# Convolution with a gaussian G(t)

1. Jan 6, 2005

### 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.

2. Jan 6, 2005

### dextercioby

$$T(x,t)\ast G(t)=:\int_{0}^{t} T(x,\tau)G(t-\tau) d\tau$$
,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. Jan 6, 2005

### Gonzolo

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. Jan 6, 2005

### Gonzolo

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. Jan 6, 2005

### dextercioby

That [itex] \tau [/tex] is viewed as a variable of integration and,for obvious reasons,it has the same dimension as "t".
This function
$$G(\tau)=\exp(-A\tau^{2})$$
is an example of dimensionless function defined everywhere.For obvious reasons,the constant 'A' is dimensional and it has the dimensions of
$$<A>_{SI} =(<\tau>_{SI})^{-2}$$

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. Jan 8, 2005

### Gonzolo

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.