1. A second-order system with a zero at -2 has its poles located at -3 + j4 and -3 – j4 in the s-plane. The undamped natural frequency and the damping factor of the system respectively are
- 5 rad/s and 0.6
- 3 rad/s and 0.6
- 5 rad/s and 0.8
- 3 rad/s and 0.8
2. The transient response of the second-order underdamped system starting from rest is given by
c(t) = Ae-6t sin(8t + θ), t ≥ 0.
The natural frequency of the system is
- 100
- 10
- 9
- 8
3. The steady state response c(t) for an input r(t) = sin2t to a system transfer function $\frac{1}{s + 4}$ is
- 0.632 cos2t
- 0.316 sin(2t – 26.5°)
- sin(2t – 45°)
- 0.25 sin2t
4. For a critically damped system, the closed-loop poles are
- real, unequal and negative
- complex conjugate with negative real part
- real, equal and negative
- purely imaginary
5. The open-loop transfer function of a unity feedback system is $\frac{K}{s(s+4)}$. For a damping factor of 0.5, the value of the gain K must be set to
- 16
- 4
- 2
- 1
6. A system has a transfer function
$\frac{C(s)}{R(s)}=\frac{4}{s^2 + 1.6s + 4}$
For a unit step response and 2% tolerance band, the settling time will be
- 2 seconds
- 3 seconds
- 4 seconds
- 5 seconds
7. How many poles does the following function have?
$R(s)=\frac{s^3 + 2s + 1}{s^2 + 3s + 2}$
- 3
- 2
- 1
- 0
8. In a system, the damping coefficient is -2. The system response will be
- Undamped
- Oscillations with decreasing magnitude
- Oscillations with increasing magnitude
- Critically damped
9. Consider the following statements regarding the effect of adding a pole in the open-loop transfer function on the closed-loop step response:
- It increases the maximum overshoot.
- It increases the rise time.
- It reduces bandwidth.
Which of the above statements are correct
- i, ii and iii
- i and ii only
- i and iii only
- ii and iii only
10. The derivative of a parabolic function becomes
- a unit-impulse function
- a ramp function
- a gate function
- a triangular function
11. In the time-domain specification, decay ratio is the ratio of the
- amplitude of the first peak and the steady-state value
- amplitudes of the first two successive peaks
- peak value to the steady-state value
- none of the above
12. Consider the time response of a second-order system with damping coefficient less than 1 to a unit step input:
- It is overdamped.
- It is a periodic function.
- Time duration between any two consecutive values of 1 is the same.
Which of the above statements is/are correct?
- i, ii and iii
- i only
- ii only
- iii only
13. A sensor requires 30 s to indicate 90% of the response to a step input. If the sensor is a first-order system, the time constant is
[given, loge (0.1) = -2.3]- 15 s
- 13 s
- 21 s
- 28 s
14. Consider the following input and system types:
Input type | System type |
Unit step | Type ‘0’ |
Unit ramp | Type ‘1’ |
Unit parabolic | Type ‘2’ |
Which of the following statements are correct?
- Unit step input is acceptable to all three types of system.
- Type ‘0’ system cannot accept unit parabolic input.
- Unit ramp input is acceptable to Type ‘2’ system only.
- i and ii only
- i and iii only
- ii and iii only
- i, ii and iii
15. The characteristic equation of a closed-loop system is s2 + 4s+ 16 = 0. The natural frequency of oscillation and damping constant respectively are
- 2 rad/s and 0.5
- $2\sqrt{3} rad/s$ and $\frac{1}{\sqrt{3}}$
- 4 rad/s and 0.5
- 4 rad/s and 0.707
16. The unit impulse response of a system is given as c(t) = -4e-t + 6e-2t. The step response of the same system for t ≥ 0 is equal to
- 3e-2t – 4e-t + 1
- -3e-2t + 4e-t + 1
- -3e-2t + 4e-t – 1
- 3e-2t + 4e-t + 1
17. A unity feedback second-order control system characterized by the open-loop transfer function
$G(s) =\frac{K}{s(Js+ B)}$
J = moment of inertia, B = damping constant and K = system gain.
The transient response specification which is not affected by system gain variation is
- Peak overshoot
- Rise time
- Settling time
- Time to peak overshoot
18. For a unit step input, a system with forward path transfer function G(s) = 20/s2 and feedback path transfer function H(s) = (s+5) has a steady-state output of
- 2
- 0.5
- 1
- 0.2
19. Consider the open-loop transfer function:
$G(s)H(s) =\frac{5(s+1)}{s^2(s+5)(s + 12)}$
The steady state error due to a ramp input is
- 0
- 5
- 12
- ∞
20. The position and velocity error coefficients for the system of transfer function,
$G(s)=\frac{50}{(1+0. 1s)(1+2s)}$ are respectively
- zero and zero
- zero and infinity
- 50 and zero
- 50 and infinity