在matlab中绘制函数

Question

我需要将原始输入转换为函数

dt = 0.01; 
t1 = 0:dt:1;
t2 = 1+dt:dt:2.5;
t3 = 2.5+dt:dt:3.5;
t4 = 3.5+dt:dt:4;
t5 = 4+dt:dt:4.5;

y1 = ones(size(t1))*-1; 
y2 = 2*(t2-2.5);
y3 = 1*sin(2*pi*(t3-1));
y4 = ones(size(t4))*0;
y5 = rand( size(t5) )-0.5;;

plot(t1, y1, t2, y2, t3, y3, t4, y4, t5 ,y5)

“这个工作正常^^^^。”

我已经完成了，但脚本中出了问题。

我在哪里弄错了？

现在看起来如何

function [t1, y1, t2, y2, t3, y3, t4, y4, t5 ,y5]=funct(t)
t = 0.01; % Its every step (0<=t) = (0<=0.01)


t1 =t((0<=t)&(t<1));
y1 = ones(size(t1))*-1;


t2 =t((1<=t)&(t<2.5));
y2 = 2*(t2-2.5);


t3 = t((2.5<=t)&(t<3.5));
y3 = 1*sin(2*pi*(t3-1)); %you wrote sn here


t4 =t((3.5<=t)&(t<4));
y4 = ones(size(t4))*0;


t5=t((4<=t)&(t<4.5));
y5 = rand( size(t5) )-0.5;


plot(t1, y1, t2, y2, t3, y3, t4, y4, t5 ,y5)

Answer 1

我获得了相同的结果。 首先我运行你的代码：

dt = 0.01; 
t1 = 0:dt:1;
t2 = 1+dt:dt:2.5;
t3 = 2.5+dt:dt:3.5;
t4 = 3.5+dt:dt:4;
t5 = 4+dt:dt:4.5;

y1 = ones(size(t1))*-1; 
y2 = 2*(t2-2.5);
y3 = 1*sin(2*pi*(t3-1));
y4 = ones(size(t4))*0;
y5 = rand( size(t5) )-0.5;;

plot(t1, y1, t2, y2, t3, y3, t4, y4, t5 ,y5)

并获得了这个：

我尝试运行你的功能后：

t = 0:dt:5;
funct(t)

，其中

function y=funct(t)
%dt = 0.01; %this is not necessary I think

%t1 = 0:dt:1;
t1 =t((0<=t)&(t<1));
y1 = ones(size(t1))*-1;

%t2 = 1+dt:dt:2.5;
t2 =t((1<=t)&(t<2.5));
y2 = 2*(t2-2.5);

%t3 = 2.5+dt:dt:3.5;
t3 = t((2.5<=t)&(t<3.5));
y3 = 1*sin(2*pi*(t3-1)); %you wrote sn here

%t4 = 3.5+dt:dt:4;
t4 =t((3.5<=t)&(t<4));
y4 = ones(size(t4))*0;

%t5 = 4+dt:dt:4.5;
t5=t((4<=t)&(t<4.5));
y5 = rand( size(t5) )-0.5;

y=[y1 y2 y3 y4 y5];

figure
plot(t1, y1, t2, y2, t3, y3, t4, y4, t5 ,y5)

我刚刚在你的代码中纠正了sn-＆gt; sin。 如果你试图在函数外部执行plot(t1, y1, t2, y2, t3, y3, t4, y4, t5 ,y5)它也无法工作，因为它的功能没有足够的输出。

您应该按以下方式修改第一行：

function [t1, y1, t2, y2, t3, y3, t4, y4, t5 ,y5]=funct(t)

希望这会有所帮助。这些错误是不同的吗？