A signal f(t) has the Laplace transform F(s) = (3s + 19)/(s² + 11s + 28). 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) = (3s + 19)/(s² + 11s + 28)
f(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (−4s + 35)/(s² + 8s + 5). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (−4s + 35)/(s² + 8s + 5)
f′(0⁺) =
3.Medium
Find f(t) for F(s) = (6s + 30)/(s² + 8s + 16) and evaluate f(0.25).
F(s) = (6s + 30)/(s² + 8s + 16)t = 0.25 s
f(0.25) =
4.Medium
Find f(t) for F(s) = 34/(s² + 2s + 10) by completing the square, and evaluate f(0.5).
F(s) = 34/(s² + 2s + 10)t = 0.5 s
f(0.5) =
5.Hard
A system obeys y″ + 4y′ + 3y = 3·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(0.25).
y″ + 4y′ + 3y = 3·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.25 s
y(0.25) =
6.Hard
A system obeys y″ + 7y′ + 6y = 12·u(t), with y(0) = 3 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″ + 7y′ + 6y = 12·u(t)y(0) = 3y′(0) = 0u(t) = 5t = 0.2 s
y(0.2) =
Answer key
- 1. f(0⁺) = 3
- 2. f′(0⁺) = 67
- 3. f(0.25) = 2.759
- 4. f(0.5) = 6.857
- 5. y(0.25) = 0.06798
- 6. y(0.2) = 3.544
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