Signal and systems miscellaneous

Signal and systems miscellaneous

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Signals and Systems

Signal and systems miscellaneous
  1. The system represented by equation
    y[n] = 8y2[n – 1] – nx[n] + 4x[n – 1] – 2x[n + 1] is—
    (i) linear
    (ii) time variant
    (iii) non-causal
    The correct statements is/are—








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    (i) Linearity: The given system
    y(n)= 3y2(n – 1) – nx(n) + 4x(n – 1) – 2x(n + 1)
    will be linear if
    F[ax(n)] = aF[x(n)]
    F[ax(n)] = ay(n) = 3a2y2(n – 1) – anx(n) + 4x(n – 1) – 2ax(n + 1)
    aF[x(n)] = a[y(n)] = 3ay2(n – 1) – an x(n) + 4x(n – 1) – 2ax(n + 1)
    Since here,
    F[ax(n)] ≠ aF[x(n)]
    Therefore, the system is non-linear.
    (ii) Shift invariance: The given system equation will be time invariant if
    y[n – n0] = F[x(n – n0)]
    The response of delayed excitation is
    y[x(n – n0)] = 3y2[(n – n0) y(n – n0)]
    = F[x(n – n0 – n)]
    = 3ay2(n – 1) – nx(n – n0) + 4x(n – n0 – 1) – 2x(n – n0 + 1)
    and the delayed response is
    y(n – n0)= 3y2[(n –n0 – 1) – (n – n0) x(n – n0) + 4x(n – n0 – 1) – 2x(n – n0 + 1)]
    Here, y(n – n0) ≠ F[x(n – n0)]
    Hence, system is not time-invariant. (iii) Causality: In the given equation
    y(n)= 3y2[(n – 1) – 1] – nx(n) x(n – n0) + 4x(n – 1) – 2x(n + 1)
    output y(n) is depends on a future input sample value x(n + 1).
    Hence, the system is non-causal. Hence, alternative (C) is the correct choice.

    Correct Option: C

    (i) Linearity: The given system
    y(n)= 3y2(n – 1) – nx(n) + 4x(n – 1) – 2x(n + 1)
    will be linear if
    F[ax(n)] = aF[x(n)]
    F[ax(n)] = ay(n) = 3a2y2(n – 1) – anx(n) + 4x(n – 1) – 2ax(n + 1)
    aF[x(n)] = a[y(n)] = 3ay2(n – 1) – an x(n) + 4x(n – 1) – 2ax(n + 1)
    Since here,
    F[ax(n)] ≠ aF[x(n)]
    Therefore, the system is non-linear.
    (ii) Shift invariance: The given system equation will be time invariant if
    y[n – n0] = F[x(n – n0)]
    The response of delayed excitation is
    y[x(n – n0)] = 3y2[(n – n0) y(n – n0)]
    = F[x(n – n0 – n)]
    = 3ay2(n – 1) – nx(n – n0) + 4x(n – n0 – 1) – 2x(n – n0 + 1)
    and the delayed response is
    y(n – n0)= 3y2[(n –n0 – 1) – (n – n0) x(n – n0) + 4x(n – n0 – 1) – 2x(n – n0 + 1)]
    Here, y(n – n0) ≠ F[x(n – n0)]
    Hence, system is not time-invariant. (iii) Causality: In the given equation
    y(n)= 3y2[(n – 1) – 1] – nx(n) x(n – n0) + 4x(n – 1) – 2x(n + 1)
    output y(n) is depends on a future input sample value x(n + 1).
    Hence, the system is non-causal. Hence, alternative (C) is the correct choice.

