power of an integral and the laplace transform

Discussion in 'Math' started by kokkie_d, Jun 8, 2010.

  1. kokkie_d

    Thread Starter Active Member

    Jan 12, 2009
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    Hi,

    I have got the following equation:

    <br />
y(t) = a_3*x^3<br />
    with <br />
<br />
x = \frac{\int{i}{dt}}{Q_p}<br />
<br />
<br />

    If I work this out then I get:
    <br />
y(t) = \frac{a_3}{Q_p^3} * \int{i}{dt}^3<br />

    If I then take the laplace of this, will this be:
    <br />
Y(s) = \frac{a_3}{Q_p^3} * \frac{I^3(s)}{s^3}<br />

    My question is: Am I following the right steps and am I arriving at the right solution?
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    Yes, it looks right.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I'm not convinced about this solution.

    Suppose we take a simple expression for the function i(t) and let

    i(t)=constant=K

    x(t) would then be given by

    x(t)=\frac{\int{Kdt}}{Q_{p}}

    x(t)=\frac{1}{Q_{p}}Kt

    So y(t) would be

    y(t)=a_{3}(\frac{1}{Q_{p}}Kt)^3=a_{3}\frac{K^3t^{3}}{Q_{p}^{3}} \ \ \ eqn(1)

    Using the Laplace transform approach - if i(t)=constant=K then

    I(s)=\frac{K}{s}

    Hence given the solution

    Y(s)=a_{3}\frac{I(s)^3}{Q_{p}^{3}s^{3}}

    Substitute I(s)=K/s

    Y(s)=a_{3}\frac{(\frac{K}{s})^3}{Q_{p}^{3}s^{3}}

    Y(s)=a_{3}\frac{K^3}{Q_{p}^{3}s^{6}}

    The inverse transform gives

    y(t)=a_{3}\frac{K^3}{Q_{p}^{3}}\frac{t^{5}}{5!} \ \ \ eqn(2)

    which does not agree with the time domain solution eqn (1) above for y(t)
     
  4. kokkie_d

    Thread Starter Active Member

    Jan 12, 2009
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    I am a bit confused here: Should the Laplace transform of i(t) not be I(s)?
    Also, with your replacement of K for i; how does this make it simpler?
    The function K(t) is still i(t).

    Could you explain your steps a bit further perhaps.

    btw I think we can leave a_3 out of the equation since it does not affect the outcome, makes writing it easier.
     
    Last edited: Jun 10, 2010
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Yes the Laplace transform of i(t) is I(s).

    I was attempting to show the inconsistency in the solution you gave by way of solving the problem for a specific case of i(t) rather than the general term i(t). The proof of the pudding is in the eating - as they say.

    If the solution doesn't work for the simple case of i(t)=constant then it's probably not going to work for more complex functions of i(t).

    So I attempted to test your solution by making i(t)=K [a constant] for which I(s)=K/s. Using two approaches - one of which is the Laplace method using your formula - gave me two different answers. That's the dilemma from my perspective.
     
  6. Ghar

    Active Member

    Mar 8, 2010
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    Weird problem... I've never done this approach before but I suspect convolution would have to factor into this.

    \mathcal{L}\{f(t)g(t)\} = F(s)*G(s)\\
    There might be some scaling factor in there too;
    http://www.esr.ruhr-uni-bochum.de/rt1/syscontrol/node9.html
    Look for item h), convolution in the frequency domain


    And since convolution and multiplication is associative...

    \mathcal{L}\{f(t)g(t)h(t)\} = F(s)*G(s)*H(s)\\

    So I guess you would end up with something like this:
    \mathcal{L}\{(\int i dt)^3\} = \frac{I(s)}{s}*\frac{I(s)}{s}*\frac{I(s)}{s}\\

    Don't quote me on this I know I'm fishing.
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    So one would end up with the same result as the OP obtained in post#1

    It still doesn't resolve my dilemma regarding the solution you get if you plug in an actual function for i(t) [or I(s)] and solve for y(t). It's easy enough to do it directly in the time domain - the expectation is that the Laplace method would yield the same solution for Y(s) <---> y(t) as that obtained via the time domain approach.

    I'm happy to be found wanting on this - it would be interesting to resolve it.

    Probably like you Ghar, I've not seen a problem quite like this before.
     
  8. Ghar

    Active Member

    Mar 8, 2010
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    I meant those as convolution symbols, not multiplication.

    A signal convolved with itself is not equal to the square.
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    OK - understood
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I would further suggest that no solution exists as the function is non-linear. If one cannot define a unique linear transformation between output Y(s) and stimulus I(s), then a solution for the Laplace transform of the given function does not exist.
     
  11. kokkie_d

    Thread Starter Active Member

    Jan 12, 2009
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    I have to agree with that. What I was trying to do is solve the mentioned equation for state space, which needs to be broken down (if at all possible) into separate linear equations and can not just be turned into state space by just switching to laplace. I am still getting to grips with the principles. If there is someone in the North East of England who can tutor me, please PM me.

    Also, what I realize from above is that the mentioned equation does not converge and thus has to be given a defined limit where it does but otherwise can not be turned into laplace transform.

    I think I am correct here. :confused:
     
  12. Georacer

    Moderator

    Nov 25, 2009
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    Let me get this straight:
    In the time-domain equation, you meant to use multiplication instead of convolution, right?
    And inside the integral, do you really wish to integrate the function i=constant against time, or you actually wanted to write
    \int t \cdot dt
    ?
    Please clarify. Even if we aren't able to extract a ss model, I would like to see the transformation completed.
     
  13. Georacer

    Moderator

    Nov 25, 2009
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