chapter 8: response of linear systems to random inputsbazuinb/ece3800cmcg/notes8...0 rxy rxx 1 h 1 d...

Notes and figures are based on or taken from materials in the course textbook: Probabilistic Methods of Signal and System Analysis (3rd ed.) by George R. Cooper and Clare D. McGillem; Oxford Press, 1999. ISBN: 0-19-512354-9.

B.J. Bazuin, Spring 2015 1 of 24 ECE 3800

Chapter 8: Response of Linear Systems to Random Inputs

8-1 Introduction

8-2 Analysis in the Time Domain

8-3 Mean and Variance Value of System Output

8-4 Autocorrelation Function of System Output

8-5 Crosscorrelation between Input and Output

8-6 Example of Time-Domain System Analysis

8-7 Analysis in the Frequency Domain

8-8 Spectral Density at the System Output

8-9 Cross-Spectral Densities between Input and Output

8-10 Examples of Frequency-Domain Analysis

8-11 Numerical Computation of System Output

Concepts:

Linear Systems Analysis in the Time Domain Mean and Variance Value of System Output Autocorrelation Function of System Output Crosscorrelation between Input and Output Example of Time-Domain System Analysis Analysis in the Frequency Domain Spectral Density at the System Output Frequency Limited Filtering Bandlimited White Noise SIMO Systems Cross-Spectral Densities between Input and Output Examples of Frequency-Domain Analysis Numerical Computation of System Output Simulating Stochastic System Responses



Chapter 8: Response of Linear Systems to Random Inputs

Linear transformation of signals: convolution in the time domain

txthty

th ty

System

tx

Linear transformation of signals: multiplication in the Laplace domain

sXsHsY

sX sH sY

The convolution Integrals (applying a causal filter)

0

dhtxty or

t

dxthty

Comment: Dr. Bazuin will use causal filters, defined as

tth

tthth

0,0,0

and when doing so typically prefers doing convolutions as

0

dhtxty



The Mean Value at a System Output

0

dhtXEtYE

Using the expected value as an operator (moving inside the intergral)

0

dhtXEtYE

For a wide-sense stationary process, this result in

00

dhXEdhXEtYE

The coherent gain of a filter is defined as:

0

dtthhgain

Therefore, gainhXEtYE

The expected value of the output is the expected value of the input times the coherent gain!

The Mean Square Value at a System Output

2

0

2 dhtXEtYE

0

2 dhRhtYE XX

Therefore, take the convolution of the autocorrelation function and then sum the filter-weighted result from 0 to infinity.



The Autocorrelation at a System Output txthtxthEtYtYERYY

The expected values of the product of two distinct convolutions:

0222

0111 dhtxdhtxERYY

021212

01 ddhhtxtxERYY

Identifying and forming the input autocorrelation

02121

021 ddhhtxtxERYY

02121

012 ddhhRR XXYY

0122

0211 ddhRhR XXYY

01111 dhRhR XXYY

or

0

1111 dhRhR XXYY

Notice that first, you convolve the input autocorrelation function with the filter function and then you convolve the result with a time inversed version of the filter!



Example: White Noise passed through an RC filter

0122

0211 ddhRhR XXYY

White Noise Signal: tNtRXX 20

0122

021

01 2

ddhNhRYY

For 0for and 01

0,2 0

1110

fordhhNRYY

For 0for and 02

0,2 0

2220

fordhhNRYY This is a correlation function of the impulse response of the filter with itself!

Now for and RC filter, the computation can continue …

tuCRt

CRth

exp1

0,expexp1

2 02

0

fordCRCRCR

NRYY

0,2expexp1

2 02

0

fordCRCRCR

NRYY

0,exp

41

0

forCRCR

NRYY

The complete autocorrelation of white noise in an RC filter is

0,expexp12

0

fordCRCRCR

NRYY

Leading to

0,exp4

10

for

CRCRNRYY

and finally

CRCRNRYY

exp

41

0



The Crosscorrelation of Input to Output txthtxEtYtXERXY

0111 dhtxtxERXY

0

111 dhtxtxERXY

0

111 dhRR XXXY

This is the convolution of the autocorrelation with the filter.

