsation, so the natural frequency \$f_0=\frac{\omega_0}{2\pi}\$
• \$\zeta\$ is the damping factor

Now just do the math and profit.

note the \$\tau=3\$s guess might be more right than you think.