chapter 7 generating and processing random signals

Post on 05-Jan-2016

73 Views

Category:

Documents

2 Downloads

Preview:

Click to see full reader

DESCRIPTION

Chapter 7 Generating and Processing Random Signals. 第一組 電機四 B93902016 蔡馭理 資工四 B93902076 林宜鴻. Outline. Outline. Stationary and Ergodic Process Uniform Random Number Generator Mapping Uniform RVs to an Arbitrary pdf Generating Uncorrelated Gaussian RV Generating correlated Gaussian RV - PowerPoint PPT Presentation

TRANSCRIPT

1

Chapter 7Generating and Processing

Random Signals

第一組電機四 B93902016 蔡馭理資工四 B93902076 林宜鴻

2

Outline

Stationary and Ergodic ProcessUniform Random Number GeneratorMapping Uniform RVs to an Arbitrary pdfGenerating Uncorrelated Gaussian RVGenerating correlated Gaussian RVPN Sequence GeneratorsSignal processing

Outline

3

Random Number Generator

Noise, interferenceRandom Number Generator- computation

al or physical device designed to generate a sequence of numbers or symbols that lack any pattern, i.e. appear random, pseudo-random sequence

MATLAB - rand(m,n) , randn(m,n)

4

Stationary and Ergodic Process

strict-sense stationary (SSS)wide-sense stationary (WSS) Gaussian

SSS =>WSS ; WSS=>SSSTime average v.s ensemble average The ergodicity requirement is that the ensemble

average coincide with the time averageSample function generated to represent signals,

noise, interference should be ergodic

5

Time average v.s ensemble average

Time average ensemble average

6

Example 7.1 (N=100)

0 0.5 1 1.5 2-1

0

1

x(t)

0 0.5 1 1.5 2-0.5

0

0.5

x ensem

ble-

avar

age(

t)0 0.5 1 1.5 2

-1

0

1

y(t)

0 0.5 1 1.5 2-1

0

1

y ensem

ble-

avar

ag(t

)

0 0.5 1 1.5 2-2

0

2

z(t)

0 0.5 1 1.5 2-2

0

2z ens

embl

e-av

arag

(t)

)2cos()(),( iii φπftμ1Aξtx

)2cos(),( ii φπftAξtx

7

Uniform Random Number Genrator

Generate a random variable that is uniformly distributed on the interval (0,1)

Generate a sequence of numbers (integer) between 0 and M and the divide each element of the sequence by M

The most common technique is linear congruence genrator (LCG)

8

Linear Congruence

LCG is defined by the operation:

xi+1=[axi+c]mod(m)

x0 is seed number of the generator

a, c, m, x0 are integer

Desirable property- full period

9

Technique A: The Mixed Congruence Algorithm

The mixed linear algorithm takes the form:

xi+1=[axi+c]mod(m)

- c≠0 and relative prime to m

- a-1 is a multiple of p, where p is the

prime factors of m

- a-1 is a multiple of 4 if m is a

multiple of 4

10

Example 7.4

m=5000=(23)(54)c=(33)(72)=1323a-1=k1‧2 or k2‧5 or 4‧k3 so, a-1=4‧2‧5‧k =40kWith k=6, we have a=241

xi+1=[241xi+ 1323]mod(5000)We can verify the period is 5000, so it’s full

period

11

Technique B: The Multiplication Algorithm With Prime Modulus

The multiplicative generator defined as :

xi+1=[axi]mod(m)

- m is prime (usaually large)

- a is a primitive element mod(m)

am-1/m = k =interger

ai-1/m ≠ k, i=1, 2, 3,…, m-2

12

Technique C: The Multiplication Algorithm With Nonprime Modulus

The most important case of this generator having m equal to a power of two :

xi+1=[axi]mod(2n)

