Signal and systems miscellaneous

Signal and systems miscellaneous

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Signals and Systems

Signal and systems miscellaneous
  1. Of the following transfer function of second order linear time-invariant systems the under damped system is represented by—








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    Here, we will calculate the value of ξ for each case. For the system to be underdamped ξ should be less than one, i.e.,
    ξ < 1

    (A) Given H(s) =
    1
    s2 + 4s + 4

    C.E. = s2 + 4s + 4
    On comparing this C.E. with standard equation
    s2 + 2ξωn + ω2n = 0
    we get, 2ξ ωn = 4
    and ω2n = 4
    or ωn = ± 2
    or 2ξ2=4
    or ξ = 1
    (B) H(s) =
    1
    s2 + 5s + 4

    Here, 2ξ2=5
    and ω2n = 4
    or ωn = ± 2
    or 2ξ2=5
    or ξ =
    5
    		= 1·25
    4

    (C) H(s) =
    1
    s2 + 4·5s + 4

    Here, 2ξωn = 4·5
    and ω2n = 4
    or ωn = ± 2
    or 2ξωn = 4·5
    or ξ =
    4·5
    		= 1·125
    4

    (D) H(s) =
    1
    s2 + 3s + 4

    Here, 2ξωn = 3,
    ω2n = 4
    or ωn = ± 2
    or 2ξ2=3
    ξ =
    3
    		= ·75 < 1
    4

    Hence, alternative (D) is the correct choice.

    Correct Option: D

    Here, we will calculate the value of ξ for each case. For the system to be underdamped ξ should be less than one, i.e.,
    ξ < 1

    (A) Given H(s) =
    1
    s2 + 4s + 4

    C.E. = s2 + 4s + 4
    On comparing this C.E. with standard equation
    s2 + 2ξωn + ω2n = 0
    we get, 2ξ ωn = 4
    and ω2n = 4
    or ωn = ± 2
    or 2ξ2=4
    or ξ = 1
    (B) H(s) =
    1
    s2 + 5s + 4

    Here, 2ξ2=5
    and ω2n = 4
    or ωn = ± 2
    or 2ξ2=5
    or ξ =
    5
    		= 1·25
    4

    (C) H(s) =
    1
    s2 + 4·5s + 4

    Here, 2ξωn = 4·5
    and ω2n = 4
    or ωn = ± 2
    or 2ξωn = 4·5
    or ξ =
    4·5
    		= 1·125
    4

    (D) H(s) =
    1
    s2 + 3s + 4

    Here, 2ξωn = 3,
    ω2n = 4
    or ωn = ± 2
    or 2ξ2=3
    ξ =
    3
    		= ·75 < 1
    4

    Hence, alternative (D) is the correct choice.

  1. Let F(ω) be the Fourier transform of a functions f(t), then F(0) is—








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    We know Fourier transform of function f(t) is given as
    F(jω) = ∫∞– ∞ f(t)·e– jωt dt
    F(0) = ∫∞– ∞ f(t)·e– j.0.t dt
    F(0) = ∫∞– ∞ f(t).dt
    Hence, alternative (A) is the correct choice.

    Correct Option: B

    We know Fourier transform of function f(t) is given as
    F(jω) = ∫∞– ∞ f(t)·e– jωt dt
    F(0) = ∫∞– ∞ f(t)·e– j.0.t dt
    F(0) = ∫∞– ∞ f(t).dt
    Hence, alternative (A) is the correct choice.

  1. The impulse response of a single-pole system would approach a non-zero constant as t → ∞ if and only if the pole is located in the s-plane—








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    NA

    Correct Option: B

    NA

  1. The impulse response of a causal, linear, time-invariant, continuous-time system is h(t). The output y(t) of the same system to an input x(t), where x(t) = 0 for t < – 2, is—








  1. View Hint View Answer Discuss in Forum

    In the given question all the alternative are linear and time invariant, but not causal all, for t < – 2. Hence, for the given system to be causal lower limit should be – 2 and upper limit should be t.
    i.e., y(t) = ∫t– 2 h(τ) x(t – τ )dt
    Hence, alternative (B) is the correct choice.

    Correct Option: B

    In the given question all the alternative are linear and time invariant, but not causal all, for t < – 2. Hence, for the given system to be causal lower limit should be – 2 and upper limit should be t.
    i.e., y(t) = ∫t– 2 h(τ) x(t – τ )dt
    Hence, alternative (B) is the correct choice.

  1. If a(n) is the response of a linear, time-invariant, discrete time system to a unit step input, then the response of the same system to a unit impulse input is—








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    NA

    Correct Option: A

    NA