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Thread: Laplace to time

  1. #1
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    Laplace to time

    Is this correct:? $$\frac{A}{C+B+sD} = A \frac{1} {s \frac{C+B}{D}}$$ using $$1 = \frac{1}{s}$$ so: $$f(t) = \frac{A}{\frac{C+B}{D}}$$?
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  2. #2
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    Re: Laplace to time

    Quote Originally Posted by Nforce View Post
    Is this correct:? $$\frac{A}{C+B+sD} = A \frac{1} {s \frac{C+B}{D}}$$ using $$1 = \frac{1}{s}$$ so: $$f(t) = \frac{A}{\frac{C+B}{D}}$$?
    No...

    $\dfrac{A}{C + B + sD} = \dfrac{A}{C+B} \cdot \dfrac{1}{1+\frac{D}{C+B}s}$

    Then use

    $\mathscr{L}(f(c t) ) = \dfrac 1 c F\left(\dfrac s c \right)$

    and

    $\mathscr{L}(a f(t)) = a F(s)$

    with $F(s) = \mathscr{L}(f(t))$

    to complete the solution
    Thanks from HallsofIvy and Nforce
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  3. #3
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    Re: Laplace to time

    Sorry, wrong problem. No wonder it get's really strange.

    The right expression is:

    $$\frac{A}{s(B+C+sD)} $$

    I should get something like this: K1e^something + K2e^something ...

    Because physics never lie. Thanks.
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  4. #4
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    Re: Laplace to time

    Quote Originally Posted by Nforce View Post
    Sorry, wrong problem. No wonder it get's really strange.

    The right expression is:

    $$\frac{A}{s(B+C+sD)} $$

    I should get something like this: K1e^something + K2e^something ...

    Because physics never lie. Thanks.
    it's time to teach yourself about partial fractions.

    $\dfrac{A}{s(B+C+sD)} = \dfrac{\alpha}{s} + \dfrac{\beta}{B+C+sD}$

    and there is a method for determining $\alpha$ and $\beta$

    Once separated into partial fractions it's much easier to compute the inverse transforms of the terms.
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  5. #5
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    Re: Laplace to time

    We can rewrite the fraction like this:

    $$\frac{A}{s(B+C+sD)} = $$


    $$\frac{A}{s}\frac{1}{(B+C+sD)}$$
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  6. #6
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    Re: Laplace to time

    Quote Originally Posted by Nforce View Post
    We can rewrite the fraction like this:

    $$\frac{A}{s(B+C+sD)} = $$


    $$\frac{A}{s}\frac{1}{(B+C+sD)}$$
    and what does this get you?
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  7. #7
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    Re: Laplace to time

    $$ \frac{A}{s} = A \frac{1}{s} = A $$

    $$ \frac{1}{B+C+sD} = \frac{1}{\frac{B+C}{D}+ s} = e^{-\frac{B+C}{D}t} $$

    =?
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  8. #8
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    Re: Laplace to time

    Quote Originally Posted by Nforce View Post
    $$ \frac{A}{s} = A \frac{1}{s} = A $$

    $$ \frac{1}{B+C+sD} = \frac{1}{\frac{B+C}{D}+ s} = e^{-\frac{B+C}{D}t} $$

    =?
    that would work if the two terms were being added but the way you have it they are being multiplied.

    What the method of partial fractions gets you is

    $\dfrac{A}{s (B+C+D s)}=\dfrac{A}{s (B+C)}-\dfrac{A D}{(B+C) (B+C+D s)}$

    Now you can mangle these terms to match the common transform templates and invert them as usual.
    Thanks from Nforce
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  9. #9
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    Re: Laplace to time

    Sorry, maybe you are right about learning the partial fraction decomposition.

    I made to here:

    $$ \alpha (B+C+sD) + s\beta = A$$

    Now I solve for $\alpha$ and $\beta$. But how, there is just A on the right hand side?
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  10. #10
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    Re: Laplace to time

    Quote Originally Posted by Nforce View Post
    Sorry, maybe you are right about learning the partial fraction decomposition.

    I made to here:

    $$ \alpha (B+C+sD) + s\beta = A$$

    Now I solve for $\alpha$ and $\beta$. But how, there is just A on the right hand side?
    I'm not checking that what you've done to reach above is correct. But assuming it is what you do is break the equation up into powers of $s$

    $\alpha(B+C) = A$ (equation for constant terms)

    $(\alpha D + \beta) = 0$ (equation for terms of $s$ )

    solve this system for $\alpha$ and $\beta$



    IMPORTANT NOTE FOR THE FUTURE

    The rules for the partial fraction decomposition get more complicated than this as powers of $s$ go up.
    Please take a moment to read about it in your textbook or on the web somewhere.
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