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 locate the new poles of the system, we need to use the following formula for the second-order system:
Poles = -ζω_n ± jω_n√(1 - ζ^2)
Where ζ is the damping factor, ω_n is the un-damped natural frequency, and j is the imaginary unit.
Given the damping factor ζ = 0.3 and the new un-damped natural frequency ω_n = 20 rad/s, we can plug these values into the formula:
Poles = -0.3 * 20 ± j20√(1 - 0.3^2)
Poles = -6 ± j20√(1 - 0.09)
Poles = -6 ± j20√(0.91)
Poles = -6 ± j18.248
Therefore, the new poles of the system are -6 ± j18.248.
Now, let's analyze the response of the new system. Since the damping factor (0.3) remains the same, the system is still considered an underdamped system. This means that the system will have oscillatory behavior and a certain amount of overshoot before settling to its steady state.
However, by increasing the un-damped natural frequency from 10 rad/s to 20 rad/s, the system becomes more oscillatory and faster in response. This is because the higher the natural frequency, the faster the system responds to changes or disturbances.
In summary, the new system will still have an underdamped response, with more oscillations and faster response time compared to the original system.