anonymous
  • anonymous
Find inverse Laplace transform by using convolution. Please step by step in integral part: 1/s^2+5s+6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
first factor s^2+5s+6 and post what you got
anonymous
  • anonymous
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)
anonymous
  • anonymous
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)

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anonymous
  • anonymous
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)
anonymous
  • anonymous
according to convolution, product of two functions in the angular frequency domain can be represented as convolution of their corresponding time domain functions.
anonymous
  • anonymous
so, h(t) = f(t)*g(t) = \[\int\limits_{0}^{t}f(\tau)g(t-\tau) d \tau\]
anonymous
  • anonymous
if F(s) = 1/(s+2), then inverse laplace transform of it is f(t) = e^(-2t)
anonymous
  • anonymous
you have all the information you need to solve this.
anonymous
  • anonymous
so final answer after integrating would be 3^(-2t)-e^(-3t)
anonymous
  • anonymous
I ment to say e^(-2t)-e^-3t
anonymous
  • anonymous
http://www.youtube.com/watch?v=TJgBEI3drUc&feature=BFa&list=SP96AE8D9C68FEB902&index=44
anonymous
  • anonymous
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
anonymous
  • anonymous
thank you just checked inm solution and it is
anonymous
  • anonymous
okay great!
anonymous
  • anonymous
\[H(s)=1/s(s^2+9)=A/s+Bs+C/(s^2+9) by \partial fraction I get (1/9s) -(s/9(s^2+9)\]
anonymous
  • anonymous
for you to use convolution, you have to express your H(s) as a product of two or more functions in s- domain
anonymous
  • anonymous
you can't express them as a summation.
anonymous
  • anonymous
\[so \Im \left with 1/9 L^-1 (1/s) -1/9 L^-1(s/s^2+9)\]
anonymous
  • anonymous
so F(s)=1/9 and G(s)=-1/9cos3t
anonymous
  • anonymous
\[\int\limits_{0}^{t}1/9-1/9\cos3rhod \rho= ?\]
anonymous
  • anonymous
I got answer 1/3sin3t and book answer is (1-cos3t)/9
anonymous
  • anonymous
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
anonymous
  • anonymous
cool

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