A signal f(t) has the Laplace transform F(s) = (−4s² + 13s + 54)/(s(s + 4)(s + 8)). 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) = (−4s² + 13s + 54)/(s(s + 4)(s + 8))
f(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (−4s² + 12s + 7)/(s(s + 1)(s + 4)). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
F(s) = (−4s² + 12s + 7)/(s(s + 1)(s + 4))
f(0⁺) =
3.Medium
Find f(t) for F(s) = (s + 37)/(s² + 2s + 1) and evaluate f(0.8).
F(s) = (s + 37)/(s² + 2s + 1)t = 0.8 s
f(0.8) =
4.Medium
F(s) = (6s + 27)/(s² + 8s + 12). Its inverse transform has the form f(t) = A·e−2t + B·e−6t. Find the coefficient A by partial fractions.
F(s) = (6s + 27)/(s² + 8s + 12)
A =
5.Hard
A system obeys y″ + 9y′ + 18y = 54·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(0.25).
y″ + 9y′ + 18y = 54·u(t)u(t) = 2y(0) = 0, y′(0) = 0t = 0.25 s
y(0.25) =
6.Hard
Solve y″ + 7y′ + 10y = 0 with y(0) = 3 and y′(0) = −1 using the Laplace transform, and find y(0.25).
y″ + 7y′ + 10y = 0y(0) = 3y′(0) = −1t = 0.25 s
y(0.25) =
Answer key
- 1. f(0⁺) = −4
- 2. f(0⁺) = −4
- 3. f(0.8) = 13.39
- 4. A = 3.75
- 5. y(0.25) = 1.67
- 6. y(0.25) = 2.353
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