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I have a function:

$$ u_1(\mathbf{x})= \begin{cases} 1, & \text{if} \;\;|x_1|\le r_1, |x_2| \le r_2 \\ 0, & \text{otherwise} \end{cases} ,\mathbf{x}=[x_1,x_2]^T \in \mathbb{R}^2 $$

And I'm told that if $r_{1}=r_{2}=5$, then plotting it in Matlab with stem3 should look like this:

enter image description here

Maybe I'm just too exhausted at this point, but I don't see how this is the same function. It looks like it's equal to 1 between 10 and 20 rather than between -5 and 5? Can someone please explain what's going on here?

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2 Answers 2

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Formally, you are correct. But the question says "should look like this". And not "should be equal too". Apparently, the graph drawn is just a hint of the solution, and one should correctly set the $x$- and $y$-axis markers to get the actual solution, only a shift away with the vector $(-15,-15)$, that could replace this graph based on a $[0,30]^2$ grid. Try:

x = [-15:1:15];
y = x;
[X,Y] = meshgrid(x,y);
z = (abs(X)<=5) .* (abs(Y)<=5);
figure(1);
subplot(1,2,1)
stem3(z);axis tight
subplot(1,2,2)
stem3(x,y,z);axis tight

enter image description here

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The axes in your plot seem to be indexes into an array, as if you were using stem3(Z) instead of stem3(X,Y,Z).

Check the example in the documentation.

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