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 \(f(t)\).