[Scilab-users] Computing Performance and debugging in Scilab.

[Rafael Guerra]

Hi,

In my Win7 Scilab 5.4 your code outputs the time of execution of the fft
command:

   -->printf("%.20f\n",t);
   0.14100000000000001000

What else do you need to know?

Rgds
Rafael

-----Original Message-----
From: users-bounces at lists.scilab.org [mailto:users-bounces at lists.scilab.org]
On Behalf Of TViT
Sent: Wednesday, March 20, 2013 5:16 PM
To: users at lists.scilab.org
Subject: Re: [Scilab-users] Computing Performance and debugging in Scilab.

Excuse, but I can not measure time of performance of function. What I not
correctly make?

clc;
clear;
Fs = 11025;
t = 0:1/Fs:1;
N=size(t,'*');

Signal=sin(2*%pi*500*t)+sin(2*%pi*1500*t)+sin(2*%pi*2500*t)+sin(2*%pi*3500*t
)+sin(2*%pi*4500*t)+grand(1,N,'nor',0,1);
disp(N);

tic();
y=fft(Signal);

t=toc();
printf("%.20f\n",t);

f=Fs*(0:(N/2))/N;
n=size(f,'*');
 bar(f,abs(y(1:n)),0.01,'blue')
//sleep(2000);close();



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