<p>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. </p>

<p>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 <span class="math-container">$\int T_{b}(\lambda)d\lambda$</span> is just a normalization term. </p>

CC BY-SA 4.0
2020-02-24T09:34:45.100Z
It is a bit of a misnomer, but I suppose the justification is that the [convolution theorem](https://en.wikipedia.org/wiki/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.
PM 2Ring
CC BY-SA 4.0
2020-02-24T09:39:15.453Z
@PM2Ring I never thought of it that way, but it would make sense !
usernumber
CC BY-SA 4.0
2020-02-24T11:33:28.337Z
@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
CC BY-SA 4.0
2020-02-24T13:05:28.443Z
I don't know about the physical relevance of switching to time domain for an SED...
usernumber
CC BY-SA 4.0
2020-02-24T15:12:57.053Z
Thank you. I was just trying to reason further why is this operation called "convolution" in astronomy.
Catarina Alves
CC BY-SA 4.0
2021-05-11T06:32:36.253Z
@PM2Ring could you please confirm your suggestion? The convolution theorem says indeed that convolution in one domain (either time or frequency) is point-wise multiplication in the other.  However, the equation in the OP is not just a point-wise multiplication, but an integral which sort of looks like a convolution with a slight deviation. Can somebody please help with this?
ngiann
CC BY-SA 4.0
2020-02-24T09:19:31.060Z
2020-02-24T09:19:31.060Z
usernumber
<p>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 
<span class="math-container">$$F_{b}=\dfrac{\int f(\lambda) T_{b}(\lambda)d\lambda}{\int T_{b}(\lambda)d\lambda}\,,$$</span> 
where <span class="math-container">$f(\lambda)$</span> is the SED, and <span class="math-container">$T_b$</span> is the filter response in band <span class="math-container">$b$</span>.</p>

<p>This formula seems very different from a <a href="https://en.wikipedia.org/wiki/Convolution" rel="nofollow noreferrer">mathematical convolution</a> that I would write as 
<span class="math-container">$$f*g(\lambda)=\int_{-\infty}^{\infty} f(x) g(\lambda-x)dx\,.$$</span></p>

<p>Are these two convolutions the same thing? Or is the "astronomy convolution" a different thing (e.g. <a href="https://sedfitter.readthedocs.io/en/stable/convolution.html" rel="nofollow noreferrer">SED Fitter python package</a>)?</p>

<spectroscopy><luminosity><photometry><mathematics>
Convolve a SED with a filter. Is convoluting the mathematical operation?
Catarina Alves
# Convolve a SED with a filter. Is convoluting the mathematical operation?

(Asked by: username_0 on 2020-02-23)

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](https://en.wikipedia.org/wiki/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](https://sedfitter.readthedocs.io/en/stable/convolution.html))?

**Tags:** spectroscopy, luminosity, photometry, mathematics


---

## 1 answers

### Answer by username_1 on 2020-02-24. Score: 2 | Accepted 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.

> **Comments:**
> **username_2**: It is a bit of a misnomer, but I suppose the justification is that the [convolution theorem](https://en.wikipedia.org/wiki/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.  
> **username_1**: @PM2Ring I never thought of it that way, but it would make sense !  
> **username_0**: @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?  
> **username_1**: I don't know about the physical relevance of switching to time domain for an SED...  
> **username_0**: Thank you. I was just trying to reason further why is this operation called "convolution" in astronomy.  
> **username_3**: @PM2Ring could you please confirm your suggestion? The convolution theorem says indeed that convolution in one domain (either time or frequency) is point-wise multiplication in the other.  However, the equation in the OP is not just a point-wise multiplication, but an integral which sort of looks like a convolution with a slight deviation. Can somebody please help with this?  

---

