Answered: (8 – 4)e-3s s2 – 4s + 5 Determine L-1

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Expert Solution

Concept used:

The inverse laplace transform rule.

If $L^{- 1} \{F (s)\} = f (t)$ then

1. $L^{- 1} \{e^{- a s} F (s)\} = H (t - a) f (t - a)$ , Here, H(t) is Heaviside step function.

2. $L^{- 1} \{F (s - a)\} = e^{a t} f (t)$

3. $L^{- 1} \{s F (s)\} = f^{'} (t) + f (0)$

The linearity property of the inverse laplace transform is

$L^{- 1} \{a \cdot f (s) + b \cdot g (s)\} = a \cdot L^{- 1} \{f (s)\} + b \cdot L^{- 1} \{g (s)\}$

From the Inverse Laplace transform table.

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

Calculation

Apply the inverse transform rule "If $L^{- 1} \{F (s)\} = f (t)$ then $L^{- 1} \{e^{- a s} F (s)\} = H (t - a) f (t - a)$ ".

For the function $\frac{(s - 4) e^{- 3 s}}{s^{2} - 4 s + 5}$ ,

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

$L^{- 1} \{\frac{e^{- 3 s} s - 4}{s^{2} - 4 s + 5}\} = H (t - 3) L^{- 1} \{F (s - 3)\}$ ...... (1)

First find $L^{- 1} \{F (s)\} = f (t)$ where $F (s) = \frac{(s - 4)}{s^{2} - 4 s + 5}$ .

$L^{- 1} \{\frac{s - 4}{s^{2} - 4 s + 5}\} = L^{- 1} \{\frac{s - 2}{{(s - 2)}^{2} + 1} - 2 \cdot \frac{1}{{(s - 2)}^{2} + 1}\}$

Step 3

Now apply the linearity property of inverse Laplace transform.

$\begin{array}{rcl} L^{- 1} \{\frac{s - 4}{s^{2} - 4 s + 5}\} & = & L^{- 1} \{\frac{s - 2}{{(s - 2)}^{2} + 1} - 2 \cdot \frac{1}{{(s - 2)}^{2} + 1}\} \\ = & L^{- 1} \{\frac{s - 2}{{(s - 2)}^{2} + 1}\} - 2 L^{- 1} \{\frac{1}{{(s - 2)}^{2} + 1}\} \end{array}$

Now apply the inverse transform rule $L^{- 1} \{F (s - a)\} = e^{a t} L^{- 1} (F (s))$ .

$\begin{array}{rcl} L \end{array} \begin{array}{rcl} ^{- 1} \end{array} \begin{array}{rcl} \{\frac{s - 2}{{(s - 2)}^{2} + 1}\} \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} L \end{array} \begin{array}{rcl} ^{- 1} \end{array} \begin{array}{rcl} \{\frac{1}{{(s - 2)}^{2} + 1}\} \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} e \end{array} \begin{array}{rcl} ^{2 t} \end{array} \begin{array}{rcl} L \end{array} \begin{array}{rcl} ^{- 1} \end{array} \begin{array}{rcl} \{\frac{s}{s^{2} + 1}\} \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} 2 \end{array} \begin{array}{rcl} e \end{array} \begin{array}{rcl} ^{2 t} \end{array} \begin{array}{rcl} L \end{array} \begin{array}{rcl} ^{- 1} \end{array} \begin{array}{rcl} \{\{\frac{1}{s^{2} + 1}\}\} \end{array}$ ..... (2)