Re: [Scilab-users] Scan file backward from the end?
On Wednesday, March 20, 2013 23.43 CET, Serge Steer serge.st...@inria.fr wrote: If you know the approximate size of the footer in number of bytes (assume N greater than it) you can try to set the pointer N bytes before the end of file using the u=mopen(myfile); mseek(-N,u,'end') then read the rest with mgetl(-1,u) Serge Thanks for the tip, I'll give it a try today. Cheers, Antoine ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] Computing Performance and debugging in Scilab.
Strange, at me houses my computer gives out always 0.. Has tested on the working computer 0.047. Question I think I shall be closed to understand what a problem of a house. -- View this message in context: http://mailinglists.scilab.org/Computing-Performance-and-debugging-in-Scilab-tp4026287p4026315.html Sent from the Scilab users - Mailing Lists Archives mailing list archive at Nabble.com. ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
[Scilab-users] problem loading enviroment
Hello, sometimes, when I try to load an eviroment, I got a message like this: loading initial environment Warning : redefining function: %_sodload . Use funcprot(0) to avoid this message inside function: createMacro . = (pos,V,R,calc_D_superp_dobcel(N, N2)) !--error 41 Incompatible output argument. at line 16 of function coord called by : at line 13 of function createMacro called by : at line 8 of function %__convertVariable__ called by : at line 836 of function %_sodload called by : then load(%fileToLoad); disp(msprintf(gettext while executing a callback and it does´t load anything. I had no problem saving it, or runing it, or manipulating the variables, or any kind of problem before or after saving, but it does not load... there seems to be some problem with some of the functions saved in the enviroment but, I repeat, I had no problem saving or running them. I can load the variables if I list them (as in load(filename,var_1,var_2,...var_n)), but it is really stupid I have to do this instead of just loading everything... ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
[Scilab-users] discrete Fourier transform
hello,
for signalanalysis I would like to use discrete Fourier transform (dft). To
see, how it works, I use the simple example below. Why is *XfA =
abs(Xf)*2/n* respectively why is XfA = abs(Xf) wrong?
clear; clc; xdel;
function y = f(x);
y = sin(x);
endfunction;
n = 200;
x=linspace(0,2*%pi,n);
y=f(x);
mat = [x',y'];
//disp(mat);
//plot2d(x , y);
//xtitle('DATA','n''','y''');
Xf=dft(y,-1);
XfA = abs(Xf)*2/n;
plot2d3([1:n/2]',XfA(1:n/2));
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[Scilab-users] Help to set parameters in an Xcos Diagram
Hello everybody,
I'd like to set parameters of my scs_m structure for an Xcos Diagram (see
attached file) using command lines.
To be specific I'd like to set the following imulation parameters :
- final integration time
- max integration time interval
I want them to be equal to a dt that I already define in my program. I am
using this code so far :
loadXcosLibs() importXcosDiagram('C:\STAMPE_3.5\STAMPE_SRC\essai.xcos');
typeof(scs_m) // Diagram structure
scs_m.props.tf = dt;// Final integration time
xcos_simulate(scs_m,4);
But I don't know what command line to use to set the max step of integration
time for my simulation.
Also, I'd like to set to dt in the same way the parameters (period,
iniialization time) of the block CLOCK_c which is in my xcos diagram.
I think I must use something like scs_m.objs.something = dt.
I would be very thankful for any help :) essai.xcos
http://mailinglists.scilab.org/file/n4026319/essai.xcos
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Re: [Scilab-users] discrete Fourier transform
Le 21/03/2013 16:41, haasejos a écrit :
hello,
for signalanalysis I would like to use discrete Fourier transform (dft). To
see, how it works, I use the simple example below. Why is *XfA =
abs(Xf)*2/n* respectively why is XfA = abs(Xf) wrong?
Why do you say that XfA = abs(Xf) is wrong
Note however. It is much more efficient using fft instead of dft.
