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Chapter 7: Problem 18

Fill in the blanks or answer true/false. \(\mathscr{^}{-1}\left\\{\frac{e^{-4}}{s^{2}}\right\\}=\)_____

Short Answer

Expert verified
The inverse Laplace transform is \( (t-4)u(t-4) \).

Step by step solution

01

Identify the Problem Type

This problem involves finding the inverse Laplace transform of a given expression. The expression provided is \( \frac{e^{-4}}{s^2} \).
02

Review Necessary Formulas

The Laplace transform that we are concerned with is for constant \( a \) and \( n \), which is given by the general formula \( \mathscr{L}^{-1}\left\{\frac{1}{s^n}\right\} = \frac{t^{n-1}}{(n-1)!} \). For expressions like \( e^{-as}\mathscr{L}(f(t))\), the inverse Laplace transform involves a time-shifting property resulting in \( f(t-a)u(t-a) \) where \( u \) is the Heaviside step function.
03

Substitute and Simplify

In our problem, \( a = 4 \) and we need to find \( \mathscr{L}^{-1}\left\{\frac{1}{s^2}\right\} \). We substitute \( n = 2 \) into the formula: \( \frac{t^{2-1}}{(2-1)!} \), which simplifies to \( t \). Thus, the inverse Laplace transform of \( \frac{1}{s^2} \) is \( t \).
04

Apply the Time-Shift Property

Now, apply the time shift property due to \( e^{-4s} \). This yields \( (t-4) \) when the time shift \( a = 4 \), placed in the Heaviside unit step function, giving the final solution as \((t-4)u(t-4)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Laplace Transform
The Laplace Transform is a powerful mathematical tool used to convert functions of a real variable, often in the time domain, into functions of a complex variable, usually in the frequency domain. This is exceptionally useful in solving differential equations and analyzing linear time-invariant systems. The transform is defined as: \[ \mathscr{L}\{ f(t) \} = F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt \]
  • Purpose: It simplifies complex differential equations into algebraic equations, making them easier to manipulate and solve.
  • Frequency Domain: By moving a problem into the frequency domain, it becomes easier to solve, especially for engineers and scientists dealing with systems' behavior over time.
Understanding the Laplace Transform is crucial because it forms the basis of the later concepts we will discuss, such as the time-shifting property and the Heaviside step function.
Time-Shifting Property
The Time-Shifting Property of the Laplace Transform allows us to manipulate functions by shifting them in time. This property is particularly noticeable when you deal with expressions that include a term like \( e^{-as} \). It suggests a delay or advancement of a signal in the time domain. Mathematically, the property is expressed as: \[ \mathscr{L}^{-1}\{ e^{-as}F(s) \} = f(t-a)u(t-a) \] In our problem, with \( e^{-4s} \), this translates to shifting the function by 4 units to the right. Key points include:
  • Expression: The term \( e^{-as} \) indicates the function \( f(t) \) is shifted by \( a \).
  • Result: The inverse transform results in \( f(t-a) \), with the Heaviside step function \( u(t-a) \) ensuring the function only begins at \( t=a \).
Applying this property helps you understand how systems respond over time when a delay or phase shift is introduced.
Heaviside Step Function
The Heaviside Step Function, often denoted as \( u(t-a) \), plays a crucial role in controlling when a function begins to have an effect in time-shifting operations in Laplace Transforms. It's essentially a switch that turns "on" the function at a specific time \( t = a \). The function can be represented as: \[ u(t-a) = \begin{cases} 0, & t < a \ 1, & t \geq a \end{cases} \] This means:
  • Control: It delays the start of a function to a specific time \( t = a \).
  • Application: In our given exercise, it ensures that the function \( (t-4) \) only starts at \( t = 4 \), aligning with the physical intuition of a delayed or phase-shifted response.
Understanding the Heaviside step function is fundamental to applying the time-shifting property, making it indispensable when working with Laplace Transforms.

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Most popular questions from this chapter

Find the given inverse transform. \(\mathscr{L}^{-1}\left\\{\frac{0.9 s}{(s-0.1)(s+0.2)}\right\\}\)

Suppose a 32-pound weight stretches a spring 2 feet. If the weight is released from rest at the equilibrium position, determine the equation of motion if an impressed force \(f(t)=\sin t\) acts on the system for \(0 \leq t<2 \pi\) and is then removed. Ignore any damping forces.

Use the Laplace transform to solve the given integral equation or integrodifferential equation. $$ \frac{d y}{d t}+6 y(t)+9 \int_{0}^{t} y(\tau) d \tau=1, \quad y(0)=0 $$

Find the given inverse transform. \(\mathscr{L}^{-1}\left\\{\frac{4}{s}+\frac{6}{s^{5}}-\frac{1}{s+8}\right\\}\)

Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions. $$ y^{\prime}+2 y=f(t), \quad y(0)=0, \text { where } f(t)= \begin{cases}t, & 0 \leq t<1 \\ 0, & t \geq 1\end{cases} $$
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