# 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)?

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.
• @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? – Catarina Alves Feb 24 at 11:33