The maximum period is 2n/4= 2n-2

the period is achieved if

- The multiplier a is 3 or 5

- The seed x0 is odd

13

Example of Multiplication Algorithm With Nonprime Modulus

a=3

c=0

m=16

x0=1

0 5 10 15 20 25 30 351

2

3

4

5

6

7

8

9

10

11

14

Testing Random Number Generator

Chi-square test, spectral test……Testing the randomness of a given sequen

ceScatterplots

- a plot of xi+1 as a function of xi

Durbin-Watson Test

-

N

n

N

n

nXN

nXnXN

2

2

2

2

][)/1(

])1[][()/1(D

15

ScatterplotsExample 7.5

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(i) rand(1,2048)

(ii)xi+1=[65xi+1]mod(2048)

(iii)xi+1=[1229xi+1]mod(2048)

16

Durbin-Watson Test (1)

N

n

N

n

nXN

nXnXND

2

2

2

2

][)/1(

])1[][()/1(

}({1

}{

}({ 22x

2

2

Y)-XEXE

Y)-XED

Let X = X[n] & Y = X[n-1]

ZρρXY 21 11 ρ

Let

Assume X[n] and X[n-1] are correlated and X[n] is an ergodic process

17

Durbin-Watson Test (2)

222222

)1()1()1(2)1(1

ZρXZρρXρEσ

D

)1(2)1()1(

2

2222

ρσ

σρσρD

X and Z are uncorrelated and zero mean

D>2 – negative correlation

D=2 –- uncorrelation (most desired)

D<2 – positive correlation

18

Example 7.6

rand(1,2048) - The value of D is 2.0081 and ρ is 0.0041.

xi+1=[65xi+1]mod(2048) - The value of D is 1.9925 and ρ is 0.0037273.

xi+1=[1229xi+1]mod(2048) - The value of D is 1.6037 and ρ is 0.19814.

19

Minimum Standards

Full period Passes all applicable statistical tests for

randomness.Easily transportable from one computer to

anotherLewis, Goodman, and Miller Minimum

Standard (prior to MATLAB 5)xi+1=[16807xi]mod(231-1)

20

Mapping Uniform RVs to an Arbitrary pdf

The cumulative distribution for the target random variable is known in closed form – Inverse Transform Method

The pdf of target random variable is known in closed form but the CDF is not known in closed form – Rejection Method

Neither the pdf nor CDF are known in closed form – Histogram Method

21

Inverse Transform Method

CDF FX(X) are known in closed form

U = FX (X) = Pr { X ≦ x }

X = FX-1

(U)

FX (X) = Pr { FX-1

(U) ≦ x } = Pr {U ≦ FX (x) }= FX (x)

FX(x)

1

U

FX-1(U) x

22

Example 7.8 (1)

Rayleigh random variable with pdf –

Setting FR(R) = U

)(2

exp)(2

2

2ru

σ

r

σ

rrf R

2

2

2

2

0 2 2exp1)(

σ

rdy

yexp

σ

yrF

r

R

r

2

2

2exp1

23

Example 7.8 (2)

∵ RV 1-U is equivalent to U (have same pdf) ∴

Solving for R gives

[n,xout] = hist(Y,nbins) - bar(xout,n) - plot the histogram

r

2

2

2exp

)ln(2R 2 Uσ

24

Example 7.8 (3)

0 1 2 3 4 5 6 7 8 90

500

1000

1500

Num

ber

of S

ampl

es

Independent Variable - x

0 1 2 3 4 5 6 7 80

0.1

0.2

0.3

0.4

Pro

babi

lity

Den

sity

Independent Variable - x

true pdf

samples from histogram

25

The Histogram Method

CDF and pdf are unknownPi = Pr{xi-1 < x < xi} = ci(xi-xi-1)

FX(x) = Fi-1 + ci(xi-xi-1)

FX(X) = U = Fi-1 + ci(X-xi) more samples

more accuracy!

