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# Ezplot Color

## Contents

I remember one time a friend was trying to find the roots of a complicated polynomial (too complicated to solve analytically) and decided to use MATLATB. EZPLOT requires that symbolic expressions are functions of only 2 symbolic variables. Please tell if this is the plot you need. –Nemesis Oct 20 '14 at 11:06 it works as a dream thanks again! :) –Sergio Haram Oct 20 '14 at What does a well diversified self-managed investment portfolio look like?

Developing web applications for long lifespan (20+ years) Are "ŝati" and "plaĉi al" interchangeable? Enviar a un amigoOtras secciones relacionadas con MatlabCursosBiblioteca de TemasCódigo FuenteChatGeneral RSS del foroExpertosTus mensajesRecomendar Seguir a @lwp_ Información y RecursosCursos y ManualesBiblioteca de TemasCódigo FuenteNoticias/ArtículosForos y ConsultasForos de ConsultaChats de For example, the MATLAB® syntax for a plot of the expressionx.^2 - y.^2which represents an implicitly defined function, is written asezplot('x^2 - y^2')That is, x^2 is interpreted as x.^2 in the However, for the sake of the following example, let's assume it's an independent variable. their explanation

## Error Using Ezplot At 70

The equation: y^2 + x*y + x*y^3 + y1 = 0 would be solved for y1 to get the following: y1 = -y^2 - x*y - x*y^3 and y1 would be Error in ==> inline.feval at 34 INLINE_OUT_ = inlineeval(INLINE_INPUTS_, INLINE_OBJ_.inputExpr, INLINE_OBJ_.expr); Error in ==> specgraph\private\ezplotfeval at 54 z = feval(f,x(1)); Error in ==> ezplot>ezplot1 at 448 [y,f,loopflag] = ezplotfeval(f,x); Error in Plot output of FourierTransform in mathematica MX record security Project going on longer than expected - how to bring it up to client? d = x^2 –6*x – 12; >> ezplot(d) Error using inlineeval (line 15) Error in inline expression ==> x.^2 –6.*x – 12 Error: The input character is not valid in MATLAB

I hope this doesn't run afoul of the homework rules here, but this is a small aspect of a larger project. La operación es la siguiente: 1234567891011121314151617181920s=tf('s')Gr=0.762+0.0434/s;G=2.5/(50*s^3+112.5*s^2+25.5*s+1);tfin=20;Go=series(Gr,G)Gw=feedback(Go,1)syms s[numGr,denGr]=tfdata(Gr);Gr_sym=poly2sym(numGr{1,1},'s')/poly2sym(denGr{1,1},'s')[numG,denG]=tfdata(G);G_sym=poly2sym(numG{1,1},'s')/poly2sym(denG{1,1},'s')Go_sym=Gr_sym*G_symGw_sym=Go_sym/(1+Go_sym)gw_sym=ilaplace(Gw_sym);figureezplot(Gw,[0 tfin]);title('Pulse characteristics of the closed loop’)axis auto Agradecería mucho un poquito de ayuda :)Valora esta pregunta0ResponderOtras secciones de LWP con contenido de Matlab- Código Sum of neighbours What are Imperial officers wearing here? Ezplot Line Style Technically, y1 is a dependent variable the way you have defined it (since it depends on the variable x).

was in office.Me too. Let me give an example of a small number of terms and it can be generalized to a large number of terms. Survey tool to ask questions on individual pages - what are they called? a fantastic read Thus you can call either new_p1x1(30, 70) # 30 is passed to f1_p1 and 70 to f2_p2 or new_p1x1([30,70], [30,70]) # The matrix [30,70] is passed to both function.

The code looks like this: k=10; G=k/((2*s+1)*(3*s+1)); T=G/(1+G); C=ilaplace(T/s); figure (4) ezplot(C) This works perfectly. Ezplot 3d But if it isn't being used as part an education, one needs to purchase the commercial license. I have tried the same as before but it doesn't work....(fplot(new_p1x1, [30,70])) I get: Error using @(xq1,xq2)f1_p1(xq1).*f2_p1(xq2) Not enough input arguments. Even though I eventually solved my problem with a different approach, I'm going to bookmark this thread because I've had very similar problems before.

## Ezplot Matlab

matlab symbolic-math share|improve this question edited Jul 2 '15 at 8:13 hbaderts 8,17721737 asked Jul 2 '15 at 6:02 user55531 253 add a comment| 1 Answer 1 active oldest votes up Reload the page to see its updated state. How to clean Car's HVAC and AC system Is intelligence the "natural" product of evolution? I'm going to do some research into Maple, because I know MATLAB can use it and it probably will come in handy in the future. Ezplot Legend

Muchisimas gracias! Please try the request again. Please tell me how can I correct the error. There's a closed-form analytic expression for both of them, and in this case the roots are all real (in fact, they're the poles of the original function you're transforming.) DrCube03-04-2009, 02:42

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## Is it possible to use this feature (splitting the equation or function into a number of terms)?

Error: The input character is not valid in MATLAB statements or expressions. Related Content Join the 15-year community celebration. By default ezplot(fun) plots the expression fun(x) over the default domain -2π < x < 2π, where fun(x) is an explicit function of only x. Matlab Ezplot Y Range Error using ==> inlineeval Error in inline expression ==> -1000./121524455857.*sum((27047643+61502320.*_alpha+612012.*_alpha.^2).*exp(_alpha.*t),_alpha = RootOf(605.*_Z.^2+501.*_Z+1100+6.*_Z.^3)) ???

I was a teaching assistant for a self-paced MATLAB course all through college, which was also when Bush Sr. Apply Today MATLAB Academy New to MATLAB? Related Content 4 Answers Geoff Hayes (view profile) 0 questions 2,000 answers 1,074 accepted answers Reputation: 5,624 Vote0 Link Direct link to this answer: https://www.mathworks.com/matlabcentral/answers/132703#answer_139646 Answer by Geoff Hayes Geoff Hayes Error in inline/feval (line 34) INLINE_OUT_ = inlineeval(INLINE_INPUTS_, INLINE_OBJ_.inputExpr, INLINE_OBJ_.expr); Error in ezplotfeval (line 52) z = feval(f,x(1)); Error in ezplot>ezplot1 (line 469) [y, f, loopflag] = ezplotfeval(f, x); Error in