Inverse Laplace transform

In summary, the conversation was about finding the inverse laplace of the function 1/s(s^2+w^2) using partial fraction method. The expert summarizer explains that the correct answers for A and C were found, but there was a discrepancy with B. After further discussion and calculations, the expert concludes that the correct answer for B is -1. The final result for the inverse laplace is 1/w^2[1-coswt).
  • #1
engnrshyckh
51
2
TL;DR 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.
 
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  • #2
engnrshyckh said:
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
Mark44 said:
I agree with your answers for A and C, but I get something different for B.
Tell me?
 
  • #4
engnrshyckh said:
Tell me?
Comparing the Coefficient of s^2 i get (1+B)/w^2 =0 which give me B=-1
 
  • #5
engnrshyckh said:
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
Mark44 said:
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
 
  • #7
engnrshyckh said:
1=A(s^2+w^2)+BS^2+Cs
For finding B
As^2+Bs^2=0
 
  • #8
engnrshyckh said:
For finding B
As^2+Bs^2=0
Right, so A + B = 0, and also Aw^2 = 1, so what do you now get for B?
 
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Likes engnrshyckh
  • #9
Mark44 said:
Right, so A + B = 0, and also Aw^2 = 1, so what do you now get for B?
Thanks - 1/w2
 
  • #10
engnrshyckh said:
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)
 

What is an inverse Laplace transform?

An inverse Laplace transform is a mathematical operation that takes a function in the complex frequency domain and transforms it back into the time domain. It is the reverse of the Laplace transform and is used to solve differential equations and analyze systems in engineering and physics.

How is an inverse Laplace transform calculated?

The inverse Laplace transform is calculated using a table of known transforms or through the use of integration techniques. It involves finding the original function from its Laplace transform by taking the integral of the transform with respect to the complex variable s.

What are the properties of an inverse Laplace transform?

The inverse Laplace transform has several properties, including linearity, time shifting, and frequency shifting. It also has a convolution property that allows for the transformation of complex functions into simpler ones. Additionally, it has a uniqueness property, meaning that a function cannot have more than one inverse Laplace transform.

What are the applications of inverse Laplace transform?

Inverse Laplace transform has various applications in engineering, physics, and mathematics. It is used to solve differential equations and analyze systems in control theory, signal processing, and circuit analysis. It is also used in the study of heat transfer, fluid dynamics, and quantum mechanics.

What are the limitations of inverse Laplace transform?

The inverse Laplace transform can only be applied to functions that have a Laplace transform. It also requires advanced mathematical skills and can be time-consuming to calculate. Additionally, it may not always give a closed-form solution, and numerical methods may be required to obtain an approximate answer.

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