1

1111 }Pr{

i

jiii PXXF

)(1

11 ii

i FUc

xX

26

Rejection Methods (1)

Having a target pdf MgX(x) ≧ fX(x), all x

otherwise

0

,0

a/)(

axMb xMg X

}max{ (x)fa

Mb X

axx+dx

M/a=b

1/a

0

0

MgX(x)

fX(x)

gX(x)

27

Rejection Methods (2)

Generate U1 and U2 uniform in (0,1)

Generate V1 uniform in (0,a), where a is the maximum value of X

Generate V2 uniform in (0,b), where b is at least the maximum value of fX(x)

If V2 ≦ fX(V1), set X= V1. If the inequality is not satisfied, V1 and V2 are discarded and the process is repeated from step 1

28

Example 7.9 (1)

R0

0

MgX(x)

fX(x)

gX(x)

πRR

M 4

R

1

otherwise0,

Rx0xRπRxf X

222

4)(

29

Example 7.9 (2)

0 1 2 3 4 5 6 70

50

100

150

Num

ber

of S

ampl

es

Independent Variable - x

0 1 2 3 4 5 6 70

0.05

0.1

0.15

0.2

Pro

babi

lity

Den

sity

Independent Variable - x

true pdf

samples from histogram

30

Generating Uncorrelated Gaussian RV

Its CDF can’t be written in closed form , so Inverse method can’t be used and rejection method are not efficient

Other techniques

1.The sum of uniform method

2.Mapping a Rayleigh to Gaussian RV

3.The polar method

31

The Sum of Uniforms Method(1)

1.Central limit theorem2.See next

.

3.

0

1( )

2

N

ii

Y B U

iU 1,2..,i N represent independent uniform R.V

B is a constant that decides the var of Y

N Y converges to a Gaussian R.V.

32

The Sum of Uniforms Method(2)

Expectation and Variance

We can set to any desired valueNonzero at

1{ }

2iE U 0

1{ } ( { } ) 0

2

N

ii

E Y B E U

1/ 2 2

1/ 2

1 1var{ }

2 12iU x dx

2

2 2

1

1var{ }

2 12

N

y ii

NBB U

12yB

N

123

2y y

NN

N

33

The Sum of Uniforms Method(3)

Approximate GaussianMaybe not a realistic situation.

34

Mapping a Rayleigh to Gaussian RV(1)

Rayleigh can be generated by

U is the uniform RV in [0,1] Assume X and Y are indep. Gaussian RV

and their joint pdf

22 lnR U

2 2

2 2

1 1( , ) exp( ) exp( )

2 22 2XY

x xf x y

2 2

2 2

1exp( )

2 2

x y

35

Mapping a Rayleigh to Gaussian RV(2)

Transform

let and

and

cosx r siny r 2 2 2x y r 1tan ( )

y

x

( , ) ( , )R R XY XYf r dA f x y dA

/ /( , )

/ /( , )XY

R

dx dr dx ddA x yr

dy dr dy ddA r

2

2 2( , ) exp( )

2 2R

r rf r

36

Mapping a Rayleigh to Gaussian RV(3)

Examine the marginal pdf

R is Rayleigh RV and is uniform RV

2 22

2 2 2 20( ) exp( ) exp( )

2 2 2R

r r r rf r d

0 r

2

2 20

1( ) exp( )

2 2 2

r rf dr

0 2

cosX R 2

1 22 ln( ) cos 2X U U

sinY R 21 22 ln( ) sin 2Y U U

37

The Polar Method

From previous

We may transform

21 22 ln( ) cos 2X U U 2

1 22 ln( ) sin 2Y U U

2 2 2 ( )s R u v R s

1cos 2 cosu u

UR s

2sin 2 sinv v

UR s

22 2

1 2

2 ln( )2 ln( ) cos 2 2 ln( )( )

u sX U U s u

ss

22 2

1 2

2 ln( )2 ln( ) sin 2 2 ln( )( )

v sY U U s v

ss

38

The Polar Method Alothgrithm

1.Generate two uniform RV , and and they are all on the interval (0,1) 2.Let and , so they are independent and uniform on (-1,1)3.Let if continue , else back to step24.Form 5.Set and

1U 2U

1 12 1V U 2 22 1V U

2 21 2S V V 1S

2( ) ( 2 ln ) /A S S S

1( )X A S V 2( )Y A S V

39

Establishing a Given Correlation Coefficient(1)

