plot graph for nested if

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liyana nadirah
liyana nadirah el 27 de En. de 2020
Comentada: liyana nadirah el 28 de En. de 2020
l=input('l=');
for t=0:100
if t>2
g=((3/(309*0.0014))*((((0.0000126*(t-2))/l)*80.4)-(l/6)));
elseif t>80
g=l-((80.4/309)*exp(((((-3*0.0000126)/(0.0014*l))*(t-80)))));
else
g=0;
end
disp (g);
end
t=0:100;
plot (t,g)
can someone explain how to plot graph using this coding because the don't get the graph.

Respuesta aceptada

Bhaskar R
Bhaskar R el 27 de En. de 2020
l=input('l=');
g = zeros(1, length(0:100));% initialize g with 0's of length t
c = 1; % counter variable
for t=0:100
if t>2
g(c)=((3/(309*0.0014))*((((0.0000126*(t-2))/l)*80.4)-(l/6)));
elseif t>80
g(c)=l-((80.4/309)*exp(((((-3*0.0000126)/(0.0014*l))*(t-80)))));
else
g(c)=0;
end
c = c+1; % increment counter variable
disp (g);
end
t=0:100;
plot (t,g)
  7 comentarios
Walter Roberson
Walter Roberson el 28 de En. de 2020
for t=0:100
t is a scalar
if (t(d)>2) && (t(d)<=80)
d is undefined. If d does not happen to be 1, then because t is a scalar, t(d) would be out of range.
for l=0.0002:0.001
every iteration of that loop, every different value of l, you are ovewriting all of g. The result would be the same as if you had only done the last of them, 0.0002 .
Notice by the way that because the default increment is 1, your code
for l=0.0002:0.001
is the same as
for l=0.0002 : 1 : 0.001
so 0.0002 would be used, then 0.0002+1 would be examined and found to be greater than 0.001 so the loop would stop.
liyana nadirah
liyana nadirah el 28 de En. de 2020
emmm sorry idon't get it, so
first i should declare d as 1?

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Walter Roberson
Walter Roberson el 27 de En. de 2020
Editada: Walter Roberson el 27 de En. de 2020
l = input('l=');
t = 0:100;
g = zeros(size(t));
mask = t>2 & t <= 80
g(mask) = ((3/(309*0.0014))*((((0.0000126*(t(mask)-2))/l)*80.4)-(l/6)));
mask = t>80;
g(mask) = l-((80.4/309)*exp(((((-3*0.0000126)/(0.0014*l))*(t(mask)-80)))));
plot (t,g)

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