标签:col plot inf label abs sub fun amp 图片
1.插值函数
%%分段线性插值 function PLI = Piecewise_linear_interpolation(X,f,precision) [m,n] = size(X);a = min(X);b = max(X); X = sort(X); F = subs(f,X); for k = 1:n-1 B = Basic_fun(X,k); I = B(1)*F(k)+B(2)*F(k+1); PLI{1,k} = [X(k),X(k+1)]; PLI{2,k} = I; t{k} = X(k):(X(k+1)-X(k))/precision:X(k+1); T{k} = subs(I,t{k}); Y_real{k} = subs(f,t{k}); end for k = 1:n-1 t_((precision+1)*(k-1)+1:(precision+1)*k) = t{k}; T_((precision+1)*(k-1)+1:(precision+1)*k) = T{k}; Y_real_((precision+1)*(k-1)+1:(precision+1)*k)= Y_real{k}; end h = figure; set(h,‘color‘,‘w‘); plot(X,F,‘r*‘,t_,T_,‘g‘,t_,Y_real_,‘b‘); xlabel(‘x shaft‘);ylabel(‘y shaft‘); legend(‘F:节点对应函数值‘,‘T:分段线性插值函数图像‘,‘Y_real:真实函数图像‘); title(‘分段线性插值‘); grid on end
2.基函数
%%基函数,max(X)>k>0 function BF = Basic_fun(X,k) X = sort(X); syms x; BF(1) = (x-X(k+1))/(X(k)-X(k+1)); BF(2) = (x-X(k))/(X(k+1)-X(k)); end
3.拟合值函数
%%线性插值拟合值 function LIV = Linear_interpolation_value(X,f,precision,x_value) [m,n] = size(X);a = min(X);b = max(X); X = sort(X); Answer = Piecewise_linear_interpolation(X,f,precision); for i = 1:n-1 if x_value >= X(i) && x_value <= X(i+1) s = i; end end LIV{1,1} = ‘线性插值拟合值‘; LIV{2,1} = vpa(subs(Answer{2,s},x_value),6); LIV{1,2} = ‘真实值‘; LIV{2,2} = vpa(subs(f,x_value),6); LIV{1,3} = ‘误差‘; LIV{2,3} = abs(LIV{2,1}-LIV{2,2}); end
4.例子
clear all clc X = -5:1:5; syms x; f = - 0.08858*x^8 + 3.694*x^7 - 64.7*x^6 + 617.8*x^5 - 3490.0*x^4 + 11820.0*x^3 - 23150.0*x^2 + 23580.0*x - 9319.0; precision = 200; %%分段线性插值 disp(‘分段线性插值‘); Piecewise_linear_interpolation(X,f,precision)
结果
分段线性插值 S = 2×10 cell 数组 列 1 至 4 {1×2 double} {1×2 double} {1×2 double} {1×2 double} {1×1 sym } {1×1 sym } {1×1 sym } {1×1 sym } 列 5 至 8 {1×2 double} {1×2 double} {1×2 double} {1×2 double} {1×1 sym } {1×1 sym } {1×1 sym } {1×1 sym } 列 9 至 10 {1×2 double} {1×2 double} {1×1 sym } {1×1 sym } >> S{2,:} ans = (227077586881*x)/50000 + 37695704689/2500 ans = (3983468847*x)/2000 + 60987657739/12500 ans = (7723057429*x)/10000 + 30518164433/25000 ans = (2518396259*x)/10000 + 4494858583/25000 ans = (3136314129*x)/50000 - 9319 ans = (465835271*x)/50000 - 9319 ans = (422501*x)/10000 - 1113617/25000 ans = 4111433/25000 - (622509*x)/10000 ans = - (271*x)/80 - 151661/12500 ans = 2072089/2500 - (10681481*x)/50000
图像如下
标签:col plot inf label abs sub fun amp 图片
原文地址:https://www.cnblogs.com/guliangt/p/12112829.html