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#### Help with computing the Crame Rao Lower Bound for Time Delay Estimation

```Dear all,

I am trying to retrieve the theoretical value of the time delay estimate when using a chirp signal.

According to theory (see http://www.ws.binghamton.edu/fowler/fowler%20personal%20page/EE522_files/EECE%20522%20Notes_08%20Ch_3%20CRLB%20Examples%20in%20Book.pdf slide 4/19), the Cramer Rao Lower Bound should be given by:

var(tau_est^2)>=sigma^2/energy_of_derivative_of_chirp

This is the theoretical lower bound for the variance of any unbiased estimator of the time delay.

Now, the ML estimate of the time delay, according also to theory (see http://www.ws.binghamton.edu/fowler/fowler%20personal%20page/EE522_files/EECE%20522%20Notes_14%20Ch_7C.pdf slide 5), is obtained as the argmax of the corss-correlation function.

ML estimate should be asymptotically unbiased and efficient, so its variance should get close to the theoretical Cramer Rao bound.

Basically, the code (see below) tries to obtain the variance of the ML estimate through Monte Carlo simulation, then compare with the theoretical value.

However, somewhere, even of this schoollike example, I make a mistake, I thing, since the results are very far away.

For example, I obtain a theoretical variance of 0.0109 and a simulated variance of 2.9543e+05. The scale factor between the two (see the MAGIC value in the code), which I was expecting to be around 1, is... 2.7214e+07

Where do I make a mistake?

This is the code:

clearvars;
%using ft_tfrs;
N=1000; %chirp signal length
fstart=0.1; %chirp start frequency
fstop=0.2; %chirp end frequency
tx=0:(N-1);
x=cos(2*pi*(fstart*tx+(fstop-fstart)/2/N*tx.^2));
figure; plot(tx,x); title('x');

%noiseless delayed copy of the chirp
M=5000;
ty=0:(M-1);
D=2000;
y=[zeros(1,D),x,zeros(1,M-D-N)];
figure; plot(ty,y); title('y');

%noising the delayed copy of the chirp
stddev=sqrt(5);
tz=ty;
z=y+(stddev^2)*randn(size(y));
figure; plot(tz,z); title('z');

%estimating the delay time (single example)
[xc,xlags]=xcorr(z,x); %order matters
[~,idxmax]=max(xc);
delay_est=xlags(idxmax);
figure; plot(xlags,xc); title('xcorr');
delay_est

%now perform a Monte-Carlo simulation
NRealisations=5000;
delay_est_v=zeros(1,NRealisations);
FIG_BASE=get(0,'CurrentFigure'); if isempty(FIG_BASE) FIG_BASE=0; end;
DISPLAY=0;
for idx=1:NRealisations
[idx NRealisations]
z=y+(stddev^2)*randn(size(y));
[xc,xlags]=xcorr(z,x); %order matters
[~,idxmax]=max(xc);
delay_est=xlags(idxmax);
delay_est_v(idx)=delay_est;
if DISPLAY==1
figure(FIG_BASE+1); plot(xlags,xc);
end
end
figure; plot(delay_est_v); title('estimated delay');
mean_estim=mean(delay_est_v);
var_estim=var(delay_est_v);
mean_estim
var_estim
stddev_estim=sqrt(var_estim);
stddev_estim

%we display convergence of the variance
var_convergence=zeros(1,NRealisations);
for idx=1:NRealisations
var_convergence(idx)=var(delay_est_v(1:idx));
end
stddev_convergence=sqrt(var_convergence);

%finally, theoretical bound of chirp
th=tx; %time bins for h
h=-sin(2*pi*(fstart*tx+(fstop-fstart)/2/N*tx.^2)).*(2*pi*(fstart+(fstop-fstart)/N*tx)); %h is the derivative of the chirp
%(alternatively) h=[0,diff(x)];
figure; plot(th,h); title('h');

var_theo=(stddev^2)/(sum(h.^2));
stddev_theo=sqrt(var_theo);
var_theo
stddev_theo

figure; plot(1:NRealisations,var_convergence,'b-'); hold on; plot(1:NRealisations,var_theo,'r-'); title('var convergence'); legend('simulation','theoretical');
figure; plot(1:NRealisations,stddev_convergence,'b-'); hold on; plot(1:NRealisations,stddev_theo,'r-'); title('stddev convergence'); legend('simulation','theoretical');

MAGIC=var_estim/var_theo;
MAGIC

%why MAGIC is not (almost) 1?
``` 0 1/19/2012 11:32:10 AM comp.soft-sys.matlab  211266 articles. 23 followers. 4 Replies 385 Views Similar Articles

