anonymous
  • anonymous
Find the Laplace Transform F(s) of: f(x)=6sin3x+4cos5x-2e^-7x+8xe^2x+7x^3
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
you're not expected to find them using the definition (integrating to infinity) you can use a table of transforms: \[F(s) = 6*\frac{ 3 }{ s^2 + 3^2 } + 5*\frac{ s^2 }{ s^2 +5^2 } + \frac{ 2 }{ s + 7 } + \frac{ 8 }{ (s -2)^2 } + \frac{ 7 * 3! }{ s^4 }\]
anonymous
  • anonymous
um can you explain what you did there?
anonymous
  • anonymous
$$\mathcal{L}\{6\sin3x+4\cos5x-2e^{-7x}+8xe^{2x}+7x^3\}$$Take it term by term:$$\mathcal{L}\{6\sin3x\}=6\cdot\mathcal{L}\{\sin3x\}=\frac{18}{s^2+9}\\\mathcal{L}\{4\cos5x\}=\frac{5s}{s^2+25}\\\mathcal{L}\{-2e^{-7x}\}=-\frac2{s+7}\\\mathcal{L}\{8xe^{2x}\}=\frac8{(s-2)^2}\\\mathcal{L}\{7x^3\}=\frac{7\cdot3!}{s^4}=\frac{42}{s^4}$$... putting it all together we have:$$F(s)=\frac{18}s^2+\frac{5s}{s^2+25}-\frac2{s+7}+\frac8{(s-2)^2}+\frac{42}{s^4}$$Here's a useful table: http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms

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anonymous
  • anonymous
oops the first term came out broken! $$F(s)=\frac{18}{s^2+9}+\dots$$
anonymous
  • anonymous
simply used the table which defines what the transformation is. you can also use the definition; but the results of all continuous functions are well known. i personally used this table right now; i use the one in my textbook for homework: https://controls.engin.umich.edu/wiki/images/9/97/Table_5_3_Single_sided_Laplace_Transforms.jpg
anonymous
  • anonymous
i made a mistake in mine too. the second term is s/(s^2 + 5^2) not s^2. sorry
anonymous
  • anonymous
My second term should be \(\dfrac{4s}{s^2+25}\)... woops.
anonymous
  • anonymous
hmm am trying to understand it, but thanks oldrin
anonymous
  • anonymous
No problem. Since the Laplace transform is a linear transformation (like differentiation and integration), you can take it term by term and constant factors pass through.
anonymous
  • anonymous
ok but may i know what is the "s" is? am kind of a beginner here
anonymous
  • anonymous
oops sorry for the lack of response. \(s\) is a "complex frequency" -- I wouldn't worry about it to much. The idea is that our Laplace transform puts us in a new space, the \(s\)-domain, in which differential equations reduce to algebraic ones.

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