A signal f(t) has the Laplace transform F(s) = (4s + 28)/(s² + 6s + 3). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (4s + 28)/(s² + 6s + 3)
f′(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = 20/(s(s² + 8s + 24)). Use the final value theorem to find the value f(∞) that f(t) settles to.
F(s) = 20/(s(s² + 8s + 24))
f(∞) =
3.Medium
Find f(t) for F(s) = (4s + 17)/(s² + 8s + 16) and evaluate f(0.2).
F(s) = (4s + 17)/(s² + 8s + 16)t = 0.2 s
f(0.2) =
4.Medium
Find f(t) for F(s) = 55/(s(s + 1)(s + 3)) by partial fractions, and evaluate f(1.5).
F(s) = 55/(s(s + 1)(s + 3))t = 1.5 s
f(1.5) =
5.Hard
A system obeys y″ + 5y′ + 4y = 12·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″ + 5y′ + 4y = 12·u(t)u(t) = 10y(0) = 0, y′(0) = 0t = 1 s
y(1) =
6.Hard
Solve y″ + 7y′ + 10y = 0 with y(0) = 9 and y′(0) = 0 using the Laplace transform, and find y(0.6).
y″ + 7y′ + 10y = 0y(0) = 9y′(0) = 0t = 0.6 s
y(0.6) =
Answer key
- 1. f′(0⁺) = 4
- 2. f(∞) = 0.8333
- 3. f(0.2) = 1.887
- 4. f(1.5) = 12.3
- 5. y(1) = 15.47
- 6. y(0.6) = 4.219
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