# Damping ratio and poles for transfer function

#### janine

Joined Jul 17, 2009
5
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

#### t_n_k

Joined Mar 6, 2009
5,447
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)

#### janine

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

Last edited:

#### janine

Joined Jul 17, 2009
5
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?

#### t_n_k

Joined Mar 6, 2009
5,447
thank you! But does it not matter that this formular hasn't been used?
Kωn2 / s(s2 + 2ςωns + ωn2 )
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.

#### t_n_k

Joined Mar 6, 2009
5,447
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?
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.

#### t_n_k

Joined Mar 6, 2009
5,447
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.
I guess I meant square roots of negative numbers take the complex operator.

#### janine

Joined Jul 17, 2009
5
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?

#### t_n_k

Joined Mar 6, 2009
5,447
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?
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.