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Writing this Mathcad equation in Matlab

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Jason 2018-3-22

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评论： Star Strider 2018-3-23

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hi, Im trying to convert an equation from mathcad into matlab but am struggling. the equation is:

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Star Strider 2018-3-22

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Try this:

psi = 0.142;
Io = 1;
f_vir2 = @(r) 4*Io * abs(integral(@(rho)besselj(0, pi*rho*r*psi).*rho, 0, 1)).^2;
r = pi;
test_call = f_vir2(r)

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Thankyou Jason

Star Strider 2018-3-22

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As always, my pleasure.

If you are new to MATLAB, see the documentation on Anonymous Functions (link) to understand those I used here.

Jason 2018-3-23

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Hello, Im trying to plot the functional form as follows:

psi = 0.142;
Io = 1;
f_vir = @(r) 4*Io * abs(integral(@(rho)besselj(0, pi*rho*r*psi).*rho, 0, 1)).^2;
r = linspace(-20,20);
axes(handles.axes1); cla reset
plot(f_vir(r))

But get the following error:

Error using  * 
Inner matrix dimensions must agree.
Error in PSF>@(rho)besselj(0,pi*rho*r*psi).*rho (line 167)
f_vir = @(r) 4*Io * abs(integral(@(rho)besselj(0, pi*rho*r*psi).*rho, 0, 1)).^2;
Error in integralCalc/iterateScalarValued (line 314)
                fx = FUN(t);
Error in integralCalc/vadapt (line 132)
            [q,errbnd] = iterateScalarValued(u,tinterval,pathlen);

Torsten 2018-3-23

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编辑：Torsten 2018-3-23

在 MATLAB Online 中打开

f_vir = @(r) 4*Io * abs(integral(@(rho)besselj(0, pi*rho*r*psi).*rho, 0, 1,'ArrayValued',true)).^2;
plot(r,f_vir(r))

Jason 2018-3-23

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Thankyou. whats the reason for "ArrayValued", "true" Jason

Torsten 2018-3-23

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You call "f_vir" with an array of values for r, "integral" usually calls "f_vir" with an array of values for rho. This would usually make multiplications of the form "r*rho" within "f_vir" incompatible. To avoid this conflict in lengths, the "ArrayValued=true" option tells "integral" to call "f_vir" only with a single scalar for "rho".

You could also use

r=linspace(-20,20);
for i=1:numel(r)
  f_vir_array(i)=f_vir(r(i));
end
plot(r,f_vir_array)

without the "ArrayValued=true" option.

Best wishes

Torsten.

Jason 2018-3-23

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Thankyou again

Star Strider 2018-3-23

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@Torsten — Thank you.

@Jason — Our pleasure.

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