# Questions tagged [fourier-analysis]

Fourier analysis, also known as spectral analysis, encompasses all sorts of Fourier expansions, including Fourier series, Fourier transform and the discrete Fourier transform (and relatives). The non-commutative analog is (representation-theory).

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### Linear Transformation from one function to another.

Let's say we have two functions $f:[-1,1]\rightarrow \mathbb{R}$ and $g:[-1,1]\rightarrow \mathbb{R}$. Suppose furthermore that one can write down $f,g$ in terms of a linear sum of basis functions ...
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### Derivative of the Fejer Kernel?

I'm looking to calculate the derivative of the Fejer kernel: $$F_n(x) = \frac1{n+1}\left(\frac{\sin\left(\frac{n+1}{2}x\right)}{\sin\left(\frac x2\right)}\right)^2$$ However, I'm running into some ...
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### Bessel decomposition from Fourier Transform

I am looking to decompose a signal in terms of Bessel functions. I'm aware of Hankel transforms; however, for computational reasons, I have to use Fourier transforms. Essentially, my question is: ...
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### $\int_{-\infty}^\infty \exp(iqy')dy' \int_{-\infty}^\infty |k|\exp(ik(y-y'))dk$ in 2 way give different results

I have integration whose result change depending on the way of calculation. I want to compute the integration below $$I=\int_{-\infty}^\infty \exp(iqy')dy' \int_{-\infty}^\infty |k|\exp(ik(y-y'))dk$$ ...
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### Fourier transform of function defined on finite interval

Let $f(t)$ be a function defined on the finite interval $[t_1,\, t_2]$. Is the Fourier transform of such a function uniquely defined? In the sense that there exists only one function $\hat{f}(\omega)$ ...
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### Energy spectral density in an LTI system

Consider the LTI system: $$x\left( t \right) \to LTI\, system \to y\left( t \right)$$ The graph of the $x(t)$ signal is the following: Remark: The graph when $t>0$ is a quarter of a circle. Q1:...
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### Fourier transform of matrix over polynomial field?

I know we can do (Discrete) Fourier Transform of vectors of polynomial coefficients. This is useful for example when multiplying polynomials, since convolution turns into multiplication in Fourier ...
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