is damp function got any limitation?

Hi,

fs =

        3
  -------------
  s^2 + 3 s + 2

Continuous-time transfer function.

   Pole        Damping       Frequency      Time Constant  
                           (rad/seconds)      (seconds)    

 -1.00e+00     1.00e+00       1.00e+00         1.00e+00    
 -2.00e+00     1.00e+00       2.00e+00         5.00e-01

above are the result of my damp line, why my wn not √2 and damping not (3√2)/2 ? is there any limitation for damp function?

Answer 1

[Birdman]

I don't understand clearly your question, but the damping ratio and natural frequency can be found mathematically as follows:

It is known that an ideal second order transfer function is as follows:

           K*Wn^2
  ------------------------
  s^2 + 2*zeta*Wn*s + Wn^2

If we extract the coefficients of both transfer functions' denominator and solve for zeta and Wn,

 2*zeta*wn=3
 Wn^2=2;

From this, Wn is found as sqrt(2) and zeta(damping ratio) is found as 3/(2*sqrt(2)). This means zeta is greater than 1, which is normal since both poles are real, which will result in an overdamped step response. Hope this helps.

[llca llasd]

what i want to ask is why the result i get from [wn,zeta]=damp(ds), the ans for wn and zeta is not correct?

[Birdman]

Oh, that is because damp function finds the damping ratio and frequency of the poles, not the system. What you want to find and what I showed you above is the damping ratio and natural frequency of the second order system. Please note the difference.

[llca llasd]

Okay, thanks for clearing my doubt.