What about the other Crosscorrelation?

txtxthEtXtYERYX

0111 dhtxtxERYX

0

111 dhtxtxERYX

0

111 dhRR XXYX

This is the correlation of the autocorrelation with the filter, inherently different than the previous, but equal at =0.

For a white noise process:

tNtRXX 20

0,0

0,2

2

0

0111

0

hNdh

NRXY

0,2

0,0

2 00111

0

hNdh

NRYX



Exercise Examples Exercise 8-2.1 Filter: tuttth 2exp

Random Process: 120,121, MforfMtX M (a) Write an expression for the output sample function. (b) Find the mean value of the output. (c) Find the variance of the output.

First:

62

12 MEtXE

4861212 222222 MMMEtXE

482 MERtXtXE XX Coherent gain

00

2exp dtttdtthhgain

1expexp 2

xaaxadxxax

411

4112

42exp

0

tthgain

Output Signal

400

MdhMdhtXtY

Therefore, 23

46

41

XEtYE

2

0

2 dhtXEtYE

2

0

2

2

0

2

dhMEdhMEtYE

34148

22

tYE

And finally 43

4912

233

2222

YY tYE



Exercise 8-2.2

Filter:

else

ttth

,010,35

Random Process: 20,21,2cos2 forfttX (a) Write an expression for the output sample function. (b) Find the mean value of the output. (c) Find the variance of the output.

First: 02cos2 tEtXE

224

2222cos142cos2 222

tEtEtXE

ttERtXtXE XX 2cos22cos2

2cos2

2222cos2cos4 tERXX

Coherent gain

835351

00

dttdtthhgain

Output Signal

0

dhtXtY

1

0

352cos2 dttY

1

0

2cos62cos10 dtttY

1

022sin62cos10

tttY

2

2sin22sin62cos10

ttttY

ttY 2cos10



Then, 0 gaingain hXEhXEtYE

22 2cos10 tEtYE 50

2100

2222cos11002

tEtYE

or

0122

0211

2 ddhRhtYE XX

0122

1

0211

2 352cos235 ddtYE

0

1112 2cos1035 dtYE

502cos30501

011

2 dtYE

And finally 50050 2222 YY tYE



Exercise 8-3.1

Filter: tuttth 2exp

Random Process: wwSnoisetX XX 42,

Therefore: 4242

,22

00 N

RN

XX

(a) Find the mean value of the output. (b) Find the variance of the output. (c) Find the mean square of the output.

First: 24 XXRtXE

642222 XXXXRtXE 22 X

Coherent gain

00

2exp dtttdtthhgain

1expexp 2

xaaxadxxax

411

4112

42exp

0

tthgain

Output Signal

0

dhtXtY

Therefore, 21

412 gainhXEtYE

2

0

2 dhtXEtYE

0122

0211

2 ddhRhtYE XX



01222

02111

2 2exp422exp ddtYE

01

02221111

2 2exp42exp22exp ddtYE

011111

2

4142exp22exp dtYE

0

11112

12 2exp4exp2 dtYE

0111

0

12

12 4exp422

44exp2

41 dtYE

0

112 14

164exp1

41 tYE

165

161

412 tYE

And finally

161

1645

21

165 2222

YY tYE

Exercise 8-3.2

Filter: 5.010 tututh

Random Process: 5, wSnoisetX XX

Therefore: 52

,52

00 NRN

XX

(a) Find the mean value of the output. (b) Find the mean square of the output.

