why definition of sine transform is not consistent with FFT?
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/149389/image.jpeg)
See the picture.
Let's suppose: G(f) = DST(g(t)) and G(f) = FFT(g(t))
There are inconsistency in 2 things:
1. sine transform uses sin(pi*f*t) while fft uses exp(i*2*pi*f*t) --- *2 is the difference
The sine transform I learn from books uses sin(2*pi*f*t)
The definition of sine transform in Matlab actually gives a result of G(f/2)
2. fft starts with (j-1) and (k-1) in the exponential while sine transform starts with n and k ---- missed 0 frequency term
Anybody know why it's defined like this? Any ideas are appreciated. Thanks~
2 Kommentare
Amith Kamath
am 29 Aug. 2013
If you look at the source for dst,
edit dst
you can see that it uses the real part of the FFT computation. Can you share some examples of why you think the documentation is inaccurate? This may be useful for someone to verify that it is indeed incorrect.
Antworten (0)
Siehe auch
Kategorien
Mehr zu Fourier Analysis and Filtering finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!