A signal f(t) has the Laplace transform F(s) = (s + 17)/(s² + 11s + 30). 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) = (s + 17)/(s² + 11s + 30)
f(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (2s + 37)/(s² + 4s + 25). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (2s + 37)/(s² + 4s + 25)
f′(0⁺) =
3.Medium
Find f(t) for F(s) = 40/(s(s + 2)(s + 6)) by partial fractions, and evaluate f(0.2).
F(s) = 40/(s(s + 2)(s + 6))t = 0.2 s
f(0.2) =
4.Medium
Find f(t) for F(s) = 40/(s(s + 4)(s + 5)) by partial fractions, and evaluate f(0.25).
F(s) = 40/(s(s + 4)(s + 5))t = 0.25 s
f(0.25) =
5.Hard
Solve y″ + 5y′ + 4y = 0 with y(0) = 4 and y′(0) = 11 using the Laplace transform, and find y(0.3).
y″ + 5y′ + 4y = 0y(0) = 4y′(0) = 11t = 0.3 s
y(0.3) =
6.Hard
A system obeys y″ + 9y′ + 18y = 36·u(t), with y(0) = 3 and y′(0) = 0. At t = 0 a step input u(t) = 10 is applied. Use the Laplace transform to find y(0.1).
y″ + 9y′ + 18y = 36·u(t)y(0) = 3y′(0) = 0u(t) = 10t = 0.1 s
y(0.1) =
Answer key
- 1. f(0⁺) = 1
- 2. f′(0⁺) = 29
- 3. f(0.2) = 0.4837
- 4. f(0.25) = 0.6132
- 5. y(0.3) = 5.161
- 6. y(0.1) = 4.142
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