Question

Variable 'u' is not defined in this context. Find the inverse Laplace transform of F(s) = 1/(s^2 + 1). (Use step(t-c) for t > c.)

a) sin(t - c)u(t - c)
b) cos(t - c)u(t - c)
c) tan(t - c)u(t - c)
d) e^(-t)u(t)

Answer 1

The inverse Laplace transform of F(s) = 1/(s^2 + 1) is sin(t), but with the condition to use step(t-c) for t > c, the correct modified inverse is sin(t - c)u(t - c), which includes a time shift and Heaviside step function.

Step-by-step explanation:

The inverse Laplace transform of F(s) = 1/(s2 + 1) can be determined by recognizing that it matches the standard form of the inverse Laplace transform for the sine function. The standard form of the Laplace transform of sin(at) is L{sin(at)} = a/(s2 + a2). In our case, a = 1, meaning the inverse Laplace transform of F(s) is simply sin(t). However, with the additional piece of information that we should use step(t-c) for t > c, we must multiply our result by the Heaviside step function.

The correct answer is therefore option (a) sin(t - c)u(t - c), which is the sine function shifted by c to the right, and it only starts at time t = c.