Plotting piecewise function self-dependent

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Eugenio Cataldo
Eugenio Cataldo 2017-5-20
评论： Eugenio Cataldo 2017-5-20
Hello,
I'm new to MATLAB and I have to plot a piecewise function and I can't manage to get it. The function is the (20) in the image, is a piecewise function which initial condition depends on the function itselft in the previous interval. It comes from the solution of the differential equation (18) with condition (19). The parameter m is equal to 0.02 and T, the "period of pulse" (from t_(n) to t_(n+1)) is 2. The "S^+" in the first of (20) stands for the value of the function at a time immediately before t_(n), when the second of (20) is applied.
I tried to write the intervals manually but there is some concept error, it already stucks at line 2 saying: "Subscript indices must either be real positive integers or logicals.
Error in Vaccini (line 2) y(0.1)=0.055;"
my code is (the first two rows are not displayed in the window, I don't know why...):
t=0:.1:8; y(0)=0.055;
for i=1:lenght(t)
if(0<=t(i))&(t(i)<=2)
y(t)=1+(y(0.1)-1)*exp(-0.02*(t));
elseif(2<=t(i))&(t(i)<=4)
y(t)=1+((1-0.05)*(y(2)-1))*exp(-0.02*(t-2));
elseif(4<=t(i))&(t(i)<=6)
y(t)=1+((1-0.05)*(y(4)-1))*exp(-0.02*(t-4));
elseif(6<=t(i))&(t(i)<=8)
y(t)=1+((1-0.05)*(y(6)-1))*exp(-0.02*(t-6));
end
end
plot(t,y);
If you have a better implementation it's welcome, with my knowledge I couldn't find anything better...
Thank you in advance

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回答（1 个）

John D'Errico
John D'Errico 2017-5-20
编辑：John D'Errico 2017-5-20
Matlab does NOT use zero based indexing. So this is an illegal expression:
y(0)=0.055;
Read the error message again. It told you exactly that.
Expressions like this will also cause failure:
y(t)=1+(y(0.1)-1)*exp(-0.02*(t));
Here you are trying to index what is apparently a vector y, with the floating point number t, which takes on non-integer values. You also try to access y(0.1). Where does the 0.1 element of a vector lie in memory? I know, it should lie somewhere between the zero'th and the first element, but that would seem to be a real problem.
Again, a vector or matrix index MUST be an integer that is greater than zero. You are creating and using y as if it is a vector.
  1 个评论
Eugenio Cataldo
Eugenio Cataldo 2017-5-20
Ok, I tried to correct that problem and I managed to plot a function. What I don't understand is the first "pulse", the one at x=2 (the abscissa, the horizontal axis). It does not perform a reduction by 1/2 like the others, but the code is the same as for x=4 and x=6! Here is the new code:
t=0:.1:8;
for i=1:length(t)
if(0<=t(i))&(t(i)<=2)
y(i)=1+(0.0556-1)*exp(-0.02*(t(i)));
elseif(2<=t(i))&(t(i)<=4)
y(i)=1+((1-0.5)*y(2)-1)*exp(-0.02*(t(i)-2));
elseif(4<=t(i))&(t(i)<=6)
y(i)=1+((1-0.5)*y(4)-1)*exp(-0.02*(t(i)-4));
elseif(6<=t(i))&(t(i)<=8)
y(i)=1+((1-0.5)*y(6)-1)*exp(-0.02*(t(i)-6));
end
end
plot(t,y);

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