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Imagine we are observing a star. The light coming from a star enters an optical instrument that will give us some observed data, such as the spectrum of light say.

What we observe is not the true signal coming from the star but the signal after it has been affected by the optical instrument (because the instrument is not perfect, no instrument is)

The true signal, call it T(x) is related to the instrument function, call it I(x), and the observed signal, call it O(y) through the following relation

$O(y)=\int T(x) I(y-x) dx$

This is nothing but the definition of a convolution of 2 functions.

My question is, why those 3 functions are related in that particular way, I mean can we prove that the relation between them must be the convolution relation?

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It is because optics is almost exactly linear. –  Ron Maimon Dec 21 '11 at 3:46
    
@RonMaimon Yeah I know that optics is almost linear. I just did not know why the convolution appears there. And it was not explained anywhere or derived. –  Revo Dec 21 '11 at 5:30
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2 Answers

There are several ways to understand this result. One is to think how $I(x,y)$ is defined. It's the signal your instrument gives you at point $y$ when your source is a delta function (that is, a point source) at point x. In mathspeak, $O_\delta(y) = \int \delta(x) I(x,y) \mathrm{d}x$.

If your source is composed of many different points, you need to sum over all those, so you have $O(y) = \int T(x) I(x,y) \mathrm{d}x$. Notice that $I$ is Green's function for your system (you might have studied it at some point). Now, in many cases $I$ depends on $x$ and $y$ only through the difference, so you get the convolution. This last step depends on your instrument, but I'm fairly certain that it's the case for telescopes with good lenses/mirrors in the paraxial approximation.

Another way to understand this result is to think about the Fourier transform. For simplicity's sake, I won't be absolutely rigorous here.

Light at the star has an amplitude profile given by $T(x)$, and when it reaches the objective of your telescope it has diffracted, so you have its Fourier transform ($\mathcal{F}[T]$). The objective acts as a filter that lets some of this light through, and has a transmission profile $\mathcal{F}[I]$, so light just after the objective is the product of both functions: $\mathcal{F}[T]\,\mathcal{F}[I]$.

Finally, the image formed at the focal plane of the objective is the inverse transform of that product: $O(y)=\mathcal{F}^{-1}\{\mathcal{F}[T]\,\mathcal{F}[I]\}$ Using the convolution theorem, $O(y)=T\otimes I$

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think how I(x,y) is defined This really is the critical point here. –  dmckee Dec 20 '11 at 20:53
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For ideal device the observed signal is identical to the true signal.
So if the true signal is a single point $$ T_\text{point}(x) = T_0 \; \delta(x-x_0) $$ then the observed signal is also single point $$ O_\text{ideal}(y)\bigl[T_\text{point}(x)\bigr] = \alpha T_0 \; \delta(y-x_0), $$ where $\alpha$ is the sensitivity.

Non-ideal device blurs the point and we get some function: $$ O(y)\bigl[T_\text{point}(x)\bigr] = T_0 I(y - x_0). \qquad (1) $$ This function is the instrument function.

Real signal is not a single point. It is a function, but this function can be represented as a sum of point signals: $$ T(x) = \int T(x_0)\delta(x-x_0) \; dx_0 \qquad (2) $$

We can apply (1) to (2): $$ O(y)\bigl[T(x)\bigr] = \int O(y)\bigl[T(x_0)\delta(x-x_0)\bigr] \; dx_0 = \int T(x_0)I(y - x_0) \; dx_0 $$

So $$ O(y)\bigl[T(x)\bigr] = \int T(x)I(y - x) \; dx $$

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Sorry, what is the physical meaning of the sensitivity $\alpha$? –  Revo Dec 21 '11 at 8:12
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@Revo, usually device does not give directly the value you measure. If you use say photoelectric detector the intensity of the light is converted to electric current. So the observed value and the true signal can even have different dimensions but they are still proportional. The sensitivity is the rate between the observed and measured values. Usually the device supports different values of sensitivity to be useful for experiments with different level of the signal. –  Maksim Zholudev Dec 21 '11 at 9:01
    
I see, excellent explanation. Thank you. +1 –  Revo Dec 22 '11 at 13:44
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