A second-order system has a damping factor of 0.3 (underdamped system) and an un�damped natural frequency of 10 rad s-1
. Keeping the damping factor the same, if the un�damped natural frequency is changed to 20 rad s-1
, locate the new poles of the system?
What can you say about the response of the new system?
To find the poles of a second-order system, we can use the general formula for the poles of the system:
p = -σ ± jωₐ
where:
p is the pole
σ = ζωₙ (damping coefficient)
ωₐ = ωₙ * sqrt(1 - ζ²) (damped natural frequency)
ζ is the damping factor
ωₙ is the undamped natural frequency
Given the system's damping factor, ζ = 0.3 and the undamped natural frequency, ωₙ = 10 rad/s, we can calculate the system's poles:
σ = ζωₙ = 0.3 * 10 = 3
ωₐ = 10 * sqrt(1 - 0.3²) = 10 * sqrt(1 - 0.09) = 10 * sqrt(0.91) ≈ 9.54
The poles of the original system: p = -3 ± j9.54
Now, we are changing the undamped natural frequency to ωₙ = 20 rad/s, keeping the damping factor the same:
σ = ζωₙ = 0.3 * 20 = 6
ωₐ = 20 * sqrt(1 - 0.3²) = 20 * sqrt(1 - 0.09) = 20 * sqrt(0.91) ≈ 19.08
The poles of the new system: p = -6 ± j19.08
As the undamped natural frequency has increased while keeping the same damping factor, the system will show an increased speed of response, and have higher oscillatory component (due to larger damped natural frequency). The system will still be underdamped, as the damping factor remains below 1, and have an oscillatory response, but the oscillations will be faster and the system will reach a steady state quicker than the original system.