Vector in a piecewise function

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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
  1 个评论
Image Analyst
Image Analyst 2013-9-10
编辑:Image Analyst 2013-9-10
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.

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采纳的回答

Roger Stafford
Roger Stafford 2013-9-10
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.)
  3 个评论
Roger Stafford
Roger Stafford 2013-9-11
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
Daniel 2013-9-11
编辑:Daniel 2013-9-11
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.

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更多回答(2 个)

Image Analyst
Image Analyst 2013-9-10
Try using .^ instead of ^ so that you square all elements, element by element.
  2 个评论
Daniel
Daniel 2013-9-10
I had tried that, didn't work. I think because of the reason below.
Image Analyst
Image Analyst 2013-9-11
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.

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Sally Al Khamees
Sally Al Khamees 2016-12-23
编辑:Sally Al Khamees 2017-2-21
If you have R2016b and the Symbolic Math Toolbox, you can just use the piecewise function with a vector
:

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