A signal f(t) has the Laplace transform F(s) = (−3s + 21)/(s² + 8s + 3). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (−3s + 21)/(s² + 8s + 3)
f′(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (−7s + 22)/(s² + 4s + 17). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (−7s + 22)/(s² + 4s + 17)
f′(0⁺) =
3.Medium
Find f(t) for F(s) = (6s + 38)/(s² + 8s + 16) and evaluate f(0.1).
F(s) = (6s + 38)/(s² + 8s + 16)t = 0.1 s
f(0.1) =
4.Medium
Find f(t) for F(s) = 45/(s(s + 1)(s + 5)) by partial fractions, and evaluate f(1).
F(s) = 45/(s(s + 1)(s + 5))t = 1 s
f(1) =
5.Hard
A system obeys y″ + 3y′ + 2y = 2·u(t), with y(0) = 2 and y′(0) = 0. At t = 0 a step input u(t) = 4 is applied. Use the Laplace transform to find y(0.25).
y″ + 3y′ + 2y = 2·u(t)y(0) = 2y′(0) = 0u(t) = 4t = 0.25 s
y(0.25) =
6.Hard
A system obeys y″ + 6y′ + 5y = 5·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 10 for t ≥ 0. Use the Laplace transform to find the output y(0.6).
y″ + 6y′ + 5y = 5·u(t)u(t) = 10y(0) = 0, y′(0) = 0t = 0.6 s
y(0.6) =
Answer key
- 1. f′(0⁺) = 45
- 2. f′(0⁺) = 50
- 3. f(0.1) = 4.96
- 4. f(1) = 4.877
- 5. y(0.25) = 2.098
- 6. y(0.6) = 3.264
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