Let's go through this problem step by step.

Part (a): Represent $f(t)$ using Heaviside step functions

The function $f(t)$ is defined in pieces:

1. $f(t) = 3$ for $t < 6$
2. $f(t) = t - 6$ for $6 \leq t < 7$
3. $f(t) = 2$ for $t \geq 7$

To express $f(t)$ using Heaviside functions, $h(t - c)$, we can set it up by noting the following:

• Start with the base value for each interval.
• Use Heaviside functions to "switch" the function at each discontinuity.

From the given solution: $f(t) = 3 - 3h(t - 6) + (t - 6)(h(t - 6) - h(t - 7)) + 2h(t - 7)$

Here's the breakdown:

1. $3$ is the value of $f(t)$ before $t = 6$.
2. $-3h(t - 6)$: At $t = 6$, subtracting $3$ sets up the change to the next interval.
3. $(t - 6)(h(t - 6) - h(t - 7))$: Adds $t - 6$ for $6 \leq t < 7$, due to the difference between $h(t - 6)$ and $h(t - 7)$, which is 1 in this range.
4. $2h(t - 7)$: Finally, adding 2 at $t = 7$ captures the constant value for $t \geq 7$.

Part (b): Finding the Laplace transform $F(s) = \mathcal{L}\{f(t)\}$

To find the Laplace transform of $f(t)$, we'll take the Laplace transform of each term separately.

1. Transform of $3$: $\mathcal{L}\{3\} = \frac{3}{s}$

2. Transform of $-3h(t - 6)$: $-3 \cdot \frac{e^{-6s}}{s}$

3. Transform of $(t - 6)(h(t - 6) - h(t - 7))$:

   • For $(t - 6) h(t - 6)$: Shifted by 6, so $\mathcal{L}\{(t - 6)h(t - 6)\} = \frac{e^{-6s}}{s^2}$
   • For $(t - 6) h(t - 7)$: Shifted by 7, so $\mathcal{L}\{(t - 6)h(t - 7)\} = \frac{e^{-7s}}{s^2}$
   • Therefore, the term becomes: $\frac{e^{-6s}}{s^2} - \frac{e^{-7s}}{s^2}$

4. Transform of $2h(t - 7)$: $2 \cdot \frac{e^{-7s}}{s}$

Combining these results gives: $F(s) = \frac{3}{s} - \frac{3e^{-6s}}{s} + \left( \frac{e^{-6s}}{s^2} - \frac{e^{-7s}}{s^2} \right) + \frac{2e^{-7s}}{s}$

Would you like further details or have any questions on specific parts of this solution?

Here are some related questions:

1. How do Heaviside step functions work in piecewise-defined functions?
2. What is the Laplace transform of a constant?
3. How can we apply shifting in the Laplace domain using Heaviside functions?
4. What is the significance of using $h(t - c)$ for modeling discontinuities?
5. How does the Laplace transform help in solving differential equations involving piecewise functions?

Tip: In Laplace transforms, the shift theorem $\mathcal{L}\{f(t - a)h(t - a)\} = e^{-as}F(s)$ is crucial for handling piecewise functions effectively.