A signal f(t) has the Laplace transform F(s) = (−4s² + 16s + 1)/(s(s + 4)(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) = (−4s² + 16s + 1)/(s(s + 4)(s + 7))
f(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (3s² + 16s + 21)/(s(s + 1)(s + 3)). Use the initial value theorem to find f(0⁺), the value of f(t) just after t = 0.
F(s) = (3s² + 16s + 21)/(s(s + 1)(s + 3))
f(0⁺) =
3.Medium
F(s) = 22/(s² + 13s + 40). Its inverse transform has the form f(t) = A·e−5t + B·e−8t. Find the coefficient A by partial fractions.
F(s) = 22/(s² + 13s + 40)
A =
4.Medium
Find f(t) for F(s) = 55/(s(s + 1)(s + 2)) by partial fractions, and evaluate f(0.8).
F(s) = 55/(s(s + 1)(s + 2))t = 0.8 s
f(0.8) =
5.Hard
A system obeys y″ + 7y′ + 12y = 36·u(t), with y(0) = 5 and y′(0) = 0. At t = 0 a step input u(t) = 5 is applied. Use the Laplace transform to find y(0.3).
y″ + 7y′ + 12y = 36·u(t)y(0) = 5y′(0) = 0u(t) = 5t = 0.3 s
y(0.3) =
6.Hard
Solve y″ + 7y′ + 6y = 0 with y(0) = −5 and y′(0) = −8 using the Laplace transform, and find y(0.2).
y″ + 7y′ + 6y = 0y(0) = −5y′(0) = −8t = 0.2 s
y(0.2) =
Answer key
- 1. f(0⁺) = −4
- 2. f(0⁺) = 3
- 3. A = 7.333
- 4. f(0.8) = 8.339
- 5. y(0.3) = 7.773
- 6. y(0.2) = −5.439
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