Question

Find the Laplace transform of the given function. $$ f(t)=\left\\{\begin{array}{ll}{0,} & {t<1} \\ {t^{2}-2 t+2,} & {t \geq 1}\end{array}\right. $$

Short answer

Answer: The Laplace Transform of the given function f(t) is L{f(t)} = e^(-s)[(2/s^2) - (2/s^3) + (1/s) + (1/s^2)].

Step-by-step solution

Break down the function into two separate Laplace transforms

For the given function f(t), we have two cases to consider. Find the Laplace transform for each case separately. For t < 1: f(t) = 0 For t >= 1: f(t) = t^2 - 2t + 2

Find the Laplace Transform for f(t) = 0 when t < 1

For the first case, where f(t) = 0, the integral becomes: $$ L\{0\} = \int_0^\infty e^{-st} (0) dt = 0 $$

Find the Laplace Transform for f(t) = t^2 - 2t + 2 when t >= 1

For the second case, where f(t) = t^2 - 2t + 2, we must multiply the function by a unit step function u(t - 1) in order to shift the function to the right by one unit. This gives us: $$ g(t) = (t^2 - 2t + 2)u(t - 1) $$ Now compute the Laplace Transform using the integration formula: $$ L\{g(t)\} = \int_1^\infty e^{-st} (t^2 - 2t + 2) dt $$ To simplify this integral, we can use tabular integration by parts. After applying integration by parts twice, we will get the following expression: $$ L\{g(t)\} = e^{-s}\left[(\frac{2}{s^2}-\frac{2}{s^3}+\frac{1}{s}+\frac{1}{s^2})\right] $$

Combine the two Laplace transforms

Now, we combine the Laplace Transforms for both cases to obtain the final expression for the Laplace Transform of the given function f(t): $$ L\{f(t)\} = L\{0\} + L\{g(t)\} = 0 + e^{-s}\left[(\frac{2}{s^2}-\frac{2}{s^3}+\frac{1}{s}+\frac{1}{s^2})\right] $$ The Laplace Transform of the given function f(t) is: $$ L\{f(t)\} = e^{-s}\left[(\frac{2}{s^2}-\frac{2}{s^3}+\frac{1}{s}+\frac{1}{s^2})\right] $$

Key concepts

Piecewise Functions

In mathematics, functions can be defined on different sections of their domain. These are known as piecewise functions. They allow us to handle different behavior in specific intervals.
For the exercise, the function \( f(t) \) is defined piecewise:

• When \( t < 1 \), the function is zero \( f(t) = 0 \).
• When \( t \geq 1 \), the function is \( f(t) = t^2 - 2t + 2 \).

This means the function changes its rule or expression depending on the value of \( t \). Analyzing each piece independently is essential for solving the problem, especially when applying operations like the Laplace Transform. Understanding this can help break down complex mathematical expressions into simpler, manageable parts, which is crucial for solving many mathematical problems.

Unit Step Function

The unit step function, often denoted as \( u(t) \), is a simple but powerful mathematical tool. It helps to shift functions and is very useful when dealing with Laplace Transforms. For a given threshold \( a \), the unit step function is defined as follows:

• \( u(t-a) = 0 \) if \( t < a \)
• \( u(t-a) = 1 \) if \( t \geq a \)

In the exercise, the function \( g(t) = (t^2 - 2t + 2)u(t - 1) \) uses the unit step function \( u(t - 1) \) to change its value starting at \( t = 1 \).
This effectively shifts the original function \( t^2 - 2t + 2 \) to the right by one unit, making it active only from \( t = 1 \) onward. The unit step function is an essential part of modeling and solving problems involving abrupt changes or shifts in a system's behavior.

Integration by Parts

Integration by parts is a technique from calculus used to integrate products of functions. It is based on the product rule for differentiation and is vital for solving integral expressions that arise in Laplace Transforms.
The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]To apply integration by parts, you choose which part of the integral to set as \( u \) and \( dv \). In our exercise, solving the integral of \( e^{-st}(t^2 - 2t + 2) \) required using this method twice.

• First, you choose \( u \) and \( dv \) wisely to simplify the integration.
• Use the formula to differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \).

Repeat the process if needed, until the integrals are manageable. Mastering this method is key to tackling complex integrals efficiently in calculus.

Laplace Transform Properties

The Laplace Transform is a crucial tool in engineering and physics. It transforms a time-domain function into a complex frequency-domain representation, simplifying the analysis of systems. For beginners, it's essential to understand some of its core properties:

• Linearity: The Laplace Transform of a sum is the sum of individual transforms.
• Shifting: If \( L\{f(t)\} = F(s) \), then \( L\{f(t-a)u(t-a)\} = e^{-as}F(s) \), which helps manage piecewise functions with unit step functions.
• Scaling: If you scale the time variable, \( L\{f(at)\} \) becomes \( \frac{1}{a}F(\frac{s}{a}) \).

In our exercise, these properties allow us to manage the function's breaking point at \( t = 1 \), apply the correct transforms, and combine them efficiently. Understanding and utilizing these properties are invaluable for solving complex differential equations and analyzing system behavior across various applications.