# Damping ratio and poles for transfer function

Discussion in 'Homework Help' started by janine, Jul 17, 2009.

1. ### janine Thread Starter New Member

Jul 17, 2009
5
0
Transfer function: 5 / s2 +2s +1
These are the steps Ive done:
Kωn2 / s(s2 + 2ςωns + ωn2 )
Y(s) = 1 / s + A1 / s-P1 + A2 / s-P2
y(s) = 1 + A1expP1t + A2expP2t
P1 + P2 = poles are equal to the roots
S2 +2ςωns + ωn2 = 0
P1, P2 = -2ςωn +- √ 4ς2ωn2 - 4ωn2 / 2 = ςωn +- ωn √ ς2  1
A1 = -(kς / 2√ ς2 1)  k / 2
A2 = (kς / 2√ ς2 1)  k / 2

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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S^2 + 2s + 1 has two equal roots at s=-1.
There are two poles, each at -1 +j0 on the complex plane.
ω0^2=1

So ω0=1

2*ζ*ω0=2

2*ζ=2

Damping factor ζ=1 (critically damped)

3. ### janine Thread Starter New Member

Jul 17, 2009
5
0
thank you! But does it not matter that this formular hasn't been used?
Kωn2 / s(s2 + 2ςωns + ωn2 )

Last edited: Jul 17, 2009
4. ### janine Thread Starter New Member

Jul 17, 2009
5
0
Also just another quick question... this transfer function is confusing me: 16/4s^2 + 4s +8
as it has complex roots I don't know how to begin solving it?

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Not necessarily - You just need to know how to equate the actual coefficients in the particular example with the general coefficients in the standard form of the equation you include above.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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16/(4s^2+4s+8) = 4/(s^2+s+2)

the roots of s^2+s+2=0 are -1/2 ± j(√7)/2

Use the normal method for solving a quadratic equation - where negative roots have the imaginary operator.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I guess I meant square roots of negative numbers take the complex operator.

8. ### janine Thread Starter New Member

Jul 17, 2009
5
0
Thanks that's really helpful! I've worked them all out now, but then the next stage was to find the peak time, rise time, settling time and overshoot. I managed to do this for all of them except this one: 1 / s^2 +8s +4, the poles are -0.54 and -7.5, zeta= 2 and ωn=2, I know it has no overshoot as it's overdamped, I worked out the settling time is 1second, but I have these formulas for rise time: pi-arctan((√1-ζ^2)/ζ) and for peak time: pi/ωn√1-ζ^2, but zeta is too big therefore you get j again but I didn't think this could be used in peak and rise time?

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Rather than just plugging numbers into a formula, you could write down the time domain solution for the example in question.

Then either plot the time domain solution or directly solve for the conditions that define a particular parameter.

Consider rise time.

This is normally considered as the interval over which the function goes from 10% to 90% of its final value. Given the time domain function, you should be able to solve for the times at which you have 10% & 90% conditions - T10 & T90. Hence Trise=T90-T10.