Find the inverse Laplace transform f(t) of F(s) = 17/(s² + 36) with a table of transform pairs, and evaluate f(1.2).
Laplace transforms worksheet with answers
1.Easy
F(s) = 17/(s² + 36)t = 1.2 s
f(1.2) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (7s + 35)/(s² + 3s + 6). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (7s + 35)/(s² + 3s + 6)
f′(0⁺) =
3.Medium
F(s) = (s + 38)/(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) = (s + 38)/(s² + 3s + 2)
A =
4.Medium
F(s) = (s + 27)/(s² + 15s + 56). Its inverse transform has the form f(t) = A·e−7t + B·e−8t. Find the coefficient A by partial fractions.
F(s) = (s + 27)/(s² + 15s + 56)
A =
5.Hard
A system obeys y″ + 7y′ + 10y = 10·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 2 for t ≥ 0. Use the Laplace transform to find the output y(0.6).
y″ + 7y′ + 10y = 10·u(t)u(t) = 2y(0) = 0, y′(0) = 0t = 0.6 s
y(0.6) =
6.Hard
Solve y″ + 10y′ + 24y = 0 with y(0) = 5 and y′(0) = −1 using the Laplace transform, and find y(0.25).
y″ + 10y′ + 24y = 0y(0) = 5y′(0) = −1t = 0.25 s
y(0.25) =
Answer key
- 1. f(1.2) = 2.249
- 2. f′(0⁺) = 14
- 3. A = 37
- 4. A = 20
- 5. y(0.6) = 1.062
- 6. y(0.25) = 3.215
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