How to make this integral

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Peppe 2017-9-21

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评论： Walter Roberson 2017-9-22

Hi Suppose we have following equation:

S(x,t) = p(x,t) + k * int_0^t h(\tau) d\tau.

What is the difference between this and S(x,t) = p(x,t) + k * int_0^t h(t) dt ?

How ould the simulink block differs? I appritiate any help. Thank you

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Answer 1

Walter Roberson 2017-9-21

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The

S(x,t) = p(x,t) + k * int_0^t h(t) dt

version is at least confusing, as it is using t both as a parameter to S and as the name of the variable to integrate over in the int() part.

S(x,t) = p(x,t) + k * int_0^t h(\tau) d\tau

can be written as

S = @(x, t) p(x,t) + k * integral(@(tau) h(tau), 0, t)

and

S(x,t) = p(x,t) + k * int_0^t h(t) dt

can potentially be written as

S = @(x, t) p(x,t) + k * integral(@(t) h(t), 0, t)

which would evaluate to the same. However, consider that if someone wrote

int_0^x sin(x)

they would normally mean the indefinite integral int( sin(x) ) -- integration of sin(x) with upperbound x. That is not the same thing that is happening with

S = @(x, t) p(x,t) + k * integral(@(t) h(t), 0, t)

which is first substituting the current specific value of t into the upper bound and then happening to use a lexically-different t as the name of the anonymous variable.

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Peppe 2017-9-22

编辑：Peppe 2017-9-22

Thankbyou for your respond How would the simulink block look like? Is same integrator used for both and same iniial condition?

Walter Roberson 2017-9-22

As outlined above, there are two potential interpretations of int_0^t h(t) dt .

In one of them, the t that is the upper bound is to be interpreted as the t that is the parameter to the function, with the t of h(t) dt being interpreted as a different t. That interpretation has exactly the same implementation as the version using tau, because the careful programmer would choose a different variable of integration to eliminate confusion... oh, say, tau perhaps.

In the other interpretation, the t that is the upper bound is to be interpreted as the t of h(t) dt. In that situation, this is an indefinite integration that would result in a symbolic formula. Simulink cannot really calculate symbolic formulae through its blocks, and you would need a quite different implementation to work around the limitation that Simulink does not like to carry around anything other than numeric and logical values.

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