Taking derivative of a time based function?
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I'm new to MatLab and want to plot out some time based functions.
As an example, I have a function y = ((t/T).^n).*exp(-t/T). I want to plot the function vs time (t), along with the derivative of the function vs t.
I'm trying to use the diff function, but it's not working right. Any suggestions? Below is my actual code:
T = 8;
n = 1;
t = 0:0.1:100;
y = ((t/T).^n).*exp(-t/T);
x = diff(y,t);
plot(t,y,t,x)
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Respuestas (3)
Azzi Abdelmalek
el 6 de Dic. de 2013
Editada: Azzi Abdelmalek
el 6 de Dic. de 2013
T = 8;
n = 1;
t = 0:0.1:100;
y = ((t/T).^n).*exp(-t/T);
x = diff(y)./diff(t);
plot(t,y,t(1:end-1),x)
% or use gradient
T = 8;
n = 1;
t = 0:0.1:100;
y = ((t/T).^n).*exp(-t/T);
x=gradient(y,t)
plot(t,y,t,x)
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Wayne King
el 6 de Dic. de 2013
Editada: Wayne King
el 6 de Dic. de 2013
Are you trying to differentiate symbolically or numerically?
T = 8;
n = 1;
t = 0:0.1:100;
y = ((t/T).^n).*exp(-t/T);
x = diff(y)./diff(t);
subplot(211)
plot(t,y); title('y(t)');
subplot(212)
plot(t(2:end),x); title('dy/dt')
Symbolically
syms t; % requires symbolic toolbox
g(t) = t/8;
h(t) = exp(-t/8);
y = g(t)*h(t);
x = diff(y,t);
subplot(211)
ezplot(y,[0 100])
subplot(212)
ezplot(x,[0 100])
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Roger Stafford
el 6 de Dic. de 2013
The 'diff' function serves two purposes, one to take derivatives and the other to take finite differences. To make it take derivatives you have to declare the variables involved as type 'sym'. To use the numeric form as you have as an approximation you will need to divide the 'diff' output by the length of the 't' interval which in your case is 0.1 .
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