Calculating Contact Time and Damping Coefficient for a Second Order System

[huma]

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

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

 

[Hesch]

2y**+4y*+8y=8x

( 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 + 2ξω_ns + ω_n² = 0

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

 

[huma]

( 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 + 2ξω_ns + ω_n² = 0

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

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

 

[Hesch]

the answer given in the book is damping coefficient is 1 .

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.

 

[huma]

thankyou :)