How do I 2D-plot my own function with two arguments? - matlab

Question

I've started out with MATLAB a few hours ago and began with plotting some simple functions, e.g.:
k = -20:20;
a = 0.01;
x_k = exp(-a * k.^2)
f_1 = figure(1);
plot(k, x_k)
This works fine and plots each value of k on the x-axis, with its corresponding function response at the y-axis. Now I've defined my own function:
function [y] = triangle(t, a)
if (a <= t) & (t <= 0)
y = (1/a * t + 1) * a
elseif (0 <= t) & (t <= a)
y = (-1/a * t + 1) * a
else
error('Function not defined for these values')
end
end
I now want to plot triangle for a range of t with a specific parameter a. So I call:
a = 1;
int = (2 * a) / 20;
t = -a:int:a;
f_1 = figure(1);
plot(t, triangle(t, a))
… but my function errors out because the entire t array (vector, I suppose?) is passed to it instead of each element of t.
How do I plot my triangle function for a specific a, over a defined range?
I've also tried fplot(triangle(a), t) (since fplot accepts a function as an argument) but that again complains about triangle not receiving enough input arguments—rightly so, but I need to pass a to it as well.

Answer 1

The problem of this is the that you pass a vector to your function triangle but you want to check every single if-statement for each single value of the vetor t. So either you use a for-loop within the function triangle or you could also do it by using this:
function [y] = triangle(t, a)
y = ((a <= t) & (t <= 0)) .* ((1/a * t + 1) * a) + ((0 <= t) & (t <= a)) .* ((-1/a * t + 1) * a);
end
I know it looks somehow weird, but if you need me to explain it a bif more in detail, I can do this for you ;-)
EDIT: Let's shortly examine the problem: You pass a vector t to the function, which means that the first if statement if (a <= t) & (t <= 0) splitted looks like this:
(a <= t) % resulting in:
ans =
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
(t <= 0) % resulting in:
ans =
1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0
so combining them via the AND operator (& for vectors), this obviously gives you
ans =
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
So this branch wont be executed. The elseif gives you:
(0 <= t)
ans =
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1
(t <= a)
ans =
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
((0 <= t) & (t <= a))
ans =
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1
so this won't be executed either even though you'd want it to be executed for the second half of your values because you'd like to check the two statements for each value-pair on it's own.
So by leaving away the statements completely you could do it like I posted above: simply multiply the result of the logical statements elementwise with the piecewise function which will yield the result you want to calculate :-)

Answer 2

The best approach is to define the triangle function so that it admits vector inputs. You and do it with logical indexing, which selects a range of values of t depending on a logical condition:
function [y] = triangle(t, a)
y = NaN(size(t)); %// pre-allocate and define NaN as default return value
ind = (a <= t) & (t <= 0); %// select first range
y(ind) = (1/a * t(ind) + 1) * a; %//compute output for first range
ind = (0 <= t) & (t <= a); %// select second range
y(ind) = (-1/a * t(ind) + 1) * a; %//compute output for second range
end
This returns NaN for out-of-range values. As an example, for your t,
>> triangle(t, a)
ans =
Columns 1 through 13
NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN 1.0000 0.9000 0.8000
Columns 14 through 21
0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0
Once you define the function like this, you can do plot(t, triangle(t, a)).

Answer 3

To pass your function to fplot, you need to use an anonymous function (like what is called a lambda in many other programming languages).
This is done with the # syntax in matlab (see help function_handle for how to use this).
So your fplot try should be:
fplot(#(x) triangle(x,a), t([1 end]))
Note that the t input for fplot is the bounds in which you want to evaluate your function, not the values to evaluate itself (you can use bounds in combination with fplot's third input argument for some control over that).
That said, as other posters also noted, the part of your example input where t<0 end up right in your else branch and yields an error.