Find the inverse Laplace transform f(t) of F(s) = 10s/(s² + 9) with a table of transform pairs, and evaluate f(0.25).
Laplace transforms worksheet with answers
1.Easy
F(s) = 10s/(s² + 9)t = 0.25 s
f(0.25) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (2s + 17)/(s² + 8s + 7). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
F(s) = (2s + 17)/(s² + 8s + 7)
f(0⁺) =
3.Medium
Find f(t) for F(s) = 28/(s² + 6s + 25) by completing the square, and evaluate f(0.1).
F(s) = 28/(s² + 6s + 25)t = 0.1 s
f(0.1) =
4.Medium
Find f(t) for F(s) = (4s + 36)/(s² + 6s + 9) and evaluate f(0.3).
F(s) = (4s + 36)/(s² + 6s + 9)t = 0.3 s
f(0.3) =
5.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) = 1 for t ≥ 0. Use the Laplace transform to find the output y(1.2).
y″ + 7y′ + 6y = 6·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 1.2 s
y(1.2) =
6.Hard
A system obeys y″ + 9y′ + 20y = 40·u(t), with y(0) = 2 and y′(0) = 0. At t = 0 a step input u(t) = 10 is applied. Use the Laplace transform to find y(0.25).
y″ + 9y′ + 20y = 40·u(t)y(0) = 2y′(0) = 0u(t) = 10t = 0.25 s
y(0.25) =
Answer key
- 1. f(0.25) = 7.317
- 2. f(0⁺) = 2
- 3. f(0.1) = 2.019
- 4. f(0.3) = 4.554
- 5. y(1.2) = 0.6387
- 6. y(0.25) = 7.519
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