Problem 18
Question
Find either \(F(s)\) or \(f(t)\), as indicated. $$ \mathscr{L}^{-1}\left\\{\frac{5 s}{(s-2)^{2}}\right\\} $$
Step-by-Step Solution
Verified Answer
The inverse Laplace transform is \( 5t e^{2t} \).
1Step 1: Identify the Inverse Laplace Transform Form
The given expression is \( \mathscr{L}^{-1}\left\{\frac{5s}{(s-2)^2}\right\} \). This expression needs to be matched to known inverse Laplace transform forms. Here, it resembles a function of the form \( \mathscr{L}^{-1}\{\frac{s-a}{(s-a)^2}\} \).
2Step 2: Use Shifting Theorem
Recognize that the expression \( \frac{5s}{(s-2)^2} \) suggests a first order derivative of an exponential function due to the \( (s-a) \) shift factor. This is aligned with the derivative of \( e^{at} \) in the time domain. Here, \( a = 2 \).
3Step 3: Apply the Derivative Property
The inverse Laplace transform of \( \frac{s-a}{(s-a)^2} \) is \( t e^{at} \). Therefore, for \( a = 2 \), the transform becomes \( t e^{2t} \).
4Step 4: Adjust for Coefficient 5
The presence of the coefficient 5 in the numerator means that the expression should be multiplied by 5. Thus, the transform becomes \( 5t e^{2t} \).
Key Concepts
Laplace transform propertiesshifting theoremderivative property
Laplace transform properties
Laplace transforms are powerful tools for solving differential equations and analyzing linear time-invariant systems. The core idea is to transform a function of time, such as \(f(t)\), into a function of a complex variable, \(s\), denoted as \(F(s)\). This process simplifies calculations, especially when dealing with linear systems.
The main properties of the Laplace transform that make it so effective include:
The main properties of the Laplace transform that make it so effective include:
- Linearity: For functions \(f(t)\) and \(g(t)\) and constants \(a\) and \(b\), the transform \(\mathscr{L}\{af(t) + bg(t)\} = aF(s) + bG(s)\).
- Time Shifting: If \(\mathscr{L}\{f(t)\} = F(s)\), then \(\mathscr{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)\), where \(u(t)\) is the unit step function.
- Frequency Shifting: The transform of \(e^{at}f(t)\) is \(F(s-a)\), useful for including exponential growth or decay effects in the analysis.
shifting theorem
The shifting theorem is one of the most crucial concepts in the context of Laplace transforms. It allows one to account for the effects of exponential functions, like \(e^{at}\), in the time domain.
For the given problem, the shifting theorem helps in translating the complex function \( \frac{s-a}{(s-a)^{2}} \) in the \(s\)-domain back into the time domain. Specifically, if you have a Laplace transform in the form of \(F(s-a)\), it signals the presence of an exponential function \(e^{at}\) in \(f(t)\).
Using this theorem, when we look at \( \frac{5s}{(s-2)^{2}} \), the fact that we have \((s-a)\) with \(a = 2\) gives us the hint that the corresponding time function involves \(e^{2t}\). This shifting property simplifies handling terms involving exponential growth or decay, making it easier to revert back to the time domain while dealing with periodic or decaying functions.
For the given problem, the shifting theorem helps in translating the complex function \( \frac{s-a}{(s-a)^{2}} \) in the \(s\)-domain back into the time domain. Specifically, if you have a Laplace transform in the form of \(F(s-a)\), it signals the presence of an exponential function \(e^{at}\) in \(f(t)\).
Using this theorem, when we look at \( \frac{5s}{(s-2)^{2}} \), the fact that we have \((s-a)\) with \(a = 2\) gives us the hint that the corresponding time function involves \(e^{2t}\). This shifting property simplifies handling terms involving exponential growth or decay, making it easier to revert back to the time domain while dealing with periodic or decaying functions.
derivative property
The derivative property of the Laplace transform involves the relationship between differentiation in the time domain and algebraic manipulation in the \(s\)-domain. This property is valuable when dealing with differential equations.
When applying the inverse Laplace transform to expressions involving derivatives, the property becomes essential. In our exercise, the expression \( \frac{5s}{(s-2)^{2}} \) suggests the use of the derivative property. Here, \( \frac{s-a}{(s-a)^{2}} \) resembles the derivative of a transformed function related to \(t \cdot f(t)\).
When applying the inverse Laplace transform to expressions involving derivatives, the property becomes essential. In our exercise, the expression \( \frac{5s}{(s-2)^{2}} \) suggests the use of the derivative property. Here, \( \frac{s-a}{(s-a)^{2}} \) resembles the derivative of a transformed function related to \(t \cdot f(t)\).
- Specifically, this indicates that the inverse transform relates to the function multiplied by time \(t\).
- This relation is crucial because it tells us that \( \frac{s-a}{(s-a)^{2}} \) in the \(s\)-domain corresponds to \(t \cdot e^{at}\) in the time domain. When \(a = 2\), this becomes \(t \cdot e^{2t}\).
Other exercises in this chapter
Problem 18
Fill in the blanks or answer true/false. $$ \mathscr{L}^{-1}\left\\{\frac{e^{-5 s}}{s^{2}}\right\\}= $$____
View solution Problem 18
In some instances the Laplace transform can be used to solve linear differential equations with variable monomial coefficients. Use Theorem 4.4.1 to reduce the
View solution Problem 19
In Problems, find either \(F(s)\) or \(f(t)\), as indicated. $$ \mathscr{L}^{-1}\left\\{\frac{2 s-1}{s^{2}(s+1)^{3}}\right\\} $$
View solution Problem 19
Fill in the blanks or answer true/false. $$ \mathscr{L}^{-1}\left\\{\frac{s+\pi}{s^{2}+\pi^{2}} e^{-s}\right\\}= $$____
View solution