Question

Find \(\mathcal{L}\\{f(t)\\}\) for each of the functions \(f\) defined. $$ f(t)=\left\\{\begin{array}{ll} 0, & 02 \end{array}\right. $$

Short answer

The Laplace transform of the given piecewise function is: \(\mathcal{L}\\{f(t)\\} = -\frac{e^{-2s}}{s^2} + \frac{1 - e^{-2s}}{s^3}\).

Step-by-step solution

Understanding the Laplace Transform formula

The Laplace Transform of a function \(f(t)\) is given by the formula: $$ \mathcal{L}\\{f(t)\\} = F(s) = \int_0^{\infty} e^{-st} f(t) dt $$ Where \(s\) is a complex variable and \(f(t)\) is the time-domain function.

Break the function into cases

The given function is a piecewise function, let's break it into two cases: 1) Case 1: \(0 < t < 2\), \(f(t) = 0\) 2) Case 2: \(t > 2\), \(f(t) = t\)

Calculate Laplace Transform for each case

We will now find the Laplace transform for each case separately and then combine them into a single result. 1) Case 1: \(0 < t < 2\), \(f(t) = 0\) When \(f(t) = 0\), the Laplace Transform becomes: $$ \mathcal{L}\\{0\\} = \int_0^{\infty} e^{-st} \cdot 0 dt $$ Since the integrand is 0, the integral becomes 0 as well: $$ \mathcal{L}\\{0\\} = 0 $$ 2) Case 2: \(t > 2\), \(f(t) = t\) In this case, we need to adjust the limits of the integral to the range for which the function is defined, i.e. from \(t=2\) to \(t=\infty\). This results in: $$ \mathcal{L}\\{t\\} = \int_2^{\infty} e^{-st} t dt $$

Integrate the second case using Integration by Parts

We will now integrate this expression using integration by parts. Let's set: $$ u = t, dv = e^{-st} dt $$ Then: $$ du = dt, v = -\frac{1}{s}e^{-st} $$ Now, apply integration by parts: $$ \int_2^{\infty} e^{-st} t dt = \left[-\frac{t}{s}e^{-st}\right]_{2}^{\infty} + \frac{1}{s}\int_2^{\infty}e^{-st} dt $$

Complete the Integration

Now, we will simplify the result and complete the integration: $$ \left[-\frac{t}{s}e^{-st}\right]_{2}^{\infty} + \frac{1}{s}\int_2^{\infty}e^{-st} dt = -\frac{e^{-2s}}{s^2} + \frac{1}{s^2}\int_0^{\infty}e^{-st} dt - \frac{1}{s^2}\int_0^{2}e^{-st} dt $$ Now, we need to find the remaining integral: $$ \frac{1}{s^2}\int_0^{\infty}e^{-st} dt - \frac{1}{s^2}\int_0^{2}e^{-st} dt = \frac{1}{s^2}\left(\int_0^{\infty}e^{-st} dt - \int_0^{2}e^{-st} dt\right) $$ Now, solve the integral and use linearity: $$ \frac{1}{s^2}\left(\frac{1}{s} - \frac{1}{s}(e^{-2s})\right) = \frac{1}{s^2}\left(\frac{1 - e^{-2s}}{s}\right) $$

Combine the cases

Now we will combine both cases of the piecewise function to find the overall Laplace Transform: $$ \mathcal{L}\\{f(t)\\} = -\frac{e^{-2s}}{s^2} + \frac{1}{s^2}\left(\frac{1 - e^{-2s}}{s}\right) $$ Therefore, the Laplace Transform of the given piecewise function is: $$ \mathcal{L}\\{f(t)\\} = -\frac{e^{-2s}}{s^2} + \frac{1 - e^{-2s}}{s^3} $$

Key concepts

Piecewise Functions

Piecewise functions are mathematical expressions defined by multiple sub-functions, each of which applies to a certain interval of the main function's domain. They can represent complex behaviors that change over different intervals of input values.

Piecewise functions often arise in real-world situations where a response or output undergoes sudden changes at specific threshold points. When applying the Laplace Transform to a piecewise function, such as in our exercise, the transformation must be applied to each piece separately. This requires careful consideration of the function's value within different ranges of time, t, as well as the modification of the integration limits to fit these ranges.

As seen in the solution, for the first interval from 0 to 2, the function is constantly zero. In the second interval, starting at t greater than 2, the function adopts a different rule, f(t) equals t. Each of these cases is treated independently before combining the results to get the full Laplace Transform of the piecewise function.

Integration by Parts

Integration by parts is a technique used to integrate the product of two functions. It is particularly handy when we are dealing with the product of a polynomial and an exponential function, as seen in the exercise.

The formula for integration by parts is derived from the product rule of differentiation and is given by: \[\int u dv = uv - \int v du\]
Where u and dv are functions of a variable (in our case, t). In our exercise, we chose to set u as t and dv as the exponential decay function, \(e^{-st} dt\). After differentiating and integrating u and dv respectively, the formula allowed us to transform the product into a form that can be integrated with respect to the integration limits imposed by the piecewise nature of the function.

Complex Variables

Complex variables are used in the Laplace Transform to extend the idea of a function beyond real number inputs. A complex variable, typically denoted by s in the context of Laplace Transforms, has both a real part and an imaginary part, written in the form s = \(\sigma + i\omega\).

In our exercise, s is a complex variable that serves as the Laplace Transform's domain. It is essential when performing the transformation because it helps determine the behavior of the function in the complex plane. When computing the Laplace Transform, we're often focused on the real part of s for convergence, while the imaginary part often relates to the frequency behavior of the transformed function.

Integration Limits

Integration limits define the range over which an integral is calculated. They are especially important in the context of piecewise functions and Laplace Transforms, as the integration limits must align with the intervals defined by the piecewise function.

In the exercise, after splitting the piecewise function into two cases, the integration limits required adjustment. For the constant zero part of the function, there is effectively no contribution to the integral, and for the linear part when \(t > 2\), the lower integration limit was changed from 0 to 2 to reflect the region where this second piece of the function applies. Such adjustments ensure that the contribution of each defined piece of the function is captured accurately during integration.