InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
For quadratic factor. In Bode log magnitude plot the high frequency asymptote is a straight line having the slope |
| Answer» For quadratic factor in denominator, slope at high frequencies is - 40 dB/decade. | |
| 52. |
For op-amp circuit of the given figure, E()/E() = |
| Answer» . | |
| 53. |
In liquid flow system the capacitance of a water tank is defined as |
| Answer» Electrical and dQ is analogous to change in amount of liquid and dv is analogous to change in head. | |
| 54. |
For a first order system having transfer function , the unit ramp response is |
| Answer» . | |
| 55. |
For a second order system, the position of poles is in the shaded region shown in the given figure. The variation in damping factor is |
| Answer» Variation in damping factor is small. | |
| 56. |
In a second order system, the time constant of exponential envelopes depends |
| Answer» | |
| 57. |
The maximum shift which can be provided by a lead compensator with transfer function is |
| Answer» The maximum phase lead is given by In this case a = 1/3. Hence sin Φm = = 0.5 or Φm = 30°. | |
| 58. |
The network of the given figure is a |
| Answer» V0 will lead Vi. Hence lead network. | |
| 59. |
For the given figure |
| Answer» . | |
| 60. |
The resistance R of a liquid flow system is defined as |
| Answer» Electrical resistance , level difference or head is analogous to v and change in flow rate is analogous to current. | |
| 61. |
For a factor in the transfer function, the log magnitude plot |
| Answer» jω term in denominator gives a straight line with a slope of - 20 dB/decade and gives a straight line with a slope of - 40 dB/decade. | |
| 62. |
The frequency at which magnitude of closed loop frequency response is 3 dB below its zero frequency value is called critical frequency. |
| Answer» It is not critical frequency. | |
| 63. |
The transfer function has the characteristics of high pass filter. |
| Answer» When ω is high, the output is low. | |
| 64. |
The transfer function can be for |
| Answer» For a lead-lag compensator β > 1 and a < 1. | |
| 65. |
The system of the given figure |
| Answer» Since there are two poles at s = -2 and s = -1 and one zero at s = +3. | |
| 66. |
If phase angle of open loop transfer function becomes - 180° at frequency ω, then gain margin is equal to |
| Answer» If gain is a, gain margin is . In this case gain is |G(jω1)|. | |
| 67. |
A system with transfer function may be approximated by the system |
| Answer» Coefficient (s + 15) gives the term e-15t and the coefficient (s + 20) gives the term e-20t. These can be neglected as compared to terms corresponding to s and (s + 1). | |
| 68. |
For an RLC series circuit Z() is of the form |
| Answer» . | |
| 69. |
If feedback factor is β the overall gain of a closed loop system is approximately equal to |
| Answer» Closed loop gain = . Since 1 << Av β, the gain is . | |
| 70. |
The functions and have |
| Answer» Magnitude are the same. i.e. . However phase angles are different because the phase angle of 1 - jωT3 is negative but that of 1 - jωT3 is positive. | |
| 71. |
For the op-amp circuit of the given figure, E()/E() = |
| Answer» | |
| 72. |
In op-amp circuit of the given figure, E()/E() = |
| Answer» | |
| 73. |
A system hasFor unit step input the settling time for 2% tolerance band is |
| Answer» ξωn - 0.8 and = 1.25 s. For 2% criterion, settling time = 4T = 5 seconds. | |
| 74. |
The primary function of lag compensator is to provide sufficient |
| Answer» It changes phase. | |
| 75. |
In the given figure, C() = |
| Answer» Use principle of superposition. Initially assume R(s) = 0 and find C(s). Then assume U(s) = 0 and find C(s). Add the two responses. | |
| 76. |
The polar plot of |
| Answer» It does not become real at any frequency. | |
| 77. |
In a second order undamped system, the poles are on |
| Answer» If poles are on imaginary axis the response is undamped and oscillatory. | |
| 78. |
For the system of the given figure, the undamped natural frequency of closed loop poles is |
| Answer» . | |
| 79. |
An open loop system has a forward path transfer function (42.25)/( + 6.5). The unit step response of the system starting from rest will have its maximum value at a time equal to |
| Answer» Unit step response is Taking inverse transform, the time domain response is - 1 + 6.5t + e-6-5t. It reaches maximum value at t = infinity. | |
| 80. |
In the given figure |
| Answer» Use block diagram algebra. | |
| 81. |
A system has transfer function. If a sinusoidal input of frequency ω is applied the steady state output is of the form |
| Answer» For sinusoidal input jω replaces s in the transfer function. | |
| 82. |
For the given figure C()/R() = |
| Answer» or . | |
| 83. |
For a quadratic factor in transfer function the phase angle is not a function of damping factor. |
| Answer» Damping ratio affects phase angle. | |
| 84. |
The current rating of dc tachogenerator is usually |
| Answer» Control systems require small current rating of components. | |
| 85. |
For the compensator in the given figure the maximum phase lead is |
| Answer» Hence Also sin φm = = 0.5 and φm = 30°. | |
| 86. |
For G(ω) = |
| Answer» When ω = 0, phase angle is 0. For high values of phase angle = - 90° - 90° = 180°. | |
| 87. |
The transient response of a second order system is given by for 5% criterion the settling time is |
| Answer» For 5% criterion settling time = 3T = . Since ξωn = 4, and 3T = 0.75 secs. | |
| 88. |
The polar plot of a transfer function passes through (-1, 0) point. The gain margin is |
| Answer» When polar plot passes through (-1, 0) the system is on limit of stability and gain margin is zero. | |
| 89. |
The gain margin for a stable system |
| Answer» For a stable system gain margin is positive. Therefore decibel value is also positive. | |
| 90. |
The signal flow graph of a system is shown in the given figure. The effect of disturbance T can be reduced by |
| Answer» If G1(s) is increased, input is amplified more while disturbance remains the same. | |
| 91. |
In the given figure, P = P sin ω. Then X() = |
| Answer» If P = Pm, sin ω . | |
| 92. |
The action of bellows is similar to that of |
| Answer» Spring and bellows convert force into displacement. | |
| 93. |
A lead compensator is basically a |
| Answer» See transfer function in the equation . | |
| 94. |
The system in the given figure |
| Answer» The relative extent of negative and positive feedbacks determines stability. | |
| 95. |
The units of thermal capacitance are |
| Answer» Heat energy (kcal) is analogous to electric charge and temperature is analogous to voltage. . | |
| 96. |
The transfer function can be for |
| Answer» For a lead compensator and a < 1 Therefore is more than . | |
| 97. |
In response to a unit step input the controller output in given figure is for |
| Answer» The intercept on y-axis is due to proportional action and inclined line is due to integral action. | |
| 98. |
For very low frequencies, log magnitude for the given figure is |
| Answer» If f very low, XC is very high and V0 ⋍ Vi. | |
| 99. |
If , initial and final values of () are |
| Answer» Use initial and final value theorems. | |
| 100. |
In the given figure the input frequency = 0.1 (1/2 RC) |
| Answer» XC = 10R. Hence V0 ⋍ Vi . | |