# Tag Info

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 ...

4

Parseval's theorem says that the the following relationship holds $$\sum_{n=1}^{N} a[n]\,a^*[n] = \frac{1}{N}\sum_{k=1}^{N} A[k]\,A^*[k]$$ where $A[k]$ is the discrete Fourier transform of $a[n]$, both assumed to be of length $N$ (no padding). This arises from the fact that the signal energy calculated from the time domain and frequency domain must be ...

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 $$\int_{-\infty}^{0}x(t) dt ... 2 Parseval's identity and Plancherel's theorem finally boil down to orthogonality. When one decomposes a data (with samples), via a scalar product, onto an orthogonal sequence (yielding coefficients), there exists a certain preservation (equality, up to a proportionality factor) of energy between samples and coefficients. There are some technical conditions, ... 2 Your first steps seem OK to me. Even if I'd put absolute values in the time domain as well (for generality), if the first three integrals exist, the fourth is valid. Integration is linear. The weak point resides in the reverse inference. The equality of energy integrals does not mean that the integrands are related. For one, PDSs are real and positive ... 2 There is, unfortunately, no simple way of getting to where you want to from what you have. As Marcus Muller's comment says, the convolutions and correlations that are associated with the DFT (FFT if you like) are cyclic or periodic whereas the autocorrelation that you seek is the linear or aperiodic autocorrelation, and it is difficult to derive the latter ... 1 There are several misconceptions in the question that have not been addressed in the existing answers. First of all, the signal x(t)=A\cos(2\pi f_0t) is a deterministic power signal (unless A or f_0 are modeled as random variables). For this reason several definitions in the question are inappropriate. First, the auto-correlation of a power signal is ... 1 Starting with from the linked question:$$S_{xx}(\omega)=\lim\limits_{T\to \infty}\mathbf{E} \left[ | \hat{x}_T(\omega) |^2 \right]  = \lim\limits_{T\to \infty}\mathbf{E} \left[ \frac{1}{T} \int\limits_0^T x^*(t) e^{i\omega t}\, dt \int\limits_0^T x(t') e^{-i\omega t'}\, dt' \right] = \lim\limits_{T\to \infty}\frac{1}{T} \int\limits_0^T \int\limits_0^T \...

1

This follows from the orthonormality of shifted versions of $\phi(x)$: $$\langle \phi(x),\phi(x+n)\rangle =\delta[n]\tag{1}$$ where $\delta[n]$ is the unit impulse. Taking the discrete-time Fourier transform (DTFT) of $(1)$ gives the equation in your question. Note that the left-hand side of $(1)$ is the auto-correlation of $\phi(x)$ evaluated at integers ...

1

There is a simple idea behind Parseval's Theorem. Unitary transform preserves the ${L}_{2}$ Norm. The Fourier Transform (All its variants) are indeed Unitary when the correct factor is in place. The problem is with the way you sum the elements and factor them. What you need to factor by $\frac{1}{N}$ is the sum of squared absolute value of the series. ...

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