If condition inside integration
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Dharma Khatiwada
le 30 Oct 2021
Commenté : Walter Roberson
le 31 Oct 2021
Hi,
I am trying to keep if condition inside integration of exponential function and solve the integral in Matlab. c(x) is 4 at x =1, 4, 7, 10, 13 otherwise zero. Any help would be appreciated.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/783968/image.png)
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Walter Roberson
le 30 Oct 2021
I am pretty sure you do not want that definition of c(x)
syms x t
c(x) = piecewise(x == 1 | x == 4 | x == 7 | x == 10 | x == 13, 4, 0)
inner = c(x) * exp(-2*(t-x))
y(t) = simplify(int(inner, x, 0, t))
string(y)
fplot(y, [-20 20])
This happens because your c(x) definition is discontinuous, and the width of the event x == 4 (or each of the other values) is 0, so the integral at those points is 0.
Compare:
c2(x) = (dirac(x-1) + dirac(x-4) + dirac(x-7) + dirac(x-10) + dirac(x-13)) * 4
inner2 = c2(x) * exp(-2*(t-x));
y2(t) = simplify(int(inner2, x, 0, t))
string(y2)
fplot(y2, [-20 20])
This defines c(x) in terms of a distribution rather than in terms of points, giving meaning to the integral at those values.
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Walter Roberson
le 31 Oct 2021
dirac(0)
syms x
int(dirac(x), x, -1, 1)
The Dirac Delta is not really a function. dirac(0) is not really inf . dirac() is defined such that the integral across 0 is 1. So what happens with int() of dirac is correct, and the Inf is not really correct.
There are different ways of defining Dirac. One of the ways is as the limit of a rectangle n units high and 1/n wide, as n approaches infinity: the area is fixed, but as the width approaches 0, the height approaches infinity. Saying that it is infinity such as dirac(0) shows, is a short-hand that is not really true.
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