Question

Find the Laplace transform of the given function. $$ f(t)=t-u_{1}(t)(t-1), \quad t \geq 0 $$

Short answer

The Laplace transform of the function f(t) is given by: $$ \mathcal{L}\{f(t)\} = \frac{1}{s^2} - e^{-1s} \left(\frac{1}{s^2} - \frac{1}{s}\right) $$

Step-by-step solution

Break down the function

The given function can be written as: $$ f(t) = t - u_1(t)(t-1), \quad t \geq 0 $$ This function can be split into two parts: \(g(t) = t\) and \(h(t) = u_1(t)(t-1)\). We will find the Laplace transforms of each of these components and then combine the results to find the Laplace transform of \(f(t)\).

Find the Laplace transform of \(g(t)\)

First, consider the first component \(g(t) = t\). The Laplace transform of this function can be found using the formula: $$ \mathcal{L}\{g(t)\} = \int_0^\infty e^{-st} g(t) dt $$ For \(g(t) = t\), this becomes: $$ \mathcal{L}\{t\} = \int_0^\infty e^{-st} t dt $$ This integral can be solved using integration by parts, with the result: $$ \mathcal{L}\{t\} = \frac{1}{s^2} $$

Find the Laplace transform of \(h(t)\)

Next, consider the second component \(h(t) = u_1(t)(t-1)\). Since this is a delayed function, we need to use the shifting property of the Laplace transform. The general shifting property is: $$ \mathcal{L}\{u_c(t)f(t-c)\} = e^{-cs}F(s) $$ For our function, we have \(c = 1\) and \(f(t-1) = t - 1\). Therefore, we need to find the Laplace transform of \(t - 1\). This is similar to the Laplace transform of \(t\) that we found in Step 2. Using the same integration by parts technique, we get: $$ \mathcal{L}\{t - 1\} = \frac{1}{s^2} - \frac{1}{s} $$ Now, we can use the shifting property to find the Laplace transform of \(h(t)\): $$ \mathcal{L}\{u_1(t)(t-1)\} = e^{-1s} \left(\frac{1}{s^2} - \frac{1}{s}\right) $$

Combine the Laplace transforms of \(g(t)\) and \(h(t)\)

Finally, we can combine the Laplace transforms of \(g(t)\) and \(h(t)\) to find the Laplace transform of \(f(t)\). Since \(f(t) = g(t) - h(t)\), the Laplace transform of \(f(t)\) is given by: $$ \mathcal{L}\{f(t)\} = \mathcal{L}\{g(t)\} - \mathcal{L}\{h(t)\} $$ Substituting the results from Steps 2 and 3, we have: $$ \mathcal{L}\{f(t)\} = \frac{1}{s^2} - e^{-1s} \left(\frac{1}{s^2} - \frac{1}{s}\right) $$ This is the Laplace transform of the given function \(f(t)\).

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are essential in modeling the behavior of systems and phenomena in various fields such as physics, engineering, and economics. In these equations, the rate of change of a quantity depends on its current state.
By solving differential equations, we often determine functions that describe physical processes, such as heat flow or motion dynamics. The Laplace transform is a powerful tool that helps solve linear differential equations by transforming them from the time domain into an algebraic equation in the frequency domain.
This transform simplifies complex problems by converting differentiation into simple algebraic manipulation. Once solved, we can transform the solution back to the time domain to interpret the results.

Step Function

The step function, often denoted as the Heaviside step function or unit step function, is a simple yet powerful tool in signal processing and control systems. It is defined as a function that jumps from zero to one at a specified point. In mathematical terms, the step function at point \(c\) is written as \(u_c(t)\), which is zero for \(t < c\) and one for \(t \geq c\).
This function is particularly useful for modeling sudden changes or switches in systems, such as the turning on or off of a signal. The Laplace transform of the step function helps in analyzing and designing systems that undergo abrupt changes.

• For example, in the exercise, the function \(u_1(t)(t-1)\) involves a step function at \(t=1\), indicating the system starts behaving differently after this time.
• This is vital to detect and handle time delays in system responses.

Integration by Parts

Integration by parts is a mathematical technique used to solve integrals by breaking them down into simpler parts. It is especially useful for integrands that are a product of two functions. 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 parts of the original integrand.
This method is particularly effective when the derivative of one part and the integral of another leads to a simpler expression. For example, to find the Laplace transform of the time function \(t\), we use integration by parts to manage the integral's complexity. By setting \(u = t\) and \(dv = e^{-st} dt\), the resulting integral is more manageable:

• After applying, it substantially reduces the difficulty in handling time-domain functions with polynomial elements.
• The Laplace transform of \(t\) thus becomes \(\frac{1}{s^2}\), simplifying the analysis in the frequency domain.

Shifting Property

The shifting property of the Laplace transform allows us to handle functions that are delayed in time. When a function starts at \(t = c\) rather than \(t = 0\), the shifting property helps transform it into a more straightforward algebraic form. The general formula for the shifting property is:\[\mathcal{L}\{u_c(t)f(t-c)\} = e^{-cs}F(s)\]This converts a time delay of \(c\) into an exponential term in the frequency domain.
The exercise uses this shifting property extensively for \(h(t) = u_1(t)(t-1)\). Here, \(c = 1\) indicates the function is delayed by one unit of time. By applying the shifting property, we multiply the usual Laplace transform by \(e^{-1s}\), which reflects this delay in the frequency spectrum.
The property simplifies calculations for systems with time-dependent behaviors, making it easier to handle delays and extract desired outcomes efficiently.