A signal f(t) has the Laplace transform F(s) = (7s + 25)/(s² + 7s + 6). 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) = (7s + 25)/(s² + 7s + 6)
f(0⁺) =
2.Easy
Find the inverse Laplace transform f(t) of F(s) = 2/(s + 2)² with a table of transform pairs, and evaluate f(0.3).
F(s) = 2/(s + 2)²t = 0.3 s
f(0.3) =
3.Medium
Find f(t) for F(s) = (5s + 38)/(s² + 6s + 9) and evaluate f(0.2).
F(s) = (5s + 38)/(s² + 6s + 9)t = 0.2 s
f(0.2) =
4.Medium
F(s) = 39/(s² + 10s + 16). Its inverse transform has the form f(t) = A·e−2t + B·e−8t. Find the coefficient A by partial fractions.
F(s) = 39/(s² + 10s + 16)
A =
5.Hard
Solve y″ + 6y′ + 8y = 0 with y(0) = 2 and y′(0) = −2 using the Laplace transform, and find y(0.25).
y″ + 6y′ + 8y = 0y(0) = 2y′(0) = −2t = 0.25 s
y(0.25) =
6.Hard
A system obeys y″ + 8y′ + 12y = 36·u(t), with y(0) = 2 and y′(0) = 0. At t = 0 a step input u(t) = 4 is applied. Use the Laplace transform to find y(0.6).
y″ + 8y′ + 12y = 36·u(t)y(0) = 2y′(0) = 0u(t) = 4t = 0.6 s
y(0.6) =
Answer key
- 1. f(0⁺) = 7
- 2. f(0.3) = 0.3293
- 3. f(0.2) = 5.269
- 4. A = 6.5
- 5. y(0.25) = 1.452
- 6. y(0.6) = 7.619
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