A signal f(t) has the Laplace transform F(s) = (7s + 40)/(s² + 3s + 25). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (7s + 40)/(s² + 3s + 25)
f′(0⁺) =
2.Easy
Find the inverse Laplace transform f(t) of F(s) = 10/(s + 4) − 7/s with a table of transform pairs, and evaluate f(0.25).
F(s) = 10/(s + 4) − 7/st = 0.25 s
f(0.25) =
3.Medium
Find f(t) for F(s) = (3s + 9)/(s² + 2s + 1) and evaluate f(1.5).
F(s) = (3s + 9)/(s² + 2s + 1)t = 1.5 s
f(1.5) =
4.Medium
Find f(t) for F(s) = 35/(s² + 2s + 10) by completing the square, and evaluate f(0.6).
F(s) = 35/(s² + 2s + 10)t = 0.6 s
f(0.6) =
5.Hard
A system obeys y″ + 11y′ + 30y = 60·u(t), with y(0) = 4 and y′(0) = 0. At t = 0 a step input u(t) = 2 is applied. Use the Laplace transform to find y(0.2).
y″ + 11y′ + 30y = 60·u(t)y(0) = 4y′(0) = 0u(t) = 2t = 0.2 s
y(0.2) =
6.Hard
A system obeys y″ + 8y′ + 15y = 30·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(0.3).
y″ + 8y′ + 15y = 30·u(t)u(t) = 10y(0) = 0, y′(0) = 0t = 0.3 s
y(0.3) =
Answer key
- 1. f′(0⁺) = 19
- 2. f(0.25) = −3.321
- 3. f(1.5) = 2.678
- 4. f(0.6) = 6.235
- 5. y(0.2) = 4
- 6. y(0.3) = 6.365
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