Vectorized parametric step function

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Pablo C.
Pablo C. 2021-9-10
评论： Pablo C. 2021-9-10
I have implemented a parametric step function using object orientation and vectorization code but there is an inconstancy in some results. The code is as follows:
classdef A6
properties
eps
end
methods
function Y = value(obj,X)
Y=double(((X>=0) & (X<0.3)))...
+double((X>=0.3) & (X<0.4)).*obj.eps...
+double((X>=0.4) & (X<0.6))...
+double((X>=0.6) & (X<0.7)).*obj.eps...
+double((X>=0.7) & (X<=1));
end
end
end
I tested the function using, eps=0.8, dx=1/20 and X=(dx:dx:1-dx). When calculating Y=value(a,X), X(12)=0.6 but Y(12)=1 when I should expect the result to be 0.8 according to the previous inequalities. Any help will be appreciated. Thanks.

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回答（1 个）

Abolfazl Chaman Motlagh
Ran in:
it is not about your code.
i test it here :
X = 0.05:0.05:1-0.05
X = 1×19
0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 0.5500 0.6000 0.6500 0.7000 0.7500 0.8000 0.8500 0.9000 0.9500
X(12)
ans = 0.6000
X(12)<0.6
ans = logical
1
this result is wrong.
because :
X(12) - 0.6
ans = -1.1102e-16
so the way you create the X have a small telorance.
do this :
X = round( (0.05:0.05:1-0.05) * 100) / 100 ;
Y=double(((X>=0) & (X<0.3)))...
+double((X>=0.3) & (X<0.4)).*0.8...
+double((X>=0.4) & (X<0.6))...
+double((X>=0.6) & (X<0.7)).*0.8...
+double((X>=0.7) & (X<=1));
plot(X,Y)
Y(X==0.6)
ans = 0.8000
-----------------------------------------------
it is strange because if you use linspace function, other part of code goes wrong:
X = linspace(0.05,1-0.05,19)
X = 1×19
0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 0.5500 0.6000 0.6500 0.7000 0.7500 0.8000 0.8500 0.9000 0.9500
Y = double(((X>=0) & (X<0.3)))...
+double((X>=0.3) & (X<0.4)).*0.8...
+double((X>=0.4) & (X<0.6))...
+double((X>=0.6) & (X<0.7)).*0.8...
+double((X>=0.7) & (X<=1));
plot(X,Y)
X(8) , Y(8)
ans = 0.4000
ans = 0.8000
which is wrong, it should be 1.
i think it's error in the way matlab create vectors.

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