Math Problem Statement

Using convolution theorem find laplace inverse of [s/(s^2+a^2)^3]

Solution

To find the inverse Laplace transform of the given function F(s)=s(s2+a2)3F(s) = \frac{s}{(s^2 + a^2)^3} using the convolution theorem, we'll follow these steps:

  1. Decompose the given function into simpler parts.
  2. Find the inverse Laplace transform of each part.
  3. Apply the convolution theorem to find the inverse Laplace transform of the product.

Step 1: Decompose the Function

We can decompose F(s)F(s) as follows:

F(s)=s(s2+a2)3=s(s2+a2)5/21(s2+a2)1/2F(s) = \frac{s}{(s^2 + a^2)^3} = \frac{s}{(s^2 + a^2)^{5/2}} \cdot \frac{1}{(s^2 + a^2)^{1/2}}

This suggests that we can write F(s)F(s) as a product of two simpler functions G(s)G(s) and H(s)H(s), where:

G(s)=s(s2+a2)5/2andH(s)=1(s2+a2)1/2G(s) = \frac{s}{(s^2 + a^2)^{5/2}} \quad \text{and} \quad H(s) = \frac{1}{(s^2 + a^2)^{1/2}}

Step 2: Find the Inverse Laplace Transforms

We now find the inverse Laplace transforms of G(s)G(s) and H(s)H(s).

1. Inverse Laplace of G(s)G(s)

For G(s)=s(s2+a2)5/2G(s) = \frac{s}{(s^2 + a^2)^{5/2}}, the inverse Laplace transform is a known result and is given by:

L1{s(s2+a2)5/2}=38a3(cos(at)+atsin(at)2)\mathcal{L}^{-1}\left\{\frac{s}{(s^2 + a^2)^{5/2}}\right\} = \frac{3}{8a^3} \left(\cos(at) + \frac{at\sin(at)}{2}\right)

2. Inverse Laplace of H(s)H(s)

For H(s)=1(s2+a2)1/2H(s) = \frac{1}{(s^2 + a^2)^{1/2}}, the inverse Laplace transform is given by:

L1{1(s2+a2)1/2}=J0(at)\mathcal{L}^{-1}\left\{\frac{1}{(s^2 + a^2)^{1/2}}\right\} = J_0(at)

where J0(at)J_0(at) is the Bessel function of the first kind of order zero.

Step 3: Apply the Convolution Theorem

The convolution theorem states that the inverse Laplace transform of a product of two Laplace transforms is the convolution of their respective inverse transforms. Therefore, the inverse Laplace transform of F(s)F(s) is given by:

L1{F(s)}=(gh)(t)=0tg(tτ)h(τ)dτ\mathcal{L}^{-1}\{F(s)\} = (g * h)(t) = \int_0^t g(t - \tau) h(\tau) \, d\tau

where g(t)g(t) and h(t)h(t) are the inverse Laplace transforms of G(s)G(s) and H(s)H(s), respectively.

Thus, the inverse Laplace transform f(t)f(t) of F(s)F(s) is:

f(t)=0t38a3[cos(a(tτ))+a(tτ)sin(a(tτ))2]J0(aτ)dτf(t) = \int_0^t \frac{3}{8a^3} \left[\cos(a(t - \tau)) + \frac{a(t - \tau)\sin(a(t - \tau))}{2}\right] J_0(a\tau) \, d\tau

This integral represents the convolution of the functions g(t)g(t) and h(t)h(t), giving the desired inverse Laplace transform.

Would you like a more detailed explanation or have any questions?


Related Questions:

  1. What is the Laplace transform of the convolution of two functions?
  2. How does the convolution theorem help in solving differential equations?
  3. Can we use other methods besides the convolution theorem to find inverse Laplace transforms?
  4. What are the properties of the Bessel function J0(x)J_0(x)?
  5. How do we handle improper integrals in the context of the convolution theorem?
  6. What are some practical applications of the Laplace transform in engineering?
  7. What is the significance of the cosine and sine functions in the solution of inverse Laplace transforms?
  8. Can the convolution theorem be extended to more than two functions?

Tip: When using the convolution theorem, ensure that the functions involved are well-behaved over the interval of integration to avoid complications with the integral's convergence.

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Math Problem Analysis

Mathematical Concepts

Laplace Transform
Inverse Laplace Transform
Convolution Theorem

Formulas

Inverse Laplace transform formula

Theorems

Convolution theorem

Suitable Grade Level

Advanced Mathematics