Question

The transfer function of a linear system is defined as the ratio of the Laplace transform of the output function y(t) over the ratio of the Laplace transform of the input function g(t) when all initial conditions are zero. Determine the transfer function H(s)=Y(s)G(s)H(s)=Y(s)G(s)for the system y′′(t)+8y′(t)+6y(t)=g(t),t>0

Answer 1

H(s) = Y(s)/G(s) = 1/(s² + 8s + 6)

Step-by-step 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)