A signal f(t) has the Laplace transform F(s) = 40/(s(s² + 3s + 30)). Use the final value theorem to find the value f(∞) that f(t) settles to.
Laplace transforms worksheet with answers
1.Easy
F(s) = 40/(s(s² + 3s + 30))
f(∞) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (−s + 11)/(s² + 4s + 28). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (−s + 11)/(s² + 4s + 28)
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
F(s) = (3s + 38)/(s² + 8s + 12). Its inverse transform has the form f(t) = A·e−2t + B·e−6t. Find the coefficient A by partial fractions.
F(s) = (3s + 38)/(s² + 8s + 12)
A =
5.Hard
A system obeys y″ + 4y′ + 3y = 9·u(t), with y(0) = 1 and y′(0) = 0. At t = 0 a step input u(t) = 10 is applied. Use the Laplace transform to find y(0.5).
y″ + 4y′ + 3y = 9·u(t)y(0) = 1y′(0) = 0u(t) = 10t = 0.5 s
y(0.5) =
6.Hard
A system obeys y″ + 10y′ + 24y = 72·u(t), with y(0) = 6 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″ + 10y′ + 24y = 72·u(t)y(0) = 6y′(0) = 0u(t) = 5t = 0.3 s
y(0.3) =
Answer key
- 1. f(∞) = 1.333
- 2. f′(0⁺) = 15
- 3. f(0.2) = 0.4837
- 4. A = 8
- 5. y(0.5) = 6.851
- 6. y(0.3) = 9.843
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