Question

clc  

clear all

x= -2:0.1:4;

y= (3*x.^3)-(26*x)+10;  

y1= diff(y)./diff(x);    

y2= diff(y1)./diff(x);      

plot(x,y,'R')              

hold on                  

plot(x,y1,'G')           

hold on     

plot(x,y2,'B')

这是找到二阶导数的正确方法吗？它没有找到二阶导数并给出一个矩阵维必须在第6行中达成一致的错误。

Answer 1

您的值y对应于值x。在计算y1的一阶导数y的近似值时，使用有限差分。这些有限差分在x之间的点处最接近导数。这意味着您计算二阶导数的近似值应该包含(x(1:end-1)+x(2:end))/2：

x1 = (x(1:end-1) + x(2:end))/2;

y2 = diff(y1) / diff(x1);

请注意，这些近似值y2最接近位置x2 = (x1(1:end-1) + x1(2:end))/2的二阶导数。

总而言之，您的代码可能如下所示：

%// Clear command window, delete all variables, close all figures
clc;  
clear all;
close all;

%// Define function
f = @(x) (3*x.^3)-(26*x)+10;

%// Define positions
x = -2:0.1:4;

%// Calculate f(x)                       
y= f(x);  

%// Calculate approximation to first derivative and their positions
x1 = (x(1:end-1) + x(2:end))/2;
y1 = diff(y) ./ diff(x);

%// Calculate approximation to second derivative and their positions
x2 = (x1(1:end-1) + x1(2:end))/2; 
y2 = diff(y1) ./ diff(x1);      

%// Plot
plot(x, y, 'R',...
     x1,y1,'G',...
     x2,y2,'B');              
legend('f','f''', 'f''''');

找到并绘制信号的二阶导数

1 个答案: