Math help

[Drizzt]

I'm playing a bit with Discreet Fourier Transform, but I'm already stuck. Problem is that I've never learn how to deal with imaginary numbers...


Now, for the infos I've found the DFT for
s={x_1, x_2, ... x_N}

is
S={y_1, y_2, ... y_N}


with:
N
y_k = sum x_h*exp(-i*2*pi*(k-1)*(h-1)/N), 1<= k <= N
h=1


Now, porting the above expression to code is pretty simple. Only, I'm stuck on that damned -i !

I mean, if
x_1 = 10337497
and
N = 196608

going for y_2, at the first iteration is:
10337497 * exp(-i*2*pi*(2-1)*(1-1)/196608)

How can I solve it?

[Drizzt]

(I've gone for the wrong forum, can someone move this to the lounge? This problem is neither a hardware nor a software one)

[Chrono_Wanderer]

LOL, sure it does!

Its scieitific calculations that can hardly be solved with today's computers. Therefore we need faster hardware

edit: sorry dunno how to do that. (still in hs hahaha, shame on me)

[the main lobe]

FFT or Euler identities may help

Fast Fourier Transforms are available for many software packages...those work well. If this isn't what your after, it is unlikely you'll be able to get an entirely real solution. Some identities that are often useful when dealing with imaginary numbers are the Euler Identities.
exp(i*x)=cos(x)+i*sin(x)
exp(-i*x)=cos(x)-i*sin(x)
also:
cos(x)=(exp(i*x)+exp(-i*x))/2
sin(x)=(exp(i*x)-exp(-i*x))/(2*i)

[Umfriend]

I was just going to say that. No, really!

[jonnylane]

These can also be useful:

sinh(ix) = isin(x)
cosh(ix) = cos(x)

Use Matlab!

Jonathan