# Relationship of Fourier Transform and Spatial Directivity Pattern

I am reading over and over the Section 1.2 of most understandable mic array tutorial for me .

A sound wave is dependent on time and space which can be observed in the great link.

But I can't understand why the Fourier transform of the spatially sampled data gives the directivity pattern of the microphone array (which can be seen in the section 1.3.2 of the first document)?

Is there any one who can describe this in more computer sciencist way?

• Your first link is not working any more. I'm interested in the tutorial, would you still have either the link or the PDF? – felipeduque Apr 9 '17 at 21:56
• @felipeduque I update the link . Also If you are curious of the answer of this question after working 2 years I figured out that relation between directivity pattern and fft not exact and similarity just luck and (kinda luck since mic array is just summing sinusodials that is what fft does as well ). – Kadir Erdem Demir Apr 28 '17 at 20:18

## 2 Answers

Have a look at the scilab code where I answered your other question.

That question is about a discrete aperture, but it is analogous to the question you are asking here.

For that question, the beam pattern is just given by: $$D(\theta) = \sum_{n=0}^{N-1} w_n \exp\left(j\frac{2\pi n d}{\lambda} \sin(\theta) \right).$$ With judicious choice of parameters, you can see how this looks somewhat like the discrete Fourier transform of the weights $w_n$ which are analogous to $A_R$ in the paper.

To obtain the formulae from the paper, you need to think of the continuous aperture rather than discrete sensors.

In your second link, the time and space dependence of a simple plane wave is given by

Ψ ( x , t ) = A sin ( ω t ± k x )

Just as a Fourier transform of the equation above can be taken from the time domain $t$ to the frequency domain $\omega$, so also can a Fourier transform be taken from the spatial $x$ domain to the wave vector $k$ domain. For the very simple one dimensional plane wave the wave vector reduces to the wavenumber $k= \frac{2\pi}{\lambda}$ in that first equation.

In general both the position and wave vector are vector quantities and the dependence of a wave on both is given by

$$e^{j\vec k\cdot\vec r}$$

The aperture of a microphone array depends on the position of the microphones (a positional quantity), and so the Fourier transform depends on wave vector.

In Section 1.3.2 of your first link, they are doing exactly that - taking the Fourier transform of the aperture from position to wave vector space, which yields the pattern of the beam in terms of the wave vector.

In the example given in the other answer by @Peter K, think of a plane wave propagating away from the microphone array at an angle $\theta$ .The term inside the exponential, is the dot product of the wave vector and the spacing of the microphone array.