# How to plot step functions in Matlab

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Sangani Prithvi on 31 Oct 2020
Commented: VBBV on 31 Oct 2020
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();

VBBV on 31 Oct 2020
Edited: VBBV on 31 Oct 2020
%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 CommentsShow 1 older commentHide 1 older comment
Sangani Prithvi on 31 Oct 2020
y(i) = 100*exp(-x(i))*cos(x(i)*pi/180
can you please explain why did you mutliply with 100 in the above equation?
VBBV on 31 Oct 2020
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

Vladimir Sovkov on 31 Oct 2020
Edited: Vladimir Sovkov on 31 Oct 2020
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));
Sangani Prithvi on 31 Oct 2020
thank you
Vladimir Sovkov on 31 Oct 2020
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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