Re: [Scilab-users] Define numerator and denominator of transfer function as variables in Xcos (DLR block)

Hi, Another way to represent filter in Xcos is to use the Discrete State-Space (DLSS) block. The matrices can be obtained directly from the syslin representation of the system: [A1 B1 C1 D1] = abcd(H1); [A2 B2 C2 D2] = abcd(H2); (This transformation can also be set in Xcos context) Yours 2017

Re: [Scilab-users] simple plot function

Thanks Pierre and Lamy. Now I have another question: I draw a simple function and in same pic, also draw differents points (this data in a vector) also I want to put a label for each point "1", "2".."12" how can I do it in scilab?, matlab's way doesn't work in my case thanks x=0:0.1:1

Re: [Scilab-users] contour : z level use

Several tests of contour's zlev parameter failed to produce contours at the request depth level. This looks like a bug too. -- View this message in context: http://mailinglists.scilab.org/Scilab-users-contour-z-level-use-tp4036103p4036114.html Sent from the Scilab users - Mailing Lists Archive

Re: [Scilab-users] contour : z level use

Le 02/04/2017 à 14:49, Rafael Guerra a écrit : // NOTE: the previous contour code line should read: contour(X,Y,Z,[Zc Zc],flag=[0 0 4]); Right. When Zc is not a strictly positive integer, it should be taken as a single Z level. But it is currently not the case. It is a bug : http://bugzilla.sc

Re: [Scilab-users] contour : z level use

Le 02/04/2017 à 14:27, paul.carr...@free.fr a écrit : Hi Samuel Even when using mode = 1 (as the other in fact) in the flag, the result remains identical Indeed. I was misled by the zlev option. It is just the single level of the plane on which contours must be plotted (instead of ?) in addi

Re: [Scilab-users] Define numerator and denominator of transfer function as variables in Xcos (DLR block)

you're welcome. anyway it should be a better way to construct the denominator, do let me know if you find a better way. :) rgds, CL On 2/4/2017 12:22 AM, Nikolay Strelkov wrote: Dear Tan Chin Luh! Thank you very much. It works as expected! The problem is solved. -- /With best regards, Ph.D

Re: [Scilab-users] contour : z level use

amazing ... once again many thanks Rafael (need time to have a look on the code and to catch how to modify some features) Thanks Paul Le 2017-04-02 14:49, Rafael Guerra a écrit : > // NOTE: the previous contour code line should read: > contour(X,Y,Z,[Zc Zc],flag=[0 0 4]); > > -- > View t

Re: [Scilab-users] contour : z level use

// NOTE: the previous contour code line should read: contour(X,Y,Z,[Zc Zc],flag=[0 0 4]); -- View this message in context: http://mailinglists.scilab.org/Scilab-users-contour-z-level-use-tp4036103p4036109.html Sent from the Scilab users - Mailing Lists Archives mailing list archive at Nabble

Re: [Scilab-users] contour : z level use

Hi, Try code below, which does not uses zlev but nz = [0,0] instead. Attached picture is produced. // START OF CODE X = linspace(-3,3,100); Y = linspace(-3,3,100); Z = sin(X'*Y)-0.3; //+ X'*ones(X); clf() xset("colorm

Re: [Scilab-users] contour : z level use

Hi Samuel Even when using mode = 1 (as the other in fact) in the flag, the result remains identical _Nb_: I probably pasted the latest trial I did :-) Paul // flag=[mode,type,box] // mode : //string representation mode. //mode=0: the level curves are drawn on the surface defined by

Re: [Scilab-users] Fundamental problem with "histc"

Found out myself: Scilab help pages say histc computes an histogram Description This function computes a histogram of the data vector using the classes x. When the number n of classes is provided instead of x, the classes are chosen equally spaced and x(1) = min(data) < x(2) = x(1) + dx < ... < x(

Re: [Scilab-users] contour : z level use

Hello, contour(X,Y,Z,1,flag=[0 1 4],zlev=0.); You are setting an inappropriate mode in flag. Samuel Le 02/04/2017 à 09:16, paul.carr...@free.fr a écrit : Hi I've been thinking that Scilab uses the closest zlev than the one I'm specifying, based on the points I provided ... am I correct?

Re: [Scilab-users] contour : z level use

Hi I've been thinking that Scilab uses the closest zlev than the one I'm specifying, based on the points I provided ... am I correct? I've to find a way to create the points I need with z=0 Paul Le 2017-04-01 23:42, paul.carr...@free.fr a écrit : > Hi > > In the following example,