DFT

How do i calculate DFT in matlab? I do not want FFT.

[Bahloul Derradji]

use KSSOLVE a matlab package for density functional calculations

Answer 1

[Wayne King]

Hi Lisa, you had a couple problems with your code:

N= 4; 
x=1:4;
for k=0:3
    for n = 0:3;
y(n+1) = x(n+1).*exp(-(1j*2*pi*k*n)/N);
    end
    xdft(k+1)= sum(y);
end

compare to

fft(x)

[Lisa Justin]

thanks, i will check

[Lisa Justin]

yes it is same but i do not think it will be same with real vibration time series. check this http://www.dataq.com/applicat/articles/an11.htm

[Amilton Pensamento]

Thanks, Wayne. I tried to use the same code as starting point to come up with the IDFT. Still using x as my input sequence.

N = 4;

x = 1:4;

for n = 0:3

for k = 0:3;

y(k+1) = x(k+1).*exp((1j*2*pi*k*n)/N);

end

xdft(n+1)= (1./N).*sum(y);

end

Answer 2

[Wayne King]

FFT() is just an efficient algorithm (actually a family of algorithms) for computing the DFT. The DFT is the mathmatical concept, the FFT is just an algorithm. You can form a matrix to compute the DFT by brute force, but the result will be identical to the output of fft().

[Lisa Justin]

i want to avoid using any window function, that is the reason i need DFT. How do i represent an interval of 0:N-1 summation in matlab? I am just experimenting with DFT to see what i get. I want to compare the results with windowing(FFT) and without windowing (DFT).

Answer 3

[Wayne King]

The DFT can be written as a matrix multiplication of a Nx1 vector, your signal, with a NxN matrix -- the DFT matrix. But that will involve N^2 multiplications and N additions. You can see that if your signal gets even reasonably large that is going to be a huge computational effort. The FFT() exploits symmetries in the DFT to reduce the number of computations greatly.

For example, here is the brute force way for N=4

    x = (1:4)';  % the signal
    W = -1j*2*pi/4;
    W = repmat(W,4,4);
    k = (0:3)';
    k = repmat(k,1,4);
    n = 0:3;
    n = repmat(n,4,1);
    W = exp(W.*k.*n);
    % W is the DFT matrix, now to get the DFT
    xdft1 = W*x

    % but that is exactly the same as
    xdft2 = fft(x)

[Lisa Justin]

N=1024

k = (1:N/2-1);

x=1:1024

for n=0:1024

xl(n+1) = (x.*exp(-i.*2.*pi.*(k./N).*n));

end

xs=sum(xl)

please can you make correction on the loop, it is easier for me to understand. i do not understand repmat on your code. thanks

Answer 4

[Wayne King]

With all due respect to that author, I think she is overstating her point. The DFT takes a N-point periodic vector (the N-point periodicity is implicit in the DFT) and projects it onto N discrete-time complex exponentials with period N. Those complex exponentials are a basis for vectors (a vector space) with period N.

Now, in reality, the DFT is most often used for sampled data, data sampled from a continuous-time process, which may or may not be periodic, and even if it is periodic, most likely does not have period N.

The problems that motivate using a window with the DFT, come from this "translation". You're taking a process which is continous, may or may not be periodic, and may not have an abrupt on and off transition, and you are creating a N-point vector out of it, which has those qualities.

So I think it is wholly artificial to draw a line between the DFT and FFT.

I think you still have to window in many cases whether you compute the DFT by brute force or use an FFT implementation.

[Lisa Justin]

Thanks!

Answer 5

[Dr. Seis]

Here is another (inefficient) way of saying the same thing as Wayne by way of example:

Fs = 100; % samples per second
dt = 1/Fs;
N = 128; % Number of samples
time = (0:1:(N-1))*dt;
timedata = sin(2*pi*time);

figure;
plot(time,timedata);

df = 1/(N*dt); % frequency increment
Nyq = 1/(dt*2); % Nyquist Frequency
freq = -Nyq:df:Nyq-df;
freqdata = zeros(size(timedata));
for i = 1 : N
for j = 1 : N
    freqdata(i) = freqdata(i) + timedata(j)*exp(-1i*2*pi*freq(i)*time(j));
end
end

figure;
plot(freq,real(freqdata),freq,imag(freqdata));

And both (actually all 3) implementations should give the same results as FFT. I can't see any reason why they wouldn't work with real data either.

[Lisa Justin]

Thanks guys!