Question

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

The Laplace transform of (a) is \( s \frac{\omega}{(s+a)^2 + \omega^2} \); (b) is \( \frac{1}{s} \cdot \frac{s+a}{(s+a)^2 + \omega^2} \). Verification involves calculating derivatives/integrals before transforming.

Step-by-step solution

Recognize the Form

For part (a), we are finding the Laplace transform of a derivative. The function is given by \( f(t) = \frac{d}{dt}(e^{-at} \sin \omega t) \). For derivatives, the Laplace transform \( \mathcal{L}\{f'(t)\} = s \mathcal{L}\{f(t)\} - f(0) \) can be used. Identify this form and note that we need to work out \( e^{-at} \sin \omega t \) first.

Compute the Function and Its Transform

Calculate \( f(t) = e^{-at} \sin \omega t \). Its Laplace transform is \( \frac{\omega}{(s+a)^2 + \omega^2} \). Since we need the derivative, we differentiate and calculate for \( f'(t) \), then apply the transformation rules. For \( f'(t) \), apply it directly: \( \mathcal{L}\{f'(t) \} = s \frac{\omega}{(s+a)^2 + \omega^2} + a\sin \omega t|_0 \) gives us the transform.

Apply Integration and Transform for Part (b)

For part (b), the function \( f(t) = \int_0^t e^{-ax} \cos \omega x \, dx \) needs to be derived. Recognize this as a convolution problem. Use the Laplace property for integration and transform using \( \mathcal{L}\{\int_0^t f(x) \, dx\} = \frac{1}{s} \mathcal{L}\{f(t)\} \). For \( e^{-ax} \cos \omega x \), the transform is \( \frac{s+a}{(s+a)^2 + \omega^2} \). Apply integration property: \( \frac{1}{s} \cdot \frac{s+a}{(s+a)^2 + \omega^2} \).

Verification for (a) by Calculating

Verify (a) by first differentiating \( e^{-at} \sin \omega t \) \[ f'(t) = -ae^{-at} \sin \omega t + e^{-at} \omega \cos \omega t \]. Then find the Laplace transform individually for the terms using standard transforms: \( -a \frac{\omega}{(s+a)^2 + \omega^2} \) and \( \frac{\omega^2}{(s+a)^2 + \omega^2} \). Combine using linearity.

Verification for (b) by Calculating

Verify (b) by carrying out the integration directly. Calculate an antiderivative of \( e^{-ax} \cos \omega x \). Check the result by taking its derivative. Apply the Laplace transform to the antiderivative and verify against the transform obtained through convolution.

Key concepts

Differentiation in Laplace Transform

Differentiation plays a crucial role when working with the Laplace transform, especially when the focus is on functions that are derivatives of another function. In essence, the Laplace transform of a derivative provides a way to analyze how changing a specific function influences the system it's modeling.
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

Integration in the context of the Laplace transform often appears in problems involving convolution or when dealing with definite integrals of functions. The transform link provides a direct way to transition from a time domain description to ample data in the Laplace domain.
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

Convolution is a pivotal idea when diving into the Laplace transform, mainly concerning two functions combined into one output. Convolution is essentially used to understand how one function modifies another, and it's particularly handy in signal processing and systems analysis.
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

The Laplace transform boasts several key properties that simplify the analysis of differential equations and system dynamics. Understanding these can greatly ease the transformation of time-domain problems into solvable algebraic equations in the frequency domain.
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.

Leveraging these properties simplifies the study of complex systems using the Laplace transform, offering a clear advantage in both theoretical and practical applications.