[PageSpeed] 28

```On Jan 19, 6:32=A0am, "Felix Totir" <n...@utopia.com> wrote:
> Dear all,
>
> =A0I am trying to retrieve the theoretical value of the time delay estima=
te when using a chirp signal.
>
> =A0According to theory (seehttp://www.ws.binghamton.edu/fowler/fowler%20p=
ersonal%20page/EE522_fi...slide 4/19), the Cramer Rao Lower Bound should be=
given by:
>
> var(tau_est^2)>=3Dsigma^2/energy_of_derivative_of_chirp
>
> =A0This is the theoretical lower bound for the variance of any unbiased e=
stimator of the time delay.
>
> Now, the ML estimate of the time delay, according also to theory (seehttp=
://www.ws.binghamton.edu/fowler/fowler%20personal%20page/EE522_fi...slide 5=
), is obtained as the argmax of the corss-correlation function.
>
> ML estimate should be asymptotically unbiased and efficient, so its varia=
nce should get close to the theoretical Cramer Rao bound.
>
> Basically, the code (see below) tries to obtain the variance of the ML es=
timate through Monte Carlo simulation, then compare with the theoretical va=
lue.
>
> However, somewhere, even of this schoollike example, I make a mistake, I =
thing, since the results are very far away.
>
> For example, I obtain a theoretical variance of 0.0109 and a simulated va=
riance of 2.9543e+05. The scale factor between the two (see the MAGIC value=
in the code), which I was expecting to be around 1, is... 2.7214e+07
>
> Where do I make a mistake?
>
> This is the code:
>
> clearvars;
> %using ft_tfrs;
> N=3D1000; %chirp signal length
> fstart=3D0.1; %chirp start frequency
> fstop=3D0.2; %chirp end frequency
> tx=3D0:(N-1);
> x=3Dcos(2*pi*(fstart*tx+(fstop-fstart)/2/N*tx.^2));
> figure; plot(tx,x); title('x');
>
> %noiseless delayed copy of the chirp
> M=3D5000;
> ty=3D0:(M-1);
> D=3D2000;
> y=3D[zeros(1,D),x,zeros(1,M-D-N)];
> figure; plot(ty,y); title('y');
>
> %noising the delayed copy of the chirp
> stddev=3Dsqrt(5);
> tz=3Dty;
> z=3Dy+(stddev^2)*randn(size(y));
> figure; plot(tz,z); title('z');
>
> %estimating the delay time (single example)
> [xc,xlags]=3Dxcorr(z,x); %order matters
> [~,idxmax]=3Dmax(xc);
> delay_est=3Dxlags(idxmax);
> figure; plot(xlags,xc); title('xcorr');
> delay_est
>
> %now perform a Monte-Carlo simulation
> NRealisations=3D5000;
> delay_est_v=3Dzeros(1,NRealisations);
> FIG_BASE=3Dget(0,'CurrentFigure'); if isempty(FIG_BASE) FIG_BASE=3D0; end=
;
> DISPLAY=3D0;
> for idx=3D1:NRealisations
> =A0 =A0 [idx NRealisations]
> =A0 =A0 z=3Dy+(stddev^2)*randn(size(y));
> =A0 =A0 [xc,xlags]=3Dxcorr(z,x); %order matters
> =A0 =A0 [~,idxmax]=3Dmax(xc);
> =A0 =A0 delay_est=3Dxlags(idxmax);
> =A0 =A0 delay_est_v(idx)=3Ddelay_est;
> =A0 =A0 if DISPLAY=3D=3D1
> =A0 =A0 =A0 =A0 figure(FIG_BASE+1); plot(xlags,xc);
> =A0 =A0 end
> end
> figure; plot(delay_est_v); title('estimated delay');
> mean_estim=3Dmean(delay_est_v);
> var_estim=3Dvar(delay_est_v);
> mean_estim
> var_estim
> stddev_estim=3Dsqrt(var_estim);
> stddev_estim
>
> %we display convergence of the variance
> var_convergence=3Dzeros(1,NRealisations);
> for idx=3D1:NRealisations
> =A0 =A0 var_convergence(idx)=3Dvar(delay_est_v(1:idx));
> end
> stddev_convergence=3Dsqrt(var_convergence);
>
> %finally, theoretical bound of chirp
> th=3Dtx; %time bins for h
> h=3D-sin(2*pi*(fstart*tx+(fstop-fstart)/2/N*tx.^2)).*(2*pi*(fstart+(fstop=
-fstart)/N*tx)); %h is the derivative of the chirp
> %(alternatively) h=3D[0,diff(x)];
