A signal f(t) has the Laplace transform F(s) = (−7s² + 12s + 60)/(s(s + 6)(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) = (−7s² + 12s + 60)/(s(s + 6)(s + 8))
f(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (7s + 35)/(s² + 7s + 18). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (7s + 35)/(s² + 7s + 18)
f′(0⁺) =
3.Medium
Find f(t) for F(s) = (3s + 11)/(s² + 8s + 17) by completing the square, and evaluate f(0.5).
F(s) = (3s + 11)/(s² + 8s + 17)t = 0.5 s
f(0.5) =
4.Medium
F(s) = (6s + 11)/(s² + 14s + 48). Its inverse transform has the form f(t) = A·e−6t + B·e−8t. Find the coefficient A by partial fractions.
F(s) = (6s + 11)/(s² + 14s + 48)
A =
5.Hard
Solve y″ + 11y′ + 30y = 0 with y(0) = −10 and y′(0) = 2 using the Laplace transform, and find y(0.1).
y″ + 11y′ + 30y = 0y(0) = −10y′(0) = 2t = 0.1 s
y(0.1) =
6.Hard
A system obeys y″ + 5y′ + 4y = 8·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 1 for t ≥ 0. Use the Laplace transform to find the output y(0.4).
y″ + 5y′ + 4y = 8·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.4 s
y(0.4) =
Answer key
- 1. f(0⁺) = −7
- 2. f′(0⁺) = −14
- 3. f(0.5) = 0.2914
- 4. A = −12.5
- 5. y(0.1) = −8.836
- 6. y(0.4) = 0.3471
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