A signal f(t) has the Laplace transform F(s) = (3s² + 12s + 9)/(s(s + 1)(s + 3)). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
Laplace transforms worksheet with answers
1.Easy
F(s) = (3s² + 12s + 9)/(s(s + 1)(s + 3))
f(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (−6s² + 10s + 24)/(s(s + 5)(s + 7)). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
F(s) = (−6s² + 10s + 24)/(s(s + 5)(s + 7))
f(0⁺) =
3.Medium
Find f(t) for F(s) = (6s + 32)/(s² + 4s + 4) and evaluate f(0.25).
F(s) = (6s + 32)/(s² + 4s + 4)t = 0.25 s
f(0.25) =
4.Medium
Find f(t) for F(s) = (6s + 8)/(s² + 4s + 4) and evaluate f(0.8).
F(s) = (6s + 8)/(s² + 4s + 4)t = 0.8 s
f(0.8) =
5.Hard
Solve y″ + 11y′ + 30y = 0 with y(0) = −8 and y′(0) = 4 using the Laplace transform, and find y(0.1).
y″ + 11y′ + 30y = 0y(0) = −8y′(0) = 4t = 0.1 s
y(0.1) =
6.Hard
A system is described by the differential equation y″ + 4y′ + 13y = 2u′ + 15u, with input u(t) and output y(t). Find its transfer function G(s) = Y(s)/U(s) (zero initial conditions) and evaluate the impulse response g(t) at t = 0.5 s. Find g(0.5).
y″ + 4y′ + 13y = 2u′ + 15ut = 0.5 s
g(0.5) =
Answer key
- 1. f(0⁺) = 3
- 2. f(0⁺) = −6
- 3. f(0.25) = 6.672
- 4. f(0.8) = 0.5653
- 5. y(0.1) = −6.93
- 6. g(0.5) = 1.398
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