Inverse Laplace transform

  • #1

Summary:

I want to find inverse laplace of function 1/s(s^2+w^2)
I used partial fraction method first as:
1/s(s^2+w^2)=A/s+Bs+C/(s^2+w^2)
I found A=1/w^2
B=-1
C=0
1/s(s^2+w^2)=1/sw^2- s/s^2 +w^2
Taking invers laplace i get
1/w2 - coswt
But the ans is not correct kindly help.
 

Answers and Replies

  • #2
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6,260
Summary:: I want to find inverse laplace of function 1/s(s^2+w^2)

I used partial fraction method first as:
1/s(s^2+w^2)=A/s+Bs+C/(s^2+w^2)
I found A=1/w^2
B=-1
C=0
I agree with your answers for A and C, but I get something different for B.
engnrshyckh said:
1/s(s^2+w^2)=1/sw^2- s/s^2 +w^2
Taking invers laplace i get
1/w2 - coswt
But the ans is not correct kindly help.
 
  • #3
I agree with your answers for A and C, but I get something different for B.
Tell me?
 
  • #5
34,549
6,260
Summary:: I want to find inverse laplace of function 1/s(s^2+w^2)

1/s(s^2+w^2)=A/s+Bs+C/(s^2+w^2)
Multiply both sides by ##s(s^2 + w^2)##. What do you get in the next step?
 
  • #6
Multiply both sides by ##s(s^2 + w^2)##. What do you get in the next step?
1=A(s^2+w^2)+BS^2+Cs
 
  • #9
Right, so A + B = 0, and also Aw^2 = 1, so what do you now get for B?
Thanks - 1/w2
 
  • #10
Thanks - 1/w2
1/s(s^2+w^2)=A/s+Bs+C/(s^2+w^2)
=1/w^2(1/s-s/s^2+w^2) taking inverse laplace ibget
=1/w^2[1-coswt)
 

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