How can I obtain the Fourier Transformation as a product of matrices?

14 Ansichten (letzte 30 Tage)
Hello everyone!
Before I being, I'd like to say I already know about the fft() command. My goal is to compare the resulting FT matrix obtained by it with the one obtained via Matrix Multiplication. I have the following code:
N = [0:255];
X = (2*cos(((15.8*pi)/128)*N)) + (3*sin(((27.3*pi)/128)*N));
fX = fft(X);
How can I write the values for 'X' as a vector and the other vector 'x' so that I may multiply them to get the Fourier Transformation?

Akzeptierte Antwort

Matt J
Matt J am 15 Sep. 2021
Bearbeitet: Matt J am 15 Sep. 2021
N = (0:255).';
X = (2*cos(((15.8*pi)/128)*N)) + (3*sin(((27.3*pi)/128)*N));
F=fft( eye(numel(X)), [],1);
fX = fft(X); %function implementation
FX = F*X; %matrix implementation
Difference = norm(fX-FX)
Difference = 4.8302e-13

Weitere Antworten (1)

Bjorn Gustavsson
Bjorn Gustavsson am 15 Sep. 2021
Have a look at the help and documentation for dftmtx. That function should return the discrete Fourier-transform matrix and give you the matrix you need. If you need to figure out how the Fourier-transform-component is an inner product between a function and a basis-function, you might gain some understanding by looking at that matrix (or its real and imaginary components) and have a think about what the matrix-multiplication does. Also just write down the integral for that Fourier-component beside the inner-product and look at how they are "similar enough". (Caveat: written by a physicist not a mathematician...)
  1 Kommentar
Adrian Lomeli Martin
Adrian Lomeli Martin am 15 Sep. 2021
Thank you! I checked the function you mentioned and it'll surely be useful down the line. Thank you again for taking the time to write this down.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Fourier Analysis and Filtering finden Sie in Help Center und File Exchange

Produkte


Version

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by