Solution of an uncontrolled LTI problem

June 12, 2023

Solution of an uncontrolled LTI problem

The uncontrolled linear time invariant system is defined by the following expression

$$ \begin{equation} \label{eq:main1} \dot{x}(t) = A x(t) \end{equation} $$

The solution of $\eqref{eq:main1}$ is defined as follows,

$$\boxed{x(t) = e^{At}x_0}$$

Where $e^{At}$ is the matrix exponential and $x_0$ is the initial state.

Solution of a controlled LTI problem

The controlled linear time invariant system is defined as follows.

$$ \begin{equation}\label{eq:main2} \dot{x}(t) = Ax(t) + Bu(t) \end{equation} $$

Since $\eqref{eq:main2}$ is a non-homogenous system of differential equations, we should first find the homogenous solution,

$$ x_h(t) = c e^{At} $$

Using the method of variation of parameters, we should change the constant $c$ to a function of $t$ to find the solution to the non-homogenous system, or

$$ \begin{equation}\label{eq:3} x(t) = v(t) e^{At} \end{equation} $$

Differentiating $\eqref{eq:3}$, we get

$$ \begin{equation}\label{eq:4} \dot{x}(t) = \dot{v}(t)e^{At} + Av(t)e^{At} \end{equation} $$

Substituting by $\eqref{eq:3}$ and $\eqref{eq:4}$ into $\eqref{eq:main2}$,

$$\dot{v}(t) e^{At} + Av(t)e^{At} = Av(t)e^{At}+Bu(t)$$

By simplifying, we get

$$ \begin{equation} \label{eq:5} \dot{v}(t) =e^{-At}Bu(t) \end{equation} $$

By integrating $\eqref{eq:5}$, we get

$$v(t) = a + \int_0^t e^{-A\tau}Bu(\tau) ~d\tau$$

where $a$ is the constant of integration. Since,

$$v(0) = v_0$$

Then

$$ \begin{equation}\label{eq:6} v(t) = v_0 + \int_0^t e^{-A\tau}Bu(\tau) ~d\tau \end{equation} $$

By substituting $\eqref{eq:6}$ into $\eqref{eq:3}$, we get

$$ \begin{equation}\label{eq:7} x(t) = v_0e^{At} + \int_0^t e^{A(t-\tau)}Bu(\tau) ~d\tau \end{equation} $$

But since,

$$x(0) = v(0)=x_0=v_0$$

Equation $\eqref{eq:7}$ can be written as follows

$$ \boxed{x(t) = x_0e^{At} + \int_0^t e^{A(t-\tau)}Bu(\tau) ~d\tau} $$