A signal f(t) has the Laplace transform F(s) = (2s² + 4s + 24)/(s(s + 5)(s + 7)). 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) = (2s² + 4s + 24)/(s(s + 5)(s + 7))
f(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (5s + 27)/(s² + 6s + 2). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (5s + 27)/(s² + 6s + 2)
f′(0⁺) =
3.Medium
Find f(t) for F(s) = (3s + 1)/(s² + 2s + 1) and evaluate f(1).
F(s) = (3s + 1)/(s² + 2s + 1)t = 1 s
f(1) =
4.Medium
Find f(t) for F(s) = (3s + 32)/(s² + 4s + 5) by completing the square, and evaluate f(0.1).
F(s) = (3s + 32)/(s² + 4s + 5)t = 0.1 s
f(0.1) =
5.Hard
A system obeys y″ + 5y′ + 6y = 24·u(t), with y(0) = 6 and y′(0) = 0. At t = 0 a step input u(t) = 5 is applied. Use the Laplace transform to find y(0.4).
y″ + 5y′ + 6y = 24·u(t)y(0) = 6y′(0) = 0u(t) = 5t = 0.4 s
y(0.4) =
6.Hard
Solve y″ + 7y′ + 6y = 0 with y(0) = −5 and y′(0) = −14 using the Laplace transform, and find y(0.2).
y″ + 7y′ + 6y = 0y(0) = −5y′(0) = −14t = 0.2 s
y(0.2) =
Answer key
- 1. f(0⁺) = 2
- 2. f′(0⁺) = −3
- 3. f(1) = 0.3679
- 4. f(0.1) = 4.569
- 5. y(0.4) = 9.562
- 6. y(0.2) = −6.06
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