## 数学代写|傅里叶分析代写Fourier analysis代考|MATH2100

2023年1月6日

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## 数学代写|傅里叶分析代写Fourier analysis代考|Fourier Boundary Descriptor

In image processing, objects in an image have to be identified. For that purpose, the image is segmented using the properties of the objects. The segmented objects have to be compactly represented. One way of characterizing an object is by its boundary representation. A boundary can be described by its coordinates. The objective is to minimize the storage requirements in representing it. Fourier boundary descriptor is one of the effective methods to represent the boundary of an object.

A closed boundary is represented by a set of its coordinates in the spatial domain. At each point on the boundary, a complex number is formed with its real part being the $x$-coordinate and the imaginary part being the $y$-coordinate. The set of the complex numbers is a periodic complex data, the period being the number of points on the boundary. Let the $N$ boundary coordinates be $${(x(0), y(0)),(x(1), y(1)), \ldots,(x(N-1), y(N-1))}$$
The complex number representation of the boundary becomes
$$b(n)={(x(0)+j y(0)),(x(1)+j y(1)), \ldots,(x(N-1)+j y(N-1))}$$
The 1-D DFT of this set, $B(k)$, is the Fourier descriptor of the boundary with significant advantages. The 2-D data is represented by a 1-D data.

It is desirable that the descriptor is as much insensitive as possible for scaling, translation, and rotation of the boundary. The properties of the DFT make it to relate the Fourier descriptors of a boundary and its modified versions. Consider the $4 \times 4$ binary image $x(m, n)$ and its shifted version $x(m-1, n-1)$
$$x(m, n)=\left[\begin{array}{llll} 1 & 1 & 1 & 0 \ 1 & 0 & 1 & 0 \ 1 & 1 & 1 & 0 \ 0 & 0 & 0 & 0 \end{array}\right] \quad x(m-1, n-1)=\left[\begin{array}{llll} 0 & 0 & 0 & 0 \ 0 & 1 & 1 & 1 \ 0 & 1 & 0 & 1 \ 0 & 1 & 1 & 1 \end{array}\right]$$

## 数学代写|傅里叶分析代写Fourier analysis代考|Periodicity

Let $x(n) \leftrightarrow X(k)$ with period $N$. Then,
$$x(n+a N)=x(n) \text {, for all } n \text { and } X(k+a N)=X(k) \text {, for all } k$$
where $a$ is an arbitrary integer. If a signal $x(n)$ is periodic with period $N$ samples, then its values over any successive $N$ samples are the same. Let
$$x(n)={\check{3}, 1,2,4} \leftrightarrow X(k)={\check{10}, 1+j 3,0,1-j 3}$$
As $-4 \bmod 4=0, x(-4)=x(0)=3$. Similarly, $X(5)=X(1)=1+j 3$.
With the exponential $e^{j \frac{2 \pi}{N} n k_0}$ also periodic with period $N, x(n) e^{j \frac{2 \pi}{N} n k_0}$ is also periodic with period $N$. Therefore, the DFT coefficients are periodic and can be determined using the expression
$$X(k)=\sum_{n=a}^{a+N-1} x(n) e^{-j k \frac{2 \pi}{N} n}, \quad k=0,1,2, \ldots, N-1$$
where $a$ is an arbitrary integer. Therefore, given a set of samples starting with index other than zero, Eq. (3.1) can be used. Alternately, the given $N$-point sequence can be periodically extended and the samples starting with index zero can be obtained. Then, Eq. (3.1), with $a=0$, can be used. For example,
$$x(n)={2 \check{2}, 1,2,4} \leftrightarrow X(k)={9 \check{9}, j 3,-1,-j 3}$$
The periodic extension of $x(n)$ is
$${\ldots, 2,1,2,4,2,1,2,4, \check{2}, 1,2,4,2,1,2,4,2,1,2,4, \ldots}$$
Then, for example,
$$x(n)={2,4,2 \check{2}, 1} \leftrightarrow X(k)={9 \check{9}, j 3,-1,-j 3}$$
using Eq. (3.1) with $a=-2$. Similarly, the IDFT can be computed using any successive $N$ samples of the periodic DFT spectrum.

# 傅里叶分析代写

## 数学代写|傅里叶分析代写Fourier analysis代考|Fourier Boundary Descriptor

$$(x(0), y(0)),(x(1), y(1)), \ldots,(x(N-1), y(N-1))$$

$$b(n)=(x(0)+j y(0)),(x(1)+j y(1)), \ldots,(x(N-1$$

## 数学代写|傅里叶分析代写Fourier analysis代考|Periodicity

$$X(k)=\sum_{n=a}^{a+N-1} x(n) e^{-j k \frac{2 \pi}{N} n}, \quad k=0,1,2, \ldots, N-$$

$\ldots, 2,1,2,4,2,1,2,4, \check{2}, 1,2,4,2,1,2,4,2,1,2,4$,

$x(n)=2,4,2 \check{2}, 1 \leftrightarrow X(k)=9 \check{9}, j 3,-1,-j 3$

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## MATLAB代写

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