A signal f(t) has the Laplace transform F(s) = (7s + 25)/(s² + 7s + 7). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (7s + 25)/(s² + 7s + 7)
f′(0⁺) =
2.Easy
A signal f(t) has the Laplace transform F(s) = 12/(s(s² + 3s + 16)). Use the final value theorem to find the value f(∞) that f(t) settles to.
F(s) = 12/(s(s² + 3s + 16))
f(∞) =
3.Medium
F(s) = (5s + 34)/(s² + 14s + 48). Its inverse transform has the form f(t) = A·e−6t + B·e−8t. Find the coefficient A by partial fractions.
F(s) = (5s + 34)/(s² + 14s + 48)
A =
4.Medium
Find f(t) for F(s) = (6s + 14)/(s² + 4s + 4) and evaluate f(1).
F(s) = (6s + 14)/(s² + 4s + 4)t = 1 s
f(1) =
5.Hard
A system obeys y″ + 9y′ + 20y = 40·u(t), with y(0) = 1 and y′(0) = 0. At t = 0 a step input u(t) = 4 is applied. Use the Laplace transform to find y(0.3).
y″ + 9y′ + 20y = 40·u(t)y(0) = 1y′(0) = 0u(t) = 4t = 0.3 s
y(0.3) =
6.Hard
Solve y″ + 3y′ + 2y = 0 with y(0) = 7 and y′(0) = 8 using the Laplace transform, and find y(0.3).
y″ + 3y′ + 2y = 0y(0) = 7y′(0) = 8t = 0.3 s
y(0.3) =
Answer key
- 1. f′(0⁺) = −24
- 2. f(∞) = 0.75
- 3. A = 2
- 4. f(1) = 1.083
- 5. y(0.3) = 3.706
- 6. y(0.3) = 8.066
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