Chapter 7: Q2RP (page 415)

In Problems 1 and 2, use the definition of the Laplace transform to determine $L {f}$ .
$f (t) = {\begin{cases} e^{- t}, 0 \leq t \leq 5 \\ - 1, 5 < t \end{cases}$

Short Answer

Expert verified

$L {f (t)} (s) = \frac{1}{s + 1} (1 - e^{- 5 (s + 1)}) - \frac{1}{s} e^{- 5 s}$

Step by step solution

Step 1:Given Information

The given function is $f (t) = {\begin{cases} e^{- t}, 0 \leq t \leq 5 \\ - 1, 5 < t \end{cases}$ .

Determining the

Using the Laplace transform definition, we get

$\begin{matrix} L {f (t)} (s) = \int_{0}^{\infty} e^{- s t} f (t) d t \\ = \int_{0}^{5} e^{- s t} e^{- t} d t + \int_{5}^{\infty} (- 1) e^{- s t} d t \\ = \int_{0}^{5} e^{- t (s + 1)} d t - \lim_{N \to \infty} \int_{5}^{N} e^{- s t} d t \end{matrix}$

Simplify further as:

$\begin{matrix} L {f (t)} = {[- \frac{e^{- t (s + 1)}}{s + 1}]}_{0}^{5} - \lim_{N \to \infty} {[- \frac{e^{- s t}}{s}]}_{5}^{N} \\ = \frac{1}{s + 1} (1 - e^{- 5 (s + 1)}) - \lim_{N \to \infty} (\frac{1}{s} (e^{- 5 s} - e^{- N s})) \\ = \frac{1}{s + 1} (1 - e^{- 5 (s + 1)}) - \frac{1}{s} e^{- 5 s} \end{matrix}$

Determining the Result

Thus, the required Laplace transform is $L {f (t)} (s) = \frac{1}{s + 1} (1 - e^{- 5 (s + 1)}) - \frac{1}{s} e^{- 5 s}$

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In Problems 1 and 2, use the definition of the Laplace transform to determine $L {f}$ .
$f (t) = {\begin{cases} e^{- t}, 0 \leq t \leq 5 \\ - 1, 5 < t \end{cases}$

Short Answer

Step by step solution

Step 1:Given Information

Determining the

Determining the Result

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In Problems 1 and 2, use the definition of the Laplace transform to determine L{f}.f(t)={e-t,0≤t≤5-1,5<t

Short Answer

Step by step solution

Step 1:Given Information

Determining the

Determining the Result

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Discrete Mathematics

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Study anywhere. Anytime. Across all devices.

In Problems 1 and 2, use the definition of the Laplace transform to determine $L {f}$ .
$f (t) = {\begin{cases} e^{- t}, 0 \leq t \leq 5 \\ - 1, 5 < t \end{cases}$