A signal f(t) has the Laplace transform F(s) = (4s + 32)/(s² + 2s + 28). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (4s + 32)/(s² + 2s + 28)
f′(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (s² + 24s + 5)/(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) = (s² + 24s + 5)/(s(s + 5)(s + 7))
f(0⁺) =
3.Medium
F(s) = (3s + 27)/(s² + 8s + 12). Its inverse transform has the form f(t) = A·e−2t + B·e−6t. Find the coefficient A by partial fractions.
F(s) = (3s + 27)/(s² + 8s + 12)
A =
4.Medium
Find f(t) for F(s) = (2s + 33)/(s² + 10s + 25) and evaluate f(0.25).
F(s) = (2s + 33)/(s² + 10s + 25)t = 0.25 s
f(0.25) =
5.Hard
A system obeys y″ + 8y′ + 12y = 12·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(0.1).
y″ + 8y′ + 12y = 12·u(t)y(0) = 5y′(0) = 0u(t) = 4t = 0.1 s
y(0.1) =
6.Hard
Solve y″ + 10y′ + 24y = 0 with y(0) = 9 and y′(0) = −11 using the Laplace transform, and find y(0.2).
y″ + 10y′ + 24y = 0y(0) = 9y′(0) = −11t = 0.2 s
y(0.2) =
Answer key
- 1. f′(0⁺) = 24
- 2. f(0⁺) = 1
- 3. A = 5.25
- 4. f(0.25) = 2.22
- 5. y(0.1) = 4.954
- 6. y(0.2) = 5.896
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