Undefined function or variable 'v'.??

2 Ansichten (letzte 30 Tage)
Libby Goodes
Libby Goodes am 3 Jun. 2018
Kommentiert: Rik am 6 Jun. 2018
t=-5:50;
if t<=8
v=10.*(t.^2)-(5.*t);
elseif (t>=8) & (t<=16)
v=624-(5.*t);
elseif (t>=16) & (t<=26)
v=(36.*t)+12.*(t-16).^2;
elseif (t>26)
v=2136.*exp(-0.1.*(t-26));
elseif (t<0)
v=0;
end
plot(t,v)

Antworten (1)

Rik
Rik am 3 Jun. 2018
You made the classic mistake of thinking that this set of logical statements would be handled for each element of your vector. But Matlab doesn't work that way, unless you tell it with a for-loop.
The code below uses logical indexing to get the result.
t=-5:50;
v=zeros(size(t));
L=t<=8;
v(L)=10.*(t(L).^2)-(5.*t(L));
L=(t>=8) & (t<=16);
v(L)=624-(5.*t(L));
L=(t>=16) & (t<=26);
v(L)=(36.*t(L))+12.*(t(L)-16).^2;
L=t>26;
v(L)=2136.*exp(-0.1.*(t(L)-26));
L=t<0;
v(L)=0;
plot(t,v)
  3 Kommentare
Rik
Rik am 3 Jun. 2018
The second line pre-allocates the vector v by filling it with zeros. This way we can use the same logical vector for v as for t.
Have you tried reading the documentation for zeros and size?
Rik
Rik am 6 Jun. 2018
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