DFT

41 次查看（过去 30 天）
Lisa Justin
Lisa Justin 2012-2-22
评论： Amilton Pensamento 2022-7-22
How do i calculate DFT in matlab? I do not want FFT.
  1 个评论
Bahloul Derradji
Bahloul Derradji 2019-1-11
use KSSOLVE a matlab package for density functional calculations

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Wayne King
Wayne King 2012-2-22
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)
  3 个评论
显示 1更早的评论
Lisa Justin
Lisa Justin 2012-2-22
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
Amilton Pensamento 2022-7-22
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

请先登录，再进行评论。

更多回答（4 个）

Wayne King
Wayne King 2012-2-22
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().
  1 个评论
Lisa Justin
Lisa Justin 2012-2-22
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).

请先登录，再进行评论。

Wayne King
Wayne King 2012-2-22
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)
  1 个评论
Lisa Justin
Lisa Justin 2012-2-22
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

请先登录，再进行评论。

Wayne King
Wayne King 2012-2-22
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.
Dr. Seis
Dr. Seis 2012-2-22
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.

类别

在 Help Center 和 File Exchange 中查找有关 Transforms 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by