Question

Change the below differential equation to transfer function
y'''+3y''+2y'-5y=sin2t

Answer 1

$$y'''(t)+3y''(t)+2y'(t)-5y(t)=x(t), x(t)=sin2t$$

To find the transfer function, first take the Laplace Transform of the differential equation (with zero initial conditions). Recall that differentiation in the time domain is equivalent to multiplication by "s" in the Laplace domain.

$$s^3 Y(s)+3s^2Y(s)+2sY(s)-5Y(s)=X(s), X(s)={2\over s^2+4}$$

The transfer function is then the ratio of output to input and is often called H(s).

$$H(s)={Y(s) \over X(s)}={1 \over s^3+3s^2+2s-5}$$