First: 0 XXRtXE

5222 XXXXRtXE 52 X



Coherent gain

00

5.010 dttutudtthhgain

52

10105.0

0

dthgain Output Signal

0

dhtXtY

Therefore, 050 gainhXEtYE

2

0

2 dhtXEtYE

0122

0211

2 ddhRhtYE XX

0122

0211

2 5 ddhhtYE

0

12

12 5 dhtYE

0

222 5.0105 dttututYE

250

2500500

5.0

0

2 dttYE

And finally 2500250 2222 YY tYE



Example section 8-4: An Exponential Autocorrelation signal through an RC filter

Signal: tStRXX exp20

The mean value is 0exp2

0

SRXXX

and 2

0 0222S

R XXXXX

Filter: CRbwheretutbbth 1,exp The autocorrealtion

0122

0211 ddhRhR XXYY

0122

021

01 exp2

ddhShRYY

For 0for and 01

012221

01

0122

021

01

1

1

exp2

exp2

ddhSh

ddhShRYY

012221

01

0122

021

01

expexp2

exp

expexp2

exp

1

1

ddbbSbb

ddbbSbbRYY

012211

20

01

02211

20

1

1

expexpexp2

expexpexp2

ddbbbS

ddbbbSRYY

01

111

02

01

111

02

expexpexp2

1expexpexp2

db

bbSb

dbb

bbSbRYY



01

111

02

01

111

02

expexpexp2

exp1expexp2

db

bbSb

db

bb

bSbRYY

011

02

01111

02

2exp2

2expexp2

dbb

Sb

dbbb

SbRYY

011

02

0111

02

2expexp2

2expexpexpexp2

dbbb

Sb

dbbbb

SbRYY

202expexp

2

202expexp0expexp

2

02

02

bbb

bSb

bbb

bb

bSbRYY

b

bSbb

bSb

bbSbRYY exp4

exp4

exp2

0002

b

bbSb

bbSbRYY exp2

exp2

02

02

bbb

SbRYY expexp2 220

Based on symmetry, we have

bbb

SbRYY expexp2 220

Conversion of input to a white noise case To compare this result to the white noise result, let 200 NS and . Then

tNtRXX 20

and

bNbbb

Sbb

SbRYY exp4exp

2exp

20

220

2

220

2



For the input signal approached white noise

bbb

SbRYY expexp2 220

becomes (isolating noise based terms prior to the others)

222

220 expexp

2

bb

bbbSbRYY

bbb

bSbRYY exp1exp2 222

0

bbb

bSbRYY exp11

1exp2

22

0

The result consists of a filtered white noise term, a constant based on b and beta, and a term in the exponential.

This suggests that for “wide bandwidth” input noise, it is often possible to approximate the noise as filtered white noise as opposed to filtering an already band-limited noise.



The Power Spectral Density at a System Output

The power spectral density is the Fourier Transform of the autocorrelation:

diwtXtXERwS XXXX exp

diwtYtYERwS YYYY exp

Where

txthty For a WSS, ergodic process,

tYtYERYY

0222

0111 dhtxdhtxERYY

021212

01 hhtxtxddERYY

021212

01 txtxEhhddRYY

021212

01 XXYY RhhddR

This is where we have gotten before. Now we need to perform the Fourier transform to find the PSD!

diwRhhddRwS XXYYYY exp

021212

01



Taking the Power Spectral Density

diwRhhddRwS XXYYYY exp

021212

01

Isolating the transform to function in time (tau)

021212

01 exp diwRhhddRwS XXYYYY

021212

01 exp jwwShhddRwS XXYYYY

Separating dependent and independent elements based on the integration

022211

01 expexp jwhdjwhdwSRwS XXYYYY

wHjwhdwSRwS XXYYYY

110

1 exp

110

1 exp jwhdwHwSRwS XXYYYY

Therefore

wHwHwSRwS XXYYYY or

2wHwSRwS XXYYYY

Relation of Spectral Density to the Crosscorrelation Function

As you might expect, computation of the cross-spectral only has a single “filter PSD term”.

wHwSwS XXXY and wHwSwS XYYY

wHwSwS XXYX and wHwSwS YXYY



Revisiting Equivalent Noise Bandwidth

In the previous notes, white noise was passed through an RC, 1st order Butterworth filter as recapped here.

Let tNtRXX 20

00

dhXEdhXEtYE

00

Hdtthhgain

0

2 dhRhtYE XX

BNdhNtYE

00

12

102

2

For the unity coherent gain filter example.

01

21

00

22

dhNBNNE EQB

0

221 dtthBEQ

This is for a normalized filter. If the filter is not a gain of 1 and DC, it must first be normalized before performing the operations.

Note: any filter gain (or loss) equally magnifies the signal and noise. Therefore, it has no effect on the signal to noise ratio (SNR).



