A signal f(t) has the Laplace transform F(s) = (−9s² + 8s + 13)/(s(s + 2)(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) = (−9s² + 8s + 13)/(s(s + 2)(s + 6))
f(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (6s² + 17s + 57)/(s(s + 1)(s + 7)). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
F(s) = (6s² + 17s + 57)/(s(s + 1)(s + 7))
f(0⁺) =
3.Medium
Find f(t) for F(s) = (4s + 36)/(s² + 6s + 34) by completing the square, and evaluate f(0.2).
F(s) = (4s + 36)/(s² + 6s + 34)t = 0.2 s
f(0.2) =
4.Medium
F(s) = (5s + 18)/(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) = (5s + 18)/(s² + 10s + 21)
A =
5.Hard
Solve y″ + 11y′ + 30y = 0 with y(0) = 7 and y′(0) = −5 using the Laplace transform, and find y(0.2).
y″ + 11y′ + 30y = 0y(0) = 7y′(0) = −5t = 0.2 s
y(0.2) =
6.Hard
A system obeys y″ + 7y′ + 10y = 10·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 2 for t ≥ 0. Use the Laplace transform to find the output y(0.4).
y″ + 7y′ + 10y = 10·u(t)u(t) = 2y(0) = 0, y′(0) = 0t = 0.4 s
y(0.4) =
Answer key
- 1. f(0⁺) = −9
- 2. f(0⁺) = 6
- 3. f(0.2) = 3.403
- 4. A = 0.75
- 5. y(0.2) = 4.576
- 6. y(0.4) = 0.6827
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