Chapter 5: Problem 16
Find the inverse Laplace transform of each function in Exercises 16-25. Create a piecewise definition for your solution that doesn't use the Heaviside function. $$ F(s)=\frac{e^{-2 s}}{s+3} $$
Short Answer
Expert verified
The inverse Laplace transform is a piecewise function: \( f(t) = 0 \) for \( t < 2 \), and \( f(t) = e^{-3(t-2)} \) for \( t \geq 2 \).
Step by step solution
01
Identify the Form of the Laplace Transform
The given function is \( F(s) = \frac{e^{-2s}}{s+3} \). Notice that it can be broken down into \( e^{-as} \times G(s) \) form, where \( a = 2 \) and \( G(s) = \frac{1}{s+3} \).
02
Find the Inverse Laplace Transform of G(s)
The function \( G(s) = \frac{1}{s+3} \) is a standard form that corresponds to the inverse Laplace transform \( L^{-1} \{ \frac{1}{s+3} \} = e^{-3t} \). This transform is valid for \( t \geq 0 \).
03
Apply the Time Delay Property
The time delay property of the Laplace transform states that if \( L^{-1}\{G(s)\} = g(t) \), then \( L^{-1}\{e^{-as}G(s)\} = g(t-a)u(t-a) \). Here, \( a = 2 \), so we perform the shift: \( e^{-3(t-2)}u(t-2) \).
04
Eliminate the Heaviside Function
Instead of using the Heaviside function to express the solution, we'll convert it into a piecewise function. The piecewise function is defined as: \[ f(t) = \begin{cases} 0, & \text{for } t < 2 \ e^{-3(t-2)}, & \text{for } t \geq 2 \end{cases} \] This removes the explicit use of the Heaviside function.
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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 often used in engineering and physics. It transforms a time domain function, which depends on time \( t \), into a complex frequency domain function, which is a function of the complex variable \( s \). This transformation simplifies the analysis and solving of differential equations, as it converts differentiation and integration into algebraic operations.\
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In its essence, the Laplace Transform, denoted usually as \( L\{f(t)\} \), is defined by the integral:
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In its essence, the Laplace Transform, denoted usually as \( L\{f(t)\} \), is defined by the integral:
- \( L\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \ dt \)
Piecewise Functions
Piecewise Functions are used to define a function for different intervals of the independent variable. They are crucial when the functions you are dealing with have different expressions based on the input value range.\
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In our exercise, we express a solution not with continuous expressions but with different cases. The function is split into distinct parts:
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In our exercise, we express a solution not with continuous expressions but with different cases. The function is split into distinct parts:
- For \( t < 2 \), the function will be zero.
- For \( t \geq 2 \), the function adapts another form, such as \( e^{-3(t-2)} \).
Time Delay Property
In the realm of Laplace Transforms, the Time Delay Property is quite similar to what its name suggests. It effectively shifts a function in time. When a basic function is delayed by \( a \) units of time, the Laplace Transform of this delayed function is modified by multiplying \( G(s) \) by \( e^{-as} \).\
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This can be understood through the relationship:
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This can be understood through the relationship:
- If \( L^{-1}\{G(s)\} = g(t) \), then \( L^{-1}\{e^{-as}G(s)\} = g(t-a)u(t-a) \).
Heaviside Function
The Heaviside Function, also known as the unit step function, is a step function with a value of zero for negative arguments and a value of one for positive arguments. It is widely used in control theory, signal processing, and other fields that require modeling switching operations or sudden changes in a signal or system.\
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Mathematically, it is expressed as:
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Mathematically, it is expressed as:
- \( u(t) = \begin{cases} 0, & t < 0 \ 1, & t \geq 0 \end{cases} \)