> figure; plot(th,h); title('h');
>
> var_theo=3D(stddev^2)/(sum(h.^2));
> stddev_theo=3Dsqrt(var_theo);
> var_theo
> stddev_theo
>
> figure; plot(1:NRealisations,var_convergence,'b-'); hold on; plot(1:NReal=
isations,var_theo,'r-'); title('var convergence'); legend('simulation','the=
oretical');
> figure; plot(1:NRealisations,stddev_convergence,'b-'); hold on; plot(1:NR=
ealisations,stddev_theo,'r-'); title('stddev convergence'); legend('simulat=
ion','theoretical');
>
> MAGIC=3Dvar_estim/var_theo;
> MAGIC
>
> %why MAGIC is not (almost) 1?

From what I recall - the CRLB for time delay estimation is actually
not a good lower bound i.e. it is hard to reach the lower bound. This
is one of the reasons for the development of the Ziv-Zakai bound - see
Johnson & Dudgeon, "Array Signal Processing" section 6.3.3 (Time delay
Estimation).

Briefly they state: ... the discrete valued nature of the parameter
calls into question the use of the Cramer-Rao bound. One of the
fundamental assumptions of the bound's derivation is the
differentiability of the likelihood function with respect to the
parameter. Mathematically, a sequence cannot be differentiated with
respect to the integers.

Therefore the sampling must be fine enough to be considered
continuous.

Cheers,
David
``` 0 1/19/2012 1:57:43 PM
```> From what I recall - the CRLB for time delay estimation is actually
> not a good lower bound i.e. it is hard to reach the lower bound. This
> is one of the reasons for the development of the Ziv-Zakai bound - see
> Johnson & Dudgeon, "Array Signal Processing" section 6.3.3 (Time delay
> Estimation).
>
> Briefly they state: ... the discrete valued nature of the parameter
> calls into question the use of the Cramer-Rao bound. One of the
> fundamental assumptions of the bound's derivation is the
> differentiability of the likelihood function with respect to the
> parameter. Mathematically, a sequence cannot be differentiated with
> respect to the integers.
>
> Therefore the sampling must be fine enough to be considered
> continuous.
>
> Cheers,
> David

Hello,

Thank you very much for replying. Indeed, the derivative of the discrete sequence with respect to the estimated parameter (namely, the delay) is of concern. It was been addressed in the provided references by taking the derivative of the signal *in continuous time*, then computing it at the required discrete times.

The code that I provided follows the same approach.

I am not sure if this is the right approach, but what else to use instead?

References for this simple problem seem to be quite scarce, especially when it comes to implementation (ie. numerical examples). I am quite amazed.

The only numerical example that I found is http://www.google.com/url?sa=t&rct=j&q=dsto%E2%80%93tr%E2%80%931705&source=web&cd=1&ved=0CB8QFjAA&url=http%3A%2F%2Fdspace.dsto.defence.gov.au%2Fdspace%2Fbitstream%2F1947%2F4441%2F1%2FDSTO-TR-1705%2520PR.pdf&ei=ziYYT4-OA8axhAe7-oWxDA&usg=AFQjCNFmIau5VUb3-FKbnAXHcBhQ6ivQdQ.

Does anyone know better examples? Maybe a digital signal processing university classes textbook?

However, *which* of the two results that I obtained (theoretical variance based in CRLB or simulation variance based on Monte Carlo) is the realistic one?

Thank you once again for your attention.

Felix
``` 0 1/19/2012 2:23:09 PM
```Above reference (weblink) it is not entirely correct (there is a bad ending point).

Correct reference is:

