A signal f(t) has the Laplace transform F(s) = (9s² + 25s + 60)/(s(s + 1)(s + 7)). 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) = (9s² + 25s + 60)/(s(s + 1)(s + 7))
f(0⁺) =
2.Easy
Find the inverse Laplace transform f(t) of F(s) = 3s/(s² + 4) with a table of transform pairs, and evaluate f(1).
F(s) = 3s/(s² + 4)t = 1 s
f(1) =
3.Medium
F(s) = (2s + 24)/(s² + 10s + 21). Its inverse transform has the form f(t) = A·e−3t + B·e−7t. Find the coefficient A by partial fractions.
F(s) = (2s + 24)/(s² + 10s + 21)
A =
4.Medium
Find f(t) for F(s) = (4s + 20)/(s² + 6s + 18) by completing the square, and evaluate f(0.4).
F(s) = (4s + 20)/(s² + 6s + 18)t = 0.4 s
f(0.4) =
5.Hard
A system obeys y″ + 6y′ + 5y = 5·u(t), with y(0) = 6 and y′(0) = 0. At t = 0 a step input u(t) = 1 is applied. Use the Laplace transform to find y(0.75).
y″ + 6y′ + 5y = 5·u(t)y(0) = 6y′(0) = 0u(t) = 1t = 0.75 s
y(0.75) =
6.Hard
A system obeys y″ + 6y′ + 8y = 8·u(t), with y(0) = 1 and y′(0) = 0. At t = 0 a step input u(t) = 4 is applied. Use the Laplace transform to find y(0.5).
y″ + 6y′ + 8y = 8·u(t)y(0) = 1y′(0) = 0u(t) = 4t = 0.5 s
y(0.5) =
Answer key
- 1. f(0⁺) = 9
- 2. f(1) = −1.248
- 3. A = 4.5
- 4. f(0.4) = 1.185
- 5. y(0.75) = 3.923
- 6. y(0.5) = 2.199
Full step-by-step solutions: practise this topic on examproblems.com.