Chapter 7: Q5E (page 404)

Use the convolution theorem to find the inverse Laplace transform of the given function.
$\frac{1}{s (s^{2} + 1)}$

Short Answer

Expert verified

The inverse Laplace transform for the given function using the convolution theorem is.

$L^{- 1} \{\frac{1}{s (s^{2} + 2)}\} (t) = 1 - \cos t$

Step by step solution

Define convolution theorem

Let $f (t)$ and $g (t)$ be piecewise continuous on $[0, \infty)$ and of exponential order $α$ and set $F (s) = L {f} (s) a n d G (s) = L {g} (s)$ , then,

$L {f * g} (s) = F (s) G (s)$

$L^{- 1} {f (s) * g (s)} (t) = (f * g) (t)$

Find the inverse Laplace transform by using the convolution theorem

Consider the given function, $\frac{1}{s (s^{2} + 1)}$

Let, $F (s) \frac{1}{s}$

$G (s) \frac{1}{(s^{2} + 1)}$

Since, we know

$f (t) = L^{- 1} \{\frac{1}{s}\} (t) = 1$

$g (t) = L^{- 1} \{\frac{1}{s^{2} + 1}\} (t) = \sin t$

Apply inverse Laplace transform and use convolution theorem to obtain,

$L^{- 1} \{\frac{1}{s^{2} + 1}\} (t) = L^{- 1} \{F (s) G (s)\} (t) = f (t) \times g (t) = 1 \times \sin t$

Thus, use the convolution formula, $(f * g) (t) = \int_{0}^{t} f (t - v) g (v) d v$

$L^{- 1} (\frac{1}{s (s^{2} + 1)}) (t) = \int_{0}^{t} \sin v d v = {[- \cos v]}_{0}^{t} = 1 - \cos t$

Therefore, $L^{- 1} (\frac{1}{s (s^{2} + 1)}) (t) = 1 - \cos t$

thus the inverse Laplace transform for the given function is

L^{- 1} (\frac{1}{s (s^{2} + 1)}) (t) = 1 - \cos t

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Use the convolution theorem to find the inverse Laplace transform of the given function.
$\frac{1}{s (s^{2} + 1)}$

Short Answer

Step by step solution

Define convolution theorem

Find the inverse Laplace transform by using the convolution theorem

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Use the convolution theorem to find the inverse Laplace transform of the given function.
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