Chapter 8: Problem 13
Use the Laplace transform table to find \(f(t)=\int_{0}^{t} e^{-\tau} \sin (t-\tau) d \tau .\) Hint: In \(L 34\) let \(g(t)=e^{-t}\) and \(h(t)=\sin t,\) and find \(G(p) H(p)\) which is the Laplace transform of the integral you want. Break the result into partial fractions and look up the inverse transforms.
Short Answer
Expert verified
The function is \(f(t) = \frac{1}{2} e^{-t} + \frac{1}{2} \cos(t) - \frac{1}{2} \sin(t)\).
Step by step solution
01
- Identify the functions
Let the functions be: \(g(t) = e^{-t}\) and \(h(t) = \sin(t)\).
02
- Determine the Laplace Transforms
Use the Laplace transform table to find the transforms of the identified functions. For \(g(t) = e^{-t}\), the Laplace transform is \[G(p) = \frac{1}{p+1}\].For \(h(t) = \sin(t)\), the Laplace transform is \[H(p) = \frac{1}{p^2+1}\].
03
- Find the product of the Laplace Transforms
Multiply the Laplace transforms of \(g(t)\) and \(h(t)\):\[G(p)H(p) = \frac{1}{p+1} \cdot \frac{1}{p^2+1} = \frac{1}{(p+1)(p^2+1)}\].
04
- Perform Partial Fraction Decomposition
To simplify \(G(p)H(p)\), perform partial fraction decomposition:\[\frac{1}{(p+1)(p^2+1)} = \frac{A}{p+1} + \frac{Bp+C}{p^2+1}\].Solve for constants \(A, B,\) and \(C\) by equating coefficients:\[1 = A(p^2 + 1) + (Bp + C)(p + 1)\].
05
- Determine constants A, B and C
Solving the equation, set specific values for \(p\):\[1 = A(p^2 + 1) + (Bp + C)(p + 1)\].For \(p = -1\):\[1 = A((-1)^2+1)\implies A = \frac{1}{2}\].Substitute \(A\) back into the equation and solve for \(B\) and \(C\):\[1 = \frac{p^2 + 1}{2} + (Bp + C)(p + 1)\implies B = \frac{1}{2}, C = \frac{-1}{2}\].
06
- Final Partial Fraction Form
Substitute \(A, B,\) and \(C\) back to get:\[\frac{1}{(p+1)(p^2+1)} = \frac{1}{2}\left(\frac{1}{p+1}\right) + \frac{1}{2}\left(\frac{p}{p^2+1}\right) - \frac{1}{2}\left(\frac{1}{p^2+1}\right)\].
07
- Find the Inverse Laplace Transforms
Look up the inverse Laplace transforms from the table:\[\mathcal{L}^{-1}\left(\frac{1}{p+1}\right) = e^{-t},\mathcal{L}^{-1}\left(\frac{p}{p^2+1}\right) = \cos(t),\mathcal{L}^{-1}\left(\frac{1}{p^2+1}\right) = \sin(t)\].Thus, the inverse transform of \(G(p)H(p)\) is\[f(t) = \frac{1}{2} e^{-t} + \frac{1}{2} \cos(t) - \frac{1}{2} \sin(t)\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Laplace Transform Table
The Laplace Transform Table is a powerful tool for solving differential equations and integrals. It contains a list of functions and their corresponding Laplace transforms, making it easier to convert time-domain functions into the s-domain. This table simplifies the process of finding Laplace transforms and inverse transforms for commonly encountered functions. For instance, in our exercise, we used the Laplace transform table to find that the Laplace transform of \(e^{-t}\) is \[ \frac{1}{p+1} \]. Similarly, the transform of \( \sin(t)\) is \[ \frac{1}{p^2+1} \]. By referring to the table, we can quickly convert these time-domain functions into the s-domain, facilitating easier manipulation and solution of the problem.
Inverse Laplace Transform
Inverse Laplace Transform is the process of converting a function from the s-domain back to the time domain. This is crucial for obtaining the final time-domain expression we need after performing operations in the s-domain. In our example, after simplifying the expression \[ \frac{1}{(p+1)(p^2+1)} \], we needed to find the inverse Laplace transforms of the resulting partial fractions. We used the table to get:\( \mathcal{L}^{-1}\left(\frac{1}{p+1}\right) = e^{-t} \), \( \mathcal{L}^{-1}\left(\frac{p}{p^2+1}\right) = \cos(t) \), and \( \mathcal{L}^{-1}\left(\frac{1}{p^2+1}\right) = \sin(t) \). Combining these inverse transforms provided us with the final answer: \[ f(t) = \frac{1}{2} e^{-t} + \frac{1}{2} \cos(t) - \frac{1}{2} \sin(t) \].
Partial Fraction Decomposition
Partial Fraction Decomposition is a method used to break down complex rational expressions into simpler fractions, which are easier to invert. In our exercise, after finding the product of the Laplace transforms \[ G(p)H(p) = \frac{1}{(p+1)(p^2+1)} \], we performed partial fraction decomposition to simplify this expression. The goal was to express it as a sum of simpler fractions: \[ \frac{1}{(p+1)(p^2+1)} = \frac{A}{p+1} + \frac{Bp+C}{p^2+1} \]. Solving for the constants \( A \), \( B \), and \( C \) involved equating coefficients and substituting specific values for \( p \) to isolate the variables. We determined that \( A = \frac{1}{2} \), \( B = \frac{1}{2} \), and \( C = -\frac{1}{2} \). This method facilitated the next step of finding the inverse Laplace transforms.
Complex Functions
Complex Functions often arise in Laplace transform problems, especially when dealing with oscillatory and exponential behaviors combined. In this exercise, we dealt with the Laplace transforms of \( e^{-t} \) and \( \sin(t) \) functions. These involve complex numbers when transformed into the s-domain. The product of their Laplace transforms gave us the expression \[ \frac{1}{(p+1)(p^2+1)} \], which needed to be simplified using partial fraction decomposition. Handling these complex functions requires understanding both their time-domain behavior and s-domain representations. By using the properties of these complex functions and their transforms, we could eventually find the simplified expression and its inverse Laplace transform, resulting in the final solution.