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anonymous

  • 5 years ago

Find inverse Laplace transform by using convolution. Please step by step in integral part: 1/s^2+5s+6

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  1. anonymous
    • 5 years ago
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    first factor s^2+5s+6 and post what you got

  2. anonymous
    • 5 years ago
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    I/(s+2)(s+3) so inverse laplace of those as F(s)=1/(s+2) and g(s)=1/s+3 and taking inverse we have F(s)=e^(-2t) and G(s)=e^(-3t)

  3. anonymous
    • 5 years ago
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    if you are taking inverse, it would be in time domain. if F(s) = 1/(s+2), then F(s) cannot also be equal to e^(-2t)

  4. anonymous
    • 5 years ago
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    if H(s) = 1/s^2+5s+6 and F(s)=1/(s+2) and G(s)=1/(s+3), then H(s) = F(s)G(s)

  5. anonymous
    • 5 years ago
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    according to convolution, product of two functions in the angular frequency domain can be represented as convolution of their corresponding time domain functions.

  6. anonymous
    • 5 years ago
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    so, h(t) = f(t)*g(t) = \[\int\limits_{0}^{t}f(\tau)g(t-\tau) d \tau\]

  7. anonymous
    • 5 years ago
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    if F(s) = 1/(s+2), then inverse laplace transform of it is f(t) = e^(-2t)

  8. anonymous
    • 5 years ago
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    you have all the information you need to solve this.

  9. anonymous
    • 5 years ago
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    so final answer after integrating would be 3^(-2t)-e^(-3t)

  10. anonymous
    • 5 years ago
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    I ment to say e^(-2t)-e^-3t

  11. anonymous
    • 5 years ago
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    http://www.youtube.com/watch?v=TJgBEI3drUc&feature=BFa&list=SP96AE8D9C68FEB902&index=44

  12. anonymous
    • 5 years ago
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    I dont know what the final answer is. I didn't actually try to solve it. I showed you the method. I trust you can take it from there

  13. anonymous
    • 5 years ago
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    thank you just checked inm solution and it is

  14. anonymous
    • 5 years ago
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    okay great!

  15. anonymous
    • 5 years ago
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    \[H(s)=1/s(s^2+9)=A/s+Bs+C/(s^2+9) by \partial fraction I get (1/9s) -(s/9(s^2+9)\]

  16. anonymous
    • 5 years ago
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    for you to use convolution, you have to express your H(s) as a product of two or more functions in s- domain

  17. anonymous
    • 5 years ago
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    you can't express them as a summation.

  18. anonymous
    • 5 years ago
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    \[so \Im \left with 1/9 L^-1 (1/s) -1/9 L^-1(s/s^2+9)\]

  19. anonymous
    • 5 years ago
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    so F(s)=1/9 and G(s)=-1/9cos3t

  20. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{t}1/9-1/9\cos3rhod \rho= ?\]

  21. anonymous
    • 5 years ago
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    I got answer 1/3sin3t and book answer is (1-cos3t)/9

  22. anonymous
    • 5 years ago
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    i dont know what you are trying to ask. I cannot help you with actually finding the values to your problems. I can only show you how. Go through that video. It explains it well with an example similar to your problem

  23. anonymous
    • 5 years ago
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    cool

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