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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).

User Merle
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Answer:

Cov(X, Y) =0.029.

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.

User Don Omondi
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