If the characteristic equation of a closed loop system is S2 + 2S + 2 = 0, then the system is
underdamped
Compare the equation with the standard second-order form and read off the damping ratio.
\(s^{2}+2\zeta\omega_{n}s+\omega_{n}^{2}=0\)
Matching against \(s^{2}+2s+2=0\):
\(\omega_{n}^{2}=2\ \Rightarrow\ \omega_{n}=\sqrt{2}=1.414\ \text{rad/s}\)
\(2\zeta\omega_{n}=2\ \Rightarrow\ \zeta=\dfrac{1}{\sqrt{2}}=0.707\)
Since \(0\lt\zeta\lt1\), the system is underdamped — option 3.
| ζ | Roots | Response |
|---|---|---|
| ζ = 0 | Purely imaginary | Undamped — sustained oscillation |
| 0 < ζ < 1 | Complex conjugate | Underdamped — decaying oscillation |
| ζ = 1 | Real, repeated | Critically damped |
| ζ > 1 | Real, distinct | Overdamped |
The roots confirm it directly. Solving the quadratic,
\(s=\dfrac{-2\pm\sqrt{4-8}}{2}=-1\pm j1\)
A complex conjugate pair means the response contains \(e^{-t}\cos t\) and \(e^{-t}\sin t\) terms — it oscillates, while the negative real part makes the envelope decay. Both facts together are precisely what "underdamped" means, and the negative real part also confirms the system is stable.
Why this particular value is famous. \(\zeta=0.707\) is the standard design target, and for good reason. Overshoot is
\(M_{p}=e^{-\pi\zeta/\sqrt{1-\zeta^{2}}}=4.3\%\)
which is small, while settling time \(4/\zeta\omega_{n}=4\) s remains short. A critically damped system reaches its final value without any overshoot but takes noticeably longer; an overdamped one is slower still. In the frequency domain, \(\zeta=0.707\) is also the largest damping ratio that produces no resonant peak — the maximally flat or Butterworth response.
Hence, the system is underdamped.
Similar Questions
The step, ramp and parabolic test input signals can respectively be expressed as
Assertion (A) : In control systems, steady state response in the final requirement for calculating the efficiency of the system.
Reason (R) : The transient response is also critical for the determination of the steady state response.
consider a unity feedback control system for an open loop transfer function \(G(s)=\frac{5}{s(s+1)}\). Arrange the time constant in ascending order at different value of damping ratio of
A. \(\xi=3\)
B. \(\xi=7\)
C. \(\xi=1\)
D. \(\xi=10\)
E. \(\xi=5\)
Choose the correct answer from the options given below :
The open loop transfer function of a unity feedback control system is given by \(G(s)=\frac{25}{s(s+5)}\). The natural frequency of oscillator is fixed.
Arrange the damped frequency of oscillation for the following damping ratio in ascending order
A. \(\xi = 0.5\)
B. \(\xi = 0.1\)
C. \(\xi = 0.3\)
D. \(\xi = 0.25\)
E. \(\xi = 0.4\)
Choose the correct answer from the options given below :
The addition of a pole to the forward path transfer function of a closed loop system, generally has the effect of
Match the following lists :
List - I List - II a. Negative real and simple roots i. Sustained oscillatory b. Negative real and equal roots ii. Overdamped c. Complex conjugate roots iii. Critically damped d. Imaginary conjugate roots iv. Underdamped Correct codes are :
Important Questions from Time Response Analysis
Which of the following is correct for over-damped and under-damped system, respectively?
What will be the time response expression for a standard first order system having unit step function \(\frac{1}{s}\) as the input
A second order system has natural frequency 3rad/sec and unity damping ratio. Identify its transfer function.
Statement (I): All the systems which exhibit overshoot in transient response will also exhibit resonance peak in frequency response.
Statement (II): A large resonance peak in frequency response corresponds to a large overshoot in transient response.
The steady state error of a control system can be minimized by:
