# Of michigan fall 2012 october 3 2012 23 93 lecture

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Unformatted text preview: E , + E , − E ) √ √ √ s2 = (+ E , − E , + E ) √ √ √ s3 = (+ E , − E , − E ) √ √ √ s4 = (− E , + E , + E ) √ √ √ s5 = (− E , + E , − E ) √ √ √ s6 = (− E , − E , + E ) √ √ √ s7 = (− E , − E , − E ) EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 23 / 93 Lecture Notes 7 Example 5: Frequency Orthogonal s0 (t ) s4 (t ) T t T T t T t s5 (t ) T t s2 (t ) s6 (t ) T t s3 (t ) s7 (t ) T EECS 455 (Univ. of Michigan) t T s1 (t ) t t Fall 2012 October 3, 2012 24 / 93 Lecture Notes 7 Example 6: Walsh/Hadamard Orthogonal φ0 (t ) φ1 (t ) Tt φ2 (t ) φ3 (t ) Tt EECS 455 (Univ. of Michigan) Tt Fall 2012 Tt October 3, 2012 25 / 93 Lecture Notes 7 Decomposition of Noise For any complete orthonormal set of signals ϕ0 (t ), ϕ1 (t ), ... we can represent a noise process as random variables and deterministic orthonormal functions ∞ n(t ) = nm ϕm (t ), nm = n(t )ϕ∗ (t )dt . m m =0 The noise variables nm and nl are independent if n(t ) is white Gaussian noise. EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 26 / 93 Lecture Notes 7 Decomposition of Noise Consider the case of real white Gaussian noise with power spectral density N0 /2. E [nm nl ] = E [ n(t )ϕm (t )dt n(s)ϕl (s)ds = E [n(t )n(s)]ϕm (t )ϕl (s)dtds = N0 δ(t − s)ϕm (t )ϕl (s)dtds 2 = = N0 ϕm (t )ϕl (t )dt 2 N0 2, 0, m=l m=l Thus nm and nl are uncorrelated for m = l . Because they are also Gaussian they are independent. EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 27 / 93 Lecture Notes 7 Decomposition of Noise 3 n(t), \hat n(t) 2 1 0 −1 −2 −3 0 0.1 0.2 0.3 0.4 0.5 time 0.6 0.7 0.8 0.9 1 0 10 20 30 40 50 l 60 70 80 90 100 0.4 nl 0.2 0 −0.2 −0.4 EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 28 / 93 Lecture Notes 7 Decomposition of Signal and Noise Consider a communication system that transmits one of M signals. s0 (t ), ..., sM −1 (t ) in additive white Gaussian noise. s Then given si (t ) was transmitted the received signal is r (t ) = si (t ) + n(t ) ∞ = (si ,m + nm )ϕm (t ) m =0 Deﬁne rm = si ,m + nm . Then ∞ r (t ) = rm ϕm (t ) m =0 We can determine the (random) variable rm by rm = EECS 455 (Univ. of Michigan) r (t )ϕ∗ (t )dt m Fall 2012 October 3, 2012 29 / 93 Lecture Notes 7 Example 1 s0 (t ) s1 (t ) Let ϕl (t ) = 1 T = −ApT (t ) exp{j 2π lf0 t }pT (t ) where f0 = 1/T . Note that ϕ0 (t ) = EECS 455 (Univ. of Michigan) = ApT (t ) 1 T pT (t ). Fall 2012 October 3, 2012 30 / 93 Lecture Notes 7 Example 1 With this set of orthonormal functions we can write √ √ s0 (t ) = E ϕ0 (t ), s1 (t ) = − E ϕ0 (t ) ∞ n(t ) = nm ϕm (t ) m =0 ∞ r (t ) = rm ϕm (t ) m =0 rm = r (t )ϕ∗ (t )dt = m = si ,m + nm , EECS 455 (Univ. of Michigan) (si (t ) + n(t ))ϕ∗ (t )dt m m = 0, 1, 2, ... Fall 2012 October 3, 2012 31 / 93 Lecture Notes 7 Example 1 s0 (t ) s1 (t ) √ √ r0 = E + n0 r0 = − E + n0 r1 = n1 r1 = n1 r2 = n2 r2 = n2 EECS 455 (Univ. of Michigan) Fall 2012 October 3, 2012 32 / 93 Lecture Notes 7 Example 2: 8PSK ϕ0 (t ) = 2/T cos(2π fc t )pT (t ) ϕ1 (t ) = 2/T sin(2π fc t )...
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## This note was uploaded on 02/12/2014 for the course EECS 455 taught by Professor Stark during the Fall '08 term at University of Michigan.

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