A signal f(t) has the Laplace transform F(s) = 11(s + 6)/(s(s + 3)(s + 5)). Use the final value theorem to find the value f(∞) that f(t) settles to.
Laplace transforms worksheet with answers
1.Easy
F(s) = 11(s + 6)/(s(s + 3)(s + 5))
f(∞) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (9s² + 8s + 17)/(s(s + 5)(s + 7)). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
F(s) = (9s² + 8s + 17)/(s(s + 5)(s + 7))
f(0⁺) =
3.Medium
Find f(t) for F(s) = 20/(s(s + 1)(s + 6)) by partial fractions, and evaluate f(0.6).
F(s) = 20/(s(s + 1)(s + 6))t = 0.6 s
f(0.6) =
4.Medium
Find f(t) for F(s) = (4s + 8)/(s² + 6s + 9) and evaluate f(0.3).
F(s) = (4s + 8)/(s² + 6s + 9)t = 0.3 s
f(0.3) =
5.Hard
A system obeys y″ + 6y′ + 5y = 5·u(t), with y(0) = 5 and y′(0) = 0. At t = 0 a step input u(t) = 4 is applied. Use the Laplace transform to find y(0.75).
y″ + 6y′ + 5y = 5·u(t)y(0) = 5y′(0) = 0u(t) = 4t = 0.75 s
y(0.75) =
6.Hard
A system obeys y″ + 7y′ + 10y = 40·u(t), with y(0) = 5 and y′(0) = 0. At t = 0 a step input u(t) = 2 is applied. Use the Laplace transform to find y(0.25).
y″ + 7y′ + 10y = 40·u(t)y(0) = 5y′(0) = 0u(t) = 2t = 0.25 s
y(0.25) =
Answer key
- 1. f(∞) = 4.4
- 2. f(0⁺) = 9
- 3. f(0.6) = 1.156
- 4. f(0.3) = 1.138
- 5. y(0.75) = 4.585
- 6. y(0.25) = 5.54
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