Using for loops and an array to plot Fourier Transform's on matlab

Question

I am trying to plot the periodic functions in matlab of the following equation:

using the following code:

tn= 25;
kn= 7;

time=0:1:200;
f=0;

for t= (-tn):1:(tn) 

for k = (-kn):1:(kn)

        s1=((1j)*k*exp(-abs(k)));
        s2=exp((1j)*k*((2*pi)/50));

        f=f+(s1*s2);
end
           tval = (f*exp(t));
        fs(1,t+1) = tval;
end

I am having trouble understanding why I am not able to see the plot. Is there another way to plot complex numbers in Matlab?

Am I going about this the wrong way? I am simply trying to plot fs against time (could be from 0-200 or -100 - 100 for all I care) and see the periodic function so I can go ahead and manipulate it but I can't seem to even get the correct plot.

I tried using the symsum function in matlab but could not figure it out. I understand C and C++ and felt like this approach was more intuitive for me.

edit:

x(1:101)=0;
t(1:101)=0;

for n=0:1:100
    t(n+1)=n;

    for k=-100:1:100
        x(n+1)=x(n+1)+abs(sin((k*pi)/2))*exp(1j*k*((2*pi)/50)*n);
    end;
end;

I plotted the function using the following code. Why do our imaginary plots look different?

Answer 1

I don't have Matlab, but here the same in Python (using Numpy & Matplotlib which are very similar to Matlab):

import numpy as np
import matplotlib.pyplot as p
%matplotlib inline

t= np.arange(0,100,0.1)
s=np.zeros(len(t))


for k in range(40):
    s=s+  np.abs(np.sin (k*np.pi/2)) * np.exp( 1j*k *2*np.pi/50.0*t)   # pos k
    s=s+  np.abs(np.sin (-k*np.pi/2)) * np.exp(- 1j*k *2*np.pi/50.0*t)   # negative k 
  # the cosine is symmetric so it survives adding the negatives, 
  # the sines get cancelled as they are antisymmetric for the negative k,
  # so the imaginary part is identically zero.

p.subplot(311)    
p.plot(np.real(s))
p.subplot(312) 
p.plot(np.imag(s),'r')
p.subplot(313)
p.plot(np.abs(s))