If the filter has coherent gain, normalize the filter first. The normalized is

0

0

Hth

hth

dtth

thtggain

Then

0

2

0

2

021

21 dt

HthdttgBEQ

20

2

02 H

dtthBEQ

From Parseval’s theorem

0

2222

21 dtthdwwHdffHdtth

Frequency Domain Computation

2

2

2

2

0402 H

dwwH

H

dffHBEQ

Time Domain Computation

2

0

0

2

20

2

21

2

dtth

dtth

h

dtthB

gainEQ

Therefore, the Equivalent noise bandwidth can be calculated purely in the time domain or purely in the frequency domain, whichever is easier to compute!



Example: A rectangular filter time domain BEQ

Ttututh 1

TdtdtthhT

gain

00

1

TT

TT

d

dtth

dh

B

T

EQ

21

22

1

2

220

1

2

0

01

21

The Fourier Filter Domain BEQ

T

dttfidttfithfH0

2exp2exp

fifi

fiTfi

fitfifH

T

202exp

22exp

22exp

0

fi

Tfififi

tfifHT

22exp

21

22exp

0

22exp

22exp

22exp

21 TfiTfiTfi

fifH

22sin2

22exp

21 TfiTfi

fifH

TfTfTTfifH

sin

22exp

TfcTTfifH

sin

22exp or

2sin

2exp TwcTTwifH

TfcTfH sin or

2sin TwcTwH

gainhTH 0



Notice that the result matches the time domain.

From here, we could solve

TT

dfTfcT

H

dffH

BEQ

2

12

sin

02 2

2

2

2

Notice that the nulls of the sinc function appear at f=1/T but that the Beq=1/2T … half the frequency distance to the first null!

Note: This is easier in the discrete time-frequency domain … summing and summing squares!



8-6 Examples of Time-Domain Analysis:

It is important to remember both the time domain and frequency domain relationships in order to pick the one that is easier based on the signal type.

When the autocorrelation has a simple form over a finite time interval, the time domain can be easier.

Filter: TtutuT

th 1

Random Process:

TARXX

12

(a) Find the mean value of the output. (b) Find the mean square of the output.

First: 0tXE

2222 0 ARtXE XXXX 22 AX

Coherent gain

T

gain dtTtutuTdtthh

00

1

11

0

TTdt

Th

T

gain

Output Signal

0

dhtXtY

Therefore, 010 gainhXEtYE

0122

0211

2 ddhRhtYE XX

T T

ddTT

AT

tYE0

120

2122 111

T T

ddTT

AtYE0

120

212

22 1



To eliminate the absolute value, perform two integral for each part,

T TT T

dddT

ddTAtYE

012122

021

012

02

22

1

111

TT

dT

TTAtYE

0121

22

0

22

212

2

22

1

1

221

T

dTTT

TTAtYE

01

21

1

2212

2

22

2221

T

dTTTAAtYE

01

211

2

3

222

2

T

TTTAAtYE

0

31

21

1

2

3

222

322

322

323

3

222 TTTT

TAAtYE

2332

22

32

3AT

TAAtYE

And finally 222222320

32 AAtYE YY

The autocorrelation function (using tables)

2wHwSRwS XXYYYY

2

exp

2

2sin

TwjTw

Tw

wH

dwjT

AwST

TXX

exp12

22

2

2

2sin

Tw

TwTAwS XX

2wHwSwS XXYY

44

22

2

2

2

2

2

2sin

2

2sin

2

2sin

Tw

TwTA

Tw

Tw

Tw

TwTAwSYY



Power Spectral Density of a filtered white noise process.

0122

0211 ddhRhR XXYY

Let tNtRXX 20

0122

021

01 2

ddhN

hRYY

0

1110

2 dhhNRYY

Taking the PSD

diwRRwS YYYYYY exp

iwdhhNdwSYY exp2 0111

0

0

1110 exp

2 diwhhdNwSYY

0

110 exp

2wHiwhdNwSYY

0

110 exp

2 wihdwHNwSYY

20022

wHN

wHwHN

wSYY

chapter 8: response of linear systems to random inputsbazuinb/ece3800cmcg/notes8...0 rxy rxx 1 h 1 d...

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