Problem 2 - Advanced Engineering Mathematics

Question

Use the definition of the Laplace transform to find \(\mathscr{L}\\{f(t)\\}\). $$ f(t)=\left\\{\begin{array}{lr} 0, & 0 \leq t<2 \\ 1, & 2 \leq t<4 \\ 0, & t \geq 4 \end{array}\right. $$

Step-by-Step Solution

Answer

The Laplace transform of \( f(t) \) is \( \frac{1}{s}(e^{-2s} - e^{-4s}) \).

Step 1: Understand the Function

The function \( f(t) \) is a piecewise function defined as 0 when \( 0 \leq t < 2 \), 1 when \( 2 \leq t < 4 \), and 0 again when \( t \geq 4 \). We need to find its Laplace transform, \( \mathscr{L}\{f(t)\} \).

Step 2: Definition of the Laplace Transform

The Laplace transform of a function \( f(t) \) is given by the integral \[ \mathscr{L}\{f(t)\} = \int_0^\infty f(t) e^{-st} \, dt\]where \( s \) is a complex number.

Step 3: Applying the Laplace Transform Definition

Break the integral into parts based on the intervals of \( f(t) \). \[\mathscr{L}\{f(t)\} = \int_0^2 0 \cdot e^{-st} \, dt + \int_2^4 1 \cdot e^{-st} \, dt + \int_4^{\infty} 0 \cdot e^{-st} \, dt\] The first and third integrals are zero because \( f(t) \) = 0 in those intervals. Focus on the second integral.

Step 4: Evaluate the Non-Zero Integral

Calculate the integral \[\int_2^4 e^{-st} \, dt\]Evaluate this integral using the antiderivative of \( e^{-st} \):\[-\frac{1}{s}e^{-st} \Bigg|_2^4 = -\frac{1}{s}(e^{-4s} - e^{-2s})\]

Step 5: Simplify the Result

Simplify the expression obtained from the integral:\[\mathscr{L}\{f(t)\} = \frac{1}{s}(e^{-2s} - e^{-4s})\]This is the final expression for the Laplace transform of \( f(t) \).

Key Concepts

Piecewise Functions

To work with piecewise functions effectively, it's crucial to understand that these are functions defined over multiple segments, each with its own rule. In our original exercise, the function \( f(t) \) is termed as a piecewise function because it behaves differently within specific intervals of time (\( t \)).
Here’s a simple breakdown:

• From \( 0 \leq t < 2 \), the function outputs 0.
• From \( 2 \leq t < 4 \), the function outputs 1.
• From \( t \geq 4 \), the function reverts back to 0.

To handle such functions in problems involving Laplace transforms, the process typically involves evaluating each segment separately.
This involves breaking down integrals into intervals corresponding to each piece of the function—this is particularly essential if you are transitioning between varying values.
This makes calculations manageable, allowing us to simplify complex problems into solvable parts.

Integral Calculus

Integral Calculus is a core component when calculating the Laplace transform of a function, as it employs integration over time. In the original exercise, the Laplace transform involves finding the integral of \( f(t) \) times \( e^{-st} \), the latter being a decaying exponential that influences the function based on a complex number \( s \).
Here's a step-by-step process:

• The Laplace transform \( \mathscr{L}\{f(t)\} = \int_0^\infty f(t) e^{-st} \, dt \) essentially translates the function from the time domain to the frequency domain.
• Piecewise functions require splitting this integral into multiple segments. This helps manage the pieces of \( f(t) \) that are non-zero.
• For our specific problem, the only non-zero integral is \( \int_2^4 e^{-st} \, dt \), which simplifies into the result after removing zero-contributing intervals.

Understanding integration, especially of exponential functions, is crucial in finding such transformations. Mastery over integral calculus rules, including using antiderivatives, is key to successfully applying these concepts.

Complex Numbers

Complex numbers, represented as \( s = a + bi \), play a fundamental role in the Laplace transform, particularly influencing the decay behavior of the function \( f(t) \). In the context of our exercise, the variable \( s \) helps to adjust how \( e^{-st} \) scales and impacts the exponential decay across different segments of \( f(t) \).
The influence of \( s \) varies based on different segments of \( f(t) \):

• In technical terms, \( e^{-st} \) is a critical component as \( s \) contains both real and imaginary parts that can affect frequency and stability of solutions.
• The transformation into the complex frequency domain often provides insights into characteristics of the original signal that aren't obvious in the time domain.
• In engineering and physics, this is pivotal for system analysis and control, allowing for deeper exploration of stability and dynamics.

Thus, complex numbers facilitate the breadth of transformations and analysis that Laplace allows, making them indispensable in understanding the behavior of transformed functions.