Serge Steer
clear; clc; xdel;
function y = f(x);
y = sin(x);
endfunction;
n = 200;
x=linspace(0,2*%pi,n);
y=f(x);
mat = [x',y'];
//disp(mat);
//plot2d(x , y);
//xtitle('DATA','n''','y''');
Xf=dft(y,-1);
XfA = abs(Xf)*2/n;
plot2d3([1:n/2]',XfA(1:n/2));
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Re: [Scilab-users] problem loading enviroment
The people, I have checked up houses once again, a problem in that that if to create test script in a file for example TicToc.sci and to start this file, shows always 0.00, and if to copy the text to insert and to start in a command window directly without start of a file. sci (or sce) that shows 0.0160. How it to understand? In what the difference when is started script from SciNotes??? -- View this message in context: http://mailinglists.scilab.org/Computing-Performance-and-debugging-in-Scilab-tp4026287p4026321.html Sent from the Scilab users - Mailing Lists Archives mailing list archive at Nabble.com. ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] discrete Fourier transform
On 2013-03-21 17:13, Serge Steer wrote:
Le 21/03/2013 16:41, haasejos a écrit :
hello,
for signalanalysis I would like to use discrete Fourier transform
(dft). To
see, how it works, I use the simple example below. Why is *XfA =
abs(Xf)*2/n* respectively why is XfA = abs(Xf) wrong?
Why do you say that XfA = abs(Xf) is wrong
Note however. It is much more efficient using fft instead of dft.
Serge Steer
Maybe haasejos meant and should have used XfA == abs(Xf)
Stefan
clear; clc; xdel;
function y = f(x);
y = sin(x);
endfunction;
n = 200;
x=linspace(0,2*%pi,n);
y=f(x);
mat = [x',y'];
//disp(mat);
//plot2d(x , y);
//xtitle('DATA','n''','y''');
Xf=dft(y,-1);
XfA = abs(Xf)*2/n;
plot2d3([1:n/2]',XfA(1:n/2));
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Re: [Scilab-users] discrete Fourier transform
good evening, what I meant is, that XfA(2) should be 1. Because this value can be understood as the amplitude of sin(x). But XfA(2) beeing calculated with XfA=abs(Xf) (see example) returns 100. This is wrong, isn't it? Josef -- View this message in context: http://mailinglists.scilab.org/discrete-Fourier-transform-tp4026318p4026324.html Sent from the Scilab users - Mailing Lists Archives mailing list archive at Nabble.com. ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] discrete Fourier transform
Hi, I think this is just a question of scaling. There is no correct scaling for the FFT - just some different conventions. This is explained here (http://www.mathworks.co.uk/matlabcentral/answers/15770-scaling-the-fft-and- the-ifft). The correct answer with Scilab would be exactly 100 (N/2) in your case, but you have a small error because your waveform is not exactly periodic - the first and last sample are the same, but you should have the last sample being the one before the first sample for a complete cycle. The following code does that: --n=201; --x=linspace(0,2*%pi,n); --x=x(1:200); --y=sin(x); --Xf=dft(y,-1); --XfA = abs(Xf); --XfA(2) ans = 100. HTH, Mike. -Original Message- From: users-boun...@lists.scilab.org [mailto:users-boun...@lists.scilab.org]On Behalf Of haasejos Sent: 21 March 2013 21:14 To: users@lists.scilab.org Subject: Re: [Scilab-users] discrete Fourier transform --n = 200; --x=linspace(0,2*%pi,n); --y=sin(x); --Xf=dft(y,-1); --XfA = abs(Xf); *please look at the difference! * --XfA(2) ans = 99.745189 -- View this message in context: http://mailinglists.scilab.org/discrete-Fourier-transform-tp4026318p4026326. html Sent from the Scilab users - Mailing Lists Archives mailing list archive at Nabble.com. ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users - No virus found in this message. Checked by AVG - www.avg.com Version: 2013.0.3267 / Virus Database: 3160/6194 - Release Date: 03/21/13 ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
