A signal f(t) has the Laplace transform F(s) = (6s + 37)/(s² + 2s + 8). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (6s + 37)/(s² + 2s + 8)
f′(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = 15/(s(s² + 6s + 15)). Use the final value theorem to find the value f(∞) that f(t) settles to.
F(s) = 15/(s(s² + 6s + 15))
f(∞) =
3.Medium
Find f(t) for F(s) = (5s + 17)/(s² + 2s + 1) and evaluate f(1).
F(s) = (5s + 17)/(s² + 2s + 1)t = 1 s
f(1) =
4.Medium
Find f(t) for F(s) = 50/(s(s + 2)(s + 4)) by partial fractions, and evaluate f(0.5).
F(s) = 50/(s(s + 2)(s + 4))t = 0.5 s
f(0.5) =
5.Hard
A system obeys y″ + 7y′ + 10y = 40·u(t), with y(0) = 2 and y′(0) = 0. At t = 0 a step input u(t) = 5 is applied. Use the Laplace transform to find y(0.3).
y″ + 7y′ + 10y = 40·u(t)y(0) = 2y′(0) = 0u(t) = 5t = 0.3 s
y(0.3) =
6.Hard
A system obeys y″ + 6y′ + 5y = 2.5·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(0.8).
y″ + 6y′ + 5y = 2.5·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.8 s
y(0.8) =
Answer key
- 1. f′(0⁺) = 25
- 2. f(∞) = 1
- 3. f(1) = 6.254
- 4. f(0.5) = 2.497
- 5. y(0.3) = 6.213
- 6. y(0.8) = 0.2215
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