Question

4. Consider a function shown below: 2 f(t) 1 O 1 2 3 4 t 0 \quad 0 \le t < 1 f(t) = \begin{cases} t - 1 \quad 1 \le t < 3 \\ 1 \quad t \ge 3 \end{cases} (a) Obtain the Laplace transformation of the function given (HINT: when \mathcal{L}[f(t)] = F(s), the Laplace transform of the function with a time shift of T is given by \mathcal{L}[f(t - T) \cdot 1(t - T)] = F(s)e^{-sT}). (b) Continue (a). Apply the Initial Value Theorem to F(s) and confirm f(0) = 0. (c) Continue (a). Apply the Final Value Theorem to F(s) and confirm f(\infty) = 1. HINT: You may use L'Hospital's rule: \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}.

          4. Consider a function shown below:

2
f(t)
1
O
1
2
3
4
t
0 \quad 0 \le t < 1
f(t) = \begin{cases} t - 1 \quad 1 \le t < 3 \\ 1 \quad t \ge 3 \end{cases}
(a) Obtain the Laplace transformation of the function given (HINT: when \mathcal{L}[f(t)] = F(s),
the Laplace transform of the function with a time shift of T is given by
\mathcal{L}[f(t - T) \cdot 1(t - T)] = F(s)e^{-sT}).
(b) Continue (a). Apply the Initial Value Theorem to  F(s) and confirm f(0) = 0.
(c) Continue (a). Apply the Final Value Theorem to F(s) and confirm f(\infty) = 1. HINT:
You may use L'Hospital's rule: \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}.

4. Consider a function shown below:

2
f(t)
1
O
1
2
3
4
t
0 0 ≤t < 1
f(t) = t - 1 1 ≤t < 3
1 t ≥3
(a) Obtain the Laplace transformation of the function given (HINT: when ℒ[f(t)] = F(s),
the Laplace transform of the function with a time shift of T is given by
ℒ[f(t - T) ·1(t - T)] = F(s)e^-sT).
(b) Continue (a). Apply the Initial Value Theorem to F(s) and confirm f(0) = 0.
(c) Continue (a). Apply the Final Value Theorem to F(s) and confirm f(∞) = 1. HINT:
You may use L'Hospital's rule: limx →c (f(x))/(g(x)) = limx →c (f'(x))/(g'(x)).

Added by Linda V.

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