A signal f(t) has the Laplace transform F(s) = (−6s + 25)/(s² + 6s + 8). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
Laplace transforms worksheet with answers
1.Easy
F(s) = (−6s + 25)/(s² + 6s + 8)
f(0⁺) =
2.Easy
Find the inverse Laplace transform f(t) of F(s) = 8/(s + 2)² with a table of transform pairs, and evaluate f(1).
F(s) = 8/(s + 2)²t = 1 s
f(1) =
3.Medium
Find f(t) for F(s) = (6s + 38)/(s² + 8s + 16) and evaluate f(0.3).
F(s) = (6s + 38)/(s² + 8s + 16)t = 0.3 s
f(0.3) =
4.Medium
Find f(t) for F(s) = (4s + 11)/(s² + 8s + 16) and evaluate f(0.5).
F(s) = (4s + 11)/(s² + 8s + 16)t = 0.5 s
f(0.5) =
5.Hard
A system obeys y″ + 10y′ + 24y = 24·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 1 for t ≥ 0. Use the Laplace transform to find the output y(0.2).
y″ + 10y′ + 24y = 24·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.2 s
y(0.2) =
6.Hard
A system obeys y″ + 7y′ + 6y = 6·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 2 for t ≥ 0. Use the Laplace transform to find the output y(0.6).
y″ + 7y′ + 6y = 6·u(t)u(t) = 2y(0) = 0, y′(0) = 0t = 0.6 s
y(0.6) =
Answer key
- 1. f(0⁺) = −6
- 2. f(1) = 1.083
- 3. f(0.3) = 3.072
- 4. f(0.5) = 0.203
- 5. y(0.2) = 0.2544
- 6. y(0.6) = 0.6938
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