The problem asks for the Laplace transform of the function $f(t)$, which is defined as:

$f(t) = \frac{t}{\tau}, \quad \text{when } 0 \leq t \leq \tau,$ and $f(t) = 1, \quad \text{when } t > \tau.$

Solution:

The Laplace transform of a function $f(t)$ is given by: $\mathcal{L}[f(t)] = F(s) = \int_0^\infty f(t)e^{-st} dt$

1. Step 1: Split the function into two parts:
   • For $0 \leq t \leq \tau$, $f(t) = \frac{t}{\tau}$
   • For $t > \tau$, $f(t) = 1$

Thus, the Laplace transform is computed as: $F(s) = \int_0^\tau \frac{t}{\tau} e^{-st} dt + \int_\tau^\infty 1 \cdot e^{-st} dt$

2. Step 2: Solve the first integral: $I_1 = \int_0^\tau \frac{t}{\tau} e^{-st} dt$ Let $k = \frac{1}{\tau}$, so $f(t) = k \cdot t$. The integral becomes: $I_1 = \frac{1}{\tau} \int_0^\tau t e^{-st} dt$ Use integration by parts ($u = t$, $dv = e^{-st} dt$): $\int t e^{-st} dt = \left[ -\frac{t}{s} e^{-st} \right]_0^\tau + \frac{1}{s} \int_0^\tau e^{-st} dt$ $= \left( -\frac{\tau}{s} e^{-s\tau} + \frac{0}{s} \right) + \frac{1}{s} \left[ -\frac{1}{s} e^{-st} \right]_0^\tau$ $= -\frac{\tau}{s} e^{-s\tau} + \frac{1}{s^2} \left( 1 - e^{-s\tau} \right)$

Thus: $I_1 = \frac{1}{\tau} \left( -\frac{\tau}{s} e^{-s\tau} + \frac{1}{s^2} (1 - e^{-s\tau}) \right)$

3. Step 3: Solve the second integral: $I_2 = \int_\tau^\infty e^{-st} dt$ $I_2 = \left[ -\frac{1}{s} e^{-st} \right]_\tau^\infty = 0 - \left( -\frac{1}{s} e^{-s\tau} \right)$ $I_2 = \frac{1}{s} e^{-s\tau}$

4. Step 4: Combine results: $F(s) = I_1 + I_2$ Substitute $I_1$ and $I_2$: $F(s) = \frac{1}{\tau} \left( -\frac{\tau}{s} e^{-s\tau} + \frac{1}{s^2} (1 - e^{-s\tau}) \right) + \frac{1}{s} e^{-s\tau}$

Simplify: $F(s) = \frac{-e^{-s\tau}}{s} + \frac{1 - e^{-s\tau}}{\tau s^2} + \frac{e^{-s\tau}}{s}$

Combine terms: $F(s) = \frac{1}{\tau s^2} (1 - e^{-s\tau})$

Final Answer:

$F(s) = \frac{1}{\tau s^2} (1 - e^{-s\tau}) + \frac{e^{-s\tau}}{s}$

Would you like a detailed explanation of any step or a further breakdown?

Related Questions:

1. How does the Laplace transform handle piecewise functions like this one?
2. Why do we use integration by parts for solving the first integral?
3. How can this solution be extended for different values of $f(t)$?
4. What is the physical significance of $\tau$ in this problem?
5. How does the exponential $e^{-st}$ affect convergence of the Laplace transform?

Tip:

Always double-check continuity and limits when working with piecewise functions in Laplace transforms.