# Second order system

1. Apr 23, 2017

### huma

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

how we can find contact time of second order system ? 2y**+4y*+8y=8x I want to find damping coefficient ... howz possible

Last edited by a moderator: Apr 27, 2017
2. Apr 23, 2017

### Hesch

( 2s2 + 4s + 8 ) * y(s) = 8 * x(s)

Finding y(s)/x(s) the characteristic equation of the transfer function will be:

2s2 + 4s + 8 = 0 →

s2 + 2s + 4 = 0

which can be formulated

s2 + 2ξωns + ωn2 = 0

So ωn = 2 , the damping coefficient, ξ = 0.5

3. Apr 23, 2017

### huma

actually I got this question from a book of M.handa .I solved my answer are same but the answer given in the book is damping coefficient is 1 ..... can you tell me about time constant . how we can find it

4. Apr 23, 2017

### Hesch

The equation

2s2 + 4s + 8 = 0

has two complex roots: s = -1 ± j√3.

If the characteristic equation were to have a damping ratio = 1, it should have two real roots at the same location, for example.
s1 = -1.2 , s2 = -1.2.
In this case the characteristic equation could be written:
s2 + 2.4s + 1.44 = 0
and the transfer function would have a double time constant = 1/1.2 sec.

If the characteristic equation were
s2 + 7s + 10 = 0
the two roots would be s1 = -2 , s2 = -5 , and the transfer function would have two time constants:
τ1 = 0.5 sec. , τ2 = 0.2 sec.

When a root is complex, we cannot speak of a "time constant". The time is complex. The system will oscillate within some damping coefficient.

Last edited: Apr 23, 2017
5. Apr 24, 2017

thankyou :)