A signal f(t) has the Laplace transform F(s) = (−8s + 11)/(s² + 3s + 2). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (−8s + 11)/(s² + 3s + 2)
f′(0⁺) =
2.Easy
Find the inverse Laplace transform f(t) of F(s) = 6s/(s² + 36) with a table of transform pairs, and evaluate f(0.4).
F(s) = 6s/(s² + 36)t = 0.4 s
f(0.4) =
3.Medium
Find f(t) for F(s) = (2s + 12)/(s² + 2s + 5) by completing the square, and evaluate f(0.4).
F(s) = (2s + 12)/(s² + 2s + 5)t = 0.4 s
f(0.4) =
4.Medium
Find f(t) for F(s) = (6s + 31)/(s² + 6s + 9) and evaluate f(0.25).
F(s) = (6s + 31)/(s² + 6s + 9)t = 0.25 s
f(0.25) =
5.Hard
A system obeys y″ + 11y′ + 30y = 120·u(t), with y(0) = 3 and y′(0) = 0. At t = 0 a step input u(t) = 1 is applied. Use the Laplace transform to find y(0.2).
y″ + 11y′ + 30y = 120·u(t)y(0) = 3y′(0) = 0u(t) = 1t = 0.2 s
y(0.2) =
6.Hard
A system obeys y″ + 7y′ + 12y = 24·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″ + 7y′ + 12y = 24·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.25 s
y(0.25) =
Answer key
- 1. f′(0⁺) = 35
- 2. f(0.4) = −4.424
- 3. f(0.4) = 3.338
- 4. f(0.25) = 4.369
- 5. y(0.2) = 3.299
- 6. y(0.25) = 0.4283
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