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UnkleRhaukus Group Title

Use partial fractions or the convolution theorem and the table of Laplace transforms, to find functions of \(t\) which have the following Laplace transforms. (b) \[\frac{p+2}{2p^2+p-1}\]

  • 2 years ago
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  1. UnkleRhaukus Group Title
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    (b) partial fractions \[\begin{align*} F(p)&=\frac{p+2}{2p^2+p-1}\\ &=\frac{p+2}{(p+1)(2p-1)}&=\frac{A}{p+1}+\frac{B}{2p-1}\\ \\&&p+2=A(2p-1)+B(p+1)\\ \\&\text{for } p=-1&1=-3A\\ &&A=-\tfrac{1}3\\ \\&\text{for } p=0&2=-A+B\\ &&B=2-\tfrac 13=\tfrac53\\ \\ F(p)&=\frac{-\tfrac13}{p+1}+\frac{\tfrac53}{2p-1}\\ &=\tfrac13\left(\frac{5}{2p-1}-\frac{1}{p+1}\right)\\ \\ f(t)&=\tfrac13\mathcal L^{-1}\left\{\frac{5}{2p-1}-\frac{1}{p+1}\right\}\\ &=\tfrac13\left(\tfrac52\mathcal L^{-1}\left\{\frac{1}{p-\tfrac12}\right\}-\mathcal L^{-1}\left\{\frac1{p+1}\right\}\right)\\ &=\tfrac13\left(\tfrac52e^{t/2}-e^{-t}\right)\\ \\ &=\frac56e^{t/2}-\frac13e^{-t}\\ \end{align*} \]

    • 2 years ago
  2. malevolence19 Group Title
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    http://www.wolframalpha.com/input/?i=inverse+laplace+transform+of+%28s%2B2%29%2F%282s%5E2%2Bs-1%29

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  3. UnkleRhaukus Group Title
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    what do i do with that/??

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  4. UnkleRhaukus Group Title
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    im having trouble with deconvolution

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  5. mahmit2012 Group Title
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    |dw:1353549631736:dw|

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  6. mahmit2012 Group Title
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    |dw:1353549766387:dw|

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  7. mahmit2012 Group Title
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    |dw:1353549812275:dw|

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  8. mahmit2012 Group Title
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    |dw:1353550099053:dw|

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  9. mahmit2012 Group Title
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    |dw:1353550140216:dw|

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  10. UnkleRhaukus Group Title
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    (b) table of laplace transforms \[\begin{align*} F(p)&=\frac{p+2}{2p^2+p-1}\\ &=\frac{p+2}{(2p-1)(p+1)}\\ &=\tfrac12\times\frac{p}{(p-\tfrac12)(p+1)}+\frac{1}{(p-\tfrac12)(p+1)}\\ f(t)&=\tfrac12\mathcal L^{-1}\left\{\frac{p}{(p-\tfrac12)(p+1)}\right\}+\mathcal L^{-1}\left\{\frac{1}{(p-\tfrac12)(p+1)}\right\}\\ &=\tfrac12\times\frac{1/2e^{t/2}+e^{-t}}{1/2+1}+\frac{e^{t/2}-e^{-t}}{1/2+1}\\ &=\tfrac12\times\frac{1/2e^{t/2}+e^{-t}}{3/2}+\frac{e^{t/2}-e^{-t}}{3/2}\\ &=\tfrac13\times\left(1/2e^{t/2}+e^{-t}\right)+\tfrac23{e^{t/2}-\tfrac23e^{-t}}\\ &=\tfrac16e^{t/2}+\tfrac13e^{-t}+\tfrac23{e^{t/2}-\tfrac23e^{-t}}\\ &=\frac56e^{t/2}-\frac13e^{-t} \end{align*}\]

    • 2 years ago
  11. UnkleRhaukus Group Title
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    the derivative fo the dirac delta is not familiar to me

    • 2 years ago
  12. malevolence19 Group Title
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    Yeah, I was wondering the same thing. How is \[\delta ' (t)\] defined?

    • 2 years ago
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