E t2 2 0 si t dt si 5 then we have a three hypothesis

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: Var[ni ] = N0 /2 and {ni }∞ 0 is an independent i= identically distributed (i.i.d.) sequence of random variables with Gaussian density functions. EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 46 / 93 Lecture Notes 7 Example 2: Additive White Gaussian Noise Let s0 (t ) = ϕ0 (t ) + 2ϕ1 (t ) s1 (t ) = 2ϕ0 (t ) + ϕ1 (t ) s2 (t ) = ϕ0 (t ) − 2ϕ1 (t ) Note that the energy of each of the three signals is the same, i.e. T2 2 0 si (t )dt = ||si || = 5. Then we have a three hypothesis testing problem. ∞ H0 : r (t ) = s0 (t ) + n(t ) = (s0,i + ni )ϕi (t ) i =0 ∞ H1 : r (t ) = s1 (t ) + n(t ) = (s1,i + ni )ϕi (t ) i =0 ∞ H2 : r (t ) = s2 (t ) + n(t ) = (s2,i + ni )ϕi (t ) i =0 EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 47 / 93 Lecture Notes 7 Example 1: Additive White Gaussian Noise 3 s (t) 0 2 s (t) 1 1 φ 1 0 −1 −2 −3 −3 s2(t) −2 −1 0 φ 1 2 3 0 EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 48 / 93 Lecture Notes 7 Example 1: Additive White Gaussian Noise 3 2 1 φ (t) 1 0 −1 −2 −3 −3 EECS 455 (Univ. of Michigan) −2 −1 0 φ0(t) Fall 2012 1 2 3 October 3, 2012 49 / 93 Lecture Notes 7 Example 1: Decision Rule The decision rule to minimize the average error probability is given as follows Decide Hi if πi pi (r) = max πj pj (r) j First let us consider the first L + 1 variables and normalize each side by the density function for the noise alone. The noise density function for L + 1 variables is p (L) (r) = EECS 455 (Univ. of Michigan) 1 2π N0 /2 N exp{− Fall 2012 1 L 2 rm } N0 2 2 m =0 October 3, 2012 50 / 93 Lecture Notes 7 Example 1: Decision Rule The optimal decision rule is equivalent to Decide Hi if πi pj (r) pi (r) = max πj . p(r) p(r) j As usual assume πi = 1/M . Then L (L) p0 (r) p(L) (r) = √ 1 2π N0 /2 exp{− 1 2 N0 2 [ i =0,1 (ri − s0,i )2 + L 2 i =2 ri ]} L 1 2π N0 /2 √ exp{− 1 2 = exp{− 1 [ N0 = exp{+ N0 2 1 2 i =0 ri + L 2 i =2 ri } 1 [2r1 + 4r2 − 5]}. N0 (ri − s0,i )2 − ri2 ]} i =0,1 Now since the above doesn’t depend on L we can let L → ∞ and the result is the same. EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 51 / 93 Lecture Notes 7 Example 1 (L) p (r) 1 p0 (r) ∆ = exp{+ [2r0 + 4r1 − 5]}. = lim 0L) ( (r) p (r) L→∞ p N0 Similarly 1 p1 (r) = exp{+ [4r0 + 2r1 − 5]} p (r) N0 1 p2 (r) = exp{+ [2r0 − 4r1 − 5]}. p (r) N0 EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 52 / 93 Lecture Notes 7 Example 1 ϕ2 (t ) s0 (t ) s1 (t ) ϕ1 (t ) s2 (t ) EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 53 / 93 Lecture Notes 7 Decision Regions 8 6 4 2 0 −2 −4 −6 −8 −8 −6 EECS 455 (Univ. of Michigan) −4 −2 0 Fall 2012 2 4 6 8 October 3, 2012 54 / 93 Lecture Notes 7 Likelihood Ratio for Real Signals in AWGN Assume two signals in Gaussian noise. H0 : r (t ) = s0 (t ) + n(t ) H1 : r (t ) = s1 (t ) + n(t ) Goal: Find decision rule to minimize the average error probability. Let n(t ) autocorrelation function R ((s, t ) = N0 δ(t − s). We assume that 2 n(t ) is a zero mean white Gaussian noise rando...
View Full Document

Ask a homework question - tutors are online