express L−1[F(s)G(s)] in terms of a convolution integral. F(s) =4G(s) =s - 1s2 – 2s +5

Given: Fs=4s3, Gs=s-1s2-2s+5. To Do: Express L-1FsGs in terms of convolution.

Answered: F(s) = 4 G(s) = s - 1 s2 – 2s +5

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

express L⁻¹[F(s)G(s)] in terms of a convolution integral.

Expert Solution

Step 1

Given:

$F (s) = \frac{4}{s^{3}}$ , $G (s) = \frac{s - 1}{s^{2} - 2 s + 5}$ .

To Do:

Express $L^{- 1} [F (s) G (s)]$ in terms of convolution.

Step 2

We know,

If $L (f * g) = F (s) G (s)$ , then $L^{- 1} [F (s) G (s)] = (f * g) (t)$ .

And, $(f * g) (t) = \int_{0}^{t} f (t - u) g (u) d u$ .

$L (t^{n}) = \frac{n!}{s^{n + 1}}$

Here we have,
$F (s) = \frac{4}{s^{3}} \Rightarrow f (t) = L^{- 1} [F (s)] \Rightarrow f (t) = L^{- 1} (\frac{4}{s^{3}}) \Rightarrow f (t) = \frac{4 t^{2}}{3!} \Rightarrow f (t) = \frac{2 t^{2}}{3} \Rightarrow f (t - u) = \frac{2 {(t - u)}^{2}}{3} . . . (1)$

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