Question

From a table of integrals, $$ \begin{aligned} &\int e^{\alpha u} \sin \beta u d u=e^{\alpha u} \frac{\alpha \sin \beta u-\beta \cos \beta u}{\alpha^{2}+\beta^{2}} \\ &\int e^{\alpha u} \cos \beta u d u=e^{\alpha u} \frac{\alpha \cos \beta u+\beta \sin \beta u}{\alpha^{2}+\beta^{2}} . \end{aligned} $$ Use these integrals to find the Laplace transform of \(f(t)\), if it exists. If the Laplace transform exists, give the domain of \(F(s)\). $$ f(t)=e^{-2 t} \cos 4 t $$

Short answer

Question: Determine the Laplace transform of the function \(f(t) = e^{-2t}\cos(4t)\) and find its domain. Answer: The Laplace transform of the function is \(F(s) = \frac{-2}{(s+2)^{2}+16}\), and its domain is \(s > -2\).

Step-by-step solution

Identify the parts of the problem

We are given \(f(t) = e^{-2t}\cos(4t)\). In our integral formula, we will need \(\alpha\) and \(\beta\). Comparing \(f(t)\) with the given formula, we can see that \(\alpha=-2\) and \(\beta=4\).

Apply the definition of Laplace transform

We want to find the Laplace transform of \(f(t)\), \(F(s) = L\{f(t)\} = \int_0^\infty e^{-st}f(t)dt\). It's important to notice that we have \(f(t) = e^{\alpha t}\cos(\beta t)\), with \(\alpha=-2\) and \(\beta=4\).

Use the given integral formula and adjust for our problem

The given integral formula for \(\int e^{\alpha u} \cos \beta u du\) is \(e^{\alpha u} \frac{\alpha \cos \beta u + \beta \sin \beta u}{\alpha^{2}+\beta^{2}}\). We will adjust this formula for our problem by replacing \(\alpha=-2\) and \(\beta=4\) and replacing \(u\) with \(t\): \(e^{-2t} \frac{-2\cos(4t) + 4\sin(4t)}{(-2)^{2}+4^{2}}\).

Compute the Laplace transform

Now we compute the Laplace transform by applying the definition: \(F(s) = \int_0^\infty e^{-st} e^{-2t} \frac{-2\cos(4t) + 4\sin(4t)}{20} dt\). We can simplify the integral by combining the terms with \(t\) in the exponent: \(F(s) = \int_0^\infty e^{-(s+2)t} \frac{-2\cos(4t) + 4\sin(4t)}{20} dt\). Now, we can use the integral formula for \(\int e^{\alpha u} \sin \beta u du\) by adjusting it to fit this integral. The Laplace transform becomes: \(F(s) = e^{-(s+2)t} \frac{(-2)\cos(4t) + (s+2)\sin(4t)}{(s+2)^{2}+4^{2}} \Big|_0^\infty \).

Evaluate the integral and find the domain of F(s)

We now evaluate the integral at the limits of integration: \(F(s) = \lim_{t\to\infty} [e^{-(s+2)t} \frac{(-2)\cos(4t) + (s+2)\sin(4t)}{(s+2)^{2}+4^{2}}] - [e^{-(s+2)(0)} \frac{(-2)\cos(4(0)) + (s+2)\sin(4(0))}{(s+2)^{2}+4^{2}}]\). Since the limit as \(t\) approaches infinity exists and is zero for \(s > -2\), the Laplace transform exists: \(F(s) = \frac{-2\cos(0) + (s+2)\sin(0)}{(s+2)^{2}+4^{2}}\), \(F(s) = \frac{-2}{(s+2)^{2}+16}\). The domain of \(F(s)\) is given by \(s>-2\), therefore: \(F(s) = \frac{-2}{(s+2)^{2}+16}, \quad s> -2\).

Key concepts

Laplace Transform of Trigonometric Functions

Understanding the Laplace transform of trigonometric functions is crucial for solving differential equations and analyzing systems in engineering and physics. The Laplace transform converts a trigonometric function in the time domain into a more manageable algebraic form in the frequency domain.

For instance, the Laplace transforms of sine and cosine functions with an exponential factor are given by integral formulas that involve the parameters \( \alpha \) and \( \beta \) representing the exponential and trigonometric components, respectively. The core idea is to express the trigonometric function as a part of an integrable expression that incorporates the exponential decay or growth dictated by the \( e^{\alpha t} \) term.

To apply these formulas to a specific function like \( f(t) = e^{-2t}\cos(4t) \) requires replacing \( \alpha \) and \( \beta \) with the corresponding values from \( f(t) \) and integrating with respect to time \( t \) from 0 to infinity; this process essentially 'filters out' the trigonometric part, leaving us with an algebraic function \( F(s) \) in terms of the complex frequency variable \( s \) for further analysis.

Integral Formulas

Integral formulas are indispensable tools for transforming complex math expressions into more tractable forms. In context of the Laplace transform, such formulas are particularly important as they allow us to integrate functions over the entire range of \( t \) from 0 to infinity.

For the given exercise, the integral formulas provided in the problem statement are fundamental for calculating the Laplace transform of an exponentially decaying cosine function. The process involves identifying \( \alpha \) and \( \beta \) and substituting them into the formulas to perform the integration necessary for the transform. In simpler terms, these formulas provide a 'shortcut' for integration by presenting a pre-calculated general result that can be customized to fit a specific function. This simplifies the process of finding the Laplace transform to mere substitution and algebraic simplification, rather than performing complex integration by parts or other lengthy integration techniques.

Domain of Laplace Transform Functions

The domain of Laplace transform functions is an essential aspect to consider, since it indicates the range of values for which the transform exists and is valid. Specifically, for the Laplace transform to exist, the integral of \( e^{-st}f(t) \) must converge, which occurs when \( s \) is greater than a certain boundary value often related to the growth rate of \( f(t) \).

In the example of \( f(t) = e^{-2t}\cos(4t) \) , we calculate the domain to be \( s > -2 \) by analyzing the point at which the exponent in \( e^{-(s+2)t} \) will ensure convergence of the integral. When \( s \) is greater than \( -2 \) , the \( e^{-(s+2)t} \) term will decay to zero as \( t \) approaches infinity, ensuring that the integral does not diverge. This analysis of the domain is critical to guaranteeing that the resulting function \( F(s) \) is meaningful and applicable within the context for which it is intended, whether that be in signal processing, control theory, or other areas of applied mathematics and engineering.