A signal f(t) has the Laplace transform F(s) = (3s + 39)/(s² + 7s + 14). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (3s + 39)/(s² + 7s + 14)
f′(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (−7s + 16)/(s² + 2s + 22). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (−7s + 16)/(s² + 2s + 22)
f′(0⁺) =
3.Medium
Find f(t) for F(s) = 50/(s(s + 1)(s + 4)) by partial fractions, and evaluate f(1.2).
F(s) = 50/(s(s + 1)(s + 4))t = 1.2 s
f(1.2) =
4.Medium
F(s) = (6s + 9)/(s² + 10s + 24). Its inverse transform has the form f(t) = A·e−4t + B·e−6t. Find the coefficient A by partial fractions.
F(s) = (6s + 9)/(s² + 10s + 24)
A =
5.Hard
A system obeys y″ + 6y′ + 8y = 16·u(t), with y(0) = 2 and y′(0) = 0. At t = 0 a step input u(t) = 2 is applied. Use the Laplace transform to find y(0.4).
y″ + 6y′ + 8y = 16·u(t)y(0) = 2y′(0) = 0u(t) = 2t = 0.4 s
y(0.4) =
6.Hard
A system obeys y″ + 6y′ + 5y = 20·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(1.2).
y″ + 6y′ + 5y = 20·u(t)y(0) = 5y′(0) = 0u(t) = 4t = 1.2 s
y(1.2) =
Answer key
- 1. f′(0⁺) = 18
- 2. f′(0⁺) = 30
- 3. f(1.2) = 7.514
- 4. A = −7.5
- 5. y(0.4) = 2.606
- 6. y(1.2) = 11.87
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