Tutorial5_solution - 1 Tutorial 5 Random Signal Analysis...

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Principles of Communications 1 Tutorial 5 Random Signal Analysis
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Principles of Communications 2 Problem 1 A random process is defined by where Θ is a random variable uniformly distributed on (0, 2 π ). Determine the mean and the autocorrelation function of X ( t ). X ( t ) = A cos(2 π f 0 t + Θ ) R X ( t 1 , t 2 ) μ X ( t )
Image of page 2
Principles of Communications We note that at time t is a function of the random variable Θ . Therefore, we have By definition, the mean of X ( t ) is given by 3 Solution μ X ( t ) We know from probability theory that a function g ( X ) of a random variable X is itself a random variable. The expected value of g ( X ) is μ X ( t ) = E [ X ( t )] = E [ A cos(2 π f 0 t + Θ )] E [ g ( X )] = g ( x ) f X ( x ) dx −∞ E [ A cos(2 π f 0 t + Θ )] = A cos(2 π f 0 t + θ ) f Θ ( θ ) d θ 0 2 π A cos(2 π f 0 t + Θ )
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Principles of Communications 4 Solution Given that μ X ( t ) = A cos(2 π f 0 t + θ ) 1 2 π d θ 0 2 π = 0 We observe that in this case is independent of t. f Θ ( θ ) = 1 2 π 0 < θ < 2 π 0 otherwise we obtain μ X ( t )
Image of page 4
Principles of Communications 5 Solution R X ( t 1 , t 2 ) = E [ X ( t 1 ) X ( t 2 )]
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern