A signal f(t) has the Laplace transform F(s) = (8s + 17)/(s² + 3s + 4). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (8s + 17)/(s² + 3s + 4)
f′(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (4s + 27)/(s² + 7s + 26). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (4s + 27)/(s² + 7s + 26)
f′(0⁺) =
3.Medium
Find f(t) for F(s) = (s + 25)/(s² + 2s + 5) by completing the square, and evaluate f(0.1).
F(s) = (s + 25)/(s² + 2s + 5)t = 0.1 s
f(0.1) =
4.Medium
F(s) = (2s + 2)/(s² + 8s + 15). Its inverse transform has the form f(t) = A·e−3t + B·e−5t. Find the coefficient A by partial fractions.
F(s) = (2s + 2)/(s² + 8s + 15)
A =
5.Hard
A system obeys y″ + 6y′ + 5y = 15·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(1.2).
y″ + 6y′ + 5y = 15·u(t)u(t) = 2y(0) = 0, y′(0) = 0t = 1.2 s
y(1.2) =
6.Hard
Solve y″ + 9y′ + 20y = 0 with y(0) = 2 and y′(0) = −3 using the Laplace transform, and find y(0.25).
y″ + 9y′ + 20y = 0y(0) = 2y′(0) = −3t = 0.25 s
y(0.25) =
Answer key
- 1. f′(0⁺) = −7
- 2. f′(0⁺) = −1
- 3. f(0.1) = 3.044
- 4. A = −2
- 5. y(1.2) = 3.745
- 6. y(0.25) = 1.143
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