wss prb5

wss prb5 - Massachusetts Institute of Technology Department...

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Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.011: Introduction to Communication, Control and Signal Processing QUIZ , April 21, 2010 ANSWER BOOKLET Your Full Name: Recitation Time : o’clock This quiz is closed book , but 3 sheet of notes are allowed. Calculators will not be necessary and are not allowed. Check that this ANSWER BOOKLET has pages numbered up to 14. The booklet contains spaces for all relevant reasoning and answers. Neat work and clear explanations count; show all relevant work and reasoning! You may want to Frst work things through on scratch paper and then neatly transfer to this booklet the work you would like us to look at. Let us know if you need additional scratch paper. Only this booklet will be considered in the grading; no additional an- swer or solution written elsewhere will be considered. Absolutely no exceptions! There are 4 problems, weighted as shown . (The points indicated on the following pages for the various subparts of the problems are our best guess for now, but may be modiFed slightly when we get to grading.) Problem Your Score 1 (6 points) 2 (15 points) 3 (15 points) 4 (14 points) Total (50 points) 1 SOLUTIONS 2
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Problem 1 (6 points) For each of the following statements, specify whether the statement is true or false and give a brief (at most a few lines) justi±cation or counterexample. (1a) (1.5 points) If X and Y are independent random variables, the unconstrained MMSE estimator Y of Y given X = x is Y = µ Y . True : The unconstrained MMSE estimator takes the form E [ Y | X ] = ± yf Y | X dy . Since Y and X are independent, we know that f Y | X = f Y . This implies E [ Y | X ] = µ Y = ± yf Y dy . (1b) (1.5 points) The following random process is strict sense stationary: x ( t ) = A where A is a continuous random variable with a pdf uniform between ± 1. True : Intuitively, you can conclude that the process x ( t ) is strict sense stationary because there is no way to tell where on the time axis you are by looking at any set of samples, no matter at what times they are taken. Mathematically, this intuition is captured by saying that PDF’s of all orders are shift invariant, or f x ( t 1 ) ,x ( t 2 ) ,...,x ( t N ) = f x ( t 1 + τ ) ,x ( t 2 + τ ) ,...,x ( t N + τ ) for all N and τ . (1c) (1.5 points) The following random process is ergodic in the mean: x ( t ) = A where A is a continuous random variable with a pdf uniform between ± 1. False : The process is not ergodic in the mean because the ensemble mean does not equal the time-average of a realization of the process x ( t ). The ensemble mean of the process x ( t ) is 0. The time-average of a realization of the process x ( t ) is the particular value of A obtained in that realization 2
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(1d) (1.5 points) If the input to a stable LTI system is WSS then the output is guaranteed to be WSS. True
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wss prb5 - Massachusetts Institute of Technology Department...

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