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. A Linear Time Invariant system with an impulse response h(t) produces output y(t) when input x(t)is applied. When the input x(t - τ) is applied to a system with impulse response h(t - τ),then output will be








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    Laplace t ransform of impulse response of a system is its transfer function.
    ∴ G1 (s) = L[h(t)] = H(s)
    ⇒ y1 (s) = H(s) X(s)
    G1 (s) = L[h(t – τ)] = e– sτH(s)
    ⇒ y2 (s) = G2(s) L[x(t – τ)]
    = G2 (s) e– sτ X(s)
    = e–2sτ H(s)X(s)
    = e–2sτ Y(s)
    ∴ y2 (t) = y(t – 2τ)
    Alternately
    x (t) → h (t) → y(t) = x(t). h(t) ...(i)
    x (t – τ) → h(t – z) → y' ...(ii)
    For system 1.
    y(t) = x(t) h(t)
    Z-transform y(z) = x(z) H(z) ...(iii)
    For system 2.
    y'(t) = x(t – τ) h(t – τ)
    Z-transform y'(τ) = z–1 x(z) z–1
    H(z) = z–2 x(z) H(z) ...(putting from (iii)
    y'(z) = z–2y(z)
    ∴ Inverse Z - transform = y'(t) = y(t– 2τ)

    Correct Option: D

    Laplace t ransform of impulse response of a system is its transfer function.
    ∴ G1 (s) = L[h(t)] = H(s)
    ⇒ y1 (s) = H(s) X(s)
    G1 (s) = L[h(t – τ)] = e– sτH(s)
    ⇒ y2 (s) = G2(s) L[x(t – τ)]
    = G2 (s) e– sτ X(s)
    = e–2sτ H(s)X(s)
    = e–2sτ Y(s)
    ∴ y2 (t) = y(t – 2τ)
    Alternately
    x (t) → h (t) → y(t) = x(t). h(t) ...(i)
    x (t – τ) → h(t – z) → y' ...(ii)
    For system 1.
    y(t) = x(t) h(t)
    Z-transform y(z) = x(z) H(z) ...(iii)
    For system 2.
    y'(t) = x(t – τ) h(t – τ)
    Z-transform y'(τ) = z–1 x(z) z–1
    H(z) = z–2 x(z) H(z) ...(putting from (iii)
    y'(z) = z–2y(z)
    ∴ Inverse Z - transform = y'(t) = y(t– 2τ)

  1. x(t) is a positive rectangular pulse from t = – 1 to t = +1 with unit height as shown in the figure. The value of








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    Correct Option: D

  1. At t = 0, the function
    ƒ(t) =
    sin t
    		 has
    t








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    Correct Option: D

  1. The second harmonic component of the periodic waveform given in the figure has an amplitude of








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    Given signal is, half wave symmetry, so even harmonics will be zero.
    Hence 2nd harmonic = 0

    Correct Option: A


    Given signal is, half wave symmetry, so even harmonics will be zero.
    Hence 2nd harmonic = 0

  1. The period of the signal
    x(t) = 8 sin0.8 πt +
    π
    		 is
    4








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    Given: x(t) = 8sin0.8 πt -
    π
    4

    Compare with x(t) = A0 sin (ωt + ω), we get
    ω = 0.8π = 2π/T
    ∴ T =
    2
    		 = 2.5 sec
    0.8

    Correct Option: B

    Given: x(t) = 8sin0.8 πt -
    π
    4

    Compare with x(t) = A0 sin (ωt + ω), we get
    ω = 0.8π = 2π/T
    ∴ T =
    2
    		 = 2.5 sec
    0.8