adaptive predictor computer experiments

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 Z - 1 Σ u(n) u(n-1) ) 1 ( ) ( ˆ  n u n w ) ( ˆ  n w f(n) - + LMS Algorithm - Computer Experiments Experiment 1: Adaptive Predictor  AR(1) Process Autoregr essive process of order 1 is dened as u ( n ) =au ( n 1 ) +v ( n)  !here" 'a' is the parameter of A#$1% process 'v(n)' is &'( !ith variance σ v 2  Adaptive F irst order Predictor Fig 1: Adaptive First order Predictor 

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Matlab program for adaptive equalizer

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Page 1: Adaptive Predictor Computer Experiments

7/18/2019 Adaptive Predictor Computer Experiments

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Z-

1

Σ

u(n) u(n-1)

)1()(ˆ   −nunw

)(ˆ   nw

f(n)-

+

LMS Algorithm - ComputerExperiments

Experiment 1: Adaptive Predictor

 AR(1) Process

Autoregressive process of order 1 is dened as

u (n )=−au (n−1 )+v (n)  

!here"

'a' is the parameter of A#$1% process

'v(n)' is &'( !ith variance σ v2

 Adaptive First order Predictor 

Fig 1: Adaptive First order Predictor 

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LMS !eight update e)uation is

w (n+1 )=w (n )+ μu (n−1 ) f (n)

f  (n )=u (n )−w (n ) u(n−1)

Experiment Results

Experiment is carried out !ith the follo!ing cases

Case1:

a* -,

.ariance of u$n%* ,/02

Case2:

a* +,

.ariance of u$n%* ,3

 4he Step si5e parameter is ta6en as µ*,3 and initial condition w (0 )=0

Steps:

1, 'enerate A# process

u (n )=−au (n−1 )+v (n)  

, 7nitiali5e w (0 )=0

/, 8pdate w (n )

w (n+1 )=w (n )+ μu (n−1 ) f (n)

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f  (n )=u (n )−w (n)u(n−1)

9, #epeat step / for 3 iterations,

3, #epeat steps 1 to 9 for 1 times and compute ensem:le average of 

w (n )

;igure sho!s the transient :ehavior of w (n ) , 7t also sho!s the

 E (w (n ) )  o:tained :< the ensem:le averaging of 1 independent trials,

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 4he experimental learning curves of adaptive rst order prediction for

var<ing step si5e parameter is sho!n in ;ig,

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Problem 5.21

 AR(2) Process

Autoregressive process of order is dened as

u (n )=−a1u (n−1 )±a

2u (n−1 )+v (n)  

a1=0.1

a2=−0.8

a) Noise variance   σ v2

 such that σ u2

=1

σ v2=

(1−a2) ((1+a2 )2−a

1

2)(1+a2 )

  σ u2

σ v2=0.27

Matlab Code for different realization of u(n)

var_v=(1-a2)*((1+a2)^2-a1^2)/(1+a2);

 % initial values of u(n)u(1)=var_v*randn(1,1); %u(2)=-a1*u(1)+var_v*randn(1,1);

 for n=3:N

  u(n)=-a1*u(n-1)-a2*u(n-2)+var_v*randn(1,1);end

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