Question

What is the laplace transform of f(t)=sin2t at the parameter p=0​

Answer 1

Given:

f(t)=sin2t at the parameter p=0​

To find:

What is the laplace transform of f(t)=sin2t at the parameter p=0​

Solution:

From given, we have,

f(t)=sin2t

The Laplace Transform is given by,

L[sin 2t]

= lim (N→∞) ∫ 0 to N e^{-st} sin (2t) dt

∫ 0 to N e^{-st} sin (2t) dt = -1/s [e^{-st} sin (2t)]_0^N - 2/s² [e^{-st} cos (2t)]_0^N - 2²/s² ∫ 0 to N e^{-st} sin (2t) dt

(1 + a²/s²) ∫ 0 to N e^{-st} sin (2t) dt = -1/s [e^{-st} sin (2t)]_0^N - 2/s² [e^{-st} cos (2t)]_0^N

(s² + 2²)/s² ∫ 0 to ∞ e^{-st} sin(2t) dt = 2/s²

∴ L[sin (2t)] = 2/(s²+4)