A signal f(t) has the Laplace transform F(s) = (3s² + 4s + 56)/(s(s + 2)(s + 5)). 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) = (3s² + 4s + 56)/(s(s + 2)(s + 5))
f(0⁺) =
2.Easy
Find the inverse Laplace transform f(t) of F(s) = 4/(s² + 9) with a table of transform pairs, and evaluate f(0.25).
F(s) = 4/(s² + 9)t = 0.25 s
f(0.25) =
3.Medium
Find f(t) for F(s) = (2s + 37)/(s² + 10s + 25) and evaluate f(0.4).
F(s) = (2s + 37)/(s² + 10s + 25)t = 0.4 s
f(0.4) =
4.Medium
Find f(t) for F(s) = 40/(s(s + 2)(s + 6)) by partial fractions, and evaluate f(0.2).
F(s) = 40/(s(s + 2)(s + 6))t = 0.2 s
f(0.2) =
5.Hard
A system obeys y″ + 5y′ + 6y = 18·u(t), with y(0) = 3 and y′(0) = 0. At t = 0 a step input u(t) = 2 is applied. Use the Laplace transform to find y(0.5).
y″ + 5y′ + 6y = 18·u(t)y(0) = 3y′(0) = 0u(t) = 2t = 0.5 s
y(0.5) =
6.Hard
Solve y″ + 9y′ + 20y = 0 with y(0) = −7 and y′(0) = 13 using the Laplace transform, and find y(0.25).
y″ + 9y′ + 20y = 0y(0) = −7y′(0) = 13t = 0.25 s
y(0.25) =
Answer key
- 1. f(0⁺) = 3
- 2. f(0.25) = 0.9089
- 3. f(0.4) = 1.732
- 4. f(0.2) = 0.4837
- 5. y(0.5) = 4.028
- 6. y(0.25) = −3.796
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