If f(t) = {E, 0 <= t <= a/2, -E, a/2 <= t <= a , where f(t+a) = f(t). Show that L[f(t)] = E/s tanh(as/4)

To show that the Laplace transform \( \mathcal{L}\{f(t)\} \) of the given periodic function \( f(t) \) is \( \frac{E}{s} \tanh\left(\frac{as}{4}\right) \), we will use the definition of the Laplace transform and properties of periodic functions.

Given:
\[ f(t) = \begin{cases} 
E, & 0 \le t < \frac{a}{2} \\
-E, & \frac{a}{2} \le t < a 
\end{cases} \]
and \( f(t+a) = f(t) \).

1. **Laplace Transform of a Periodic Function:**

   For a periodic function \( f(t) \) with period \( a \):
   \[
   \mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-as}} \int_0^a e^{-st} f(t) \, dt
   \]

2. **Compute the Integral:**
   
   We need to evaluate the integral:
   \[
   \int_0^a e^{-st} f(t) \, dt
   \]
   Given the piecewise nature of \( f(t) \), this integral can be split into two parts:
   \[
   \int_0^a e^{-st} f(t) \, dt = \int_0^{a/2} e^{-st} E \, dt + \int_{a/2}^a e^{-st} (-E) \, dt
   \]

3. **Evaluate Each Part:**

   \[
   \int_0^{a/2} e^{-st} E \, dt = E \int_0^{a/2} e^{-st} \, dt = E \left[ \frac{e^{-st}}{-s} \right]_0^{a/2} = E \left( \frac{1 - e^{-sa/2}}{s} \right)
   \]

   \[
   \int_{a/2}^a e^{-st} (-E) \, dt = -E \int_{a/2}^a e^{-st} \, dt = -E \left[ \frac{e^{-st}}{-s} \right]_{a/2}^a = -E \left( \frac{e^{-sa/2} - e^{-sa}}{s} \right)
   \]

4. **Combine the Results:**

   \[
   \int_0^a e^{-st} f(t) \, dt = E \left( \frac{1 - e^{-sa/2}}{s} \right) - E \left( \frac{e^{-sa/2} - e^{-sa}}{s} \right)
   \]

   Simplify:
   \[
   \int_0^a e^{-st} f(t) \, dt = E \left( \frac{1 - e^{-sa/2} - e^{-sa/2} + e^{-sa}}{s} \right)
   \]
   \[
   \int_0^a e^{-st} f(t) \, dt = E \left( \frac{1 - 2e^{-sa/2} + e^{-sa}}{s} \right)
   \]

5. **Simplify the Expression:**

   Notice that \( 1 - 2e^{-sa/2} + e^{-sa} = (1 - e^{-sa/2})^2 \), so:
   \[
   \int_0^a e^{-st} f(t) \, dt = E \left( \frac{(1 - e^{-sa/2})^2}{s} \right)
   \]

6. **Use the Periodic Function Property:**

   \[
   \mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-as}} \int_0^a e^{-st} f(t) \, dt
   \]
   Substitute the evaluated integral:
   \[
   \mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-as}} \left( \frac{E (1 - e^{-sa/2})^2}{s} \right)
   \]

7. **Final Simplification:**

   \[
   \mathcal{L}\{f(t)\} = \frac{E (1 - e^{-sa/2})^2}{s(1 - e^{-as})}
   \]

   Notice that \( 1 - e^{-as} = (1 - e^{-sa/2})(1 + e^{-sa/2}) \):
   \[
   \mathcal{L}\{f(t)\} = \frac{E (1 - e^{-sa/2})^2}{s(1 - e^{-sa/2})(1 + e^{-sa/2})}
   \]
   \[
   \mathcal{L}\{f(t)\} = \frac{E (1 - e^{-sa/2})}{s(1 + e^{-sa/2})}
   \]

   Recognize that:
   \[
   \frac{1 - e^{-sa/2}}{1 + e^{-sa/2}} = \tanh\left(\frac{sa}{4}\right)
   \]

8. **Final Result:**

   \[
   \mathcal{L}\{f(t)\} = \frac{E}{s} \tanh\left(\frac{sa}{4}\right)
   \]

Thus, we have shown that the Laplace transform of \( f(t) \) is:
\[
\mathcal{L}\{f(t)\} = \frac{E}{s} \tanh\left(\frac{as}{4}\right)
\]

Would you like any further details or explanations on this solution?

Here are 5 additional questions you might be interested in:

1. How do you find the Laplace transform of a piecewise function?
2. What are the properties of the Laplace transform that can simplify calculations?
3. Can you explain the significance of the hyperbolic tangent function in this context?
4. How does the periodicity of a function affect its Laplace transform?
5. Can you show an example of finding the inverse Laplace transform?

**Tip:** Always break down complex integrals into manageable parts and utilize properties of functions, such as periodicity, to simplify the calculations.

Learn how to compute the Laplace transform of the periodic function f(t) = {E, 0 <= t <= a/2, -E, a/2 <= t <= a}. This detailed solution covers the use of Laplace transform properties and periodic function characteristics, providing a step-by-step integration approach. Suitable for advanced undergraduate students studying integral calculus and periodic functions.

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