``` 0 1/19/2012 3:18:10 PM
```> From what I recall - the CRLB for time delay estimation is actually
> not a good lower bound i.e. it is hard to reach the lower bound. This
> is one of the reasons for the development of the Ziv-Zakai bound - see
> Johnson & Dudgeon, "Array Signal Processing" section 6.3.3 (Time delay
> Estimation).

> Cheers,
> David

Thank you, Dave, for directing me towards the Ziv-Zakai bound (although it is a Bayesian, ie global bound).

I found this little gem:

http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.43.6786

While I am still tackling the problem, this reference grealy increased my knowledge. I put it here for the interested reader.
``` 0 1/19/2012 4:13:09 PM Similar Artilces:

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Matlab help

MATLAB-help!!!
I am trying to write code to iterate a function 50 times, however, every time i write it without help i get ()-indexing must appear last in an index expression. I'm not sure what this means or how to fix it, can someone please explain. Also, my input is this: x(i)=a*x(i-1)(1-x(i-1)+h*x(i-1)^2/2) On Jan 24, 12:45=A0pm, michelle taylor <miss.smileyf...@gmail.com> wrote: > I am trying to write code to iterate a function 50 times, however, > every time i write it without help i get ()-indexing must appear last > in an index expression. > > I'm not sure what this means or how to fix it, can someone please > explain. > > Also, my input is this: > > x(i)=3Da*x(i-1)(1-x(i-1)+h*x(i-1)^2/2) Aren't you missing an operator? x(i)=3Da*x(i-1) here (1-x(i-1)+h*x(i-1)^2/2) x(i)=3Da*x(i-1)*(1-x(i-1)+h*x(i-1)^2/2) or x(i)=3Da*x(i-1)/(1-x(i-1)+h*x(i-1)^2/2) or x(i)=3Da*x(i-1)^(1-x(i-1)+h*x(i-1)^2/2) On Jan 23, 3:58=A0pm, TideMan <mul...@gmail.com> wrote: > On Jan 24, 12:45=A0pm, michelle taylor <miss.smileyf...@gmail.com> > wrote: > > > I am trying to write code to iterate a function 50 times, however, > > every time i write it without help i get ()-indexing must appear last > > in an index expression. > > > I'm not sure what this means or how to fix it, can someone please > > explain. > > > Also, my input is this: > > > x(i)=3Da*x(i-1)(1-x(i-1)+h*x(i-1)^2/...

[JOB] Help
Hi all Sorry to advertise within the user group but I really need some help! I was hoping that people in this usergroup could help other fellow Matlab developers who may be looking for a new role, or might be interested in what I have on offer. I need to find a Application engineer who is a confident developer with Matlab and its toolboxes. Competent at explaining to people how and why they are doing things with the technology, able to solve complex problems with simple solutions. Plus, have have a burning desire for technology. Personally, I have been looking for this type of person for almost 4 months and I am almost ready to give in, but I wont be beaten! If anyone is a experienced within C++ / C or Java, and have a lot of experience with using Matlab, then please get in touch with me. My contact details are danielsheppard1983@gmail.com or call me on 0282523120 thanks for letting me post this, I just really needed the help Thanks Dan "Dan Sheppard" <danielsheppard1983@gmail.com> wrote in message news:bd137cfa-cbbb-4069-8d6b-3721cba3b17b@t8g2000pbe.googlegroups.com... > Hi all > > Sorry to advertise within the user group but I really need some help! > I was hoping that people in this usergroup could help other fellow > Matlab developers who may be looking for a new role, or might be > interested in what I have on offer. > > I need to find a Application engineer who is a confident developer >...

