plot graph for nested if

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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.

采纳的回答

Bhaskar R
Bhaskar R 2020-1-27
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 个评论
Walter Roberson
Walter Roberson 2020-1-28
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 2020-1-28
emmm sorry idon't get it, so
first i should declare d as 1?

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更多回答(1 个)

Walter Roberson
Walter Roberson 2020-1-27
编辑:Walter Roberson 2020-1-27
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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