"A function f (t) that has the given..." solved and explained

Rui Baldwin

Answered question

2020-10-21

A function f (t) that has the given Laplace transform F (s).

Cullen

Skilled2020-10-22Added 89 answers

Given:
Laplace transform F (s) of a function f (t) defined on an interval

[0, \infty) - F (s) = \frac{2}{s (s^{2} + 16)}

Calculation:
Given,

F (s) = \frac{2}{s (s^{2} + 16)}

Partial fraction decomposition,

F (s) = \frac{2}{s (s^{2} + 16)}

Approach:
The

= \frac{A}{s} + \frac{B s + C}{s^{2} + 16}

2 = A (s^{2} + 16) + s (B s + C)

= s^{2} (A + B) + C s + 16 A

Compare the Left hand side and Right hand side,

A = \frac{1}{8}

B = - \frac{1}{8}

C = 0

Substitute -

F (s) = \frac{1}{8 s} - \frac{s}{8 (s^{2} + 16)}

Laplace transform,

L [f] = \int_{0}^{\infty} e^{- s} t f (t) d t

= F (s)

= \frac{2}{s (s^{2} + 16)}

= \frac{1}{8 s} - \frac{s}{8 (s^{2} + 16)}

Inverse Laplace transform-

L^{- 1} [F] (t) = L^{- 1} [\frac{2}{s (s^{2} + 16)}]

= L^{- 1} [\frac{1}{8 s} - \frac{s}{8 (s^{2} + 16)}]

= \frac{1}{8} L^{- 1} [\frac{1}{s}] - \frac{1}{8} L^{- 1} [\frac{s}{s^{2} + 4^{2}}]

= \frac{1}{8} - \frac{1}{8} \cos (4 t)

L^{- 1} [F] (t) = \frac{1}{8} (1 - \cos (4 t))

= f (t)

Conclusion:
Hence, a function f (t) that has the given Laplace transform

F (s) = \frac{2}{s (s^{2} + 16)} i s \frac{1}{8} (1 - \cos (4 t)) .

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