A signal f(t) has the Laplace transform F(s) = (−5s² + 27s + 49)/(s(s + 3)(s + 7)). 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) = (−5s² + 27s + 49)/(s(s + 3)(s + 7))
f(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = 30/(s(s² + 3s + 23)). Use the final value theorem to find the value f(∞) that f(t) settles to.
F(s) = 30/(s(s² + 3s + 23))
f(∞) =
3.Medium
Find f(t) for F(s) = (4s + 29)/(s² + 10s + 25) and evaluate f(0.25).
F(s) = (4s + 29)/(s² + 10s + 25)t = 0.25 s
f(0.25) =
4.Medium
Find f(t) for F(s) = (5s + 14)/(s² + 10s + 25) and evaluate f(0.3).
F(s) = (5s + 14)/(s² + 10s + 25)t = 0.3 s
f(0.3) =
5.Hard
A system obeys y″ + 3y′ + 2y = 6·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(0.5).
y″ + 3y′ + 2y = 6·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.5 s
y(0.5) =
6.Hard
Solve y″ + 5y′ + 4y = 0 with y(0) = −5 and y′(0) = 4 using the Laplace transform, and find y(0.8).
y″ + 5y′ + 4y = 0y(0) = −5y′(0) = 4t = 0.8 s
y(0.8) =
Answer key
- 1. f(0⁺) = −5
- 2. f(∞) = 1.304
- 3. f(0.25) = 1.791
- 4. f(0.3) = 0.3793
- 5. y(0.5) = 0.4645
- 6. y(0.8) = −2.383
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