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Find the laplace transform of the following time function given that the final answer is 1/s^2 - e-Ts/s2 + (-T*e^-2Ts)/s

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:

\[
f(t) = \begin{cases}
\frac{t}{T}, & 0 \leq t < T \\
1, & T \leq t < 2T \\
0, & t \geq 2T
\end{cases}
\]

### 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}\

Discover how to find the Laplace Transform of a piecewise defined function step-by-step. This problem involves analyzing piecewise functions, unit step functions, and applying the time-shifting property. Suitable for advanced undergraduate and graduate students.

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