Determine the Laplace transform ofg(t)=−8sin(5t−30)u6(t)
Question
Determine the Laplace transform ofg(t)=−8sin(5t−30)u6(t)
Solution
The Laplace transform of a function g(t) is given by the integral from 0 to infinity of e^(-st)g(t) dt.
The Laplace transform of a function g(t) = u_c(t)f(t-c) is given by e^(-cs)F(s), where F(s) is the Laplace transform of f(t).
The Laplace transform of a function g(t) = sin(at) is given by a/(s^2 + a^2).
So, to find the Laplace transform of g(t) = -8sin(5t - 30)u_6(t), we first need to rewrite the function in the form u_c(t)f(t-c).
We can rewrite -8sin(5t - 30) as -8sin(5(t - 6)), so our function is now g(t) = u_6(t)(-8sin(5(t - 6))).
Now we can apply the formula for the Laplace transform of a function g(t) = u_c(t)f(t-c).
The Laplace transform of -8sin(5(t - 6)) is -8 * 5/(s^2 + 25), or -40/(s^2 + 25).
So, the Laplace transform of g(t) = u_6(t)(-8sin(5(t - 6))) is e^(-6s)(-40/(s^2 + 25)).
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