To find the Laplace Transform of the given time function $f(t)$, let's break down the function based on the piecewise definition implied by the graph.

Step 1: Analyze the function

From the graph:

• For $0 \leq t < T$, $f(t) = \frac{t}{T}$ (This is a line with slope $\frac{1}{T}$).
• For $T \leq t < 2T$, $f(t) = 1$ (This is a constant function).
• For $t \geq 2T$, $f(t) = 0$.

So, the function $f(t)$ can be represented as:

\frac{t}{T}, & 0 \leq t < T \\ 1, & T \leq t < 2T \\ 0, & t \geq 2T \end{cases}$$ ### Step 2: Express $$f(t)$$ using unit step functions We can express this function in terms of unit step functions $$u(t)$$: $$f(t) = \left( \frac{t}{T} \right) u(t) - \left( \frac{t-T}{T} \right) u(t-T) + u(t-T) - u(t-2T)$$ Simplifying the expression: $$f(t) = \frac{t}{T} u(t) - \frac{t-T}{T} u(t-T) + u(t-T) - u(t-2T)$$ ### Step 3: Find the Laplace Transform of each term Using the Laplace Transform properties: 1. **$$\frac{t}{T} u(t)$$**: $$\mathcal{L}\left\{ \frac{t}{T} u(t) \right\} = \frac{1}{T} \cdot \frac{1}{s^2}$$ 2. **$$-\frac{t-T}{T} u(t-T)$$**: Use the time-shifting property: $$\mathcal{L}\{f(t-T)u(t-T)\} = e^{-Ts} \mathcal{L}\{f(t)\}$$. $$\mathcal{L}\left\{ -\frac{t-T}{T} u(t-T) \right\} = -\frac{1}{T} e^{-Ts} \cdot \frac{1}{s^2}$$ 3. **$$u(t-T)$$**: $$\mathcal{L}\left\{ u(t-T) \right\} = \frac{e^{-Ts}}{s}$$ 4. **$$-u(t-2T)$$**: $$\mathcal{L}\left\{ -u(t-2T) \right\} = -\frac{e^{-2Ts}}{s}$$ ### Step 4: Combine the results The total Laplace Transform is: \[ \mathcal{L}\