Dirac function and matlabFunction

hello!

I'm having problems evaluating a matlab function that has dirac within. I have the following symbolic function: mf = v^2*dirac(x + 0.5) + 3*v^2*dirac(x + 0.5, 1)

I then created a matlab function by doing this: ht = matlabFunction(mf)

ht = @(v,x)(v.^2.*dirac(x + 0.5) + 3.*.v.^2.*dirac(x + 0.5, 1))

Everything works well. But when I then try to evaluate the function by doing this:

ht(2, 7)

I get an error. It says "Error using dirac. Too many input arguments." How do I evaluate that function correctly?

thanks in advance!

Answer 1

[Carlos]

The problem is this line of the code dirac(x + 0.5, 1), dirac only allows one argument. Why do you include the second parameter 1? What do you want to represent? Just a dirac function centered in -0.5? then just write dirac(x + 0.5) as in the previous command

Answer 2

[Azzi Abdelmalek]

In your expression what is dirac(x+0.5,1)? it should be dirac(x+0.5)

ht = @(v,x)(v.^2.*dirac(x + 0.5) + 3*v.^2.*dirac(x + 0.5))
ht(2,7)

Answer 3

[jim]

Well, I am using the heaviside function in an earlier expression called 'z'. When I took it's derivative by doing this:

mf = diff(z)

I got this:

mf = v^2*dirac(x + 0.5) + 3*v^2*dirac(x + 0.5, 1)

So that's where that came from.

After that I created a matlab function by doing this:

ht = matlabFunction(mf)

which gives

ht = @(v,x)(v.^2.*dirac(x + 0.5) + 3.*.v.^2.*dirac(x + 0.5, 1))

Again, everything works well. I just can't evaluate it. It gives me an error when I do h(2,7).

[Carlos]

Try with

mf = v^2*dirac(x + 0.5) + 3*v^2*dirac(x + 0.5)

Answer 4

[Walter Roberson]

Please show the z formula for which mf is the derivative.

It appears to me that your z most likely contains dirac and not heaviside and that dirac(x + 0.5,1) is an attempt to represent the first derivative of the dirac function.

Which MATLAB version are you using? Are you possibly using Maple as your symbolic engine instead of MuPAD ?