Chapter 7: Problem 3
Find the imerse Laplace transform \(f(t)\) of each fiunction given in Problems. Then sketch the graph of \(f\). $$ F(s)=\frac{e^{-s}}{s+2} $$
Short Answer
Expert verified
Step function starting at \( t = 1 \) with exponential decay, \( f(t) = e^{-2(t-1)} \) for \( t \geq 1 \).
Step by step solution
01
Identify the Inverse Laplace Transform Formula
The given function, \( F(s) = \frac{e^{-s}}{s+2} \), suggests we can use the inverse Laplace transform involving a time shift. Recall that, if \( G(s) = \frac{1}{s+a} \), then its inverse is \( g(t) = e^{-at} u(t) \), where \( u(t) \) is the Heaviside step function.
02
Apply the Time Shift Property
The function has an exponential term \( e^{-s} \) which implies a time shift. The time shift property of Laplace transforms states that if \( F(s) = e^{-cs} G(s) \), then the inverse is \( f(t) = u(t-c) g(t-c) \). Here, \( c = 1 \), and \( G(s) = \frac{1}{s+2} \).
03
Determine the Corresponding g(t) for G(s)
Given \( G(s) = \frac{1}{s+2} \), we use the standard inverse Laplace transform:\( g(t) = e^{-2t} u(t) \), where \( u(t) \) is the Heaviside step function.
04
Combine the Results for f(t)
Apply the time shift result to find \( f(t) = e^{-2(t-1)} u(t-1) \). This translates to the function being \( e^{-2(t-1)} \) for \( t \geq 1 \) and zero otherwise.
05
Sketch the Graph
Draw the graph of \( f(t) = e^{-2(t-1)} u(t-1) \). Start at \( t = 1 \) where the function initiates, as \( u(t-1) \) is zero for \( t < 1 \). The function \( e^{-2(t-1)} \) quickly decays as \( t \) increases beyond 1, starting from value 1 at \( t = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Time Shift Property
The Time Shift Property is a handy tool when working with inverse Laplace transforms, especially when dealing with functions that appear with an exponential term. A shift in time often appears in the form of an exponential factor like in our given problem. Here, with the function \( F(s) = \frac{e^{-s}}{s+2} \), the term \( e^{-s} \) implies a shift in time.
This property can simplify the process by breaking it into recognizable parts.
This property can simplify the process by breaking it into recognizable parts.
- Suppose you have \( F(s) = e^{-cs} G(s) \).
- The time shift property tells us the inverse transform \( f(t) \) is \( u(t-c) g(t-c) \).
Heaviside Step Function
The Heaviside Step Function, often denoted as \( u(t) \), is a crucial piece in many engineering and physical applications. It acts as a switch that turns a function "on" or "off" at a certain point in time. In this example, the Heaviside function is used to represent our function starting at time \( t = 1 \).
Think of \( u(t) \) as a gate:
Think of \( u(t) \) as a gate:
- For \( t < c \), the output of \( u(t-c) \) is 0, thus \( f(t) = 0 \).
- Once \( t \geq c \), the gate opens, and \( u(t-c) = 1 \), reflecting the original function \( g(t-c) \).
Sketching Graphs
Sketching the graph of a function involves a visual translation of the mathematical expression into an understandable shape on the coordinate plane. Given \( f(t) = e^{-2(t-1)} u(t-1) \), there's a clear pattern to follow.
- Start with recognizing the shift at \( t = 1 \).
- Before \( t = 1 \), the function value is 0 due to the Heaviside step function.
- From \( t = 1 \) onwards, the function begins with a peak value of 1 (since \( e^0 = 1 \)).
- Pin the start at \( (1, 1) \).
- Imagine a rapidly decreasing curve from the start point as \( t \) increases.