Chapter 7: 1RP (page 415)

In Problems 1 and 2, use the definition of the Laplace transform to determine $L {f}$ .
$f (t) = {\begin{cases} 3, 0 \leq t \leq 2 \\ 6 - t, 2 < t \end{cases}$

Short Answer

Expert verified

$L {f (t)} (s) = \frac{3}{s} + \frac{e^{- 2 s}}{s} - \frac{e^{- 2 s}}{s^{2}}$

Step by step solution

Step 1:Given Information

The given function is $f (t) = {\begin{cases} 3, 0 \leq t \leq 2 \\ 6 - t, 2 < t \end{cases}$

Determining the L{f}

Using the Laplace transform definition, we get

$\begin{matrix} L {f (t)} = \int_{0}^{\infty} e^{- s t} f (t) d t \\ = \int_{0}^{2} 3 e^{- s t} d t + \int_{2}^{\infty} (6 - t) e^{- s t} d t \\ = 3 {[- \frac{e^{- s t}}{s}]}_{0}^{2} + \lim_{N \to \infty} \int_{2}^{N} (6 - t) e^{- s t} d t \\ = \frac{3}{s} (1 - e^{- 2 s}) + \lim_{N \to \infty} \int_{2}^{N} (6 - t) e^{- s t} d t \end{matrix}$

Letrole="math" localid="1664044470656" $|\begin{matrix} \begin{array}{ll} 6 - t = u & e^{- s t} d t = d v \\ - d t = d u & v = - \frac{e^{- s t}}{s} \end{array} \end{matrix}|$ in second integral, then we can write as:

$\begin{matrix} L {f (t)} = \frac{3}{s} (1 - e^{- 2 s}) + \lim_{N \to \infty} ({[- (6 - t) \frac{e^{- s t}}{s}]}_{2}^{N} - \int_{2}^{N} \frac{e^{- s t}}{s} d t) \\ = \frac{3}{s} (1 - e^{- 2 s}) + \lim_{N \to \infty} (\frac{4 e^{- 2 s}}{s} - \frac{(6 - N) e^{- s N}}{s} + {[\frac{e^{- s t}}{s^{2}}]}_{2}^{N}) \\ = \frac{3}{s} (1 - e^{- 2 s}) + \lim_{N \to \infty} (\frac{4 e^{- 2 s}}{s} - \frac{(6 - N) e^{- s N}}{s} + \frac{e^{- s N}}{s^{2}} - \frac{e^{- 2 s}}{s^{2}}) \\ = \frac{3}{s} (1 - e^{- 2 s}) + \lim_{N \to \infty} \frac{4 e^{- 2 s}}{s} - \lim_{N \to \infty} \frac{(6 - N) e^{- s N}}{s} + \lim_{N \to \infty} \frac{e^{- s N}}{s^{2}} - \lim_{N \to \infty} \frac{4 e^{- 2 s}}{s^{2}} \end{matrix}$

Simplify further as:

$\begin{matrix} L {f (t)} = \frac{3}{s} (1 - e^{- 2 s}) + \frac{4 e^{- 2 s}}{s} - 0 + 0 - \frac{4 e^{- 2 s}}{s^{2}} \\ = \frac{3}{s} (1 - e^{- 2 s}) + \frac{4 e^{- 2 s}}{s} - \frac{4 e^{- 2 s}}{s^{2}} \\ = \frac{3}{s} + \frac{e^{- 2 s}}{s} - \frac{e^{- 2 s}}{s^{2}} \end{matrix}$

Determining the Result

Thus, the required Laplace transform is $L {f (t)} (s) = \frac{3}{s} + \frac{e^{- 2 s}}{s} - \frac{e^{- 2 s}}{s^{2}}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

In Problems 1 and 2, use the definition of the Laplace transform to determine $L {f}$ .
$f (t) = {\begin{cases} 3, 0 \leq t \leq 2 \\ 6 - t, 2 < t \end{cases}$

Short Answer

Step by step solution

Step 1:Given Information

Determining the L{f}

Determining the Result

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Mechanics Maths

Geometry

Logic and Functions

Theoretical and Mathematical Physics

Decision Maths

Study anywhere. Anytime. Across all devices.

In Problems 1 and 2, use the definition of the Laplace transform to determineL{f}.f(t)={3,0≤t≤26-t,2<t

Short Answer

Step by step solution

Step 1:Given Information

Determining the L{f}

Determining the Result

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Mechanics Maths

Geometry

Logic and Functions

Theoretical and Mathematical Physics

Decision Maths

Study anywhere. Anytime. Across all devices.

In Problems 1 and 2, use the definition of the Laplace transform to determine $L {f}$ .
$f (t) = {\begin{cases} 3, 0 \leq t \leq 2 \\ 6 - t, 2 < t \end{cases}$