A signal f(t) has the Laplace transform F(s) = (2s + 1)/(s² + 3s + 3). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (2s + 1)/(s² + 3s + 3)
f′(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (5s + 16)/(s² + 6s + 8). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (5s + 16)/(s² + 6s + 8)
f′(0⁺) =
3.Medium
F(s) = (2s + 12)/(s² + 9s + 8). Its inverse transform has the form f(t) = A·e−t + B·e−8t. Find the coefficient A by partial fractions.
F(s) = (2s + 12)/(s² + 9s + 8)
A =
4.Medium
Find f(t) for F(s) = (6s + 6)/(s² + 8s + 16) and evaluate f(0.2).
F(s) = (6s + 6)/(s² + 8s + 16)t = 0.2 s
f(0.2) =
5.Hard
A system obeys y″ + 7y′ + 6y = 24·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(0.25).
y″ + 7y′ + 6y = 24·u(t)y(0) = 4y′(0) = 0u(t) = 2t = 0.25 s
y(0.25) =
6.Hard
A system obeys y″ + 8y′ + 12y = 36·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 5 for t ≥ 0. Use the Laplace transform to find the output y(0.4).
y″ + 8y′ + 12y = 36·u(t)u(t) = 5y(0) = 0, y′(0) = 0t = 0.4 s
y(0.4) =
Answer key
- 1. f′(0⁺) = −5
- 2. f′(0⁺) = −14
- 3. A = 1.429
- 4. f(0.2) = 1.078
- 5. y(0.25) = 4.44
- 6. y(0.4) = 5.57
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