MATLAB Central Newsreader Recent Posts

Low pass filtering

Dias Papas — Sat, 07 Nov 2009 20:08:02 -0500

I am new in signal processing and I want to filter some time series.
I searched and I found the below code.
It seems to work but I am not sure if it works properly.
And I have two questions.

First: Why the first values of the output 'y' are close to 0? Is it correct? Wouldn't be more logic to be close to the input data 'x'?

And the second one: How can I plot the input data 'X' in the frequency domain with values of Hz, so I will see the frequencies of 'X' and easily decide the cut off frecuency.

If I am in completely wrong thought, please tell me.

The code:

fs = 300;
n = 0:1/fs:0.2;
x = cos(2*pi*30*n) + cos(2*pi*50*n) + sin(2*pi*60*n);
wc = pi*40 / (fs/2);
N = 21;
n = 0:1:N-1;
hd = ideal_lp(wc,N);
w_tet = (boxcar(N))';
h = hd.*w_tet;
y = filter(h,1,x);

fr = 0:0.01:pi;
X = freqz(x,1,fr);
H = freqz(h,1,fr);
Y = freqz(y,1,fr);

figure (1)
subplot(311)
plot(fr,abs(X)); axis([0 3.5 0 40])
subplot(312)
plot(fr,abs(H)); axis([0 3.5 0 2])
subplot(313)
plot(fr,abs(Y)); axis([0 3.5 0 40])

figure (2)
subplot(311)
plot(x)
subplot(312)
plot(h)
subplot(313)
plot(y)

Thanks veeeery much
Dias

Re: radial-basis network

Greg Heath — Sat, 07 Nov 2009 20:07:07 -0500

On Nov 7, 12:10 pm, banakil <bana...@list.ru> wrote:
> Does the implementation of a radial-basis network using linear equations in matlab?

The hidden layer weights are training vectors
that generated the worst errors during training.
One weight is added each epoch.

The output layer weights are obtained by solving a
linear system between the hidden nodes and the
output nodes. They are recalculated after each
new hidden node is added.

Hope this helps.

Greg.

Re: Computational complexity of fmincon algorithm

Lior Vernia — Sat, 07 Nov 2009 20:06:02 -0500

Hi Matt,

Thanks for the speedy reply. You are correct, of course. However, let's assume that since the objective function is written by me, its complexity is known to us. We can call it T(n), n being the number of variables. Let's also say that f(x) is analytical and that I'm supplying the gradients and the Hessian matrix, and thus we know their complexity as well. What should the complexity of the entire call to fmincon be? Obviously fmincon is the one who decides how many times to compute f(x), and its gradients and Hessian, so how does that depend on n?

Thanks, Lior.

Using subclass 'get' function to access inherited superclass property

Janice — Sat, 07 Nov 2009 19:20:04 -0500

Hi all,

I have read several posts about the problems with overloading a superclass 'set'/'get' function, but I'm interested in simply writing a subclass 'get' function that is able to return an inherited superclass property. It seems that the 'get' function is blind to the superclass properties, despite the fact that one could simply access this property by using dot notation with the subclass object (if the property is public?) elsewhere in the code. The code I'm using is below:

classdef ExptBasic < handle

     properties (GetAccess = 'public', SetAccess = 'protected')
          logfile_name = '';
     end

     methods
          function value = get.logfile_name(Expt)
               value = Expt.logfile_name;
          end
     end

end

classdef TrialBasic < ExptBasic
     methods
          function get.logfile_name(Trial)
               value = get.logfile_name@ExptBasic(Trial);
          end
     end
end

Am I going about this the wrong way? I tried extending the superclass 'get.logfile_name' method in the subclass TrialBasic.

Thanks in advance for the help,
Janice

Re: Computational complexity of fmincon algorithm

Matt — Sat, 07 Nov 2009 19:20:03 -0500

"Lior Vernia" <liorvern@post.tau.ac.il> wrote in message <hd4b47$mq$1@fred.mathworks.com>...
> Hello,
>
> I've wondered what the computational complexity of the fmincon algorithm is (the medium scale one, I think it's a line search). I'd especially like to know how it depends on the number of variables; is it exponential? I thought it'd be easier to ask here than to look at the m-file and try to figure out the exact algorithm myself.
======

That would really depend on your objective function. For example

f(x)=norm(x)^2

is obviously linear in the number of variables, while for a dense square matrix A

f(x)=norm(A*x)^2

is obviously quadratic in the the number of variables, because computing A*x is O(N^2)

