A convolution identification (alteration of the inquiry "Is this convolution identification known?")

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The adhering to workout is motivated by response to the inquiries this and also this, yet no expertise of probability concept is called for. The remedy is uncomplicated thinking knowledge with the pertinent theorems/properties. The evidence is intended to be casual.

Allow $\hat \varphi (\omega)$ represent the Fourier change of a function $\varphi$, and also $*$ convolution. Express $\hat g(\omega)$ in regards to $\hat f(\omega)$ and also a constant $a \in (0,1)$, to make sure that $$ (f_2 * g)(t) = at(f * g)(t), $$ where $f_2 (x) = xf(x)$.

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2019-12-02 02:51:55
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Answers: 2

I might be screwing up my use of Fourier changes and also convolutions with circulations, yet I assume this can be addressed with a couple of identifications (particularly just how changes work with items and also convolutions, and also just how convolutions collaborate with by-products and also Dirac deltas). Offered the beginning factor $(xf) * g = a t (f * g)$, take the (unitary, average regularity) change to get

$(\hat{x}*\hat{f} ) \hat{g} = a\hat{x} * (\hat{f} \hat{g}) \implies (\delta'*\hat{f} ) \hat{g} = a\delta' * (\hat{f} \hat{g}) \implies \hat{f}' \hat{g} = a (\hat{f}\hat{g})'$

If you aluminum foil and also revise you get the straightforward differential formula $a \hat{g} ' + (a-1) (\hat{f}' / \hat{f}) \hat{g} =0 $, which is very easy to address and also brings out $\hat{g} = \hat{f}^{1/a - 1}$ (I wish I did that all appropriately!).

[MODIFY - many thanks to listed below comment. ]

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2019-12-03 05:37:46
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My solution resembles chroma is solution, just that I get $$ (xf) * g = at(f * g) \Rightarrow \hat f'\hat g = a(\hat f \hat g)' $$ right from the adhering to typical building (incorporated, certainly, with the convolution theory) : if $\psi = x \varphi$, after that $\hat \psi = i \hat \varphi'$.

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2019-12-03 04:26:24
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