  1. The three statements according to Poley-Wiener criterion for an amplitude response to be realizable are:
    (i) The magnitude |H(jω)| of a realizable network can be zero at a discrete set of frequencies, but it cannot have zero magnitude over a finite band of frequencies.
    (ii) The amplitude function |H(jω)| must not fall to zero faster than the exponential order.
    (iii) Any ideal filter is non-causal. The true statements is/are—








  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: D

    NA

  1. The system represented by equations—
    (i) y[n] = anx(n) + b
    (ii) y[n] = ex[n]








  1. View Hint View Answer Discuss in Forum

    (i) y(n) = anx(n) + b
    F[x1(n) + x2(n)] = an[x1(n) + x2(n)] + b F[x1(n)] + F[x2(n)]
    = [an x1(n) + b] + [an x2(n) + b]
    = an[x1(n) + x2(n)] + 2b
    Therefore,
    F[x1(n) + x2(n)] ≠ F[x1(n)] + F[x2(n)]
    Hence, the system is non-linear.
    (ii) y(n) = ex(n)
    F[x1(n) + x2(n)] = e[x1 (n) + x2 (n)]
    = ex1(n) + ex2(n)
    F[x1(n)] + F[x2(n)] = ex1(n) + ex2(n)
    Since here,
    F[x1(n) + x2(n)] ≠ F[x1(n)] + F[x2(n)]
    Hence, the system is non-linear, therefore alternative (D) is the correct choice.

    Correct Option: D

    (i) y(n) = anx(n) + b
    F[x1(n) + x2(n)] = an[x1(n) + x2(n)] + b F[x1(n)] + F[x2(n)]
    = [an x1(n) + b] + [an x2(n) + b]
    = an[x1(n) + x2(n)] + 2b
    Therefore,
    F[x1(n) + x2(n)] ≠ F[x1(n)] + F[x2(n)]
    Hence, the system is non-linear.
    (ii) y(n) = ex(n)
    F[x1(n) + x2(n)] = e[x1 (n) + x2 (n)]
    = ex1(n) + ex2(n)
    F[x1(n)] + F[x2(n)] = ex1(n) + ex2(n)
    Since here,
    F[x1(n) + x2(n)] ≠ F[x1(n)] + F[x2(n)]
    Hence, the system is non-linear, therefore alternative (D) is the correct choice.

  1. The system represented by equations—
    (i)
    N – 1
    y(n) = 1/N
    		x(n – m)
    m = 0

    (ii) y(n) = [x(n)]2








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    N – 1
    (i) y(n) = 1/N
    		x(n – m)
    m = 0

    N – 1
    Let y1(n) = 1/N
    		x1(n – m)
    m = 0

    N – 1
    and y2(n) = 1/N
    		x2(n – m)
    m = 0

    then y(n) = y1(n) + y2(n)
    N – 1
    N – 1
    = 1/N
    		x1(n – m) + 1/N
    		x2(n – m)
    m = 0
    m = 0

    N – 1
    = 1/N
    		[x1(n – m) + x2(n – m)]
    m = 0

    N – 1
    = 1/N
    		x(n – m)
    m = 0

    Therefore, this system is linear. (ii) y[n] = [x(n)]2 is non-linear already discussed in previous question.
    Hence, alternative (A) is the correct choice.

    Correct Option: A

    N – 1
    (i) y(n) = 1/N
    		x(n – m)
    m = 0

    N – 1
    Let y1(n) = 1/N
    		x1(n – m)
    m = 0

    N – 1
    and y2(n) = 1/N
    		x2(n – m)
    m = 0

    then y(n) = y1(n) + y2(n)
    N – 1
    N – 1
    = 1/N
    		x1(n – m) + 1/N
    		x2(n – m)
    m = 0
    m = 0

    N – 1
    = 1/N
    		[x1(n – m) + x2(n – m)]
    m = 0

    N – 1
    = 1/N
    		x(n – m)
    m = 0

    Therefore, this system is linear. (ii) y[n] = [x(n)]2 is non-linear already discussed in previous question.
    Hence, alternative (A) is the correct choice.

  1. The discrete time Fourier transform (DTFT) of x[n] = 2(3)n u[– n] is equal to—








  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: C

    NA