EECE 522 Notes_08 Ch_3 CRLB Examples in Book

7/30/2019 EECE 522 Notes_08 Ch_3 CRLB Examples in Book

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3.11 CRLB Examples

1. Range Estimation sonar, radar, robotics, emitter location

2. Sinusoidal Parameter Estimation (Amp., Frequency, Phase) sonar, radar, communication receivers (recall DSB Example), etc.

3. Bearing Estimation sonar, radar, emitter location4. Autoregressive Parameter Estimation

speech processing, econometrics

Well now apply the CRLB theory to several examples of practical signal processing problems.

Well revisit these examples in Ch. 7 well derive ML

estimators that will get close to achieving the CRLB


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max,

);(

0)()()( o st s

o T T t t wt st x

o

+=+=!"!#$

Transmit Pulse: s(t ) nonzero over t [0,T s]

Receive Reflection: s(t o)

Measure Time Delay: o

C-T Signal Model

BandlimitedWhite Gaussian

t T s

s(t )

t T

s(t o)BPF& Amp

x(t )

PSD of w(t )

f B B

N o/2

Ex. 1 Range Estimation Problem


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Sample Every = 1/2 B secw[n] = w(n)

DT WhiteGaussian Noise

Var 2

= BN o1,,1,0][)(][ =+= N nnwn sn x o

f

ACF of w(t )

1/2 B

B B

1/ B 3/2 B

PSD of w(t )

N o/2 2 = BN o

+

++

=

1][

1][)(

10][

][

N n M nnw

M nnnnwn s

nnnw

n x

o

ooo

o

s[n;o] has M non-zero samples starting at no

Range Estimation D-T Signal Model


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Now apply standard CRLB result for signal + WGN:

= =

+

= =

+

=

=

=

=

=

1

0

2

2

1 2

2

1 2

2

1

0

2

2

)()(

)(];[)var(

M

nnt

M n

nn nt

M n

nn o

o N

n o

oo

t t s

t t s

n sn s

o

o o

o

o

Plug in and keepnon-zero terms

Exploit Calculus!!! Use approximation: o

= noThen do change of variables!!

Range Estimation CRLB


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s

T

o

s

T

o

T o

E

dt t t s

N E

dt t t s

N

dt t t s s s s

=

=

0

2

0

2

0

2

2

)(

2/

1

)(

2/

)(1)var(

Assume sample spacing is small approx. sum by integral

= sT s dt t s E 0

2 )(

( )

( )

=

dt f S

df f S f

N

E

E

df f S f

N E

o

s

s

T

o

s

o s

2

22

0

22

)(

)(2

2/

1

)(2

2/

1)var(

FT Theorem& Parseval

Parseval( )

=

dt f S

df f S f Brms 2

22

)(

)(2

Define a BW measure:

Brms is RMS BW (Hz)A type of SNR

Range Estimation CRLB (cont.)


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Using these ideas we arrive at the CRLB on the delay:

)(sec1

)var( 22rms

o BSNR

To get the CRLB on the range use transf. of parms result:

)(m4/

)var( 22

2

rms BSNR

c R

oCRLB

RCRLB

o R

2

= with R = c o

/ 2

CRLB is inversely proportional to: SNR Measure RMS BW Measure

CRLB is inversely proportional to: SNR Measure RMS BW Measure

So the CRLB tells us Choose signal with large B

rms Ensure that SNR is large Better on Nearby/large targets Which is better?

Double transmitted energy?

Double RMS bandwidth?

Range Estimation CRLB (cont.)


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1,,1,0][)cos(][ =++= N nnwn An x o

Given DT signal samples of a sinusoid in noise.

Estimate its amplitude, frequency, and phase

DT White Gaussian NoiseZero Mean & Variance of 2

o is DT frequency incycles/sample: 0 < o <

Multiple parameters so parameter vector: T o A ][ =

Recall SNR of sinusoid in noise is:

2

2

2

2

2

2/

A A P P

SNR

n

s ===

Ex. 2 Sinusoid Estimation CRLB Problem


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Approach : Find Fisher Info Matrix Invert to get CRLB matrix Look at diagonal elements to get bounds on parm variances

Recall : Result for FIM for general Gaussian case specialized to

signal in AWGN case:

j

N

n i

T

jiij

n sn s

=

=

=

];[];[1

1)]([

1

02

2

ssI

Sinusoid Estimation CRLB Approach


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Taking the partial derivatives and using approximations given in book (valid when o is not near 0 or ) :

( )

( )

2

21

0

22233

1

02

21

0

2223223

1

0

22

21

0

22

21

0

222222

1

02

1

023113

1

02

1

022112

2

1

02

1

0

2211

2)(sin

1)]([

2)(sin

1)]([)]([

2

)22cos(1

2

)(sin)(1

)]([

0)22sin(2

)sin()cos(1)]([)]([

0)22sin(2

)sin()cos(1

)]([)]([

2)22cos(1

2

1)(cos

1)]([

NAn A

n A

nn A

n A

nn A

nn A

n Ann A

nn A

nn An

N nn

N

no

N

n

N

no

N

n

N

n

o

N

n

o

N

no

N

noo

N

no

N

noo

N

no

N

no

+=

+==

+=+=

+

=++

==

+=++==

++=+=

=

=

=

=

=

=

=

=

=

=

=

=

I

II

I

II

II

I

T o A ][ =

Sinusoid Estimation Fisher Info Elements


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10/19

=

=

=

2

21

0

2

2

1

02

21

022

2

2

220

220

002

)(

NAn

A

n An A

N

N

n

N

n

N

nI

Fisher Info Matrix then is:

