Discrete Fourier Transform a dummy approach

I'm trying to understand how DFT works exactly. However, when experimenting around, I compared both Matlab generated FFT result with a dummy approach result and I get similar result. However, the Imaginary part of both result are negated. The code below is my implementation. The dummy approach is based on http://en.wikipedia.org/wiki/Discrete_Fourier_transform#Definition

Fs = 500; % Sampling frequency 
T = 1/Fs; % Sample time 
L = 1000; % Length of signal 
x = cos(2*pi*100*t)+randn(size(t));

% Calculate by using Matlab build-in FFT 
fdft = fft(x);

% A dummy way to calculate DFT
n = 0:L-1;
k = 0:L-1;
table = bsxfun(@times, n.', k);
dummydft = x * exp(-i * 2 * pi .* table ./ L )';

Is there anything I miss? Thank you.

Answer 1

[Wayne King]

Yes, the only problem you have is the use of the transpose operator. The transpose operator with complex-valued data returns the conjugate transpose by default, you want to use .'

   Fs = 500; % Sampling frequency 
   T = 1/Fs; % Sample time 
   L = 1000; % Length of signal 
   t = 0:T:(1000*T)-T;
   x = cos(2*pi*100*t)+randn(size(t));
   x = x(:);
   n = 0:L-1;
   k = 0:L-1;
   table = bsxfun(@times, n.', k);
   dftmatrix = exp(-i * 2 * pi .* table ./ L ).';
   dft1 = dftmatrix*x;
   dft2 = fft(x);