**Problem:**Find the Laplace transform of(a) $ x(t) = -e^{-at}u(-t) $**Explanation:**The problem involves finding the Laplace transform of a given function, $ x(t) $, where $ x(t) = -e^{-at}u(-t) $. - **$ e^{-at} $:** This is an exponential function where $ a $ is a constant.- **$ u(-t) $:** This is the unit step function, but it's applied for negative time, which essentially means it activates when time is negative.The solution approach typically involves recognizing how the unit step function $ u(-t) $ affects the transformation process and applying relevant Laplace transform properties. Make sure to consider the time-shifting property and scaling of functions when solving such problems.

The mirrored unit step function u(-t) is defined as u-t=0 for t≥01 for t<0 For any function x(t),…

Answered: Find the Laplace transform of (a) x(1)…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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**Problem:**

Find the Laplace transform of

(a) $ x(t) = -e^{-at}u(-t) $

**Explanation:**

The problem involves finding the Laplace transform of a given function, $ x(t) $, where $ x(t) = -e^{-at}u(-t) $.

- **$ e^{-at} $:** This is an exponential function where $ a $ is a constant.
- **$ u(-t) $:** This is the unit step function, but it's applied for negative time, which essentially means it activates when time is negative.

The solution approach typically involves recognizing how the unit step function $ u(-t) $ affects the transformation process and applying relevant Laplace transform properties. Make sure to consider the time-shifting property and scaling of functions when solving such problems.

Expert Solution

Step 1

The mirrored unit step function u(-t) is defined as

$u (- t) = \{\begin{cases} 0 for t \geq 0 \\ 1 for t < 0 \end{cases}$

For any function x(t), the Laplace transform of the function x(t) is defined as

$L [x (t)] = \int_{- \infty}^{+ \infty} x (t) e^{- s t} d t - - - > (1)$

Plugging the given function into equation (1), following can be obtained:

$L [x (t)] = - \int_{- \infty}^{0} e^{- a t} e^{- s t} d t = > L [x (t)] = - \int_{- \infty}^{0} e^{- (s + a) t} d t = > L [x (t)] = \frac{1}{s + a} {[e^{- (s + a) t}]}_{- \infty}^{0} = > L [x (t)] = \frac{1}{s + a} [1 - e^{- (s + a) (- \infty)}] - - - > (2)$

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