Signals and systems electrical engineering miscellaneous

Signals and systems electrical engineering miscellaneous

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Signals and Systems

Signals and systems electrical engineering miscellaneous
  1. The z - transform of a signal x[n]is given by
    4z-3 + 3z-1 + 2 - 6z² + 2z³
    It is applied to a system, with a transfer function H(z) = 3z–1 – 2.
    Let the output be y(n). Which of the following is true?








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    Given: X(z) = 4 z–3 + 3z–1 + 2 – 6z² + 2z³
    H(z) = 3z–1 – 2,
    output, y(n) = x(n) × h(n)
    Taking z-tranform both sides,
    Y(z) = H(z) X (z)
    Y(z) = X(z) · H(z)
    Y(z) = 12 z–4 – 8z–3 + 9z–2 – 4 – 18z + 18z² – 4z³
    As, y[n] ≠ 0 for n < 0
    So we see that, we have z, z², z³ which depends on future input but finite terms. So y(n) is non casual with finite support.

    Correct Option: A

    Given: X(z) = 4 z–3 + 3z–1 + 2 – 6z² + 2z³
    H(z) = 3z–1 – 2,
    output, y(n) = x(n) × h(n)
    Taking z-tranform both sides,
    Y(z) = H(z) X (z)
    Y(z) = X(z) · H(z)
    Y(z) = 12 z–4 – 8z–3 + 9z–2 – 4 – 18z + 18z² – 4z³
    As, y[n] ≠ 0 for n < 0
    So we see that, we have z, z², z³ which depends on future input but finite terms. So y(n) is non casual with finite support.

  1. The system represented by the input-output relationship








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    so that system is non-causal.

    Correct Option: B


    so that system is non-causal.

  1. Given the finite lengt h input x[n] and the corresponding finite length output y[n] of an LTI system as shown below. The impulse response h[n] of the system is








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    x[n] = {1, – 1},     N = 2
    y[n] = {1, 0, 0, 0, – 1},   N = 5
    If P is t he number of elements in impulse response h[n], then,
    ⇒ N = M + P – 1
    ⇒ 5 = 2 + P – 1
    ⇒ P = 4
    Let h[n] = {x1, x2, x3, x4}
    then, y[n] = {x1, (x2 – x2), (x3 – x2), (x4 – x3), – x1}
    Comparing with y[n] = {1, 0, 0, 0, – 1}
           ↑
    We get x1 = 1, x2 = x1 = 1, x3 = x4 =1, x4 = x3 = 1
    then, h[n] = {1, 1, 1, 1}

    Correct Option: C

    x[n] = {1, – 1},     N = 2
    y[n] = {1, 0, 0, 0, – 1},   N = 5
    If P is t he number of elements in impulse response h[n], then,
    ⇒ N = M + P – 1
    ⇒ 5 = 2 + P – 1
    ⇒ P = 4
    Let h[n] = {x1, x2, x3, x4}
    then, y[n] = {x1, (x2 – x2), (x3 – x2), (x4 – x3), – x1}
    Comparing with y[n] = {1, 0, 0, 0, – 1}
           ↑
    We get x1 = 1, x2 = x1 = 1, x3 = x4 =1, x4 = x3 = 1
    then, h[n] = {1, 1, 1, 1}

  1. A band-limited signal with a maximum frequency of 5 kHz is to be sampled. According to the sampling theorem, the sampling frequency in kHz which is not valid is








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    As ƒmax = 5 kHz
    ƒ sampling ≥ 2 ƒmax = 2 ƒmax = 10 kHz
    Not 12, 15, 20 kHz a valid samply frequency so sampling frequency ƒs, such is not valid sampling frequency.

    Correct Option: A

    As ƒmax = 5 kHz
    ƒ sampling ≥ 2 ƒmax = 2 ƒmax = 10 kHz
    Not 12, 15, 20 kHz a valid samply frequency so sampling frequency ƒs, such is not valid sampling frequency.

  1. A signal x(t) = sinc (αt) where α is a real constant
    sin c (x) =
    sin (πx)
    πx

    is the input to a Linear Time invariant system whose impulse response h(t) = sinc (βt) where β is a real constant.
    If min (α , β) denotes the minimum of α and β, and similarly max (α , β) denotes the maximum of α and β, and K is a constant, which one of the following statements is true about the output of the system ?








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    NA

    Correct Option: A

    NA