Vector in a piecewise function

So, I've trying to learn to use vectors in a piecewise function. I've got the function working properly, (code below). When I use any single number, (pwise(5), it works right. When I use a vector, say, pwise(-5:0:50), it gives me numbers I know are not correct. -5 should be -5, I'm just not sure what MatLab is doing. Any pointers?

function[v]=pwise(t)

%Figures Piecewise Function
if t<0

      v=0
  elseif t<8
      v=10*t^2-5*t
  elseif t<16
      v=624-5*t
  elseif t<26
      v=36*t+12*(t-16)^2
  else
      v=2136*exp(-0.1*(t-26))
  end

[Image Analyst]

What happens when you follow the normal debugging process and step through your code? I bet that would explain things. http://blogs.mathworks.com/videos/2012/07/03/debugging-in-matlab/ And what do you expect to happen when you say "if [-5,-4,-3,-2,-1,0,1,2,3,4,5] < 0"? Do you expect it to automatically somehow form a loop over all those numbers and do the if statement for each number? It doesn't.

Answer 1

[Roger Stafford]

The "if-elseif-etc" construct you have there doesn't work the way you want for vectors. For example if you say "if t<8", for t a vector, it comes true only in case the proposition is true for all elements of t. You need something different for your problem to work with vectors.

One way is to write a for-loop that goes through the values of t one element at a time and figures out the corresponding value of v for each one using your "if elseif ..." stuff.

A vectorized method could also be done as follows, though I think you will find that the above for-loop method is easiest to do.

 v = (10*t.^2-5*t).*((t>=0)&(t<8)) + ...
     (624-5*t).*((t>=8)&(t<16)) + ...
     (36*t+12*(t-16).^2).*((t>=16)&(t<26)) + ...
     (2136*exp(-0.1*(t-26))).*(t>=26);

(Note that the t<0 case automatically gives zero for this and doesn't have to be written in specifically.)

[Daniel]

Okay, so I got over my fear of loops and conjured up a simple one. It worked! Code below. But then, when I went to plot it, I crashed. I can't plot (t,v) because v only exists within my function. I tried t,ans, but that didn't seem to work either. Could I put v into an array within my loop??? Any thoughts?

t=-5:1:50;

for k=-5:1:50

pwise(k)

end

plot(t,ans)

[Roger Stafford]

You have to store the results of pwise(k) somewhere if you want to use it later such as in 'plot'. Remember though, its indices must be positive integers. If I were you I would write it like this where index k is known to be positive:

 t = -5:1:50;
 v = zeros(1,length(t));
 for k = 1:length(t)
  v(k) = pwise(t(k));
 end
 plot(t,v,'yo')

Of course if you use the vectorized version of pwise I gave you earlier, you can do the plot directly without the for-loop and its k index.

[Daniel]

Beautiful! That's what I wanted to do, just didn't know how to say it. Thanks so much Mr. Stafford. I've learned a lot tonight. Have a good night.

Answer 2

[Image Analyst]

Try using .^ instead of ^ so that you square all elements, element by element.

[Daniel]

I had tried that, didn't work. I think because of the reason below.

[Image Analyst]

Well it does work - it was at least part of the solution, as Roger showed, but not the whole solution. You should really learn when to use the dot in front of the operators.

Answer 3

[Sally Al Khamees]

If you have R2016b and the Symbolic Math Toolbox, you can just use the piecewise function with a vector

:

https://www.mathworks.com/help/symbolic/piecewise.html