# Tag Info

## Hot answers tagged fourier-analysis

3

There are several good reasons for using wavelets over Fourier. To name a few: The Fourier basis is nonlocal, meaning that signal features such as transients, discontinuities, and edges (in the case of images) show up in all the Fourier coefficients. This is because the functions $e^{ikt}$ are nonzero for all $t$. Consequentially, this means that with ...

3

If we substitute $x = e^u$ and write $g(u) = f(e^u)$, on the one hand, we have $$\int_0^\infty \lvert f(x)\rvert^2\, \frac{dx}{x} = \int_{-\infty}^\infty \lvert f(e^u)\rvert^2\,du = \int_{-\infty}^\infty \lvert g(u)\rvert^2\,du.$$ On the other hand, \begin{align} F(it) &= \int_0^\infty x^{-it} f(x)\,\frac{dx}{x}\\ &= \int_{-\infty}^\infty ... 2 The first thing to understand is that any periodic function can be represented as the sum of discrete "spikes" in the frequency domain, that is, by its "Fourier series". The longer the period of the function, that is, the longer our sampling interval, the closer together those spikes need to be. In that vein, one common explanation is that the Fourier ... 2 Yes, you must show that the tempered distribution (1+|\xi|^2)^{s/2}\,\hat{u} is given by integration against an L^2 function. The multiplication by (1+|\xi|^2)^{s/2} does not change whether or not it is an a.e. pointwise function, but it does affect square integrability. For example, the Fourier transform of Dirac \delta is (up to a constant) the ... 2 This might looks like a comment. First I think you understand the Fourier Transform for L^2(\mathbb{R}) is only "formally" defined in the formula you wrote. It's actually defined for L^1(\mathbb{R}) in that way. Now the above argument provided by BigM gives you the property of Fourier Transform on L^1(\mathbb{R})\cap L^2(\mathbb{R}), the only thing ... 1 The mistake on the last step: \frac{6}{\sqrt{2\pi}} \left[ \frac {\sin^2\frac{3\omega}2}{\frac{3\omega^2}2} \right] = \frac{6}{\sqrt{2\pi}} \left[ \frac {\sin^2\frac{3\omega}2}{\frac{2}{3}\left(\frac{3\omega}2\right)^2} \right] = \frac{9}{\sqrt{2\pi}} \left[ \frac {\sin\frac{3\omega}2}{\frac{3\omega}2} \right]^2 = \frac{9}{\sqrt{2\pi}} {{\rm ...

1

Although $\int_0^\pi \cos(x)\,dx = 0$, $a_0\ne 0$ because $$\int_0^{\pi/2} |\cos(x)|\,dx=\int_{\pi/2}^{\pi} |\cos(x)|\,dx.$$ We can evaluate it as follows, as can be seen in the plot below a_0 = \frac 1 \pi \int_{-\pi}^\pi |\cos(x)|\,dx=\frac 2 \pi \int_0^\pi |\cos(x)|\,dx=\frac 4 \pi \int_0^{\pi/2} |\cos(x)|\,dx = \frac 4 \pi \int_0^{\pi/2} ...

1

Hint.plug $T(F)(x)$ in the left hand side of the ...

1

Depends on what you want to achive with it. If you have a (quasi-)periodic signal and want to estimate the used frequencies you use the fourier-transform. If you just want to interpolate between two sample point use wavelets. These are only examples. The convolution product is not simplified in the wavelet domain, use FFT

Only top voted, non community-wiki answers of a minimum length are eligible