A signal f(t) has the Laplace transform F(s) = (6s + 4)/(s² + 3s + 8). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (6s + 4)/(s² + 3s + 8)
f′(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (−9s + 40)/(s² + 3s + 2). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
F(s) = (−9s + 40)/(s² + 3s + 2)
f(0⁺) =
3.Medium
Find f(t) for F(s) = (2s + 26)/(s² + 8s + 52) by completing the square, and evaluate f(0.2).
F(s) = (2s + 26)/(s² + 8s + 52)t = 0.2 s
f(0.2) =
4.Medium
Find f(t) for F(s) = (4s + 6)/(s² + 4s + 4) and evaluate f(0.5).
F(s) = (4s + 6)/(s² + 4s + 4)t = 0.5 s
f(0.5) =
5.Hard
A system obeys y″ + 7y′ + 12y = 48·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 5 for t ≥ 0. Use the Laplace transform to find the output y(0.1).
y″ + 7y′ + 12y = 48·u(t)u(t) = 5y(0) = 0, y′(0) = 0t = 0.1 s
y(0.1) =
6.Hard
A system obeys y″ + 7y′ + 10y = 30·u(t), with y(0) = 4 and y′(0) = 0. At t = 0 a step input u(t) = 2 is applied. Use the Laplace transform to find y(0.25).
y″ + 7y′ + 10y = 30·u(t)y(0) = 4y′(0) = 0u(t) = 2t = 0.25 s
y(0.25) =
Answer key
- 1. f′(0⁺) = −14
- 2. f(0⁺) = −9
- 3. f(0.2) = 1.582
- 4. f(0.5) = 1.104
- 5. y(0.1) = 0.9537
- 6. y(0.25) = 4.36
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