How to plot step functions in Matlab
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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
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VBBV
il 31 Ott 2020
Modificato: VBBV
il 31 Ott 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 Commenti
VBBV
il 31 Ott 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
Più risposte (1)
Vladimir Sovkov
il 31 Ott 2020
Modificato: Vladimir Sovkov
il 31 Ott 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));
2 Commenti
Vladimir Sovkov
il 31 Ott 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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