2

2

2

ASNR =Recall and closed form results for these sums

T

o A ][ =

Sinusoid Estimation Fisher Info Matrix


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Inverting the FIM by hand gives the CRLB matrix and thenextracting the diagonal elements gives the three bounds:

(using co-factor & detapproach helped by 0s)

)rad(4

)1()12(2

)var(

))rad/sample(()1(

12)var(

)volts(2

)var(

2

22

22

N SNR N N SNR N

N N SNR

N A

o

+

Amp. Accuracy: Decreases as 1/ N , Depends on Noise Variance (not SNR)

Freq. Accuracy: Decreases as 1/ N 3, Decreases as as 1/S NR

Phase Accuracy: Decreases as 1/ N , Decreases as as 1/S NR

To convert to Hz 2

multiply by ( F s /2)2

Sinusoid Estimation CRLBs


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The CRLB for Freq. Est. referred back to the CT is

)Hz()1()2(

12)var( 2

22

2

N N SNR

F f so

Does that mean we do worse if we sample faster than Nyquist?

NO!!!!! For a fixed duration T of signal: N = TF s

Also keep in mind that F s has effect on the noise structure:

Not in Book

f

ACF of w(t )

1/2 B

B B1/ B 3/2 B

PSD of w(t ) N o/2 2 = BN o

Frequency Estimation CRLBs and Fs


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Uniformly spaced linear array with M sensors: Sensor Spacing of d meters Bearing angle to target radians

Figure 3.8

from textbook:

Simple model

Emits or reflects

signal s(t )

)2cos()( += t f At s ot

Propagation Time to nth Sensor: 1,,1,0cos0 == M ncd

nt t n

+

+=

=

cos2cos

)()(

0 cd nt t f A

t t st s

o

nn

Signal at nth Sensor:

Ex. 3 Bearing Estimation CRLB Problem


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Now instead of sampling each sensor at lots of time instantswe just grab one snapshot of all M sensors at a single instant t s

( )

~cos

~cos

2cos

cos2cos)( 0

+=

+

=

+

+=

n And c f

A

cd

nt t f At s

so

so sn

s

s! ! ! "! ! ! #$

! ! "! ! #$

Spatial sinusoid w/spatial frequency

s

Spatial Frequencies : s is in rad/meter s is in rad/sensor

For sinusoidal transmitted signal BearingEst. reduces to Frequency Est.

And we already know its FIM & CRLB!!!

Bearing Estimation Snapshot of Sensor Signals


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][~

cos][)(][ nwn Anwt sn x s sn ++=+=

Each sample in the snapshot is corrupted by a noise sample

and these M samples make the data vector x = [ x[0] x[1] x[ M -1] ]:

Each w[n] is a noise sample that comes from a different sensor soModel as uncorrelated Gaussian RVs (same as white temporal noise)Assume each sensor has same noise variance 2

So the parameters to consider are: T s A ][ =

which get transformed to:

===

2arccos

)(d f

c

A A

o

sg

Parameter of interest!

Bearing Estimation Data and Parameters


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Using the FIM for the sinusoidal parameter problem together with the transform. of parms result (see book p. 59 for details):

( )

)rad(

sin11

)2(

12)var( 2

22

2

+

L

M M

M SNR

Bearing Accuracy:

Decreases as 1/ SNR Depends on actual bearing Decreases as 1/ M Best at = /2 (Broadside)

Decreases as 1/ L r 2 Impossible at = 0! (Endfire)

L = Array physical length in meters M = Number of array elements

= c/ f o Wavelength in meters (per cycle)

Define: L r = L / Array Length inwavelengths

Low-frequency (i.e., long wavelength) signals needvery large physical lengths to achieve good accuracy

Bearing Estimation CRLB Result


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In speech processing (and other areas) we often model the

signal as an AR random process and need to estimate the AR parameters. An AR process has a PSD given by

2

1

2

2

][1

);(

=

+

= p

m

fm j

u xx

ema

f P

AR Estimation Problem: Given data x[0], x[1], , x[N-1]estimate the AR parameter vector

[ ]T u paaa 2][]2[]1[ &=

This is a hard CRLB to find exactly but it has been published.

The difficulty comes from the fact that there is no easy directrelationship between the parameters and the data.

It is not a signal plus noise problem

Ex. 4 AR Estimation CRLB Problem


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Approach : The asymptotic result we discussed is perfect here: An AR process is WSS is required for the Asymp. Result

Gaussian is often a reasonable assumption needed for Asymp. Result

The Asymp. Result is in terms of partial derivatives of the PSD andthat is exactly the form in which the parameters are clearly displayed!

[ ] [ ] [ ]df f P f P N

j xx

i xxij

2

1

21

);(ln);(ln2)(

I

2

1

222

1

2

2][1lnln

][1

ln);(ln

=

=

+=

+

= p

m

fm ju

p

m

fm j

u xx ema

ema

f P

Recall:

AR Estimation CRLB Asymptotic Approach


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After taking these derivatives you get results that can besimplified using properties of FT and convolution.

The final result is: [ ]

N

pk N

k a

uu

kk xxu

42

12

2)var(

,,2,1])[var(

= R

Both Decreaseas 1/ N

To get a little insight look at 1 st order AR case ( p = 1):

])1[1(1

])1[var( 2a N

a

Complicateddependence on

AC Matrix!!

Improves as polegets closer to unit

circle PSDs

with sharppeaks are easier

to estimate

a[1] Re( z )

Im( z )

AR Estimation CRLB Asymptotic Result

EECE 522 Notes_08 Ch_3 CRLB Examples in Book

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