Convolution with a gaussian G(t)

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Discussion Overview

The discussion revolves around the convolution of a temperature function T(x,t) with a Gaussian function G(t), focusing on the units of G(t) and the implications for the resulting convolution in terms of physical calculations. The scope includes mathematical reasoning and conceptual clarification regarding the properties of Gaussian functions in this context.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification
  • Exploratory

Main Points Raised

  • Some participants propose that for the convolution to maintain the same units as T(x,t) (Kelvins), the Gaussian G(t) must be dimensionless.
  • One participant questions how the amplitude of the Gaussian relates to the maximum temperature T in Kelvins, suggesting that normalization may be necessary to achieve a meaningful maximum.
  • Another participant raises a concern about the units of the differential element dtau, suggesting that if G has no units, the integral would yield units of K.s, which complicates the interpretation.
  • A later reply clarifies that while dtau has units of time (seconds), the Gaussian function can still be dimensionless, with the constant A in the Gaussian function having dimensions that ensure the overall function remains dimensionless.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the units of G(t) and the role of normalization in defining the Gaussian amplitude. The discussion does not reach a consensus on these points, indicating ongoing uncertainty and exploration of the topic.

Contextual Notes

There are unresolved aspects regarding the normalization of the Gaussian function and the implications of dimensionality in the convolution process. The discussion highlights the dependence on definitions and assumptions about the properties of the functions involved.

Gonzolo
Hi. Suppose I have a function T(x,t), units are in Kelvins. I then do a convolution with a gaussian G(t), and the result is also in Kelvins. What are the units of the gaussian G(t)? Thanks.
 
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Gonzolo said:
Hi. Suppose I have a function T(x,t), units are in Kelvins. I then do a convolution with a gaussian G(t), and the result is also in Kelvins. What are the units of the gaussian G(t)? Thanks.

[tex]T(x,t)\ast G(t)=:\int_{0}^{t} T(x,\tau)G(t-\tau) d\tau[/tex]
,and if u want the convolution to have the same units as the T,this means that the 'G' must be dimensionless (a genuine exponential (Gauss-bell) is dimensionless).

Daniel.
 
Thanks, that helps a lot.

Now, the Gaussian that I have has a defined width (duration). But what about its height (amplitude, maximum etc.)?

I would expect the result to have a defined maximum T in Kelvins, that I can use for further physical calculations. How can such a known maximum exist? How must I define my gaussian amplitude? I suspect normalization is involved but I'm not sure how to do it so I have a meaningful maximum T in the end.
 
Actually, doesn't the dtau have units (say s)? So that if G has no units, the integral would have K.s as units?
 
Gonzolo said:
Actually, doesn't the dtau have units (say s)? So that if G has no units, the integral would have K.s as units?

That [itex]\tau [/tex] is viewed as a variable of integration and,for obvious reasons,it has the same dimension as "t".<br /> This function<br /> [tex]G(\tau)=\exp(-A\tau^{2})[/tex] <br /> is an example of dimensionless function defined everywhere.For obvious reasons,the constant 'A' is dimensional and it has the dimensions of <br /> [tex]<A>_{SI} =(<\tau>_{SI})^{-2}[/tex]<br /> <br /> Yes,the integral will have dimensions of K.s,and that's because the parameter is dimensional.If it wasn't,it would have been K.<br /> <br /> Daniel.[/itex]
 
Thanks for the help. I figured out what units my gaussian was in, everything came into place. Reading back the thread, everything now seems so trivial. How typicial.
 

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