A signal f(t) has the Laplace transform F(s) = 36/(s(s² + 4s + 22)). Use the final value theorem to find the value f(∞) that f(t) settles to.
Laplace transforms worksheet with answers
1.Easy
F(s) = 36/(s(s² + 4s + 22))
f(∞) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (8s + 26)/(s² + 4s + 20). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (8s + 26)/(s² + 4s + 20)
f′(0⁺) =
3.Medium
Find f(t) for F(s) = (2s + 33)/(s² + 10s + 25) and evaluate f(0.2).
F(s) = (2s + 33)/(s² + 10s + 25)t = 0.2 s
f(0.2) =
4.Medium
Find f(t) for F(s) = (5s + 28)/(s² + 4s + 13) by completing the square, and evaluate f(0.6).
F(s) = (5s + 28)/(s² + 4s + 13)t = 0.6 s
f(0.6) =
5.Hard
A system obeys y″ + 5y′ + 6y = 6·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(0.5).
y″ + 5y′ + 6y = 6·u(t)y(0) = 4y′(0) = 0u(t) = 2t = 0.5 s
y(0.5) =
6.Hard
A system obeys y″ + 11y′ + 30y = 30·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″ + 11y′ + 30y = 30·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.2 s
y(0.2) =
Answer key
- 1. f(∞) = 1.636
- 2. f′(0⁺) = −6
- 3. f(0.2) = 2.428
- 4. f(0.6) = 1.418
- 5. y(0.5) = 3.315
- 6. y(0.2) = 0.2987
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