[next] [prev] [prev-tail] [tail] [up]

Homework 9

[PDF]

ECE 460: Communication and Information Theory
Prof. B.-P. Paris
Homework 9
Solution

  1. Problem 4.35
    mX (t) = E [A + Bt] = E [A ] + E[B ]t = 0

    where the last equality follows from the fact that A, B are uniformly distributed over [-1 1] so that E[A] = E[B] = 0.

    RX (t1,t2) = E [X (t1)X (t2)] = E [(A + Bt1 )(A + Bt2)] = E [A2] + E[AB ]t2 + E[BA ]t1 + E[B2 ]t1t2
    The random variables A, B are independent so that E[AB] = E[A]E[B] = 0. Furthermore
    ∫ 1 E [A2 ] = E[B2 ] = x21-dx = 1-x3|1 = 1- - 1 2 6 -1 3

    Thus

    1 1 RX (t1,t2) = --+ -t1t2 3 3

    __________________________________________________________________________________________________________________________________________

  2. Problem 4.40
    1) f(τ) cannot be the autocorrelation function of a random process for f(0) = 0 < f(1∕4f0) = 1. Thus the maximum absolute value of f(τ) is not achieved at the origin τ = 0.

    2) f(τ) cannot be the autocorrelation function of a random process for f(0) = 0 whereas f(τ)⁄=0 for τ⁄=0. The maximum absolute value of f(τ) is not achieved at the origin.

    3) f(0) = 1 whereas f(τ) > f(0) for |τ| > 1. Thus f(τ) cannot be the autocorrelation function of a random process.

    4) f(τ) is even and the maximum is achieved at the origin (τ = 0). We can write f(τ) as

    f(τ) = 1.2Λ (τ) - Λ(τ - 1) - Λ(τ + 1)

    Taking the Fourier transform of both sides we obtain

    2 2 ( -j2πf j2πf) 2 S (f) = 1.2sinc (f ) - sinc (f) e + e = sinc (f )(1.2 - 2 cos(2 πf))

    As we observe the power spectrum S(f) can take negative values, i.e. for f = 0. Thus f(τ) can not be the autocorrelation function of a random process.

    ____________________________________________________________________________________________________

  3. Problem 4.52
    The impulse response of a delay line that introduces a delay equal to Δ is h(t) = δ(t - Δ). The output autocorrelation function is
    RY (τ) = RX (τ) ⋆ h(τ) ⋆ h(- τ )

    But,

    ∫ ∞ h(τ) ⋆ h(- τ) = -∞ δ(- (t - Δ ))δ(τ - (t - Δ ))dt ∫ ∞ = δ(t - Δ )δ(τ - (t - Δ))dt ∫ -∞∞ = δ(t′)δ(τ - t′)dt′ = δ(τ) -∞
    Hence,
    RY (τ) = RX (τ ) ⋆ δ(τ) = RX (τ)

    This is to be expected since a delay line does not alter the spectral characteristics of the input process.

    ____________________________________________________________________________________________________

[next] [prev] [prev-tail] [front] [up]