NaN Solution Problem with erfi() and erf()

So when I solve for Ff only the first column (1 through 2) is returned with values and the rest are NaN and all of the values for Zf are NaN. Is there any way that I can get a number value out of these? I need them because I'm trying to graph something.

I have tried using erf() and just making whats inside the erf() function a real number but it isn't coming out like it should that way. Any suggestions?

if true
format long
a = 1 
b = 3*10.^-7 
c = 5*10.^-8 
f0 = 4*10.^9 
sigma = 0.2 
t0 = 0 
tmax = 2*b 
t = linspace(t0, tmax)
f =  linspace(10.^6, 10.^10)
omega0 = 2*pi*f0
yt = a*exp((-(t-b).^2)/((2*c).^2))
figure(1)
plot(t,yt)
zt = yt.*(1-((sigma/2)*(1-sin(omega0*t))))
figure(2)
plot(t,zt)
G = -2.*pi.*f.*(pi.*c.^2.*f + sqrt(-1).*b)
D = (((b-(sqrt(-1)).*2.*pi.*c.^2.*f-t0)/(sqrt(2).*c)))
E = (((b-(sqrt(-1)).*2.*pi.*c.^2.*f-tmax)/(sqrt(2).*c)))
Ff = sqrt(pi/2).*a.*c.*exp(G).*((-sqrt(-1).*(erfi(-sqrt(-1).*D)))-((-sqrt(-1).*(erfi(-sqrt(-1).*E)))))
Zf = ((1-(sigma/2)).*Ff) + (sqrt(-1).*sqrt(pi/2).*((a.*c.*sigma)/4).*exp(-((2.*pi.*f+omega0).*(c.^2.*(2.*pi.*f+omega0)+2.*sqrt(-1).*b)))).*(-exp(4.*pi.*c.^2.*f.*omega0+2.*sqrt(-1).*b.*omega0).*((-sqrt(-1).*erfi(((tmax-b+sqrt(-1).*c.^2).*(2.*pi.*f-omega0))/(sqrt(2).*c)))-(-sqrt(-1).*erfi(((t0-b+sqrt(-1).*c.^2).*(2.*pi.*f-omega0))/(sqrt(2).*c))))+(-sqrt(-1).*(erfi(((tmax-b+sqrt(-1)*c.^2).*(2.*pi.*f+omega0))/(sqrt(-1).*c))))-(-sqrt(-1).*erfi(((t0-b+sqrt(-1).*c.^2).*(2.*pi.*f+omega0))/(sqrt(-1).*c))))
figure(3)
plot(f,Zf)
  % code
end

Answer 1

[Ghada Saleh]

Hi Imarquez,

The issue you encountered is not because of 'erfi', it is because you are exceeding the maximum floating number that can be represented in MATLAB. Basically you are computing 'inf-inf' which results in 'NAN'. The 'inf' is a result of calculating 'erfi(c)' where 'c' is some large number. Note that the definition of 'erfi' is an integration of exponential square. To see that, try the following command in MATLAB:

>> exp(200^2)
ans =

   Inf

Hence,

>> erfi(200)
ans =

   Inf

since, in the calculation in 'Ff' and 'Zf' you have 'inf - inf', you end up with 'NAN' value.

I hope this helps,

Ghada

[imarquez]

That makes a lot of sense now. This is a long shot, but do you know if excel will be able to return a value based on the same calculation?

[Stephen23]

Consider how large those numbers would be, and read this:

https://en.wikipedia.org/wiki/Numeric_precision_in_Microsoft_Excel