Control systems miscellaneous

Control systems miscellaneous

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Control Systems

Control systems miscellaneous
  1. A linear stable invariant system is forced with an input given below
    x (t) = A sin ωt
    x (t)⎯→ [G (s) ]⎯→y (t)
    Under steady-state conditions, the output y (t) of the system will be—








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    Given x(t) ⎯→ [G(s)] ⎯→ y(t)
    x(t) = A sin ωt
    steady state response
    Y(jω) = X(jω) .G(jω)
    where, X(jω) = A ∠sin ωt
    G(jω) = |G(jω)| < G(jω)
    or
    Y(t) = A sin [ωt + < G(jω)].|G(jω)|
    or
    y(t) = A|G(jω)| sin [ωt + < G(jω)]

    Correct Option: B

    Given x(t) ⎯→ [G(s)] ⎯→ y(t)
    x(t) = A sin ωt
    steady state response
    Y(jω) = X(jω) .G(jω)
    where, X(jω) = A ∠sin ωt
    G(jω) = |G(jω)| < G(jω)
    or
    Y(t) = A sin [ωt + < G(jω)].|G(jω)|
    or
    y(t) = A|G(jω)| sin [ωt + < G(jω)]

  1. The transfer function for the given Bode plot is–










  1. View Hint View Answer Discuss in Forum

    From figure we conclude that two pole at origin, one pole at ω = 5 and one zero at ω = 1 so the transfer function would be of the type

    T.F =
    K(s + 1)
    s2((s/5) + 1)

    Hence alternative (A) is the correct choice.


    Correct Option: A

    From figure we conclude that two pole at origin, one pole at ω = 5 and one zero at ω = 1 so the transfer function would be of the type

    T.F =
    K(s + 1)
    s2((s/5) + 1)

    Hence alternative (A) is the correct choice.


  1. A unity feedback system has an open-loop transfer function of the from
    KG(s) =
    K(s + a)
    		; b > a
    s2(s + b)

    Which of the loci system in figure can be a valid root-loci for the system?








  1. View Hint View Answer Discuss in Forum

    Given,

    KG(s) =
    K(s + a)
    		; b > a
    s2(s + b)

    Here, pole are at s = 0, 0, – b and zero are at s = – a The asymptote angles are
    =
    (2q + 1)
    		q = 0, 1.
    p – z

    = 90° and 270° degrees.
    Centroid is at,
    σ =
    – (real part of all poles – real parts of all zeros)
    p – z

    =
    – (b – a)
    3 – 1

    =
    – b + a
    2

    The characteristic equation is, 1 + KG(s) = 0
    or
    1 +
    K(s + a)
    		= 0
    s2(s + b)

    or
    s3 + bs2 + Ks + Ka = 0
    and, Routh array is
    s3 1 K
    s2 B Ka
    s1 Kb – Ka/b
    s0 Ka
    Since given that, b > a, therefore the system is always stable i.e., there is no intersection on the imaginary axis. Hence the roots locus meets the imaginary axis at s = 0. So, alternative (A) is the correct choice.

    Correct Option: A

    Given,

    KG(s) =
    K(s + a)
    		; b > a
    s2(s + b)

    Here, pole are at s = 0, 0, – b and zero are at s = – a The asymptote angles are
    =
    (2q + 1)
    		q = 0, 1.
    p – z

    = 90° and 270° degrees.
    Centroid is at,
    σ =
    – (real part of all poles – real parts of all zeros)
    p – z

    =
    – (b – a)
    3 – 1

    =
    – b + a
    2

    The characteristic equation is, 1 + KG(s) = 0
    or
    1 +
    K(s + a)
    		= 0
    s2(s + b)

    or
    s3 + bs2 + Ks + Ka = 0
    and, Routh array is
    s3 1 K
    s2 B Ka
    s1 Kb – Ka/b
    s0 Ka
    Since given that, b > a, therefore the system is always stable i.e., there is no intersection on the imaginary axis. Hence the roots locus meets the imaginary axis at s = 0. So, alternative (A) is the correct choice.

  1. A unity feedback system has the open-loop transfer function
    G(s) =
    1
    (s – 1)(s + 2)(s + 3)

    The Nyquist plot of G(s) encircles the origin—








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    No. of encirclement, N = P+ – Z+
    where P+ = no. of poles on RHS on the open loop T.F.
    Z+ = no. of zeros on RHS on the closed loop
    T.F. The given open loop T.F.
    G(s) = 1/(s – 1)(s + 2)(s + 3)
    Here
    P+ = 1
    Z+ = 0
    So, N = 1 – 0 = 1 Hence alternative (B) is the correct choice.

    Correct Option: B

    No. of encirclement, N = P+ – Z+
    where P+ = no. of poles on RHS on the open loop T.F.
    Z+ = no. of zeros on RHS on the closed loop
    T.F. The given open loop T.F.
    G(s) = 1/(s – 1)(s + 2)(s + 3)
    Here
    P+ = 1
    Z+ = 0
    So, N = 1 – 0 = 1 Hence alternative (B) is the correct choice.

  1. The open-loop transfer function of a system is given by
    G(s) H(s) =
    K(s + 1)(s + 3)
    s2 + 4s + 8

    Its root locus diagram is—








  1. View Hint View Answer Discuss in Forum

    The open-loop transfer function is

    G(s)H(s) =
    K(s + 1)(s + 3)
    s2+ 4s + 8

    Now s2 + 4s + 8 = 0
    has the roots s1, 2 = – 2 ± j2
    Number of zeros of G(s) H(s) is 2 at
    s = – 1, s = – 3.
    Number of poles of G(s) H(s) is 2 at
    s = – 2 + j2, s = – 2 – j2.
    The root locus will having two branches starting at s = – 2 + j2 and s = – 2 – j2 and terminating at s = – 1 and s = – 3.
    Number of asymptotes = α = p (Number of finite poles) – Z (Number of finite zeros)
    = 2 – 2 = 0.

    Correct Option: A

    The open-loop transfer function is

    G(s)H(s) =
    K(s + 1)(s + 3)
    s2+ 4s + 8

    Now s2 + 4s + 8 = 0
    has the roots s1, 2 = – 2 ± j2
    Number of zeros of G(s) H(s) is 2 at
    s = – 1, s = – 3.
    Number of poles of G(s) H(s) is 2 at
    s = – 2 + j2, s = – 2 – j2.
    The root locus will having two branches starting at s = – 2 + j2 and s = – 2 – j2 and terminating at s = – 1 and s = – 3.
    Number of asymptotes = α = p (Number of finite poles) – Z (Number of finite zeros)
    = 2 – 2 = 0.