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The short explanation of the DFT that I like the most is that it projects the signal vector on to the vector room with the exponential functions as basis vector
by sorenjan 3y ago
The short explanation of the DFT that I like the most is that it projects the signal vector on to the vector room with the exponential functions as basis vectors. Then you can see how much of the signal that each basis vector, corresponding to a frequency, can explain.
It's intuitive to see if you start with a 2D vector space (the regular euclidian plane) and a 2D vector, and then you can expand the definition of vector spaces to include orthogonal functions as bases.
- sampo 3y ago> vector room Is Swedish, "rum" means both space, and a room. In English, "vector space" is used.
- amelius 3y agoMinor point, it's not a projection because you don't lose dimensions.
- nh23423fefe 3y agonit: the identity matrix is a projection P^2 = P is the definition. not losing dimensions
- amelius 3y agoTrue, but missing the point as you would not normally refer to the identity matrix as a projection.
- nh23423fefe 3y agoWhat was your point? An incorrect characterization of projections?
- mrfox321 3y agoNit to your nit, which is incorrect w.r.t. parent reply: A Fourier transform is not a projection, it's a change of basis represented by a unitary transformation.
- ndriscoll 3y agoThe w Fourier coefficient F(w) is the dot product of f with an exponential function, `e_w • f`, and is in that sense a projection. The inverse Fourier transform writes the original function as a sum of the projected components: `f = sum_w (e_w • f) e_w = sum_w F(w) e_w`. This is exactly how writing an "arrow" style 2- or 3-D vector as a sum of orthogonal projections works.
- anvuong 3y agoIs it a math terminology? Because the way I think it's still a projection, just with 0 residuals?