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

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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Fourier analysis代考|MAT3105

数学代写|傅里叶分析代写傅里叶分析代考|图像低通滤波


过滤器是一个系统,从任何通过它的东西中删除一些东西。咖啡过滤器去除咖啡渣,让煎剂流过。滤水器可以去除盐和细菌。紫外线滤光片可以阻挡或吸收紫外线


在信号和图像处理中,信号的光谱按需要的方式被改变。如果允许低频分量而抑制高频分量,则称为低通滤波器。如果高频分量被允许,低频分量被抑制,它被称为高通滤波器。如果我们在一个长数据序列的一个数的邻域中找到一组数的运行平均值,序列的颠簸性就会减少。另一方面,如果我们发现了差异,则会增强凹凸性,减少平滑性。一个纯高频信号,如${1,-1}$,如果平均将变为零。如果有差异,纯低频信号(如${1,1}$)将变为零。时域的滤波操作是用卷积来模拟的。各种类型的滤波器的脉冲响应表征了滤波作用


滤波操作在频域更容易可视化,因为时域的卷积变成了频域的乘法。信号的频谱是其有序频率分量相对于其各自振幅的显示。“因此,我们可以简单地以想要的方式改变光谱的任何部分。如果我们使信号高频分量的系数小或为零,信号就会变得更平滑。如果我们使信号的低频分量的系数小或为零,信号就会变得颠簸。除了更容易可视化的滤波操作,在频域中实现滤波操作对于较长的滤波器是更快的。让我们研究一下时域和频域的滤波操作


有不同特性的低通滤波器可用。最简单和广泛使用的$3 \times 3$二维低通滤波器,称为平均滤波器,其脉冲响应为$$
\begin{aligned}
h(m, n) &=\left[\begin{array}{rrr}
h(-1,-1) & h(-1,0) & h(-1,1) \
h(0,-1) & \boldsymbol{h ( 0 . 0 )} & h(0,1) \
h(1,-1) & h(1,0) & h(1,1)
\end{array}\right] \
&=\frac{1}{9}\left[\begin{array}{lll}
1 & 1 & 1 \
1 & \mathbf{1} & 1 \
1 & 1 & 1
\end{array}\right], \quad m=-1,0,1, \quad n=-1,0,1
\end{aligned}
$$
。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
$$
\begin{gathered}
y(m, n)=\frac{1}{9}(x(m-1, n-1)+x(m-1, n)+x(m-1, n+1)+x(m, n-1)+x(m, n) \
\quad+x(m, n+1)+x(m+1, n-1)+x(m+1, n)+x(m+1, n+1))
\end{gathered}
$$
给出,输出图像变得更平滑。滤波器越大,平滑效果越好

数学代写|傅里叶分析代写傅立叶分析代考|低通滤波与DFT


对于二维信号,我们必须在两个方向上置零。图$5.9$显示了$8 \times 8$零填充图像$x z(m, n)$,它是从上一个示例中使用的$4 \times 4$图像$x(m, n)$获得的。还显示了$3 \times 3$过滤器$h(m, n)$,以及相应的补零和移位的行和列过滤器。由于卷积输出的大小是$6 \times 6$, 2的最近次幂是8,因此零填充图像的大小是$8 \times 8$ .

因为平均滤波器是可分离的。卷积运算可以使用两个1-D滤波器${h(-1)=1, h(0)=1, h(1)=1} / 3$进行。因为$x z(m, n)$的大小是$8 \times 8$,原点在左上角,所以零填充过滤器的长度必须是8,开头是$h(0)$。零填充的滤波器可以写成
$$
{h(-1)=1, h(0)=1, h(1)=1, h(2)=0, h(3)=0, h(4)=0, h(5)=0, h(6)=0} / 3
$$
我们得到零填充的列滤波器
$$
h z(m)=\frac{1}{3}{1,1,0,0,0,0,0,1}
$$
,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
$$
H(k)={3,2.4142,1,-0.4142,-1,-0.4142,1,2.4142}
$$
由于该滤波器在时域是实值且偶对称的,它的DFT也是实值且偶对称的。列滤波器$H(k)$的DFT是行滤波器$H(l)$的DFT的转置。图$5.9$中$x z(m, n)$的二维DFT $X(k, l)$如表5.6所示。频域中的部分卷积输出$3 P(k, l)=X(k, l) H(l)$如表5.7所示,是将$X(k, l)$的每一行按点乘$H(l)$得到的。在频域中的卷积输出$9 Y(k, l)=3 P(k, l) H(k)$如表$5.8$所示,是将$3 P(k, l)$的每列按点乘$H(k)$得到的

数学代写|傅里叶分析代写Fourier analysis代考

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