A signal f(t) has the Laplace transform F(s) = (−3s + 22)/(s² + 6s + 24). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (−3s + 22)/(s² + 6s + 24)
f′(0⁺) =
2.Easy
Find the inverse Laplace transform f(t) of F(s) = 19s/(s² + 25) with a table of transform pairs, and evaluate f(1).
F(s) = 19s/(s² + 25)t = 1 s
f(1) =
3.Medium
Find f(t) for F(s) = (s + 28)/(s² + 6s + 25) by completing the square, and evaluate f(0.5).
F(s) = (s + 28)/(s² + 6s + 25)t = 0.5 s
f(0.5) =
4.Medium
Find f(t) for F(s) = (2s + 33)/(s² + 4s + 29) by completing the square, and evaluate f(0.8).
F(s) = (2s + 33)/(s² + 4s + 29)t = 0.8 s
f(0.8) =
5.Hard
A system obeys y″ + 7y′ + 12y = 12·u(t), with y(0) = 5 and y′(0) = 0. At t = 0 a step input u(t) = 1 is applied. Use the Laplace transform to find y(0.25).
y″ + 7y′ + 12y = 12·u(t)y(0) = 5y′(0) = 0u(t) = 1t = 0.25 s
y(0.25) =
6.Hard
A system obeys y″ + 6y′ + 5y = 5·u(t), with y(0) = 2 and y′(0) = 0. At t = 0 a step input u(t) = 4 is applied. Use the Laplace transform to find y(0.75).
y″ + 6y′ + 5y = 5·u(t)y(0) = 2y′(0) = 0u(t) = 4t = 0.75 s
y(0.75) =
Answer key
- 1. f′(0⁺) = 40
- 2. f(1) = 5.39
- 3. f(0.5) = 1.175
- 4. f(0.8) = −1.15
- 5. y(0.25) = 4.143
- 6. y(0.75) = 2.831
Full step-by-step solutions: practise this topic on examproblems.com.