How can I use a Matlab function, whose inputs varies with time (real time function), in simulink
Hello I need your help I would llike to know how can I use a Matlab function, whose inputs varies with time (real time function), in simulink. Thank you in advance Hiba ...

help new in MATLAB
Hi Maybe a too easy problem for you but maybe you help me to solve this in MATLAB I have a vector of point let say for example sin(x) from -PI to PI and I would like to plot the -- vector -- derivative of the vector -- integral of the vector any idea how do do this in MATLAB Thks On 8/5/2012 1:15 PM, mourad wrote: > Hi > Maybe a too easy problem for you but maybe you help me to solve this in > MATLAB > I have a vector of point let say for example sin(x) from -PI to PI > and I would like to plot the > -- vector > -- derivative of the vector > -- integral of the vector > doc plot doc diff doc trapz -- ...

Help with Parallel Matlab
Long story short, here is the criteria I need. 1) I need two modules running in parallel. One function must activate these two They don't need to START at the same time but both neither return info nor end. They run pretty much infinitely 1.1) I would prefer if the One function also ran separate from the above two in it's own loop but i can deal with a 3'rd function also not ending. 2) I need a shared variable between two of the modules. 1 Module will place information there while the other just reads it, but as they are running parallel. That's it. I can do this in pretty much any other language but matlab I've been having issues with. Problem is I absolutely need to use matlab for one module. If this is impossible to accomplish all in matlab then can i instead run this in C and call one matlab module with separate thread? Sample code below. function mainParallelTest() persistent counter; counter = 0; parfor i=0:2 if (i == 0) ImageProccessorParallelTest(); elseif(i == 1) EventProcessorParallelTest(); elseif(i == 2) prevCounter = 0; tempCounter = 1; while(true) tempCounter = counter; if (tempCounter == prevCounter) %disp('hi') end prevCounter = tempCounter; end end end "Alex Cruikshank" <cruiksam@gmail.com> writes: > Long story short, here is the criteria I need. >...

Help!! Time Calculations.
Hi all, I'm creating a TimeSheet Database, I need to calculate how many hours the Employee works.The problem is that when they enter the time, it doesn't calculate the minutes, it just calculate hours.it rounds up to 6 and not 5:30. I'm using this formula: Me.Hours = (DateDiff("h", Me.Start, Me.Stop)) Ex: Start Stop Hours 08:30AM 02:00PM 6 <it should be 5 hours and half I would be grateful, if someone can help me. Thanks in advanced. Try using the minutes ("n") argument instead and then divide the result by 60 to get hours as a decimal. Mike Storr www.veraccess.com "christian" <csepulveda@partners.org> wrote in message news:e1cc3cc3.0402030742.550a0e15@posting.google.com... > Hi all, > I'm creating a TimeSheet Database, I need to calculate how many hours > the Employee works.The problem is that when they enter the time, it > doesn't calculate the minutes, it just calculate hours.it rounds up to > 6 and not 5:30. > I'm using this formula: > > Me.Hours = (DateDiff("h", Me.Start, Me.Stop)) > > Ex: > Start Stop Hours > 08:30AM 02:00PM 6 <it should be 5 hours and half > > > I would be grateful, if someone can help me. > > Thanks in advanced. csepulveda@partners.org (christian) wrote in news:e1cc3cc3.0402030742.550a0e15@posting.google.com: > Me.Hours = (DateD...

