If condition inside integration

4 次查看（过去 30 天）
Dharma Khatiwada
Dharma Khatiwada 2021-10-30
评论： Walter Roberson 2021-10-31
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.

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2021-10-30
Ran in:
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)
c(x) = 
inner = c(x) * exp(-2*(t-x))
inner = 
y(t) = simplify(int(inner, x, 0, t))
y(t) = 
string(y)
ans = "piecewise(in(t, 'real'), 0)"
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
c2(x) = 
inner2 = c2(x) * exp(-2*(t-x));
y2(t) = simplify(int(inner2, x, 0, t))
y2(t) = 
string(y2)
ans = "4*heaviside(t - 1)*exp(-2*t)*exp(2) + 4*heaviside(t - 4)*exp(-2*t)*exp(8) + 4*heaviside(t - 7)*exp(-2*t)*exp(14) + 4*heaviside(t - 10)*exp(-2*t)*exp(20) + 4*heaviside(t - 13)*exp(-2*t)*exp(26)"
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.
  6 个评论
显示 4更早的评论
Dharma Khatiwada
Dharma Khatiwada 2021-10-30
Hi Walter,
Just one more quick question with the value of c2(x) in the line:
c2(x) = (dirac(x-1) + dirac(x-4) + dirac(x-7) + dirac(x-10) + dirac(x-13)) * 4
Normally, we should expect dirac function giving the output i.e. infinity if for example, when x=1 otherwise zero. In my context, I am expecting that to be 4 i.e. c2(x)=4. I am not sure whether it is returning to that value. My concern is only on the value of c2(x).
I found something similar in other Matlab posting. Please suggest if I also need to normalize like below.
Thank you again!
x = -1:0.1:1;
y = dirac(x);
idx = y == Inf; % find Inf
y(idx) = 1; % set Inf to finite value
Walter Roberson
Walter Roberson 2021-10-31
Ran in:
dirac(0)
ans = Inf
syms x
int(dirac(x), x, -1, 1)
ans = 
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.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Numbers and Precision 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by