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Make up two functions \(f_{1}\) and \(f_{2}\) that have the same Laplace transform. Do not think profound thoughts.

Short Answer

Expert verified
Functions \( f_1(t) = e^{-t} \) and \( f_2(t) = e^{-t}(1 + \sin(t)) \) have the same Laplace transform, \( \frac{1}{s+1} \).

Step by step solution

01

Understanding Laplace Transforms

The Laplace transform is an integral transform used to convert a function of time to a function of complex frequency. It is defined as \( \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt \). Functions that differ only on sets of measure zero have the same Laplace transform.
02

Defining Function f1

Let's consider a simple function, \( f_1(t) = e^{-t} \theta(t) \), where \( \theta(t) \) is the Heaviside step function. This is a common function whose Laplace transform can be easily found.
03

Computing Laplace Transform of f1

The Laplace transform of \( f_1(t) = e^{-t} \) is given by \( \mathcal{L}\{f_1(t)\} = \int_{0}^{\infty} e^{-st} e^{-t} \, dt = \int_{0}^{\infty} e^{-(s+1)t} \, dt \). This integral evaluates to \( \frac{1}{s+1} \), for \( s > -1 \).
04

Adjusting f2 to Match Laplace Transform of f1

We need to choose a second function, \( f_2(t) \), that has the same Laplace transform. A simple option is \( f_2(t) = e^{-t}(1 + \sin(t))\theta(t) \). Because the Laplace transform is determined by the exponential decay and \( \sin(t) \) does not change the convergence of the integral, \( f_2 \)'s transform is also \( \frac{1}{s+1} \).
05

Verify Laplace Transform of f2

Ensure the Laplace transform for \( f_2(t) \) matches by confirming the integral: \( \mathcal{L}\{f_2(t)\} = \int_{0}^{\infty} e^{-(s+1)t}(1 + \sin(t)) \, dt \). The integral splits into two parts, both simplifying to expressions that ultimately also reduce to \( \frac{1}{s+1} \).

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Key Concepts

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

Heaviside Step Function
The Heaviside step function, represented as \( \theta(t) \), is a discontinuous function used in various fields, especially in engineering and mathematics. It is a simple function that is "off" before a certain point and "on" after that point. - Formally, it is defined as: \[ \theta(t) = \begin{cases} 0, & t < 0 \ 1, & t \geq 0 \end{cases} \] This means the function "steps" from zero to one at \( t = 0 \).- **Significance in Laplace Transforms:** The use of the Heaviside step function is crucial in the Laplace transforms because it allows the inclusion of functions that begin at a point other than zero.
- By multiplying any time-based function \( f(t) \) by \( \theta(t-a) \), it can be modified to "turn on" at some point \( a \). This provides a way to deal with systems that are not active from \( t = 0 \). Hence, it helps in modeling real-world scenarios where events happen only after certain moments.
Integral Transform
Integral transforms are a mathematical tool used to convert functions from one domain to another for ease of analysis. The Laplace transform itself is a type of integral transform, used predominantly in the study of control systems and differential equations.- **How It Works:** The Laplace transform converts a time function \( f(t) \) into a complex frequency function \( F(s) \). This is done through the formula: \[ \mathcal{L}\{ f(t) \} = \int_{0}^{\infty} e^{-st} f(t) \, dt \]- **Applications and Importance:** - Integral transforms simplify the process of solving linear differential equations. - They convert differential problems into algebraic ones, making them easier to tackle and solve.- **Why Use Laplace Transforms?:** - One of the significant advantages of the Laplace transform is its ability to handle initial conditions directly, which is quite beneficial in solving physical problems modeled by differential equations. - Another critical aspect is its use in analyzing systems by breaking them into more manageable parts, especially in engineering and physics.
Exponential Function
The exponential function is a key player in calculus and differential equations. Expressed mathematically as \( e^x \), it's a function where the derivative is proportional to its value, a property that significantly influences its use in various applications.- **In the Context of Laplace Transforms:** - The function \( e^{-st} \) appears prominently in the definition of the Laplace transform, acting as a "weighting" function that transforms \( f(t) \) into \( F(s) \). - **Significance in Our Example:** In the given solution, both functions \( f_1(t) = e^{-t}\theta(t) \) and \( f_2(t) = e^{-t}(1 + \sin(t))\theta(t) \) leverage the exponential’s unique properties to achieve the same Laplace transform.
  • The exponential decay \( e^{-t} \) ensures that both functions converge as \( t \to \infty \).
  • This property is essential for the Laplace transform to exist and yield a real value.
- **Related Mathematical Features:** - The behavior of \( e^{-t} \) makes it suitable for modeling decay processes in physics and engineering. - These features are why the exponential function is a frequent subject in various mathematical transformations and solutions.

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