# 数学代写|傅里叶分析代写Fourier analysis代考|MAT3105

## 数学代写|傅里叶分析代写Fourier analysis代考|Lowpass Filtering of Images

Filter is a system that removes something from whatever passes through it. A coffee filter removes the coffee grounds and allows the decoction to flow through. A water filter removes salt and bacteria. An ultraviolet filter blocks or absorbs ultraviolet light.

In signal and image processing, the spectra of signals are altered in a desired way. If the low-frequency components are allowed and high-frequency components suppressed, it is called a lowpass filter. If the high-frequency components are allowed and low-frequency components suppressed, it is called a highpass filter. If we find the running averages of a set of numbers in a neighborhood of a number of a long data sequence, the bumpiness of the sequence is reduced. On the other hand, if we find the differences, the bumpiness will be enhanced and the smoothness reduced. A purely high frequency signal such as ${1,-1}$ will become zero, if averaged. A purely low-frequency signal such as ${1,1}$ will become zero, if differenced. The filtering operation, in the time domain, is modeled by convolution. The impulse responses of the various types of filters characterize the filtering action.

The filtering operation is easier to visualize in the frequency domain, since convolution in the time domain becomes multiplication in the frequency domain. The spectrum of a signal is a display of its ordereed frequency components verrsus their respective amplitudes. ‘Iherefore, we can simply alter any part of the spectrum in a desired way. If we make the coefficients of the high-frequency components of the signal small or zero, the signal becomes smoother. If we make the coefficients of the low-frequency components of the signal small or zero, the signal becomes bumpier. In addition to easier visualization of the filtering operation, the implementation of the filtering operation in the frequency domain is faster for longer filters. Let us study the filtering operation in both the time domain and the frequency domain.

Lowpass filters with different characteristics are available. The impulse response of the simplest and widely used $3 \times 3$ 2-D lowpass filter, called the averaging filter, is \begin{aligned} h(m, n) &=\left[\begin{array}{rrr} h(-1,-1) & h(-1,0) & h(-1,1) \ h(0,-1) & \boldsymbol{h ( 0 . 0 )} & h(0,1) \ h(1,-1) & h(1,0) & h(1,1) \end{array}\right] \ &=\frac{1}{9}\left[\begin{array}{lll} 1 & 1 & 1 \ 1 & \mathbf{1} & 1 \ 1 & 1 & 1 \end{array}\right], \quad m=-1,0,1, \quad n=-1,0,1 \end{aligned}
The origin of the filter is shown in boldface. All the coefficient values are 1/9. The filter outputs the average of the pixel values of the image in each neighborhood. Variable weighting is applied in other lowpass filters. When an image is passed through this filter, the output at each point is given by
$$\begin{gathered} y(m, n)=\frac{1}{9}(x(m-1, n-1)+x(m-1, n)+x(m-1, n+1)+x(m, n-1)+x(m, n) \ \quad+x(m, n+1)+x(m+1, n-1)+x(m+1, n)+x(m+1, n+1)) \end{gathered}$$
The output image becomes smoother. The smoothing effect increases with larger filters.

## 数学代写|傅里叶分析代写Fourier analysis代考|Lowpass Filtering with the DFT

For a 2-D signal, we have to zero pad in two directions. Figure $5.9$ shows the $8 \times 8$ zero-padded image $x z(m, n)$, obtained from the $4 \times 4$ image $x(m, n)$ used in the last example. The $3 \times 3$ filter $h(m, n)$, along with the corresponding zero-padded and shifted row and column filters, is also shown. Since the convolution output is of size is $6 \times 6$ and the the nearest power of 2 is 8 , the size of the zero-padded image is $8 \times 8$.

As the averaging filter is separable. the convolution operation can be carried out using two 1-D filters, ${h(-1)=1, h(0)=1, h(1)=1} / 3$. Since $x z(m, n)$ is of size $8 \times 8$ and the origin is at the top-left corner, the zero-padded filter has to be of length 8 with $h(0)$ in the beginning. The zero-padded filter can be written as
$${h(-1)=1, h(0)=1, h(1)=1, h(2)=0, h(3)=0, h(4)=0, h(5)=0, h(6)=0} / 3$$
We get the zero-padded column filter
$$h z(m)=\frac{1}{3}{1,1,0,0,0,0,0,1}$$
with the origin at the beginning, by circularly shifting left by one position, The row filter, shown in Fig. 5.9, is the transpose of the column filter. Computing the 1-D DFT of this filter (division by 3 is deferred), we get
$$H(k)={3,2.4142,1,-0.4142,-1,-0.4142,1,2.4142}$$
Since the filter is real-valued and even-symmetric in the time domain, its DFT is also real-valued and even-symmetric, The DFT of the column filter $H(k)$ is the transpose of that of the row filter $H(l)$. The 2-D DFT, $X(k, l)$, of $x z(m, n)$ in Fig. $5.9$ is shown in Table 5.6. The partial convolution output, $3 P(k, l)=X(k, l) H(l)$ shown in Table 5.7, in the frequency domain is obtained by pointwise multiplication of each row of $X(k, l)$ by $H(l)$. The convolution output, $9 Y(k, l)=3 P(k, l) H(k)$ shown in Table $5.8$, in the frequency domain is obtained by pointwise multiplication of each column of $3 P(k, l)$ by $H(k)$.

## 数学代写|傅里叶分析代写傅里叶分析代考|图像低通滤波

。所有的系数都是1/9。过滤器输出图像在每个邻域的像素值的平均值。其他低通滤波器采用了变加权。当图像通过这个过滤器时，每个点的输出由
$$\begin{gathered} y(m, n)=\frac{1}{9}(x(m-1, n-1)+x(m-1, n)+x(m-1, n+1)+x(m, n-1)+x(m, n) \ \quad+x(m, n+1)+x(m+1, n-1)+x(m+1, n)+x(m+1, n+1)) \end{gathered}$$

## 数学代写|傅里叶分析代写傅立叶分析代考|低通滤波与DFT

$${h(-1)=1, h(0)=1, h(1)=1, h(2)=0, h(3)=0, h(4)=0, h(5)=0, h(6)=0} / 3$$

$$h z(m)=\frac{1}{3}{1,1,0,0,0,0,0,1}$$
，原点在开始，通过左圆移动一个位置，行滤波器，如图5.9所示，是列滤波器的转置。计算该滤波器的一维DFT(除以3是延迟的)，我们得到
$$H(k)={3,2.4142,1,-0.4142,-1,-0.4142,1,2.4142}$$

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