Control systems miscellaneous

Control systems miscellaneous

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Control Systems

Control systems miscellaneous
  1. For a gain constant K, the phase-lead compensator—








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    For a constant gain, the phase lead compensator reduces the slope of the magnitude curve in the entire range of frequency domain. For more detail refer synopsis.

    Correct Option: A

    For a constant gain, the phase lead compensator reduces the slope of the magnitude curve in the entire range of frequency domain. For more detail refer synopsis.

  1. Consider the following statements regarding time-domain analysis of a control system
    1. Derivative control improves system’s transient performance
    2. Integral control does not improve system’s steady state performance
    3. Integral control can convert a second order system into a third order system. Of these statements—








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    For more detail refer synopsis of integral and derivative control action.

    Correct Option: B

    For more detail refer synopsis of integral and derivative control action.

  1. In a linear system, an input of 5 sin ωt produces an output of 10 cos ωt. The output corresponding to input 10 cos ωt will be equal to—








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    Given
    it means the system is differentiator with ω = 2
    So with,
    ω = 2
    = – 10. ω sin ωt
    = – 10 × 2. sin ωt
    = – 20 sin ωt,

    Correct Option: D

    Given
    it means the system is differentiator with ω = 2
    So with,
    ω = 2
    = – 10. ω sin ωt
    = – 10 × 2. sin ωt
    = – 20 sin ωt,

  1. A linear second-order system with the transfer function
    G(s) =
    49
    s2 + 8s + 49

    is initially at rest and is subject to a step input signal. The response of the system will exhibit a peak overshoot of—








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    We know that maximum overshoot is given as
    Mp = e– ξπ/√1 – ξ2

    From given T.F. = G
    49
    s2 + 8s + 49

    C.E. = s2 + 8s + 49
    on comparing this equation with the standard equation
    ωns + ω2n = 0
    2 ξωn and ωn = 7
    or
    ξ =
    8
    		 =
    8
    		 = .57
    2ωn2 × 7

    % Mp = e
    -3.14 × (0.57)2
    		 × 100 = 2%
    √1 - (0.57)2

    Correct Option: C

    We know that maximum overshoot is given as
    Mp = e– ξπ/√1 – ξ2

    From given T.F. = G
    49
    s2 + 8s + 49

    C.E. = s2 + 8s + 49
    on comparing this equation with the standard equation
    ωns + ω2n = 0
    2 ξωn and ωn = 7
    or
    ξ =
    8
    		 =
    8
    		 = .57
    2ωn2 × 7

    % Mp = e
    -3.14 × (0.57)2
    		 × 100 = 2%
    √1 - (0.57)2

  1. NA










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    From figure
    C = GR – CH2
    C (1 + H2) = GR

    C
    		 =
    G
    R1 + H2
    From figure
    C = GR – CH2
    C (1 + H2) = GR
    C
    		 =
    G
    R1 + H2


    Correct Option: B

    From figure
    C = GR – CH2
    C (1 + H2) = GR

    C
    		 =
    G
    R1 + H2