Answer:
H(s) = Y(s)/G(s) = 1/(s² + 8s + 6)
Explanation:
Since we are required to determine the transfer function G(s) and Y(s) = G(s)H(s). So, H(s) = Y(s)/G(s).
Since our system is given as y′′(t) + 8y′(t) +6y(t) = g(t), t>0 with all initial condition zero, that is, y(0) = 0, y"(0) = 0 and g(0) = 0.
Taking the Laplace transform of both the left hand side and right hand side of the equation, we have
y′′(t) + 8y′(t) +6y(t) = g(t),
L{y′′(t) + 8y′(t) +6y(t)} = L{g(t)}
L{y′′(t)} + L{8y′(t)} + L{6y(t)} = L{g(t)}
L{y′′(t)} + 8L{y′(t)} + 6L{y(t)} = L{g(t)}
[s²Y(s) - sy(0) - y'(0)] + 8[sY(s) - y(0)] + 6Y(s) = G(s)
[s²Y(s) - s(0) - 0] + 8[sY(s) - 0] + 6Y(s) = G(s)
s²Y(s) + 8sY(s) + 6Y(s) = G(s)
(s² + 8s + 6)Y(s) = G(s)
Y(s)/G(s) = 1/(s² + 8s + 6)
So, H(s) = Y(s)/G(s) = 1/(s² + 8s + 6)
