A signal f(t) has the Laplace transform F(s) = (8s + 16)/(s² + 8s + 9). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (8s + 16)/(s² + 8s + 9)
f′(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = 2(s + 2)/(s(s + 1)(s + 7)). Use the final value theorem to find the value f(∞) that f(t) settles to.
F(s) = 2(s + 2)/(s(s + 1)(s + 7))
f(∞) =
3.Medium
Find f(t) for F(s) = 30/(s(s + 1)(s + 6)) by partial fractions, and evaluate f(0.8).
F(s) = 30/(s(s + 1)(s + 6))t = 0.8 s
f(0.8) =
4.Medium
Find f(t) for F(s) = (5s + 40)/(s² + 6s + 13) by completing the square, and evaluate f(0.6).
F(s) = (5s + 40)/(s² + 6s + 13)t = 0.6 s
f(0.6) =
5.Hard
A system obeys y″ + 10y′ + 24y = 48·u(t), with y(0) = 4 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″ + 10y′ + 24y = 48·u(t)y(0) = 4y′(0) = 0u(t) = 10t = 0.1 s
y(0.1) =
6.Hard
Solve y″ + 3y′ + 2y = 0 with y(0) = 8 and y′(0) = −4 using the Laplace transform, and find y(1.2).
y″ + 3y′ + 2y = 0y(0) = 8y′(0) = −4t = 1.2 s
y(1.2) =
Answer key
- 1. f′(0⁺) = −48
- 2. f(∞) = 0.5714
- 3. f(0.8) = 2.312
- 4. f(0.6) = 2.225
- 5. y(0.1) = 5.387
- 6. y(1.2) = 3.251
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