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Easy laplace conversion from s to t-domain gone wrong

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data

    [tex]I(s)=\frac{2}{s^2 + s}[/tex]

    Convert this to the time domain

    3. The attempt at a solution

    Step 1 - Divide by s
    [tex]I(s)=\frac{\frac{2}{s}}{s + 1}[/tex]

    Step 2 - Substract and add 1 to create new fractions
    [tex]I(s)=\frac{\frac{2}{s}-1+1}{s + 1}[/tex]

    Step 3 - Split into new fractions
    [tex]I(s)=\frac{\frac{2}{s}}{s + 1} - \frac{1}{s + 1} + \frac{1}{s + 1}[/tex]

    Step 4 - Contract the two newbie fractions
    [tex]I(s)=\frac{\frac{2}{s}}{s + 1} - \frac{2}{s + 1}[/tex]

    Step 5: Split up the first fraction
    [tex]I(s)=\frac{2}{s} \frac{1}{s + 1} - \frac{2}{s + 1}[/tex]


    Step 6 - Convert to time domain
    [tex]i(t)=2e^{-t}-e^{-t}[/tex]

    BUT, this is not correct. It should be: [itex] 2-2e^{-t}[/itex]

    What is wrong in my approach here?
     
  2. jcsd
  3. May 5, 2012 #2

    Ray Vickson

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    You started with [tex] I(s) = \frac{2}{s^2+s} = \frac{2}{s(s+1)} = \frac{2}{s}\frac{1}{s+1} [/tex] and ended up with [tex] I(s) =\frac{2}{s} \frac{1}{s + 1} - \frac{2}{s + 1}[/tex] and you see nothing wrong with this? Why did you not just apply partial fractions to
    [tex] I(s) = \frac{2}{s(s+1)}?[/tex]

    RGV
     
  4. May 5, 2012 #3

    Mark44

    Staff: Mentor

    What's wrong is your first step. Factor the denominator to s(s + 1) and then use partial fraction decomposition to rewrite 2/[s(s + 1)] in the form A/s + B/(s + 1).
     
  5. May 5, 2012 #4
    Thank you Ray.

    Can you tell me exactly what step in my first approach which was "illegal"?
     
  6. May 5, 2012 #5

    Ray Vickson

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    You don't see your claim that
    [tex] -\frac{1}{s+1} + \frac{1}{s+1} = -\frac{2}{s+1}[/tex] when going from Step3 to step 4? You don't see that the fraction
    [tex]\frac{\frac{2}{s}}{s+1}[/tex] in the first term in Step 3 is exactly the same as I(s), so you have really done nothing that will be useful?

    RGV
     
  7. May 5, 2012 #6
    OK, I'm gonna do this one more time from scratch.
    With partial fraction decomposition this time.

    [tex]
    I(s) = \frac{2}{s^2 + s}
    [/tex]


    Step 1 - Factor the denomenator
    [tex]
    I(s) = \frac{2}{s(s + 1)}
    [/tex]

    Step 2 - Set up the partial fractions stuff
    [tex]
    \frac{A}{s}+\frac{B}{s + 1}
    [/tex]

    Step 3 - Merge A and B fractions
    [tex]
    \frac{A(s+1)+B(s)}{s(s + 1)}
    [/tex]

    Step 4 - Determine A and B
    [tex](A+B)s = 0[/tex]
    [tex]A = 2[/tex]
    Using that A = 2, B must be -2 to fullfill the first equation.

    Step 5 - Replace A and B with determined values
    [tex]
    \frac{2}{s}-\frac{2}{s + 1}
    [/tex]

    Step 6 - Convert to the holy time domain
    [tex]\frac{2}{s} = 2[/tex]
    [tex]\frac{-2}{(s+1} = -2e^-t[/tex]
    [tex]i(t) = 2(1-e^-t)[/tex]
     
  8. May 5, 2012 #7

    I like Serena

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  9. May 5, 2012 #8
    Hey Serena! Long time no see ;)
    I managed this one after realizing that I had to utlize the partial fraction-stuff :)

    But I still need some help in this other thread of mine about the fifth harmonic of a square wave passing trough a HP filter. And my exam is on monday :/
    https://www.physicsforums.com/showthread.php?t=603179
     
  10. May 5, 2012 #9

    Ray Vickson

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    This is almost OK now, but it is a bad idea to write things like [tex]\frac{2}{s} = 2[/tex] because that is false, and might get you a mark of zero just for writing it. If you mean that the inverse transform of 2/s is 1, why not just say so, or maybe use a better notation, such as [itex] L^{-1}(2/s) = 1 [/itex] or [itex] 2/s \longrightarrow 1?[/itex] Also, in LaTeX, if you want [itex] e^{-t}[/itex] rather than your [itex] e^-t,[/itex], you need to use a curly bracket. To get [itex]a^{anything},[/itex] put the "anything" between curly brackets, like this: {anything}. If you don't, you will get [itex]a^anything.[/itex]

    RGV
     
  11. May 5, 2012 #10
    I didnt find the arrow in latex, but in my hand written notation i always use that.
     
  12. May 5, 2012 #11

    Ray Vickson

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    There are various types of arrows.
    [itex]\text{\rightarrow gives} \rightarrow,[/itex] [itex]\text{\longrightarrow gives} \longrightarrow,[/itex] [itex]\text{\Rightarrow gives} \Rightarrow,[/itex] [itex]\text{\Longrightarrow gives} \Longrightarrow.[/itex]
     
  13. May 5, 2012 #12

    I like Serena

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    I like \to (##\to##), which is nice and short to type.

    I also like \mathcal{L} ##\mathcal{L}^{-1}(\frac 2 s) = 2##, or more formally ##\mathcal{L}^{-1}_s[ \frac 2 s ](t) = 2##.
     
  14. May 5, 2012 #13

    Ray Vickson

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    I, too, like mathcal, but have also seen things like [tex] \frac{1}{s} \leftrightarrow 1 .[/tex]
    However, this uses the somewhat lengthy "\leftrightarrow". Is there a shorter version?

    RGV
     
  15. May 5, 2012 #14
    Thread now about latex syntax. I have a feeling I will need to use some Laplace symbols soon so it's great info. Ty
     
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