A signal f(t) has the Laplace transform F(s) = (−6s² + 13s + 19)/(s(s + 5)(s + 6)). 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) = (−6s² + 13s + 19)/(s(s + 5)(s + 6))
f(0⁺) =
2.Easy
Find the inverse Laplace transform f(t) of F(s) = 12/(s² + 9) with a table of transform pairs, and evaluate f(0.1).
F(s) = 12/(s² + 9)t = 0.1 s
f(0.1) =
3.Medium
Find f(t) for F(s) = (6s + 38)/(s² + 6s + 9) and evaluate f(0.25).
F(s) = (6s + 38)/(s² + 6s + 9)t = 0.25 s
f(0.25) =
4.Medium
F(s) = (5s + 31)/(s² + 13s + 42). Its inverse transform has the form f(t) = A·e−6t + B·e−7t. Find the coefficient A by partial fractions.
F(s) = (5s + 31)/(s² + 13s + 42)
A =
5.Hard
A system obeys y″ + 3y′ + 2y = 2·u(t), with y(0) = 4 and y′(0) = 0. At t = 0 a step input u(t) = 2 is applied. Use the Laplace transform to find y(1).
y″ + 3y′ + 2y = 2·u(t)y(0) = 4y′(0) = 0u(t) = 2t = 1 s
y(1) =
6.Hard
A system obeys y″ + 8y′ + 12y = 12·u(t), with y(0) = 6 and y′(0) = 0. At t = 0 a step input u(t) = 10 is applied. Use the Laplace transform to find y(0.4).
y″ + 8y′ + 12y = 12·u(t)y(0) = 6y′(0) = 0u(t) = 10t = 0.4 s
y(0.4) =
Answer key
- 1. f(0⁺) = −6
- 2. f(0.1) = 1.182
- 3. f(0.25) = 5.196
- 4. A = 1
- 5. y(1) = 3.201
- 6. y(0.4) = 7.485
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