Problem with using ODE solvers

Markthomas — Sat, 07 Nov 2009 19:10:04 -0500

I am trying to run an ODE case usin ode45... the scripts are seen below:
---------------------------------------------code 1------------------------------------------------------------------
function varargout = odefile_V2(t,T,flag,tau,ET,R1,rho1,L1,C1,R3,R2,rho2,L2,C2,rho3,L3,C3)

switch flag
    case ''
        varargout{1} = f(t,T,tau,ET,R1,rho1,L1,C1,R3,R2,rho2,L2,C2,rho3,L3,C3);
    case 'init'
        [varargout{1:3}] = init(tau,ET,R1,rho1,L1,C1,R3,R2,rho2,L2,C2,rho3,L3,C3);
end

%--------------------------------------------------------------------------
function dydt = f(t,T,tau,ET,R1,rho1,L1,C1,R3,R2,rho2,L2,C2,rho3,L3,C3)

dydt(1) = (-(ET*(T(1)^4-(1200*(1-exp(-t/tau)))^4)+(R1*ET*(T(1)^4-(1200*(1-exp(-t/tau)))^4)+T(1)-T2(t))/R2)/rho1/L1/C1+4*ET*R1*(1200*(1-exp(-t/tau)))^3*((1200/tau)*exp(-t/tau)))/(4*ET*R1*T(1)^3+1);
dydt(2) = (R3*ET*R1*T(1)^4+R3*T(1)+(-R3-R2)*T(2)-R3*ET*R1*(1200*(1-exp(-t/tau)))^4+R2*T(3))/R2/R3/rho2/L2/C2;
dydt(3) = (T(2)-T(3))/R3/rho3/L3/C3;

function [tspan,y0,options] = init(tau,ET,R1,rho1,L1,C1,R3,R2,rho2,L2,C2,rho3,L3,C3)
tspan = [0; 100];
y0 = [300; 300; 300];
options = [];
---------------------------------------------code 2------------------------------------------------------------------
clear; clc; close all
% Material @ Node 0
T0_0 = 300;
% Material @ Node 1
T1_0 = 300;
K1 = 60;
rho1 = 8000;
C1 = 440;
L1 = .07;
% Material @ Node 2
T2_0 = 300;
K2 = 1.4;
rho2 = 1300;
C2 = 1500;
L2 = .04;
% Material @ Node 2
T3_0 = 300;
K3 = 50;
rho3 = 7500;
C3 = 400;
L3 = .02;
%Resistances
R1 = L1/(2*K1);
R2 = R1 + (L2/(2*K2));
R3 = (L2/(2*K2)) + (L3/(2*K3));
% Other Constants
ET = 4e-8;
tau = 50;

[t,T]=ode45('odefile_V2',tau,ET,R1,rho1,L1,C1,R3,R2,rho2,L2,C2,rho3,L3,C3);
------------------------------------------------------------------------------------------------------------------------
i keep getting errors and was hoping somebody was able to figure out what is going wrong. I have rewritten the code several times and can't seem to get it to work... any help would be great!

Thanks!
MT

Re: Data cursor coordinates -- increasing precision or number of decimal places

Richard Sonnenfeld — Sat, 07 Nov 2009 19:05:03 -0500

I stumbled across the thread because I was looking for a way to change the
number of decimal digits (precision) displayed by the data cursor (or data-tip) as they
seem to call it. Four digits is not necessarily enough for fine work.

It turns out to be easy -- once you think of it. The poster
figured out his question. I'll document what he discovered when he said "nevermind --
I got it."

1) Select Data Cursor tool and click a point on your plot
2) Right click the tool and select "Edit Text Update Function"
3) On line 8 you will see an expression "output_txt = ... (pos(1),4)...
(pos(1),4)...
4) Change the ,4's to ,7's or whatever level of precision you need.
5) Save the file as "custom_cursor.m" (or whatever you like
6) Next time you need to use the cursor, repeat step 1, but then Right Click and
select "Select Text Update Function". Now select your custom_cursor.m and
you will have your customized cursor.
7) The cursor always returns to its default state. For now, I don't mind selecting custom_cursor every time I need one. I'll be happy to have a suggestion on how to permanently change the cursor if someone feels like making one.

Re: handles pass strings

ucd puri — Sat, 07 Nov 2009 19:02:02 -0500

i would say rather try
strcmp(alpha,'A') ---> strcmp is case sensitive
strcmpi(alpha,'A') ---> strcmpi is NOT case sensitive

Re: MLE - use with Discrete Weibull mean function

Erik Gadzinski — Sat, 07 Nov 2009 18:59:01 -0500

"Erik Gadzinski" <ergadz@yahoo.com> wrote in message <hd4ed6$l92$1@fred.mathworks.com>...
> I am having trouble with an MLE function that was passed along to me. I am new to MATLAB and am having trouble shooting problems. I am taking 'v' values that are randomly generated that are discrete numbers that range from 0 to a couple thousand. In the code, "v' can have a record length of 10, 50, 100, 1000, and 10,000. In the Llikef function I would like to take the log of the function, but I have been getting log zero when large numbers usually over 1,022 are entered into the function. So I took the log out. The thing I need help with is that I want my q and N (eta) values to be positive. The function sometimes outputs negative q's and eta's. Does anybody know how to do this? Thank you in advance!

