How to plot step functions in Matlab
显示 更早的评论
I have a function involving
y=o for x<o;
y=exp(-x)*cos(x) for 0<x<2pi();
y=2*exp(-x)*cos(x) for x>2pi();
please help me to plot such a type of function in xy space
采纳的回答
%if true
% code
%end
x = -2*pi:0.1:3*pi;
for i = 1:length(x);
if x(i)<0;
y(i)=0;
elseif x(i)<=2*pi & x(i)>=0 ;
y(i) = exp(-x(i))*cos(x(i)*pi/180);
elseif x(i) > 2*pi ;
y(i) = 100*exp(-x(i))*cos(x(i)*pi/180);
end;
end;
plot(x,y)
axis([-2*pi 3*pi -0.2 1])
3 个评论
Sangani Prithvi
2020-10-31
Sangani Prithvi
2020-10-31
VBBV
2020-10-31
you can use a smaller number, say 2, but
y(i) = 2*exp(-x(i))*cos(x(i)*pi/180);
gives a very small step height compared to
y(i) = exp(-x(i))*cos(x(i)*pi/180)
So in the graph it is not noticeable clearly
Thats why i used to 100 which amplifies the step height.
Remember in both cases the step nature does not vary. i,e, decreasing exponential function according to your equations
更多回答（1 个）
Vladimir Sovkov
2020-10-31
编辑：Vladimir Sovkov
2020-10-31
syms x;
y=piecewise(x<0,0, 0<=x<2*pi,exp(-x).*cos(x), x>=2*pi,2*exp(-x).*cos(x));
t=linspace(-pi,4*pi,1000);
plot(t,subs(y,x,t));
2 个评论
Sangani Prithvi
2020-10-31
Vladimir Sovkov
2020-10-31
Welcome.
By the way, in the question, you did not specify what the function is equal to at the boundary points x=0 and x=2*pi; in the code I implied the right limit but you can easily alter this convention.
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