A signal f(t) has the Laplace transform F(s) = (−5s + 37)/(s² + 5s + 24). Find the initial slope f′(0⁺) with the initial value theorem.
Laplace transforms worksheet with answers
1.Easy
F(s) = (−5s + 37)/(s² + 5s + 24)
f′(0⁺) =
2.Easy
Find the inverse Laplace transform f(t) of F(s) = 6s/(s² + 9) with a table of transform pairs, and evaluate f(0.8).
F(s) = 6s/(s² + 9)t = 0.8 s
f(0.8) =
3.Medium
Find f(t) for F(s) = 40/(s(s + 1)(s + 5)) by partial fractions, and evaluate f(1).
F(s) = 40/(s(s + 1)(s + 5))t = 1 s
f(1) =
4.Medium
Find f(t) for F(s) = (4s + 34)/(s² + 2s + 17) by completing the square, and evaluate f(0.25).
F(s) = (4s + 34)/(s² + 2s + 17)t = 0.25 s
f(0.25) =
5.Hard
A system obeys y″ + 9y′ + 18y = 90·u(t). It starts at rest (y(0) = 0, y′(0) = 0) and the input is a step u(t) = 1 for t ≥ 0. Use the Laplace transform to find the output y(0.1).
y″ + 9y′ + 18y = 90·u(t)u(t) = 1y(0) = 0, y′(0) = 0t = 0.1 s
y(0.1) =
6.Hard
Solve y″ + 7y′ + 12y = 0 with y(0) = −10 and y′(0) = 6 using the Laplace transform, and find y(0.3).
y″ + 7y′ + 12y = 0y(0) = −10y′(0) = 6t = 0.3 s
y(0.3) =
Answer key
- 1. f′(0⁺) = 62
- 2. f(0.8) = −4.424
- 3. f(1) = 4.335
- 4. f(0.25) = 6.598
- 5. y(0.1) = 0.3359
- 6. y(0.3) = −6.595
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