# Convolve a SED with a filter. Is convoluting the mathematical operation?

I know that in order to get the Flux of a star (or something else) in a particular filter from its SED (luminosity per unit wavelength), I need to convolve the spectrum (SED) with the filter response. Most of the formulas I see to do this are $$F_{b}=\dfrac{\int f(\lambda) T_{b}(\lambda)d\lambda}{\int T_{b}(\lambda)d\lambda}\,,$$ where $$f(\lambda)$$ is the SED, and $$T_b$$ is the filter response in band $$b$$.

This formula seems very different from a mathematical convolution that I would write as $$f*g(\lambda)=\int_{-\infty}^{\infty} f(x) g(\lambda-x)dx\,.$$

Are these two convolutions the same thing? Or is the "astronomy convolution" a different thing (e.g. SED Fitter python package)?

## 1 Answer

To get the flux of an SED through a particular filter, you actually multiply the the SED by the filter's response. Talking about convolution in this context is a bit of a misnomer.

Basically, for each wavelength, you look at what fraction of the light will go through the filter, and you sum up the values you get at those wavelengths. The division by $$\int T_{b}(\lambda)d\lambda$$ is just a normalization term.

• It is a bit of a misnomer, but I suppose the justification is that the convolution theorem says that convolution in the time domain equals point-wise multiplication in the frequency domain. And of course spectra & filter responses are in the frequency (or wavelength) domain. Commented Feb 24, 2020 at 9:34
• @PM2Ring I never thought of it that way, but it would make sense ! Commented Feb 24, 2020 at 9:39
• @PM2Ring, does this means I should be able to find a $T_b$ defined in time domain? Additionally, I always need to integrate over frequency/ wavelength in the Fourier space or over time in the time domain (i.e. $\int_{t^*=-\infty}^{t^*=\infty}\int_{t^=-\infty}^{t=\infty}f(t)T_b(t^*-t)dtdt^*$), right? Commented Feb 24, 2020 at 11:33
• I don't know about the physical relevance of switching to time domain for an SED... Commented Feb 24, 2020 at 13:05
• Thank you. I was just trying to reason further why is this operation called "convolution" in astronomy. Commented Feb 24, 2020 at 15:12