why definition of sine transform is not consistent with FFT?

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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~
Amith Kamath
Amith Kamath on 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.

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