Damping and natural frequency

Hello I'm trying to derive the damping ratio and the natural frequency of a second order system, but it appears that using the 'damp(sys)' function returns to me a different value from what I calculated... My transfer function is:

G = tf([18],[1 8 12.25])

I calculated the damping ratio and natural frequency using:

s^2+(2*zeta*w)s+w^2

The result was zeta = 8/7 and w is 3.5.

However I can't understand why damp(G) returns that the damping ratio is 1 when it's approximately 1.14 and additionally the w is inaccurate. If this is not the right function to use what is the right function that I could use to find damping ratio and frequency of the system? I'd greatly appreciate the help, thanks!

Answer 1

[Les Beckham]

The damping ratio that you calculated is greater than one, meaning this system is overdamped (non-oscillatory).

It has two real poles:

p = roots([1 8 12.25])

p = 2×1

-5.9365 -2.0635

Thus, the denominator of your transfer function is not really in the form s^2 + (2*zeta*w)s + w^2

but rather (s + p(1)) * (s + p(2))

Let's check

p(1)*p(2)

ans = 12.2500

Damping ratios greater than one don't really have a physical meaning, so the damp() function doesn't report them that way, it just reports 1.0.

[Arcadius]

That makes much more sense thank you for the detailed explanation!

[Sam Chak]

@Arcadius, just to add a little bit more. If you look into the documentaion,

https://www.mathworks.com/help/control/ref/lti.damp.html

you will find the algorithm to compute the damping ratio of a continuous-time system is given by

where is the pole location. It is different from the textbook formula

p = roots([1 8 12.25])
zeta = - (p(1) + p(2))/(2*sqrt(p(1)*p(2)))

because, like what @Les Beckham has explained, the overdamped systems do not oscillate and they behave just like any first-order systems. For first-order systems, the natural frequency is registered as the absolute value of the pole location itself, and the damping ratio can be , depending on location of the pole.