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

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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...
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