4.6 Java1490-2d fourier transforms using java, part 1  (Page 10/15)

Two full peaks but at different locations

If you consider the peaks at the ends of the wavenumber response for the leftmost and center images in Figure 3 to each represent only half a peak (with the other half being off the scale to the left and the right) , all three scenarios have two complete peaks in their wavenumber responses.

(You could think in terms of printing the wavenumber response on a piece of paper, cutting it out, and taping the two ends together to form acontinuous ring. As you made a complete traversal of the ring, you would encounter two peaks.)

However, the locations of the two peaks for the rightmost array are at completely different wavenumber values than are the peaks for the other twoarrays. The two peaks exhibited by the rightmost array are in the locations of the two nulls for the center array. Similarly, the null points for the rightmostarray are in the same locations as the two peaks for the center array.

What can we learn from these scenarios?

We learn that we can have a significant impact on the wavenumber response of an array by increasing the number of elements in the array. We can also have asignificant impact on the wavenumber response by applying weights, (including sign changes) , to the electrical signals produced by the array elements before adding them together.

Extending into two dimensions

Now let's complicate things a bit by extending our array analysis into two dimensions. Up to this point, we have assumed that our sensors were attached toa wire that was free to move up and down only. As such, waves impinging on the array were constrained to approach the array from one end or the other. In thiscase the wavenumber was completely determined by the wavelength of the wave.

(For our purposes, the wavelength is given by the ratio of propagation speed in meters per second to frequency in cycles per second.Canceling out the units leaves us with wavelength in meters per cycle.)

Move the array to a table top

Let's move our array of sensors from the wire to a large sheet of metal on the top of a table. For the time being, we will still place the elements in aline with uniform spacing. However, we will now assume that a wave can impinge on the array from any direction along the surface of the sheet of metal.

(For simplicity, we will assume that there is some sort of insulation between the sheet of metal and the table top to prevent waves from impinging on the array from below.)

What does a wave look like in this scenario?

Imagine a piece of corrugated sheet metal or fiber glass. (Material like this is sometimes used to build a roof on a patio.) When you look at it from one end, it looks something like the sine wave in Figure 1 . However, if you keep it at eye level and slowly turn it in the horizontal plane, the distance between the peaks willappear to become shorter and shorter until finally you don't see any peaks at all. What you see at that point is something that appears to have the samethickness from one end to the other. This is the view that one of our sensors sees as the wavefront of an impinging wave.