Time Response of a Second Order Underdamped System

  • Find the damping ratio and natural frequency of a second-order system.

  • Find the settling time, peak time, percent overshoot, and rise time for an underdamped second-order system.

A general form of a second order system is: \[\dfrac{d^{2}y(t)}{dt^{2}} + 2 \xi \omega _{n}\dfrac{dy(t)}{dt} + \omega _{n}^{2}y(t) = K \omega _{n}^{2}u(t)\]

And the transfer function is: \[\dfrac{Y(s)}{R(s)} = \dfrac{K \omega _{n}^{2}}{s^{2}+ 2\xi \omega _{n}s + \omega _{n}^{2}}\] where:
K = Gain of the system,
\(\xi\) = Damping ratio of the system,
\(\omega _{n}\) = The undamped natural frequency of the system
Depending upon the value of \(\xi\) ,there are three possibilities:

  • \(\xi\) = 1 means critically damped

  • \(\xi\) \(<\) means 1 underdamped

  • \(\xi\) \(>\) means overdamped

The maximum overshoot is the maximum peak value of the response curve measured from unity. \[\%M_{p}= e^{\dfrac{-\zeta \pi }{\sqrt{1- \zeta ^{2}}}} * 100\] Settling time \(t_s\) = It is the time required for the response curve to reach and stay within a specified percentage (usually 2% or 5%) of the final value. \[t_{s} = \dfrac{4}{\zeta \omega _{n}}\]

Solved Examples

Solved Example:

87-3-01

For the following second order control system model, the damped natural frequency in rad/s is: \[\dfrac{Y(s)}{R(s)} = \dfrac{20}{s^{2}+5s+24}\]

Solution:
\[\dfrac{Y(s)}{R(s)} = \dfrac{20}{s^{2}+5s+24}\] Also, \[\dfrac{Y(s)}{R(s)} = \dfrac{20}{24}\left [\dfrac{24}{s^{2}+5s+24}\right]\] Comparing with, \[\dfrac{Y(s)}{R(s)} = K \left [\dfrac{\omega_{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}\right]\] \[\omega _{n}^{2}= 24\] $\mathrm{Undamped\ natural\ frequency\ \ } \omega _{n} = 4.89 \ \mathrm{rad/s}$
$\mathrm{And\ Also\ }\ 2\xi \omega _{n} = 5$
$\mathrm{The\ damping \ ratio\ \ } \xi = 0.511$
$\mathrm{The \ damped \ natural \ frequency\ \ }$
$\omega _{d} = \omega_{n}\sqrt{1-\xi ^{2}} = 4.2 \ \mathrm{rad/s}$

Correct Answer: C

Solved Example:

87-3-02

Considering the unity feedback system G(s)=$\dfrac{9}{s(s+2.4)}$ , the settling time of the resulting second order system for 2% tolerance band will be:

Solution:
\[G(s)=\dfrac{9}{s(s+2.4)}\] Unity feedback system means H(s) = 1
Closed loop transfer function \[\dfrac{Y(s)}{R(s)} = \dfrac{G(s)}{1+G(s)H(s)}\] Substituting the values of G(s) and H(s), Closed loop transfer function \[ \dfrac{Y(s)}{R(s)} = \dfrac{9}{s^2+2.4s + 9}\] Comparing with, \[\dfrac{Y(s)}{R(s)} = K \left [\dfrac{\omega_{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}\right]\] \[\dfrac{9}{s^2+2.4s + 9} = K \left [\dfrac{\omega_{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}\right]\] \[K = 1, \omega_n = 3\ rad/s, \zeta = 0.4\] The settling time is given by, \[t_s = \dfrac{4}{\zeta \omega_n} = \dfrac{4}{(0.4)(3)} = 3.33\ sec\]

Correct Answer: B

Solved Example:

87-3-03

Peak overshoot of step-input response of an underdamped second-order system is explicitly indicative of: (TANGEDCO AE EE 2018)

Solution:
Peak overshoot refers to the damping of the system as if the damping id less than the peak will be more.

Correct Answer: A

Solved Example:

87-3-04

Match the transfer functions of the second order systems with the nature of the systems given below: (GATE EE 2018)

Transfer Function
P.$\dfrac{15}{s^2+5s+15}$
Q.$\dfrac{25}{s^2+10s+25}$
R.$\dfrac{35}{s^2+18s+35}$

Nature of System
I. Overdamped
II. Critically damped
III. Under damped

Solution:
Compare each equation with the standard second order system equation: \[\dfrac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}\]
For system P.$\dfrac{15}{s^2+5s+15}$
$\omega_n^2 = 15$ and $2\zeta \omega_n = 5$
$\zeta = \dfrac{5}{2\sqrt{15}}< 1$ (Underdamped)

For system Q.$\dfrac{25}{s^2+10s+25}$
$\omega_n^2 = 25$ and $2\zeta \omega_n = 10$
$\zeta = 1$ (Critically Damped)

For system R.$\dfrac{35}{s^2+18s+35}$
$\omega_n^2 = 35$ and $2\zeta \omega_n = 18$
$\zeta = \dfrac{9}{\sqrt{35}}> 1$ (Overdamped)

Correct Answer: C