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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.
 Fall '08
 Stark

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