5

Yes indeed! In theory as long as the wavelet is orthogonal, the sum of the squares of all the coefficients should be equal to the energy of the signal. In practice, one should be careful that: the decomposition is not "expansive", i.e. the number of samples and of coefficients is the same. wavelet filter coefficients are not re-scaled, as happens in some ...


3

Sure. You can can just integrate in the time domain from $0$ to $T$ since the area outside $[0, T]$ is 0. Please note that you still must integrate from $-\infty$ to $+\infty$ in the frequency domain since finite support in the time domain implies infinite support in the frequency domain (and vice versa). Things are different if the function is periodic.


3

I don't think you will find a reference for this. Your are simply stating that "integrating over a whole bunch of zeros results in 0". In other words, if $$x(t) = 0, x\not\in [0,T]$$ then $$ \int_{-\infty}^{+\infty}x(t) dt = \int_{-\infty}^{0}x(t) dt + \int_{0}^{T}x(t) dt + \int_{T}^{+\infty}x(t)dt = \int_{0}^{T}x(t) dt $$ since $$ \int_{-\infty}^{0}x(t) dt ...


2

Your first steps seem OK to me. Even if I'd put absolute values in the time domain as well (for generality), if the first three integrals exist, the fourth is valid. Integration is linear. The weak point resides in the reverse inference. The equality of energy integrals does not mean that the integrands are related. For one, PDSs are real and positive ...


2

There is, unfortunately, no simple way of getting to where you want to from what you have. As Marcus Muller's comment says, the convolutions and correlations that are associated with the DFT (FFT if you like) are cyclic or periodic whereas the autocorrelation that you seek is the linear or aperiodic autocorrelation, and it is difficult to derive the latter ...


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