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amberjordan2006

  • one year ago

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  1. zepdrix
    • one year ago
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    Hmm it looks like we can do Partial Fraction Decomposition.

  2. amberjordan2006
    • one year ago
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    And how would I do that?

  3. zepdrix
    • one year ago
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    \[\large \frac{2s-3}{s^2-3s+2}\]The denominator can be factored, giving us,\[\large \frac{2s-3}{(s-2)(s-1)}\] From here, let's try applying Partial Fraction Decomposition. The factors in the denominator will break up like so, \[\large \frac{2s-3}{(s-2)(s-1)} \qquad = \qquad \frac{A}{s-2}+\frac{B}{s-1}\] Where \(\large A\) and \(\large B\) are unknown constants that we need to solve for. The reason they're constants is because the unknown term you place on the top of each fraction, should be one degree lower than the denominator. Just something to maybe remember. :) Ok let's try solving this,

  4. zepdrix
    • one year ago
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    Hmm, yes it is :) It's good if you understand why you're doing that though. We'll start by multiplying through by the denominator on the left.\[\large \cancel{(s-2)(s-1)}\frac{2s-3}{\cancel{(s-2)(s-1)}} \qquad = \qquad \left(\frac{A}{s-2}+\frac{B}{s-1}\right)(s-2)(s-1)\] And yes, after cancelling some stuff out on the right, you should get the thing you're thinking.

  5. zepdrix
    • one year ago
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    Hopefully you're coming up with something like this, \[\large 2s-3=A(s-1)+B(s-2)\] The s-2 's cancelled on the A term, while the s-1 's cancelled on the B term.

  6. zepdrix
    • one year ago
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    There are several ways to continue from here. Since we have nice easy factors, the way I would recommend is by plugging in values for \(\large s\), (which is what I think you meant c; ). If you plug in \(\large s=1\) you should be able to find B very easily.

  7. zepdrix
    • one year ago
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    A=1, B=1? Yah that's what I'm coming up with also. Remember the initial setup that we did? You want to plug the values back into that.

  8. zepdrix
    • one year ago
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    Looks good c: Understand how to solve it from here?

  9. zepdrix
    • one year ago
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    Ummm, I'm a little rusty on my Leplace Transforms. I think we just take the transform. This step requires a bit of memorization I guess. \[\huge \mathscr{L}[e^{at}]=\frac{1}{s-a}\]Hopefully I'm remembering that correctly.

  10. zepdrix
    • one year ago
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    We'll want to be using it in this direction.\[\huge \mathscr{L}^{-1}\left[\frac{1}{s-a}\right]=e^{at}\]

  11. zepdrix
    • one year ago
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    Here's just a quick reminder of where we've gotten so far, and the proper way to break it up. \[\large \mathscr{L}^{-1}\left[\frac{2s-3}{s^2-3s+2}\right]\qquad =\qquad \mathscr{L}^{-1}\left[\frac{1}{s-1}+\frac{1}{s-2}\right]\] \[\large =\qquad \mathscr{L}^{-1}\left[\frac{1}{s-1}\right]+\mathscr{L}^{-1}\left[\frac{1}{s-2}\right]\]

  12. zepdrix
    • one year ago
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    Hmm I think you'll get two terms out of this one. When your \(\large a\) is \(\large 1\), it should give you \(\large e^{t}\). Which is what that first inverse leplace should be giving us. It looks like you have the second term correctly figured out though.

  13. zepdrix
    • one year ago
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    Cool looks like you were able to get through this one without too much trouble \c:/ Just needed to brush up on those darn Partial Fractions! heh

  14. zepdrix
    • one year ago
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    Yes looks good! :) You have to use curly brackets on the exponent so it formats correctly hehe e^{2t}

  15. zepdrix
    • one year ago
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    If we were simply given F(s) at the start, not a differential equation, then I suppose we're done :O

  16. zepdrix
    • one year ago
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    Yah we weren't given initial conditions or anything silly like that :) Luckily

  17. zepdrix
    • one year ago
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    Close this thread at the top, it's getting too long. It's going to get laggy with all of the fancy latex formatted. Open a new thread with your next question. I'll look for it. You can type @zepdrix in the comments somewhere if I'm taking too long, heh. That will send me a page and I can find yer question easier.

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