Whitening signal vs. whitening its DFT

Question

A whitening transformation (PCA) is simply a rotation into a space in which variables become uncorrelated.

Because a DFT is a transformation into a coordinate space of orthogonal frequency components, a DFT is a also just a rotation. We can obtain the DFT of a signal by creating a matrix of orthogonal basis vectors and solving for their weights. You can test this in Matlab using:

B = 1000;
b = 0:B-1;
Fs = 1000;
t = b/Fs;
y = 3*cos(2*pi*1*t) + 2*sin(2*pi*2*t + pi/2);
X = [cos(2*pi*1*Fs/B*t)' sin(2*pi*1*Fs/B*t)' cos(2*pi*2*Fs/B*t)' sin(2*pi*2*Fs/B*t)'];
Y = (X'*X)\X'*y'; % assuming y = XY, so Y gives the coefficients of the frequency components

We see that the coefficients corresponding to cosine of 1 and 2 Hz are 3 and 2, as expected (since the sin of 2 Hz has a phase of pi/2).

If both of these are just rotations of the original data, does that mean that whitening the original signal is equivalent to whitening its DFT?

Answer 1

Provided you define it appropriately, the DFT is just an orthonormal transformation: the vectors that make up the DFT matrix are orthogonal to each other and are unit vectors.

does that mean that whitening the original signal is equivalent to whitening its DFT?

Yes.

In fact, the DFT of a white noise sequence is... white noise.