Assume two Gaussian RV X and Y , they are zero mean and uncorrelated

Define a new RV We also can see Z is Gaussian RV Show is correlation coefficient relating

X and Z

21Z X Y | | 1

40

Establishing a Given Correlation Coefficient(2)

Mean , Variance , Correlation coefficient { } { } { } 0E Z E X E Y

2 2 2 2 2{ } 2 1 { } (1 ) { }E X E XY E Y

{ } { } { } 0E XY E X E Y 2 2 2 2 2 2 2{ } ( { }) { } { }X Y E X E X E X E Y

2 2 2 2 2(1 )

2 2 2{[ 1 ] }Z E X Y

41

Establishing a Given Correlation Coefficient(3)

Covariance between X and Z

as desired

{ } { [ (1 ) ]}E XZ E X X Y

2{ } (1 ) { }E X E XY

2 2{ }E X

2

2

{ }XZ

X Z

E XZ

42

Pseudonoise(PN) Sequence Genarators

PN generator produces periodic sequence that appears to be random

Generated by algorithm using initial seedAlthough not random , but can pass man

y tests of randomnessUnless algorithm and seed are known , t

he sequence is impractical to predict

43

PN Generator implementation

44

Property of Linear Feedback Shift Register(LFSR)

Nearly random with long periodMay have max period If output satisfy period , is called

max-length sequence or m-sequenceWe define generator polynomial as

The coefficient to generate m-sequence can always be found

45

Example of PN generator

46

Different seed for the PN generator

47

Family of M-sequences

48

Property of m-sequence

Has ones , zerosThe periodic autocorrelation of a m-se

quence is

If PN has a large period , autocorrelation function approaches an impulse , and PSD is approximately white as desired

1

49

PN Autocorrelation Function

50

Signal Processing

Relationship

1.mean of input and output

2.variance of input and output

3.input-output cross-correlation

4.autocorrelation and PSD

51

Input/Output Means

Assume system is linearconvolution

Assume stationarity assumption

We can getand

[ ] [ ] [ ]k

k

y n h k x n k

{ [ ]} { [ ] [ ]} [ ] { [ ]}

k k

E y n E h k x n k h k E x n k

{ [ ]} { [ ]}E x n k E x n

{ } { } [ ]k

E y E x h k

[ ] (0)

k

h k H

{ } (0) { }E y H E x

52

Input/Output Cross-Correlation

The Cross-Correlation is defined by

This use is used in the development of a number of performance estimators , which will be developed in chapter 8

{ [ ] [ ]} [ ] { [ ] [ ] [ ]}xyj

E x n y n m R m E x n h j x n j m

[ ] [ ] { [ ]}xy

j

R m h j E x n j m

[ ] [ ]xxj

h j R m j

53

Output Autocorrelation Function(1)

Autocorrelation of the output

Can’t be simplified without knowledge of the Statistics of

{ [ ] [ ]} [ ]yyE y n y n m R m

{ [ ] [ ] [ ] [ ]}j k

E h j x n j h k x n k m

[ ] [ ] [ ] { [ ] [ ]}yy

j k

R m h j h k E x n j x n m k

[ ] [ ] ( )xxj k

h j h k R m k j

[ ]x n

54

Output Autocorrelation Function(2)

If input is delta-correlated(i.e. white noise)

substitute previous equation

2

[ ] { [ ] [ ]}0

xxxR m E x n x n m

20

[ ]0 x

mm

m

[ ]yyR m

2[ ] [ ] [ ] ( )yy xj k

R m h j h k m k j

2 [ ] [ ]x

j

h j h j m

55

Input/Output Variances

By definition Let m=0 substitute into

But if is white noise sequence

2[0] { [ ]}yyR E y n

[ ]yyR m

2 [0] [ ] [ ] [ ]y yy xxj k

R h j h k R j k

[ ]x n

2 2 2[0] [ ]y yy xj

R h j

56

The EndThanks for listening

top related