WHAT TO GRAPH USING ELEMENTWISE MULTIPLCATION
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YOGESHWARI PATEL
el 5 de Jun. de 2017
Comentada: YOGESHWARI PATEL
el 5 de Jun. de 2017
i WANT TO PLOT THE GRAPH FOR
U(X,T)=EXP(-4*PI^2*0.5*y)*SIN(2*PI*x)
AS PER THE SUGGESTION GIVEN ON MATHWORK I WROTE THIS CODE BUT STILL GRAPH IS NOT CORRECT. i AM USING R2013 VERSION OF MATLAB. pROVIDE SOME MORE EXAMPLE.
x = linspace(0,1 ,51)';
y = linspace(0,1,51)';
[X, Y] = meshgrid(x, y);
Z=exp(-4.*(pi^2)*0.5*Y)*sin(2*pi*X);
surf(X, Y, Z);
ylabel('t');
xlabel('x');
zlabel('u(x,t)')
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Respuesta aceptada
Stephen23
el 5 de Jun. de 2017
Editada: Stephen23
el 5 de Jun. de 2017
x = linspace(0,1 ,51)';
y = linspace(0,1,51)';
[X,Y] = meshgrid(x, y);
Z = exp(-2*pi^2*Y).*sin(2*pi*X); % I only changed this line!
surf(X, Y, Z);
ylabel('t');
xlabel('x');
zlabel('u(x,t)')
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/176652/image.png)
Lets have a look, so you can understand why that line. You wrote this:
Z=exp(-4.*(pi^2)*0.5*Y)*sin(2*pi*X);
^ element-wise multiply, but pointless.
But what is the point of element-wise multiplication with 4? 4 is a scalar number and so is pi, it serves no purpose whatsoever to use element-wise multiplication with them. So what parts of that calculation are not scalar? Answer: X and Y. Therefore any operation involving X and Y, or any other arrays derived from them will need to use element-wise operations. In this case this means the exp(..) and sin(..) terms will need to be multiplied using element-wise multiply. And thus I showed you (twice so far) this, which uses element-wise multiply for the terms that are arrays (and I did not use it on the scalar values like you did):
exp(-2*pi^2*Y).*sin(2*pi*X)
^ element-wise multiply, required.
or if you really want to have that 4 and 0.5:
exp(-4*pi^2*0.5*Y).*sin(2*pi*X)
You have now been told this multiple times:
2 comentarios
Stephen23
el 5 de Jun. de 2017
Editada: Stephen23
el 5 de Jun. de 2017
@YOGESHWARI PATEL: no, 2*pi^2*X does not need element-wise multiplication. It is optional in that set of operations. In my answer I also explained that it is not required for the operations that you used that include only scalars and one array. You could use it there, but it does not need it (there is no harm in using either, it is up to you). In my answer I showed you working code that does not use it. Did you not notice that my code was working (it included a nice colorful figure: did you see it?)
Did you actually read my answer?
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