I'm trying to understand how to incorporate a set of initial conditions when starting from a transfer function, i.e. I know the general response of my system, and I want to reach a time-domain representation where the initial state is nonzero. I am familiar with this process for polynomial functions: take the inverse Laplace transform, then take the Laplace transform with the initial conditions included, and then take the inverse Laplace transform of the results. However, it is not clear how to do so when the impulse response is not a polynomial function.
For example, given y(0) = 1, y'(0) = 0, and the transfer function
$$
H(s) = \frac{K_0}{(s-a)(s-b)}
$$
I can decompose this into first order components to get the impulse response:
$$
h(t) = u(t)[ K_1e^{at} + K_2e^{bt} ]
$$
How do I incorporate my initial conditions? It is not clear from the definition of the transform how initial conditions would be captured for non-polynomial expressions. My first guess was to treat the input as a homogeneous solution, and then use my initial conditions and solve for a particular solution, and since this is a LTI system, I can sum the results. However, I couldn't find any text or discussion that supports this as valid.
Best Answer
The equation you are solving with $\frac{Y(s)}{X(s)}=H(s)$ is $$ y''-(a+b)y'+aby=K_0x(t), $$ $x(t)=y(t)=0$ for $t<0$.
Now translate back the homogeneous equation incorporating arbitrary initial conditions $$ [s^2Y(s)-y(0)s-y'(0)]-(a+b)[sY(s)-y(0)]+Y(s)=0 $$ so that $$ Y(s)=\frac{(s-a-b)y(0)+y'(0)}{(s-a)(s-b)}=H(s)\frac{(s-a-b)y(0)+y'(0)}{K_0}. $$ Which tells you that the input to realize wanted initial conditions is $$ x(t)=\frac{y(0)u(t)+\bigl[y'(0)-(a+b)y(0)\bigr]\delta_0(t)}{K_0}. $$
So with $y(0)=1$, $y'(0)=0$ one gets $$ Y(s)=\frac{ s-a-b }{(s-a)(s-b)}=-\frac{a}{(b-a)(s-b)}+\frac{b}{(b-a)(s-a)} \\ \implies y(t)=u(t)\frac{be^{at}-ae^{bt}}{b-a} $$