A signal f(t) has the Laplace transform F(s) = (s + 34)/(s² + 11s + 24). 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) = (s + 34)/(s² + 11s + 24)
f(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (−s² + 24s + 13)/(s(s + 2)(s + 4)). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
F(s) = (−s² + 24s + 13)/(s(s + 2)(s + 4))
f(0⁺) =
3.Medium
F(s) = 17/(s² + 10s + 16). Its inverse transform has the form f(t) = A·e−2t + B·e−8t. Find the coefficient A by partial fractions.
F(s) = 17/(s² + 10s + 16)
A =
4.Medium
Find f(t) for F(s) = (2s + 40)/(s² + 8s + 17) by completing the square, and evaluate f(0.5).
F(s) = (2s + 40)/(s² + 8s + 17)t = 0.5 s
f(0.5) =
5.Hard
A system obeys y″ + 4y′ + 3y = 4.5·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 10 for t ≥ 0. Use the Laplace transform to find the output y(1).
y″ + 4y′ + 3y = 4.5·u(t)u(t) = 10y(0) = 0, y′(0) = 0t = 1 s
y(1) =
6.Hard
Solve y″ + 9y′ + 18y = 0 with y(0) = −9 and y′(0) = 10 using the Laplace transform, and find y(0.2).
y″ + 9y′ + 18y = 0y(0) = −9y′(0) = 10t = 0.2 s
y(0.2) =
Answer key
- 1. f(0⁺) = 1
- 2. f(0⁺) = −1
- 3. A = 2.833
- 4. f(0.5) = 2.314
- 5. y(1) = 7.096
- 6. y(0.2) = −6.342
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