Laplace transform with Convolution

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SUMMARY

The discussion focuses on applying the Convolution Theorem to solve the Laplace transform of the expression \( \mathfrak{L} \left \{t\int_{0}^{t} \sin \tau d\tau \right \} \) without directly solving the integral. The key equation referenced is \( \mathfrak{L}\left \{ f*g \right \}=F(s)G(s) \), where \( f(t) = \int_0^t \sin \tau \, d\tau \). The solution provided in the textbook is \( \frac{3s^2+1}{s^2(s^2+1)^2} \), and the discussion emphasizes using the property \( \mathfrak{L}\{t f(t)\} = -F'(s) \) to derive \( F(s) \) through convolution.

PREREQUISITES
  • Understanding of Laplace transforms, specifically \( \mathfrak{L} \{ f(t) \} \)
  • Familiarity with the Convolution Theorem in Laplace transforms
  • Knowledge of integration techniques, particularly definite integrals
  • Ability to differentiate functions in the context of Laplace transforms
NEXT STEPS
  • Study the Convolution Theorem in detail, focusing on its applications in Laplace transforms
  • Learn about the differentiation property of Laplace transforms, specifically \( \mathfrak{L}\{t f(t)\} = -F'(s) \)
  • Practice solving Laplace transforms involving integrals of trigonometric functions
  • Explore partial fraction decomposition techniques for inverse Laplace transforms
USEFUL FOR

Students studying differential equations, engineers applying Laplace transforms in control systems, and anyone seeking to deepen their understanding of convolution in the context of Laplace transforms.

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Homework Statement



Use convolution theorem to solve:

\mathfrak{L} \left \{t\int_{0}^{t} \sin \tau d\tau \right \}

Do not solve the integral.

Homework Equations



"Convolution Theorem" in textbook is stated as:

\mathfrak{L}\left \{ f*g \right \}=F(s)G(s)f*g=\int_{0}^{t} \ f(\tau )g(t-\tau ) d\tau

The Attempt at a Solution



Not quite sure how to approach this one with convolution and not solving the integral.

I need a t-τ instead of a t, but I can't have τ's in my Laplace because they won't go to S-space with any meaning.

Answer in book is :

\frac{3s^2+1}{s^2(s^2+1)^2}working backwards only leads me to a partial fraction type inverse transform.

Thanks in advance for help!
 
Last edited:
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Perhaps the idea is to use the property
$$\mathfrak{L}\{t f(t)\} = -F'(s) $$ where
$$f(t) = \int_0^t \sin \tau \, d\tau, $$ and to find F(s) using the convolution theorem.
 
Last edited:
Thanks Vela, that's exactly it
 

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