# How to develop a characteristic equation if we only know zeta and w?

I am currently solving a question on pole placement method and in it we develop a required characteristic equation for the system using % overshoot and settling time. I know in a 2nd order system,the characteristic equation is $$s^2+2\cdot\zeta\cdot w\cdot s+w^2$$, and then we replace the values of $$\zeta$$ and $$w$$ from the values we get from the values which we get from % overshoot and settling time.

My question is how would we do this in a 3rd order system? My guess is that we take a general equation like $$(s+p)(s^2+2\cdot\zeta\cdot w\cdot s+w^2)$$ and then find $$p$$, $$\zeta$$ and $$w$$ but I dont know if this is correct or how would I even do that.

The question I am currently solving is an example from Norman Nise

Given the plant
G(s) = some transfer function
design the phase-variable feedback gains to yield 9.5% overshoot and a settling
time of 0.74 second.


In the solutions, it simply says

This equation must match the desired characteristic equation,
s^3 + 15.9s^2 + 136.08s + 413.1 = 0


I want to know the method of finding this equation

• My guess would have been to find $\zeta$ and $\omega$ and to set the third pole at a value $p>>\omega$; i.e., concept of dominant poles. However, the solution given by the book doesn't appear to satisfy this criteria.
– AJN
Apr 14, 2023 at 12:13
• maybe im supposed to use some matlab function to find this equation Apr 14, 2023 at 17:53
• It's a weird solution because the author is incorporating the concept of pole-zero cancellation and second-order approximation too! So @AJN's way might make more sense in general, but since the question has G(s) with a LHP zero, the choice of the 3rd pole is determined by this and the author makes an odd choice in order to make a point. p.190 in Nise discusses this point Apr 16, 2023 at 1:04
• @GeoffreyLiddell If you have access to Nise, consider making the above comment an answer with relevant quotes from Nise. Also, you comment seems to hint (LHP zero) that OP has not posted the full details of the problem required to reach the same solution as the text.
– AJN
Apr 16, 2023 at 8:23

Searching for 9.5% in Nise showed that the question you are answering seems to be Example 12.1 on p.655. The important information you have left out is what $$G(s)$$ is.

$$G(s) = \frac{20(s+5)}{s(s+1)(s+4)}$$

Since the plant has a zero and 3 poles, we can't use the formulas relating % overshoot and settling time to $$\zeta$$ and $$\omega_n$$ without also considering whether the system can be approximated by a second-order system. In section 4.7 p.182 of Nise you can read about this, but the conclusion is there must be 2 dominant complex poles, i.e. 2 poles which are much closer to the imaginary axis than all the others. Nise also notes that you should always simulate the system before going about the final design.

The solution then goes about assuming that the second-order approximation does hold, then finding the two complex poles that would fulfill the design requirements. So they have chosen probably $$\zeta = 0.6$$ for the overshoot and $$t_s \approx \frac{4}{\zeta \omega_n}$$ so that $$\omega_n = 9$$, so then you get $$p_{1,2} = - 5.4 \pm j7.2$$. You have already got this far anyway by your approach of assuming that the solution will be of the form $$(s + p)(s^2 + 2\zeta \omega_n s + \omega_n ^2)$$.

The next step is to notice that if the plant has a zero, in order for us to get a response similar to our general second-order system, we will have to cancel it out. So we could choose p = -5. Pole-zero cancellation is going to be fine in this situation since it's in the LHP, but if it were a RHP zero, there would be a problem, see 'non-minimum phase system.'

In the book, Nise has chosen p = -5.1 to show that the system will still be fine even if the cancellation is not exact, then the characteristic equation comes from expanding

$$(s + 5.1)(s + 5.4 + 7.2j)(s + 5.4 - 7.2j)$$

$$(s + 5.1)(s^2 + 10.8s + 81)$$

$$s^3 + 15.9 s^2 + 136.1 s + 413.1$$