A signal f(t) has the Laplace transform F(s) = (−4s + 31)/(s² + 8s + 24). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (−4s + 31)/(s² + 8s + 24)
f′(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (−3s² + 13s + 54)/(s(s + 5)(s + 8)). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
F(s) = (−3s² + 13s + 54)/(s(s + 5)(s + 8))
f(0⁺) =
3.Medium
Find f(t) for F(s) = (2s + 8)/(s² + 10s + 25) and evaluate f(0.2).
F(s) = (2s + 8)/(s² + 10s + 25)t = 0.2 s
f(0.2) =
4.Medium
Find f(t) for F(s) = 60/(s(s + 3)(s + 6)) by partial fractions, and evaluate f(0.3).
F(s) = 60/(s(s + 3)(s + 6))t = 0.3 s
f(0.3) =
5.Hard
A system is described by the differential equation y″ + 7y′ + 12y = 2u′ + 17u, 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 = 0.3 s. Find g(0.3).
y″ + 7y′ + 12y = 2u′ + 17ut = 0.3 s
g(0.3) =
6.Hard
Solve y″ + 11y′ + 30y = 0 with y(0) = 1 and y′(0) = 10 using the Laplace transform, and find y(0.1).
y″ + 11y′ + 30y = 0y(0) = 1y′(0) = 10t = 0.1 s
y(0.1) =
Answer key
- 1. f′(0⁺) = 63
- 2. f(0⁺) = −3
- 3. f(0.2) = 0.5886
- 4. f(0.3) = 1.174
- 5. g(0.3) = 1.762
- 6. y(0.1) = 1.472
Full step-by-step solutions: practise this topic on examproblems.com.