Damping and natural frequency
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Arcadius
am 7 Apr. 2022
Kommentiert: Sam Chak
am 9 Apr. 2022
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!
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Les Beckham
am 7 Apr. 2022
Bearbeitet: Les Beckham
am 7 Apr. 2022
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])
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)
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.
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Sam Chak
am 9 Apr. 2022
@Arcadius, just to add a little bit more. If you look into the documentaion,
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.
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