Chapter 7: Q11E (page 404)

Use the convolution theorem to find the inverse Laplace transform of the given function. $\frac{s}{(s - 1) (s + 2)} [H i n t : \frac{s}{s - 1} = 1 + \frac{1}{s - 1}]$

Short Answer

Expert verified

The inverse Laplace transform for the given function by using the convolution theorem is. $y (t) = \frac{2 e^{- 2 t}}{3} + \frac{e^{t}}{3}$

Step by step solution

Define convolution theorem

Let and be piecewise continuous on $[0, \infty)$ and of exponential orderand set, $F (s) = L {f} (s) and G (s) = L {g} (s)$ then,

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

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

Use the convolution theorem to obtain the inverse Laplace transform

Consider the given equation,

$\frac{s}{(s - 1) (s + 2)}$

Let,

$\begin{matrix} y (s) = \frac{s}{(s - 1) (s + 2)} \\ = \frac{1}{s + 2} [1 + \frac{1}{s - 1}] \\ = \frac{1}{s + 2} + \frac{1}{(s - 1) (s + 2)} \end{matrix}$

Take inverse Laplace transform on both sides,

$L^{- 1} [y (s)] = L^{- 1} [\frac{1}{s + 2}] + L^{- 1} [\frac{1}{(s - 1) (s + 2)}]$ $\to (1)$

Hence, the convolution formula is, $L^{- 1} [f (s) \cdot g (s)] = f ⋆ g = \int_{0}^{t} f (t - v) g (v) d v$ , where

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

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

Thus, the equationcan be written as,

$\begin{matrix} y (t) = e^{- 2 t} + \int_{0}^{t} e^{t - v} \cdot e^{- 2 v} d v \\ = e^{- 2 t} + e^{t} \int_{0}^{t} e^{- 3 v} d v \\ = e^{- 2 t} + e^{t} {[\frac{e^{- 3 v}}{- 3}]}_{0}^{t} \\ = e^{- 2 t} - \frac{e^{t}}{3} [e^{- 3 t} - 1] \end{matrix}$

$\begin{matrix} y (t) = e^{- 2 t} - \frac{e^{- 2 t}}{3} + \frac{e^{t}}{3} \\ = \frac{2 e^{- 2 t}}{3} + \frac{e^{t}}{3} \end{matrix}$

Therefore, the inverse Laplace transform for the given function is. $y (t) = \frac{2 e^{- 2 t}}{3} + \frac{e^{t}}{3}$

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Use the convolution theorem to find the inverse Laplace transform of the given function. $\frac{s}{(s - 1) (s + 2)} [H i n t : \frac{s}{s - 1} = 1 + \frac{1}{s - 1}]$

Short Answer

Step by step solution

Define convolution theorem

Use the convolution theorem to obtain the inverse Laplace transform

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Use the convolution theorem to find the inverse Laplace transform of the given function.s(s-1)(s+2)[Hint:ss-1=1+1s-1]

Short Answer

Step by step solution

Define convolution theorem

Use the convolution theorem to obtain the inverse Laplace transform

One App. One Place for Learning.

Most popular questions from this chapter

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Use the convolution theorem to find the inverse Laplace transform of the given function. $\frac{s}{(s - 1) (s + 2)} [H i n t : \frac{s}{s - 1} = 1 + \frac{1}{s - 1}]$