• 数字信号处理Day2-小波基与规范正交化


    我们有这么一张灰度图64*64



    我们能够定义出4096个基,各自是某一位是0其它是1,在这样的情况下,假设我们传输图片,那么就相当于传输原始数据

    如果传到一半,网络坏了。

    于是,我们得到

    我们能够计算原图像和这图像的差距

    error = I - I_approx; 
    distance = sqrt(sum(sum(error.*error))) 
    distance = 
    713.5559

    如果我们使用Haar基



    function h = haar(N)
    
        % Assume N is power of 2
        h = zeros(N,N); 
        h(1,1:N) = ones(1,N)/sqrt(N);
    
        for k = 1:N-1 
            p = fix(log(k)/log(2)); 
            q = k-(2^p); 
            k1 = 2^p;
            t1 = N/k1; 
            k2 = 2^(p+1);
            t2 = N/k2; 
            for i=1:t2 
                h(k+1,i+q*t1) = (2^(p/2))/sqrt(N); 
                h(k+1,i+q*t1+t2) = -(2^(p/2))/sqrt(N); 
            end 
        end

    然后,我们看一下仅仅接受一半的图片能够恢复成什么样子,这个东西能够用到图片压缩(PCA降维)

    clear all
    close all
    clc
    
    % Load image
    I = double(imread('camera.jpg'));
    
    % Arrange image in column vector
    I = I(:);
    
    % Generate Haar basis vector (rows of H)
    H = haar(4096);
    
    % Project image on the new basis
    I_haar = H*I;
    
    % Remove the second half of the coefficient
    I_haar(2049:4096) = 0;
    
    % Recover the image by inverting change of basis
    I_haar = H'*I_haar;
    
    % Rearrange pixels of the image
    I_haar = reshape(I_haar, 64, 64);
    
    % Display image
    figure
    imshow(I_haar,[]);


    效果还能够

    error = I - I_haar; 
    distance = sqrt(sum(sum(error.*error))) 
    distance = 
    350.6765


    Haar基是正交基

    怎样将普通基变成正交基呢

    线性代数中的拉格慕-施密特正交化即可

    function E=gs_orthonormalization(V)
    % V is a matrix where each column contains the vectors spanning the space of which we want to compute the orthonormal base
    % E is a matrix where each column contains an ortho-normal vector of the base of the space
    
    [numberLines,numberColumns]=size(V);
    
    % U is a matrix containing the orthogonal vectors (non normalized)
    U=zeros(numberLines,numberColumns);
    
    for indexColumn=1:numberColumns
        U(:,indexColumn)=V(:,indexColumn);
    
        for index=1:(indexColumn-1)
            R(index,indexColumn) =U(:,index)'*V(:,indexColumn);
            U(:,indexColumn)=U(:,indexColumn)-R(index,indexColumn)*U(:,index);
        end
    
        R(indexColumn,indexColumn)=norm(U(:,indexColumn));
        U(:,indexColumn)=U(:,indexColumn)./R(indexColumn,indexColumn);
    
    end
    
    E=U;
    
    return 


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  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/3808526.html
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