Chapter 12: Problem 23
Find the Laplace transform for (a) and (b). a) \(f(t)=\frac{d}{d t}\left(e^{-a t} \sin \omega t\right)\) b) \(f(t)=\int_{0}^{t} e^{-a x} \cos \omega x d x\) c) Verify the results obtained in (a) and (b) by first carrying out the indicated mathematical operation and then finding the Laplace transform.
Short Answer
Step by step solution
Recognize the Form
Compute the Function and Its Transform
Apply Integration and Transform for Part (b)
Verification for (a) by Calculating
Verification for (b) by Calculating
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation in Laplace Transform
For any function, say \( f(t) \), its derivative is represented by \( f'(t) \). The Laplace transform of a derivative is given by the formula \( \mathcal{L}\{f'(t)\} = s \mathcal{L}\{f(t)\} - f(0) \). This formula accounts for the initial condition \( f(0) \), which is essential in engineering and physics to determine the system's behavior at the starting point.
When applying this to a function such as \( e^{-at} \sin \omega t \), it's essential first to find its Laplace transform before applying the differentiation property. This step-by-step process ensures that all changes in the function during the differentiation are accounted for in the transformed domain.
Integration and the Laplace Transform
When a function is expressed as an integral, for instance, \( \int_{0}^{t} e^{-ax} \cos \omega x \, dx \), it's essential to recognize this as potentially involving convolution operations. The Laplace property for integration simplifies this process using the formula: \( \mathcal{L}\{\int_0^t f(t) \, dt\} = \frac{1}{s}\mathcal{L}\{f(t)\} \).
This property allows for a concise translation by practically dividing the Laplace transform of the integrated function by \( s \), simplifying the analysis. Grasping this concept enables one to manage complex integral transformations efficiently.
Understanding Convolution in Laplace Transform
The convolution of two functions \( f(t) \) and \( g(t) \), denoted \( (f * g)(t) \), can be complex to compute directly. However, when using the Laplace transform, the computation simplifies dramatically. The Laplace transform of a convolution is given by \( \mathcal{L}\{(f * g)(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\} \). This property turns the convolution operation into a straightforward multiplication of the Laplace transforms of each function.
Recognizing when a function can be treated as a convolution is key to simplifying Laplace transform applications, enabling more efficient problem-solving.
Key Laplace Properties
Some fundamental properties are:
- Linearity: This allows the transformation of linear combinations of functions. It means \( \mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\} \).
- Time Shifting: Allows shifting the function in the time domain. If \( \mathcal{L}\{f(t)\} = F(s) \), then \( \mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) \) for \( a > 0 \).
- Frequency Shifting: Modifies functions by exponential multipliers in the frequency domain. If \( \mathcal{L}\{f(t)\} = F(s) \), then \( \mathcal{L}\{e^{bt}f(t)\} = F(s-b) \).
- Initial and Final Value Theorems: These properties help ascertain the behavior of functions as \( t \to 0 \) and \( t \to \infty \), providing insights into system stability.