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    Suppose that x = ax + u (t > 0) with x(0) = 1 in which a > 0 is a constant and u = u(t) is the control. Show that the response of x to the control is given by:

     x(t) = e^{at} (1 + \int^t_0 u(s)e^{-as}ds)

    Answer:

    Integrating factor  \mu(t) = e^{\int -a \ dt} = e^{-at}
     d/dt(e^{-at}x) = e^{-at}u
     xe^{-at} = \int^t_0 e^{-as}u(s) ds + x(0)

    I'm confused about a few things on the last line...firstly how do we know to integrate between 0 and t? secondly, why has the variable suddenly changed to s, where did this come from?! And finally why has the constant x(0) been added before we've even integrated?

    If anyone could please explain these things I would very grateful...

    Thanks in advance!
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    I assume you mean dx/dt = ax + u.

    Integrating between 0 and t is a quick way of integrating directly and adding a constant c. Think about it - if you just integrate and add c, you'll then have to plug in the initial conditions where t = 0 and x = 1 anyway. What they've really done is

    \int^t_0 d/dt(e^{-at}x) \mathrm{d}t

    = e^{-at}x - e^{-a*0}*x(0) = e^{-at}x - x(0)

    This is the left hand side. So we get

    e^{-at}x - x(0) = \int^t_0 e^{-as}u(s) ds

    e^{-at}x = \int^t_0 e^{-as}u(s) ds + x(0)

    They changed the letter to s to be notationally correct. It's wrong to write (even though I've done it above for clarity)

    \int^t_0 f(t) \mathrm{d}t - you can't have t in the limit if you're integrating w.r.t t. So they introduce a "dummy" variable, s, and integrate that instead. But for all intents and purposes, it's basically just t.
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    (Original post by sexyzebra)
    Suppose that x = ax + u (t > 0) with x(0) = 1 in which a > 0 is a constant and u = u(t) is the control. Show that the response of x to the control is given by:

     x(t) = e^{at} (1 + \int^t_0 u(s)e^{-as}ds)

    Answer:

    Integrating factor  \mu(t) = e^{\int -a \ dt} = e^{-at}
     d/dt(e^{-at}x) = e^{-at}u
     xe^{-at} = \int^t_0 e^{-as}u(s) ds + x(0)

    I'm confused about a few things on the last line...firstly how do we know to integrate between 0 and t?
    It is a general result that \int_a^b \frac{d}{ds} f(s) \, ds = f(b)-f(a) (usual caveats about f being differentiable, etc).

    So if we know f(0) and we want f(t), we need to integrate between 0 and t.

    This is where the f(0) has come from as well, it's been moved over to the RHS.

    As for changing the variable to s, the 's' in "ds" is a 'dummy' variable. It doesn't matter what it is, except that it mustn't be one of the variables we're actually using. So it can't be 't', because we're using that as one of the limits to integrate, so they just picked the letter s at random. It makes no real difference if you write

    \int_a^b f(x)\, dx or \int_a^b f(t) \, dt or \int_a^b f(s) \, ds. (But \int_a^b f(a) \, da would be wrong).
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    Thank you both for such detailed explanations, it makes much more sense now
 
 
 
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Updated: November 11, 2008

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