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Plotting a function of a function with piecewise limits

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Braden Kerr
Braden Kerr on 25 Sep 2020
Commented: Walter Roberson on 26 Sep 2020
Hello,
I am trying to plot a function of a function, f(x(t)) with piecewise limits, however I keep getting an error that "Subscript indices must either be real positive integers or logicals."
Im wondeirng if I need to adjust my limits im using in my piecewise function so they are in terms of t or if I can leave it in the current form and am missing something else.
t =[0:0.1:2*pi];
x = 2*sin(t);
%% Plot 1
f_1 = piecewise(x<-1, 1.5*x+1.5, 0, -1<x<1, 1.5*x(t)-1.5, x>1);
plot (f_1, t);
Thank you

  1 Comment

Walter Roberson
Walter Roberson on 26 Sep 2020
user asked near duplicate at https://www.mathworks.com/matlabcentral/answers/600166-plotting-a-piecewise-returns-error?s_tid=srchtitle which I have Answered.

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Answers (1)

Dana
Dana on 25 Sep 2020
It's the x(t) that shows up in your definition of f_1: t is a vector of real numbers (including non-integers), so x(t) is not a sensible MATLAB expression. Did you want that to just be x rather than x(t)?

  2 Comments

Braden Kerr
Braden Kerr on 26 Sep 2020
I need to plot f(x(t)) so it should be x(t) and not just x. Seems like easiest may be to plot each piecewise part on the same graph with hold on
Dana
Dana on 26 Sep 2020
No, it shouldn't be x(t), because x(t) doesn't make sense as a MATLAB expression, and that's why you're getting the error you're getting.
For a vector y, the syntax y(j) means the j-th element of y if j is a scalar, and if j is a vector then y(j) = [y(j(1)) y(j(2)) ... ], where j(k) is the k-th element of j.
In your case, t is a vector, so x(t) would put you in the second case above. But t(k) is not typically a positive integer in your case, so there's no such thing as the t(k)-th element of x.
In addition, the use of the symbolic function piecewise is probably not the best way to go about this, not to mention you've messed up the order of your input arguments. Ignoring the x(t) issue, I think you wanted
f_1 = piecewise(x<-1, 1.5*x+1.5, -1<x<1, 0, x>1, 1.5*x(t)-1.5);
I suspect the following will do what you actually want:
t =[0:0.1:2*pi];
x = 2*sin(t);
% The following function returns 1.5*z+1.5 if z<-1,
% 1.5*z-1.5 if z>1, and 0 otherwise
f_1 = @(z) (z<-1).*(1.5*z+1.5) + (z>1).*(1.5*z-1.5);
plot (t,f_1(x));

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