GNU Astronomy Utilities


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6.3.2.3 Fourier series

In astronomical images, our variable (brightness, or number of photo-electrons, or signal to be more generic) is recorded over the 2D spatial surface of a camera pixel. However to make things easier to understand, here we will assume that the signal is recorded in 1D (assume one row of the 2D image pixels). Also for this section and the next (Fourier transform) we will be talking about the signal before it is digitized or pixelated. Let’s assume that we have the continuous function \(f(l)\) which is integrable in the interval \([l_0, l_0+L]\) (always true in practical cases like images). Take \(l_0\) as the position of the first pixel in the assumed row of the image and \(L\) as the width of the image along that row. The units of \(l_0\) and \(L\) can be in any spatial units (for example meters) or an angular unit (like radians) multiplied by a fixed distance which is more common.

To approximate \(f(l)\) over this interval, we need to find a set of frequencies and their corresponding ‘magnitude’s (see Circles and the complex plane). Therefore our aim is to show \(f(l)\) as the following sum of periodic functions:

$$f(l)=\displaystyle\sum_{n=-\infty}^{\infty}c_ne^{i{2{\pi}n\over L}l} $$

Note that the different frequencies (\(2{\pi}n/L\), in units of cycles per meters for example) are not arbitrary. They are all integer multiples of the fundamental frequency of \(\omega_0=2\pi/L\). Recall that \(L\) was the length of the signal we want to model. Therefore, we see that the smallest possible frequency (or the frequency resolution) in the end, depends on the length we observed the signal or \(L\). In the case of each dimension on an image, this is the size of the image in the respective dimension. The frequencies have been defined in this “harmonic” fashion to insure that the final sum is periodic outside of the \([l_0, l_0+L]\) interval too. At this point, you might be thinking that the sky is not periodic with the same period as my camera’s view angle. You are absolutely right! The important thing is that since your camera’s observed region is the only region we are “observing” and will be using, the rest of the sky is irrelevant; so we can safely assume the sky is periodic outside of it. However, this working assumption will haunt us later in Edges in the frequency domain.

The frequencies are thus determined by definition. So all we need to do is to find the coefficients (\(c_n\)), or magnitudes, or radii of the circles for each frequency which is identified with the integer \(n\). Fourier’s approach was to multiply both sides with a fixed term:

$$f(l)e^{-i{2{\pi}m\over L}l}=\displaystyle\sum_{n=-\infty}^{\infty}c_ne^{i{2{\pi}(n-m)\over L}l} $$

where \(m>0\)68. We can then integrate both sides over the observation period:

$$\int_{l_0}^{l_0+L}f(l)e^{-i{2{\pi}m\over L}l}dl =\int_{l_0}^{l_0+L}\displaystyle\sum_{n=-\infty}^{\infty}c_ne^{i{2{\pi}(n-m)\over L}l}dl=\displaystyle\sum_{n=-\infty}^{\infty}c_n\int_{l_0}^{l_0+L}e^{i{2{\pi}(n-m)\over L}l}dl $$

Both \(n\) and \(m\) are positive integers. Also, we know that a complex exponential is periodic so after one period (\(L\)) it comes back to its starting point. Therefore \(\int_{l_0}^{l_0+L}e^{2{\pi}k/L}dl=0\) for any \(k>0\). However, when \(k=0\), this integral becomes: \(\int_{l_0}^{l_0+T}e^0dt=\int_{l_0}^{l_0+T}dt=T\). Hence since the integral will be zero for all \(n{\neq}m\), we get:

$$\displaystyle\sum_{n=-\infty}^{\infty}c_n\int_{l_0}^{l_0+T}e^{i{2{\pi}(n-m)\over L}l}dl=Lc_m $$

The origin of the axis is fundamentally an arbitrary position. So let’s set it to the start of the image such that \(l_0=0\). So we can find the “magnitude” of the frequency \(2{\pi}m/L\) within \(f(l)\) through the relation:

$$c_m={1\over L}\int_{0}^{L}f(l)e^{-i{2{\pi}m\over L}l}dl $$


Footnotes

(68)

We could have assumed \(m<0\) and set the exponential to positive, but this is more clear.


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