discrete inverse fourier transform

I have an = (T0* (exp(-T1)*(2*n*pi/T0*sin((n*pi*(T0-2))/(2*T0 ))-cos((n*pi*(T0-2))/(2*T0 )))+1))/(1+(2*n*pi/T0)^2);

and I need to preform a ifft using this formula

an.*exp(j*n*w0*x)

while n is between -25 to 25

what am I missing? I'm trying to use symsum and ifft functions and plot?

[Matt J]

Not sure why you would use SYMSUM. Why not just IFFT?

[Rotem]

I have both n and x so I need symsum no? I can't just write ifft(an.*exp(j*n*w0*x)) because ifft doesn't accept sym

[Matt J]

OK, but then why are using symbolic variables if ultimately you just want a plot? Or, if you insist on summing symbolically, with SYMSUM, why do you think you need IFFT as well?

In any case, we probably need to see your code and more of a description from you of what isn't working in it.

[Star Strider]

The symbolic inverse Fourier transform is ifourier.

[Rotem]

I need to synthesize the temporary signal by reverse Fourier transform with a finite number of harmonies as described The following equation: Σ an.*exp(j*n*w0*x) from -N to N

while I found my an as i wrote before Im not sure about how am I supposed to find the inverse using the ifft function .. and am using the sym because its more generic that way

[Matt J]

But once you've used SYMSUM to evaluate Σ an.*exp(j*n*w0*x) then what more is there to do? The expression above is the inverse fourier transform of an, is it not? Why does IFFT have to be part of this?

[Walter Roberson]

Note that ifourier is for the continuous inverse fourier, not the discrete.

[Rotem]

yes but I got a stright line not a hrominc like I think I should get.. I am using the sum let say N = 25 and I get a symbolic answer of a vector of 41 in length using plot with a x vector eq in length to that of the answer and gets a stright line