0.16 Time domain state and output equations

Given the state equations and output equations

we would like to be able to find solutions for $x$ and $y$ in terms of the initial state of the system and the system's input. To find such solutions, we begin with an examination ofa scalar ( $n=1$ , $m=1$ ) state equation:

If we looked at a special case of this equation, one where the input $u$ was $0$ , we'd have $\frac{d^{1}x(t)}{dt^1}=ax$ . We've seen this many times before; to solve this, we need afunction whose derivative is directly proportional to itself. This function is the exponential function. Therefore, insolving the more general case presented by the state equation , we expect that the exponential function will also come into play.

Starting with the state equation , we can collect like terms, multiply through by $e^{-(at)}$ , and rewrite the left-hand side of the derivative equation in terms of the derivative. (We take this last step after noticing that theleft-hand side of the derivation equation looks like the derivative product rule has already been applied to it.)

Since we are searching for $x(t)$ instead of its derivative, we will integrate both sides from $t_0$ to $t$ .

In the left-hand side of this equation, the integral and the derivative counteract each other and we are left with the difference of the function $e^{-(at)}x(t)$ evaluated at the upper and lower integration limits. To avoid confusion, the variable of integration will be changed from $t$ (which is now a constant limit in the integral) to $$ .

We now move the $x(t_0)$ term to the other side and divide through by $e^{-(at)}$ . This leaves us with a solution for the state variable $x(t)$ in the scalar case:

What happens if we let $t_0$ go to $-$ ? The first term on the right-hand side will go to zero since $x(())$∞ 0 . Then, if we say that $h(t)=e^{at}b$ , the second term can be rewritten as

This is the convolution equation $(h, u)$ .

For the scalar case, the solution to the output equation $y(t)$ has the same basic form as the solution to the state equation:

Again, we can see the convolution in the second term.

The general matrix forms of the solutions for the state and output equations follow the same pattern. The onlydifferences are that we use the matrix exponential instead of the scalar exponential, and that we use the matrices $A$ , $B$ , $C$ , and $D$ instead of the scalars $a$ , $b$ , $c$ , and $d$ .

The convolution term is easy to see in the solution for $x(t)$ . However, it takes a little regrouping to find it in the solution for $y(t)$ . If we pull the $D$ matrix into the integral (by multiplying it by the impulse function, $$ , we once again have the integral of a function of $t-$ being multiplied by the input (convolution).