Convolution with a gaussian G(t)

In summary, the conversation discusses the use of a convolution with a gaussian function in Kelvins, and the question of what units the gaussian function should have. It is determined that the gaussian function must be dimensionless and that the integral will have units of K.s. The conversation concludes with the realization that the units of the gaussian function were already known and everything now seems trivial.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
Actually, doesn't the dtau have units (say s)? So that if G has no units, the integral would have K.s as units?
 
  • #5
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".
This function
[tex] G(\tau)=\exp(-A\tau^{2}) [/tex]
is an example of dimensionless function defined everywhere.For obvious reasons,the constant 'A' is dimensional and it has the dimensions of
[tex] <A>_{SI} =(<\tau>_{SI})^{-2} [/tex]

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.

Daniel.
 
  • #6
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.
 

What is the purpose of convolving a function with a gaussian?

The purpose of convolving a function with a gaussian is to smooth out the function and remove any high frequency noise or fluctuations. This can be useful in data analysis or signal processing to make the data easier to interpret and analyze.

How is convolution with a gaussian different from other types of convolution?

Convolution with a gaussian is different from other types of convolution because the gaussian function has a unique shape that allows it to effectively smooth out data without significantly distorting or altering the original function. Other types of convolution may result in more significant changes to the original function.

What is the mathematical formula for convolution with a gaussian?

The mathematical formula for convolution with a gaussian is given by the equation G(t)*f(t) = ∫f(τ)g(t-τ)dτ, where G(t) is the gaussian function and f(t) is the original function. This equation represents the process of sliding the gaussian function over the original function and calculating the area under the curve at each point.

Can convolution with a gaussian be applied to any type of function?

Yes, convolution with a gaussian can be applied to any type of function. However, the effectiveness of the convolution may vary depending on the shape and characteristics of the original function. In general, it is most effective for smoothing out functions with a lot of high frequency noise or fluctuations.

How can convolution with a gaussian be used in image processing?

In image processing, convolution with a gaussian is commonly used as a blurring or smoothing filter. This helps to reduce any pixelated or jagged edges in the image and create a more visually appealing result. It can also be used to remove any noise or artifacts in the image.

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