A signal f(t) has the Laplace transform F(s) = 2(s + 6)/(s(s + 1)(s + 7)). Use the final value theorem to find the value f(∞) that f(t) settles to.
Laplace transforms worksheet with answers
1.Easy
F(s) = 2(s + 6)/(s(s + 1)(s + 7))
f(∞) =
2.Easy
Find the inverse Laplace transform f(t) of F(s) = 5/(s² + 16) with a table of transform pairs, and evaluate f(1).
F(s) = 5/(s² + 16)t = 1 s
f(1) =
3.Medium
Find f(t) for F(s) = (5s + 13)/(s² + 4s + 13) by completing the square, and evaluate f(0.75).
F(s) = (5s + 13)/(s² + 4s + 13)t = 0.75 s
f(0.75) =
4.Medium
F(s) = (s + 22)/(s² + 9s + 20). Its inverse transform has the form f(t) = A·e−4t + B·e−5t. Find the coefficient A by partial fractions.
F(s) = (s + 22)/(s² + 9s + 20)
A =
5.Hard
A system obeys y″ + 9y′ + 18y = 36·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(0.3).
y″ + 9y′ + 18y = 36·u(t)u(t) = 2y(0) = 0, y′(0) = 0t = 0.3 s
y(0.3) =
6.Hard
A system obeys y″ + 7y′ + 6y = 6·u(t), with y(0) = 5 and y′(0) = 0. At t = 0 a step input u(t) = 10 is applied. Use the Laplace transform to find y(0.75).
y″ + 7y′ + 6y = 6·u(t)y(0) = 5y′(0) = 0u(t) = 10t = 0.75 s
y(0.75) =
Answer key
- 1. f(∞) = 1.714
- 2. f(1) = −0.946
- 3. f(0.75) = −0.5272
- 4. A = 18
- 5. y(0.3) = 1.409
- 6. y(0.75) = 7.177
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