Time Help #2
Hello, I have this script and it shows GMT time and I need it to show -8 GMT for PST time. Here is the code that needs changing. ( "F j, Y g:ia T", \$row[date] ) What do I change to get it to -8? Any help would be appreciated. gmstrftime() http://www.php.net/gmstrftime On Thu, 13 Dec 2007 15:29:19 -0800 (PST), macca <ptmcnally@googlemail.com> wrote: >gmstrftime() > > >http://www.php.net/gmstrftime I just need an example as I looked at the site and I'm not good with PHP. John schreef: > On Thu, 13 Dec 2007 15:29:19 -0800 (PST), macca > <ptmcnally@googlemail.com> wrote: > >> gmstrftime() >> >> >> http://www.php.net/gmstrftime > > > I just need an example as I looked at the site and I'm not good with > PHP. > The example is quite obvious, what is it that you don't understand? JW John wrote: > ( "F j, Y g:ia T", \$row[date] ) > What do I change to get it to -8? ( "F j, Y g:ia T", \$row[date] - 8*3600 ) i.e. subtract 8 hours (an hour is 3600 seconds) from the time before displaying it. There might be some funniness around daylight savings, but there's always funniness around then. This is why I think daylight savings, and perhaps even time zones as a whole, should be abolished. Everyone should just use UTC for everything all the time. -- Toby A Inkster BSc (Hons) ARCS [Geek of HTML/SQL/Perl/PHP/Python/Apache/Linux] [OS: ...

Matlab LVQ Help
Hi, I am new on neural network toolbox. I am trying to use LVQ algorithm,but it is not working at all. Right now I am trying to train the network for a simple sine curve, but it is not trained according to the target. Can you help me with your suggestion, my function is y=sin(x); x is my input and y is my target output. I classified target results in four different classes. 0.5<=value<=1.0; class 1 0.0<=value<0.5; class 2 -0.5<=value<0.0; class 3 -1<=value<-0.5; class 4 I would highly appriciate if you help me with some idea, i am stucked here for a long time. Here is the code I am using, --------------------------------------------------- t=-5:.1:5; x=t; for i=1:length(x) y(i)=sin(x(i)); end for i=1:1:length(y) if(0.5<(y(1,i)) && (y(1,i))<=1) tc1(1,i)=1; end end for i=1:1:length(y) if(0<=(y(1,i)) && (y(1,i))<0.5) tc1(1,i)=2; end end for i=1:1:length(y) if(-0.5<=(y(1,i)) && (y(1,i))<0) tc1(1,i)=3; end end for i=1:1:length(y) if(-1.0<=(y(1,i)) && (y(1,i))<-0.5) tc1(1,i)=4; end end p1=x; check(1,:)=p1; check(2,:)=tc1; checkall=sortrows(check,2); p=checkall(1,:); tc=checkall(2,:); t=ind2vec(tc); targets=full(t); sc=16; calu(1,1)=sum(targets(1,:)); calu(2,1)=sum(targets(2,:)); calu(3,1)=sum(targets(3,:)); calu(4,1)=sum(targets(4,:)); c1=calu(1,1)/(sum(calu(1,1)+calu(2,1)+calu(3,1)+calu(4,1))); c2=calu(2,1)/(sum(calu(1,1)+calu(2,1)+calu(3,1)+calu(4,1))); c3=calu...

Need Help With Matlab
Hi I am trying to simulate the low pass filtered output of a square wav (pulse train) using Matlab. The pulse varies between 0 and 1. So th filtered output looks somewhat like a sinewave. The filter is a low pas filter (FIR/raised cosine filter). Essentially what I need is to be abl to generate a look up table that will store the output waveform rise fall transition (from 0-1 and 1-0). So what i want essentially is to b able to generate a look up table of different lengths (like 100,50 etc) t store these transitions. I tried using Simulink for this, but havent bee successful so far. So my table will look something like 1 0.995 0.923 0.91 0.85 ..... to 0 Any help is highly appreciated. Thanks. Best regards Vikram ... Web resources about - Help with computing the Crame Rao Lower Bound for Time Delay Estimation - comp.soft-sys.matlab Estimation - Wikipedia, the free encyclopedia
... here. For the racehorse, see Estimate (horse) . For the card game, see Estimate (card game) . For the symbol, see Estimated sign . Estimation ...

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Evolving Estimation Process
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