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From: Walter Roberson on 17 Jun 2010 22:59
Mat Hunt wrote:

> The function in question is:
>
> 16.2668*exp(-1/4*k^2)/(144+2*k^4-12*k^2+9*abs(k))
>
> I tried ifourier and I got the result:
>
> 16.2668*ifourier(exp(-1/4*k^2)/(144+2*k^4-12*k^2+9*abs(k)),k,x)
>
> So I think I will have to use ifft. I would ideally like to plot the
> original function but I don't know how to plot it.

I'm trying to figure out what kind of function could possibly have
abs(k) as part of its fourier transform. It appears that at the very
least the answer involves complex numbers.
From: Mat Hunt on 18 Jun 2010 06:04
No, there are no problems with it, no complex numbers. The integrand is an even function and real, as a result, the complex part of the solution is zero.

Mat
From: Mat Hunt on 18 Jun 2010 08:58
So, the best way to perform this calculation is numerically. In which case how would I go about this using matlab's inbuilt functions. I wrote a small program myself:

%Start the integration
x=-8:0.01:8;
n=length(x);
y=zeros(n,1);
dk=0.001;
k=-5:dk:5;
l=length(k)-1;
for j=1:n
z=f(k).*exp(i*k*x(j));
y(j)=real(trapz(k,z));
end
plot(x,y)
xlabel('x');
ylabel('\eta_{1}(x)');

which seemed to work in some simple cases rather well. Now I want to compare it to matlabs ifft routine.

Any suggestions on what I should do?

Mat
From: Walter Roberson on 18 Jun 2010 09:37
Mat Hunt wrote:

> The function in question is:
>
> 16.2668*exp(-1/4*k^2)/(144+2*k^4-12*k^2+9*abs(k))
>
> I tried ifourier and I got the result:
>
> 16.2668*ifourier(exp(-1/4*k^2)/(144+2*k^4-12*k^2+9*abs(k)),k,x)
>
> So I think I will have to use ifft. I would ideally like to plot the
> original function but I don't know how to plot it.

I get something vaguely plausible if I use

k = -100:.1:100;
t = 16.2668*exp(-1/4*k.^2)/(144+2*k.^4-12*k.^2+9*abs(k));
ti = ifft([0 t]);
plot(ti)
From: Mat Hunt on 18 Jun 2010 10:18

>
> k = -100:.1:100;
> t = 16.2668*exp(-1/4*k.^2)/(144+2*k.^4-12*k.^2+9*abs(k));
> ti = ifft([0 t]);
> plot(ti)

I used the above code and didn't get anything remotely sensible. The code I use is:

h=1;
E_b=0.1; %0.2;
B=0.2;
rho=1;
F=0.2;
mu=0;
%Start the integration
x=-8:0.01:8;
n=length(x);
y=zeros(n,1);
dk=0.001;
k=-20:dk:20;
l=length(k)-1;
for j=1:n
z=f_1(k,F,B,E_b,h,rho,mu).*exp(i*k*x(j));
y(j)=real(trapz(k,z));
end

plot(x,y)
xlabel('x');
ylabel('\eta_{1}(x)');

Along with the function:

function eta=f_1(k,F,B,E_b,h,rho,mu)
m=length(k);
for j=1:m
eta(j)=-(sqrt(pi)*exp(-0.25*k(j)^2)/(2*rho*9.8065))./(1-F+(h*k(j))^4/90+0.5*(B-1/3)*(h*k(j))^2-0.5*E_b*h*abs(k(j))+i*mu);
end

If I don't get anything like the above function then the answer is incorrect. The code that you gave me didn't give anything like the above M file.

Mat
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