# An Aircraft Object Detection Algorithm Based on Small Samples in Optical Remote Sensing Image
## 2 Theoretical Analysis and Methods  
Generally, the grayscale, shape, size and shadow of aircraft objects in remote sensing images are used to distinguish aircraft from other interfering objects. Usually aircraft objects are brighter than the background, and most aircraft have similar shapes that they have four parts of the aircraft nose, left wing, tail and right wing. As shown in Figure 2a, the portion where the aircraft fuselage and the wingspan intersect may form a rectangle whose length and width are the widths of the fuselage and the wing, respectively. If the center of gravity of the rectangle is the center of the circle, draw a circle is with a diameter larger than the width of the fuselage and smaller than the length of the fuselage and wingspan, as shown in Figure 2b. The pixel points on the circumference are extracted clockwise, and the gray value of the pixels on the circumference will show the characteristic of \"bright-dark-bright-dark-bright-dark-bright-dark\" [32]. As shown in Figure 2c, a bar graph is made by taking 601 pixels on the circumference as the horizontal axis and the gray value at each pixel as the vertical axis. However, when the center of the circle is not at the center of the aircraft, it does not have such a characteristic. Therefore, based on this feature, aircraft and non-aircraft can be distinguished.  
Figure 2: (**a**) The rectangle of the intersection of the aircraft fuselage and the wingspan. (**b**) The circumference on the aircraft, the size of the original image is \(1024\times 958\), the radius of the circle is 200 pixels. (**c**) The gray value curve on the circumference, the horizontal axis is 601 pixels, and the vertical axis is the gray value at each pixel point.  
Let \(f_{k}\) (\(k=0\), \(1\), \(2\),..., \(N-1\)) denote the gray value on the circumference centered on \((i,j)\) and the radius is \(R\). Make a discrete Fourier transform on this one-dimension array and obtain  
\[F=\sum_{k=0}^{N-1}f_{k}e^{-j\frac{2\pi}{N}kn}(k=0,1,2,\ldots,N-1), \tag{1}\]  
Square the above equation to get the amplitude value.  
\[F^{2}=\left(\sum_{k=0}^{N-1}f_{k}\cos\frac{2\pi}{N}kn\right)^{2}+\left(\sum_{k =0}^{N-1}f_{k}\sin\frac{2\pi}{N}kn\right)^{2}, \tag{2}\]  
where \(n\) is a constant used to control the number of periods of the sine and cosine functions for the discrete Fourier transform.  
The gray value curve on the circumference has 4 peaks and 4 valleys, which is similar to the sine and cosine curves of 4 cycles. So, if four cycles of sine and cosine functions are selected in Equations (1) and (2) (that is, \(n=4\)), the amplitude value of the circumferential frequency filter [32] in the center of the aircraft is large.  
Since the non-aircraft center point usually does not have the special properties, the value of the circle frequency filter [32] is small. The amplitude value is the square of the value obtained by the Fourier transform, therefore, the amplitude values on the center of the aircraft and the non-aircraft center will vary greatly. Therefore, it is easy to distinguish between aircraft centers and non-aircraft centers.