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I want to define a function $y(t)$ that has 2 cases:
$$y(t)= \begin{cases} -1 & \text{for} & -T \lt t \leq 0 \\ 1 & \text{for} & 0 \lt t\leq T \end{cases} $$ where $T$ is an integer.
I already defined $t$ using linspace between $-T$ to $T$ in a resolution of 1000 points.
I was advised to use the command ones, but I dont know how to apply it by the function terms.
Thank you.
Well, this is Octave but should work in Matlab:
function y=Y(t)
y=ones(size(t));
y=y+(t<=0)*(-2);
.. and it is vectorised. Calling will look like:
>t=linspace(-T,T,1000);
>y=Y(t);
To make this more general we could also make the time at which the function switches a parameter also:
function y=Y(t,S)
y=ones(size(t));
y=y+(t<=S)*(-2);
y=[-ones(500,1);ones(500,1)];$\endgroup$ – Matt L. Apr 6 '15 at 18:01yhas nothing to do witht, just the lengths of the two vectors must be the same.ysimply gives you the values, andtdefines where on the t-axis these values are located. That's the way it works. $\endgroup$ – Matt L. Apr 7 '15 at 6:58signfunction and set the value fort=0to be 1? $\endgroup$ – jojek♦ Jun 6 '15 at 13:34