In short, how do I make sure that "that" comes back postive in the two lines below: (the rest of the code is below that)

that0 = [0.5 1];
[that,Llikef] = fminsearch(Llikef,that0)
>
> In the function that(1) = q and that(2) = eta.
> vsim are whole numbers that range from 0 to a few thousand
>
> %find the Mean MLE Fitted Average.
> %1) take each record length of 'v' values and plug it into the MLE routine
> %2) use the q and eta values generated from that function and plug them
> %in the DW mean function
>
>
> slength=1;
> elength=recordlength;
> meanindex=1;
> vsim;
> while meanindex<=100
> if sum(vsim(slength:elength))>0
> x=vsim(slength:elength);
>
> Llikef = @(that)-sum(max(that(1),.00001).^(x.^max(that(2),.00001))-max(that(1),.00001).^((x+1).^max(that(2),.00001)));
> %Llikef = @(that)-sum(log(max(that(1),.00001).^(x.^max(that(2),.00001))-max(that(1),.00001).^((x+1).^max(that(2),.00001))))
>
> that0 = [0.5 1];
> [that,Llikef] = fminsearch(Llikef,that0);
>
> if Llikef<=0
> that(1)=.5;
> that(2)=1;
> else end
> if that(1)>0
> q2(meanindex)=that(1);
> else q2(meanindex)=.5;
> end
> if that(2)>0;
> eta2(meanindex)=that(2);
> else eta2(meanindex)=1;
> end
>
> N=eta2(meanindex);
> q=q2(meanindex);
> M=1e5;
> vm=1:M;
> mmlefm(meanindex)= sum(q.^(vm.^N))+(gamma(1/N)*gammainc((M+1)^N*(-log(q)),1/N,'upper'))/(N*(-log(q))^(1/N));
> meanindex=meanindex+1;
> slength=elength+1;
> elength=elength+recordlength;
> else
> q2(meanindex)=.5;
> eta2(meanindex)=1;
> mmlefm(meanindex)=0;
> slength=elength+1;
> elength=elength+recordlength;
> meanindex=meanindex+1;
> end
> end

Re: matrix operations in F2

Bruno Luong — Sat, 07 Nov 2009 18:57:01 -0500

"Joren Heit" <jorenheit@gmail.com> wrote in message <hd42o0$re0$1@fred.mathworks.com>...
> Not what I meant...
> I need to solve a set of equations like
>
> | 1 1 0 0 | | x1 | | 1 |
> | 1 1 1 0 | | x2 | = | 1 |
> | 0 1 1 1 | | x3 | | 1 |
> | 0 0 1 1 | | x4 | | 1 |
>
> Under the conditions that 0-1=0 and 1+1=0, this results in
> x = (1 0 0 1)
>
> Now I need to do this for very large matrices....
>

I guess you have to write your own Gaussian pivoting, unless if there is such thing already on FEX.

This has been discussed here:
http://www.mathworks.com/matlabcentral/newsreader/view_thread/261760#684562

Bruno

Histogram

Jonas — Sat, 07 Nov 2009 18:50:04 -0500

Does anybody have a better solution for the code below? The x- and edge-vectors has a size of 5000 elements in my problem, and the sum of k-vector is equal to the length of the x- and edge-vectors, the code allocates a lot of memory unnecessarily.

Similar problem
x = (1:1:11)';
u = [2 3.5 4 5 6 5 4 4 3 2 1]';
umin = min(u);
umax = max(u);
edge = linspace(umin,umax,numel(x))';
[k bin]=histc(u,edge);

u1 = zeros(length(k),max(k));
edge = flipud(edge);
for i=1:length(k)
u1(i,1:k(i))=edge(i);
end
[m,n] = size(u1);
for i=1:m
u2((i*n+1):((i+1)*n)) = u1(i,:);
end
ind1=find(u2==0);
u2(ind1)=[];
subplot(2,1,1); plot(x,u)
subplot(2,1,2); plot(x,u2)

handles pass strings

eematic ee — Sat, 07 Nov 2009 18:39:01 -0500

Do "handles" pass strings? I have a program that enters a string using a popup box and passes it using handles to a input callback then an Evaluation callback. The proces worked for a number. In the Evaluation callback there is an error:

if alpha = 'A' then
delta = -delta;
end;

The error is in the equals statement in "if alpha = 'A' then". Does this mean that "handles" is not passing strings thru alpha to the Evaluation callback?

