hw5-sol

# hw5-sol - EE 278 Statistical Signal Processing Handout#12...

This preview shows pages 1–4. Sign up to view the full content.

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE 278 October 30, 2009 Statistical Signal Processing Handout #12 Homework #5 Solutions 1. Additive-noise channel with path gain. Consider the additive noise channel shown in the figure below, where X and Z are zero mean and uncorrelated, and a and b are constants. a b X Z Y = b ( aX + Z ) Find the MMSE linear estimate of X given Y and its MSE in terms only of σ X , σ Z , a , and b . Solution (10 points) First we find the mean and variance of Y and its covariance with X . In the following we use the notation σ 2 X = P and σ 2 Z = N . E( Y ) = E( abX + bZ ) = ab E( X ) + b E( Z ) = 0 Var( Y ) = E( abX + bZ ) 2- (E( abX + bZ )) 2 = E( a 2 b 2 X 2 + 2 ab 2 XZ + b 2 Z 2 )- E( Y ) 2 = a 2 b 2 E( X 2 ) + 2 ab 2 E( X ) E( Z ) + b 2 E( Z 2 )- E( Y ) 2 = a 2 b 2 P + 0 + b 2 N- 0 = a 2 b 2 P + b 2 N Cov( X, Y ) = E[( X- E( X ))( Y- E( Y ))] = E( XY ) = E[ X ( abX + bZ )] = ab E( X 2 ) + b E( XZ ) = abP + b E( X ) E( Z ) = abP The MMSE linear estimate of X given Y is given by ˆ X = Cov( X, Y ) σ 2 Y ( Y- E( Y )) + E( X ) = abP a 2 b 2 P + b 2 N ( Y- E( Y )) + E( X ) = aP b ( a 2 P + N ) Y The MSE of the linear estimate is the minimum MMSE: MMSE = σ 2 X- Cov 2 ( X, Y ) σ 2 Y = P- a 2 b 2 P 2 a 2 b 2 P + b 2 N = a 2 P 2 + PN- a 2 P 2 a 2 P + N = PN a 2 P + N 2. Shot Noise Channel. Consider an additive noise channel with input signal X ∼ U(0 , 1) and output signal Y = X + Z , where the noise Z |{ X = x } ∼ N (0 , ax ), for some constant a > 0, i.e., the noise variance is proportional to the signal. Observing Y , find the minimum MSE linear estimate of X . Your answer should be in terms only of a . Solution (10 points) To find the minimum MSE linear estimate we need to find the means, variances and covariance of the signal and observation. Consider E( X ) = 1 2 . E( Y ) = E( X ) + E( Z ) = 1 2 + E X (E( Z | X )) = 1 2 + 0 . Var( X ) = 1 12 . E( Y 2 ) = E( X 2 ) + 2 E( XZ ) + E( Z 2 ) = 1 3 + 2 E X ( X E( Z | X )) + E X (E( Z 2 | X )) = 1 3 + 0 + E X ( aX ) = 1 3 + a 2 . Var( Y ) = 1 3 + a 2- 1 4 = 1 12 + a 2 . E( XY ) = E( X 2 ) + E X ( X E( Z | X )) = 1 3 + 0 . Cov( X, Y ) = 1 12 . Thus ˆ X = Cov( X, Y ) Var( Y ) ( Y- E( Y )) + E( X ) = 1 1 + 6 a parenleftbigg Y- 1 2 parenrightbigg + 1 2 = Y + 3 a 1 + 6 a . Page 2 of 14 EE 278, Autumn 2009 3. Worst noise distribution. Consider an additive noise channel with signal X ∼ N (0 , P ) and noise Z with zero mean and variance N . Assume X and Z are independent and the output is Y = X + Z . Specify a distribution of Z that maximizes the minimum MSE of estimating X given Y , i.e., the distribution of the worst noise Z that has the given mean and variance. You need to justify your answer. Solution (5 points) Note that the MMSE is always less than or equal to the best MSE of the linear estimate. Thus for our setting, for any distribution on Z with average power N E(Var( X | Y )) ≤ PN P + N ....
View Full Document

## This note was uploaded on 11/28/2009 for the course EE 278 taught by Professor Balajiprabhakar during the Fall '09 term at Stanford.

### Page1 / 14

hw5-sol - EE 278 Statistical Signal Processing Handout#12...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online