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5 March, 22:26

In electrical engineering, the unwanted "noise" in voltage or current signals is often modeled by a Gaussian (i. e., normal) distribution. Suppose that the noise in a particular voltage signal has a constant mean of 0.9 V, and that two noise instances sampled τ seconds apart have a bivariate normal distribution with covariance equal to 0.04e-jτj/10. Let X and Y denote the noise at times 3 s and 8 s, respectively. (a) Determine Cov (X, Y).

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  1. 5 March, 22:30
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    Cov (X, Y) = 0.029.

    Step-by-step explanation:

    Given that:

    The noise in a particular voltage signal has a constant mean of 0.9 V. that is μ = 0.9V ... (1)

    Also, the two noise instances sampled τ seconds apart have a bivariate normal distribution with covariance.

    0.04e-jτj/10 ... (2)

    Having X and Y denoting the noise at times 3 s and 8 s, respectively, the difference of time = 8-3 = 5seconds.

    That is, they are 5 seconds apart,

    τ = 5 seconds ... (3)

    Thus,

    Cov (X, Y), for τ = 5seconds = 0.04e-5/10

    = 0.04e-0.5 = 0.04/√e

    = 0.04/1.6487

    = 0.0292

    Thus, Cov (X, Y) = 0.029.
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