MLE - use with Discrete Weibull mean function

Erik Gadzinski — Sat, 07 Nov 2009 18:31:02 -0500

I am having trouble with an MLE function that was passed along to me. I am new to MATLAB and am having trouble shooting problems. I am taking 'v' values that are randomly generated that are discrete numbers that range from 0 to a couple thousand. In the code, "v' can have a record length of 10, 50, 100, 1000, and 10,000. In the Llikef function I would like to take the log of the function, but I have been getting log zero when large numbers usually over 1,022 are entered into the function. So I took the log out. The thing I need help with is that I want my q and N (eta) values to be positive. The function sometimes outputs negative q's and eta's. Does anybody know how to do this? Thank you in advance!

In the function that(1) = q and that(2) = eta.
vsim are whole numbers that range from 0 to a few thousand

%find the Mean MLE Fitted Average.
%1) take each record length of 'v' values and plug it into the MLE routine
%2) use the q and eta values generated from that function and plug them
%in the DW mean function

slength=1;
elength=recordlength;
meanindex=1;
vsim;
while meanindex<=100
    if sum(vsim(slength:elength))>0
        x=vsim(slength:elength);

        Llikef = @(that)-sum(max(that(1),.00001).^(x.^max(that(2),.00001))-max(that(1),.00001).^((x+1).^max(that(2),.00001)));
        %Llikef = @(that)-sum(log(max(that(1),.00001).^(x.^max(that(2),.00001))-max(that(1),.00001).^((x+1).^max(that(2),.00001))))

        that0 = [0.5 1];
        [that,Llikef] = fminsearch(Llikef,that0);

        if Llikef<=0
            that(1)=.5;
            that(2)=1;
        else end
            if that(1)>0
                  q2(meanindex)=that(1);
               else q2(meanindex)=.5;
               end
               if that(2)>0;
                  eta2(meanindex)=that(2);
               else eta2(meanindex)=1;
               end

        N=eta2(meanindex);
        q=q2(meanindex);
        M=1e5;
        vm=1:M;
        mmlefm(meanindex)= sum(q.^(vm.^N))+(gamma(1/N)*gammainc((M+1)^N*(-log(q)),1/N,'upper'))/(N*(-log(q))^(1/N));
        meanindex=meanindex+1;
        slength=elength+1;
        elength=elength+recordlength;
    else
        q2(meanindex)=.5;
        eta2(meanindex)=1;
        mmlefm(meanindex)=0;
        slength=elength+1;
        elength=elength+recordlength;
        meanindex=meanindex+1;
    end
end

Volume Of Interest

ucd puri — Sat, 07 Nov 2009 18:30:06 -0500

Hello All,

I am using matlab for my project, i am wondering if it is possible to draw a volume of interest on the 3D dataset.

For Example, i can use functions imrect, imellipse, etc... to draw a 2D region of interest in 2D image/slices. but is there a way i can draw a cuboid/cube i.e. a 3D volume of interest on a 3D volumetric dataset formed by the multiple 2D slices?

Kindly suggest
Thanks a lot.

Doubly Linked Lists of objects

rebecca donnelly — Sat, 07 Nov 2009 18:28:01 -0500

Hi,
I am a first year PhD student and am implementing the image processing algorithm "Seeded Region Growing" as an exercise to help me get to grips with Matlab and to gain a deeper understanding of the algorithm.

The paper asks for a datastructure they call a "Sequentially Sorted List", it needs to hold an object which contains the x and y coordinates of a pixel and another value called Sigma which the list is sorted by.

I attempted to do this using the following code:

% Calculate sigma and create a new pxladrs object
        sigma_value = abs(seg_img(x+1, y) - mean_gl);
        p = pxladrs(sigma_value);
        % Add x and y coords to pxladrs object
        p.Xvalue = x+1;
        p.Yvalue = y;
        % Add object to SSL
        if counter==0
            ssl_start = dlnode(p);
        else
            cur = ssl_start;
            % Insert ordered by sigma value
            while cur ~= 0
                if p.Sigma < cur.Data.Sigma
                    ssl_insert = dlnode(p);
                    ssl_insert.insertBefore(cur);
                    break
                end
         end
                cur = cur.Next;

However this didn't work as I couldn't store the pxladrs object in the doubly linked list.

Error:
??? Undefined function or method 'dlnode' for input arguments of
type 'pxladrs'.
Error in ==> SeededRegionGrowing at 58
            ssl_start = dlnode(p);

So my question is; How do I modify the doubly linked list so it can handle objects? Or am I going about this totally the wrong way?

Any comments appriciated,
Rebecca