Time Response Analysis Questions and Answers

Answer: D

The damping ratio \(\zeta\) of a second order system decides the nature of the response. For \(\zeta = 0\) the response is undamped, for \(0 < \zeta < 1\) it is underdamped (oscillatory with decaying amplitude), for \(\zeta = 1\) it is critically damped, and for \(\zeta > 1\) the system is overdamped.

In the overdamped case the characteristic roots are real and unequal, and the response creeps to its final value without overshoot. So the damping ratio of an overdamped system is greater than unity, option D.

Answer: A

The characteristic equation of the closed loop system is \(s^2 + 4s + K = 0\), with roots

\[ s = \frac{-4 \pm \sqrt{16 - 4K}}{2} \]

For \(K = 4\) the discriminant is zero, so

\[ s_1 = s_2 = -2 \]

The roots are real and equal (repeated), at the same point \(-2\) on the negative real axis, and the system is critically damped. The roots are not on the imaginary axis, which would need no \(s\) term, and the value \(-6\) in the last option is wrong. The answer is option A.

Answer: A

The settling time decreases.

For an underdamped second-order system the response settles within an envelope \(e^{-\zeta\omega_{n}t}\), and the settling time on the 2% criterion is about

\[ t_{s} \approx \frac{4}{\zeta\omega_{n}} \]

For a fixed natural frequency \(\omega_{n}\), a larger damping ratio \(\zeta\) makes the envelope decay faster, so \(t_{s}\) falls. It varies inversely with \(\zeta\), not with its square, and it is certainly affected by damping.

Answer: C

The damping ratio is 0.5.

With unity negative feedback, the closed loop characteristic equation is \(1 + G(s) = 0\):

\[ \begin{aligned} s(s + 2) + 4 &= 0 \\ s^{2} + 2s + 4 &= 0 \end{aligned} \]

Comparing with the standard form \(s^{2} + 2\zeta\omega_{n}s + \omega_{n}^{2} = 0\):

\[ \begin{aligned} \omega_{n}^{2} &= 4, \quad \omega_{n} = 2 \\ 2\zeta\omega_{n} &= 2, \quad \zeta = 0.5 \end{aligned} \]

The system is underdamped. A value of 1 would mean critical damping, 1.5 overdamping, and 0.25 comes from taking \(\omega_{n}\) as 4, the value of \(\omega_{n}^{2}\), instead of its square root.

Answer: C

The keyed meaning is analysing the system with an infinite frequency spectrum.

Transient analysis studies how a control system responds to sudden inputs, such as a unit step or impulse, before it settles to steady state: rise time, overshoot and settling time. Such inputs change abruptly, and their spectrum extends over all frequencies; a unit impulse has a flat spectrum to infinite frequency. Testing with them therefore excites the system across its whole frequency range at once.

Subjecting the system to a short period or a momentary high voltage describes a stress test, not transient analysis, and calling it merely 'a mathematical tool' does not say what is analysed.

Answer: C

It is the natural frequency.

For a second-order system with poles at \(s = -\zeta\omega_{n} \pm j\omega_{n}\sqrt{1 - \zeta^{2}}\), the undamped natural frequency \(\omega_{n}\) is the frequency of oscillation that would occur if the damping ratio \(\zeta\) were zero; the poles then lie on the imaginary axis at \(\pm j\omega_{n}\). With damping, the system oscillates at the lower damped frequency \(\omega_{d} = \omega_{n}\sqrt{1 - \zeta^{2}}\).

'Forced damping frequency', 'critical damping frequency' and 'conditional frequency' are not standard names for this quantity.

Answer: B

The feature not in the list is a simple system.

The standard requirements of a good control system are performance requirements: good stability, so that it settles rather than oscillates; good accuracy, a small steady-state error; adequate speed of response; and sufficient power handling, so that it can drive its load.

Simplicity is welcome in any design, but it is not one of the performance features that define a good control system; meeting the others often makes the system more complex, through feedback, compensators and sensors.