Find the inverse Laplace transform f(t) of F(s) = 12/(s² + 4) with a table of transform pairs, and evaluate f(0.6).
Laplace transforms worksheet with answers
1.Easy
F(s) = 12/(s² + 4)t = 0.6 s
f(0.6) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (5s² + 27s + 34)/(s(s + 2)(s + 4)). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
F(s) = (5s² + 27s + 34)/(s(s + 2)(s + 4))
f(0⁺) =
3.Medium
Find f(t) for F(s) = (5s + 34)/(s² + 4s + 4) and evaluate f(0.5).
F(s) = (5s + 34)/(s² + 4s + 4)t = 0.5 s
f(0.5) =
4.Medium
Find f(t) for F(s) = (2s + 32)/(s² + 8s + 16) and evaluate f(0.25).
F(s) = (2s + 32)/(s² + 8s + 16)t = 0.25 s
f(0.25) =
5.Hard
A system obeys y″ + 11y′ + 30y = 15·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(0.1).
y″ + 11y′ + 30y = 15·u(t)u(t) = 10y(0) = 0, y′(0) = 0t = 0.1 s
y(0.1) =
6.Hard
A system obeys y″ + 6y′ + 8y = 48·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 5 for t ≥ 0. Use the Laplace transform to find the output y(0.5).
y″ + 6y′ + 8y = 48·u(t)u(t) = 5y(0) = 0, y′(0) = 0t = 0.5 s
y(0.5) =
Answer key
- 1. f(0.6) = 5.592
- 2. f(0⁺) = 5
- 3. f(0.5) = 6.254
- 4. f(0.25) = 2.943
- 5. y(0.1) = 0.5244
- 6. y(0.5) = 11.99
Full step-by-step solutions: practise this topic on examproblems.com.