The given function is G(s) = 6s * e^(-s) / (s^2 + 36). This function is a product of two functions: 6s / (s^2 + 36) and e^(-s). The first function 6s / (s^2 + 36) is a standard Laplace transform of a function, which is 6 * cos(6t). The second function e^(-s) corresponds to a shift in the time domain by 1 unit. So, the inverse Laplace transform of G(s) is 6 * cos(6(t-1)) * u(t-1), where u(t-1) is the unit step function. This function is zero for t = 1.

Question

The given function is G(s) = 6s * e^(-s) / (s^2 + 36).

This function is a product of two functions: 6s / (s^2 + 36) and e^(-s).

The first function 6s / (s^2 + 36) is a standard Laplace transform of a function, which is 6 * cos(6t).

The second function e^(-s) corresponds to a shift in the time domain by 1 unit.

So, the inverse Laplace transform of G(s) is 6 * cos(6(t-1)) * u(t-1), where u(t-1) is the unit step function.

This function is zero for t < 1 and equals 6 * cos(6(t-1)) for t >= 1.

Knowee AI · Accepted Answer

The given function is G(s) = 6s * e^(-s) / (s^2 + 36).

This function is a product of two functions: 6s / (s^2 + 36) and e^(-s).

The first function 6s / (s^2 + 36) is a standard Laplace transform of a function, which is 6 * cos(6t).

The second function e^(-s) corresponds to a shift in the time domain by 1 unit.

So, the inverse Laplace transform of G(s) is 6 * cos(6(t-1)) * u(t-1), where u(t-1) is the unit step function.

This function is zero for t < 1 and equals 6 * cos(6(t-1)) for t >= 1.

Determine the inverse Laplace transform ofG(s)=6se−ss2+36𝐺(𝑠)=6𝑠𝑒−𝑠𝑠2+36Note: You must use the notation u(t−c)𝑢(𝑡−𝑐) rather than uc(t)𝑢𝑐(𝑡) in order for your answer to be accepted by SOWISO.