A signal f(t) has the Laplace transform F(s) = (8s + 17)/(s² + 8s + 12). 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) = (8s + 17)/(s² + 8s + 12)
f(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (−4s + 7)/(s² + 10s + 16). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
F(s) = (−4s + 7)/(s² + 10s + 16)
f(0⁺) =
3.Medium
Find f(t) for F(s) = (s + 34)/(s² + 8s + 25) by completing the square, and evaluate f(0.1).
F(s) = (s + 34)/(s² + 8s + 25)t = 0.1 s
f(0.1) =
4.Medium
Find f(t) for F(s) = (2s + 11)/(s² + 4s + 40) by completing the square, and evaluate f(0.3).
F(s) = (2s + 11)/(s² + 4s + 40)t = 0.3 s
f(0.3) =
5.Hard
Solve y″ + 8y′ + 15y = 0 with y(0) = −6 and y′(0) = −13 using the Laplace transform, and find y(0.3).
y″ + 8y′ + 15y = 0y(0) = −6y′(0) = −13t = 0.3 s
y(0.3) =
6.Hard
A system obeys y″ + 11y′ + 30y = 60·u(t), with y(0) = 1 and y′(0) = 0. At t = 0 a step input u(t) = 5 is applied. Use the Laplace transform to find y(0.2).
y″ + 11y′ + 30y = 60·u(t)y(0) = 1y′(0) = 0u(t) = 5t = 0.2 s
y(0.2) =
Answer key
- 1. f(0⁺) = 8
- 2. f(0⁺) = −4
- 3. f(0.1) = 2.621
- 4. f(0.3) = 0.3742
- 5. y(0.3) = −5.283
- 6. y(0.2) = 3.688
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