Find the inverse Laplace transform f(t) of F(s) = 15/(s² + 16) with a table of transform pairs, and evaluate f(0.3).
Laplace transforms worksheet with answers
1.Easy
F(s) = 15/(s² + 16)t = 0.3 s
f(0.3) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (−6s + 39)/(s² + 2s + 18). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (−6s + 39)/(s² + 2s + 18)
f′(0⁺) =
3.Medium
F(s) = (6s + 33)/(s² + 3s + 2). Its inverse transform has the form f(t) = A·e−t + B·e−2t. Find the coefficient A by partial fractions.
F(s) = (6s + 33)/(s² + 3s + 2)
A =
4.Medium
Find f(t) for F(s) = (5s + 40)/(s² + 8s + 52) by completing the square, and evaluate f(0.3).
F(s) = (5s + 40)/(s² + 8s + 52)t = 0.3 s
f(0.3) =
5.Hard
A system obeys y″ + 3y′ + 2y = 4·u(t), with y(0) = 1 and y′(0) = 0. At t = 0 a step input u(t) = 1 is applied. Use the Laplace transform to find y(0.3).
y″ + 3y′ + 2y = 4·u(t)y(0) = 1y′(0) = 0u(t) = 1t = 0.3 s
y(0.3) =
6.Hard
A system obeys y″ + 7y′ + 12y = 36·u(t), with y(0) = 4 and y′(0) = 0. At t = 0 a step input u(t) = 2 is applied. Use the Laplace transform to find y(0.3).
y″ + 7y′ + 12y = 36·u(t)y(0) = 4y′(0) = 0u(t) = 2t = 0.3 s
y(0.3) =
Answer key
- 1. f(0.3) = 3.495
- 2. f′(0⁺) = 51
- 3. A = 27
- 4. f(0.3) = 0.6356
- 5. y(0.3) = 1.067
- 6. y(0.3) = 4.555
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