A signal f(t) has the Laplace transform F(s) = 7(s + 5)/(s(s + 1)(s + 6)). Use the final value theorem to find the value f(∞) that f(t) settles to.
Laplace transforms worksheet with answers
1.Easy
F(s) = 7(s + 5)/(s(s + 1)(s + 6))
f(∞) =
2.Easy
A signal f(t) has the Laplace transform F(s) = (5s + 21)/(s² + 6s + 19). Find the initial slope f′(0⁺) with the initial value theorem.
F(s) = (5s + 21)/(s² + 6s + 19)
f′(0⁺) =
3.Medium
F(s) = (3s + 5)/(s² + 9s + 18). Its inverse transform has the form f(t) = A·e−3t + B·e−6t. Find the coefficient A by partial fractions.
F(s) = (3s + 5)/(s² + 9s + 18)
A =
4.Medium
Find f(t) for F(s) = 35/(s(s + 1)(s + 5)) by partial fractions, and evaluate f(1).
F(s) = 35/(s(s + 1)(s + 5))t = 1 s
f(1) =
5.Hard
A system obeys y″ + 8y′ + 15y = 37.5·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 10 for t ≥ 0. Use the Laplace transform to find the output y(0.2).
y″ + 8y′ + 15y = 37.5·u(t)u(t) = 10y(0) = 0, y′(0) = 0t = 0.2 s
y(0.2) =
6.Hard
Solve y″ + 11y′ + 30y = 0 with y(0) = −7 and y′(0) = 10 using the Laplace transform, and find y(0.2).
y″ + 11y′ + 30y = 0y(0) = −7y′(0) = 10t = 0.2 s
y(0.2) =
Answer key
- 1. f(∞) = 5.833
- 2. f′(0⁺) = −9
- 3. A = −1.333
- 4. f(1) = 3.793
- 5. y(0.2) = 4.495
- 6. y(0.2) = −4.242
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