Laplace Transform : sin^2(wt) question

Replies

• 

  Sagar07

  Didn't get your question exactly.......
• 

  Brad Grillakis

  Sagar07

  Didn't get your question exactly.......

  My question is, from this function 32/(s(s[SUP]2[/SUP]+64)), how is the answer sin[SUP]2[/SUP](4t)? (I have been trying to work out how my professor got this answer but I can not seem to figure this out. I tried doing partial fractions, but u cant because of the s[SUP]2[/SUP].)
  
  The 2 formula I gave is from the Laplace Transform Table.
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  ritesh05

  Here as we can see that in the denominator there is 's' term so we can directly use the division by s formula that is
  
  so after using this formula you will get -.5(cos8t-1)=-.5(1-2sin^(4t)-1)=sin^2(4t)...............{cos2t=1-2sin^2(t)}
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  rashidsohail

  i dont understand that
  Here as we can see that in the denominator there is 's' term so we can directly use the division by s formula that is
• 

  simplycoder

  Brad Grillakis

  My question is, from this function 32/(s(s[SUP]2[/SUP]+64)), how is the answer sin[SUP]2[/SUP](4t)? (I have been trying to work out how my professor got this answer but I can not seem to figure this out. I tried doing partial fractions, but u cant because of the s[SUP]2[/SUP].)
  
  The 2 formula I gave is from the Laplace Transform Table.

  This can be solved by partial fractions. All you need is proper substitution.
  
  Substitute it as (A/s)+(Bs+C)/(s^2+64).
  
  A=0.5,B=-0.5,C=0.
  
  so you get 0.5/s- 0.5/(s^2+64)
  
  Laplace inverse of the first term is 1, and that of second term is cos(8t).
  So you get 0.5*(1-cos8t).By trignometry, cos8t=1-2sin^2 (4t).
  So you get 0.5*(1-(1-2sin^2(4t))).
  =0.5*2sin^2(4t).
  =sin^2(4t).//
• 

  silverscorpion

  Write the fraction 32/(s(s2+64) as follows..
  
  32/(s(s2+64) = (2*4[sup]2[/sup])/(s*(s[sup]2[/sup] + 4*4[sup]2[/sup]))
  
  So, you get w[sub]0[/sub] to be 4. Hence, the inverse Laplace transform is, from the formula you have given, sin[sup]2[/sup] 4t