State Transition Matrix & Forced Response

The state equation is local: it gives the instantaneous velocity $\dot x$. The state transition matrix turns that local rule into a finite-time map. It is the bridge between differential equations, modal behavior, sampled models, controllability, and prediction.

Autonomous LTI propagation #

For

$$\dot x=Ax,\qquad x(t_0)=x_0,$$

the solution is

$$x(t)=\Phi(t,t_0)x_0.$$

For constant $A$,

$$\Phi(t,t_0)=e^{A(t-t_0)},\qquad e^{A\tau}=I+A\tau+\frac{(A\tau)^2}{2!}+\cdots.$$

This matrix has the defining properties

$$\Phi(t_0,t_0)=I,\qquad \Phi(t_2,t_0)=\Phi(t_2,t_1)\Phi(t_1,t_0),\qquad \frac{\partial\Phi}{\partial t}=A\Phi.$$

The composition law says that state is sufficient: propagating from $t_0$ to $t_2$ directly gives the same result as stopping at $t_1$ and continuing from the state there.

Inputs create a forced response #

For

$$\dot x=Ax+Bu,$$

variation of constants gives

$$x(t)=e^{A(t-t_0)}x_0+ \int_{t_0}^{t}e^{A(t-\tau)}B\,u(\tau)\,d\tau.$$

The first term is the zero-input or homogeneous response. The integral is the zero-state or forced response. A disturbance channel $Ed$ adds another integral with $E d(\tau)$. Output is then $y(t)=Cx(t)+Du(t)$.

Worked double-integrator example #

Let position $q$, velocity $v$, and commanded acceleration $u$ satisfy

$$x=\begin{bmatrix}q\\v\end{bmatrix},\quad A=\begin{bmatrix}0&1\\0&0\end{bmatrix},\quad B=\begin{bmatrix}0\\1\end{bmatrix}.$$

Because $A^2=0$, the exponential series terminates:

$$e^{At}=I+At=\begin{bmatrix}1&t\\0&1\end{bmatrix}.$$

Therefore

$$v(t)=v_0+\int_0^t u(\tau)\,d\tau,$$

$$q(t)=q_0+t v_0+\int_0^t(t-\tau)u(\tau)\,d\tau.$$

The weighting $(t-\tau)$ has a physical explanation: an early acceleration changes velocity for longer, so it accumulates more position change than the same acceleration applied near the final time.

Modes, Jordan structure, and stability #

If $A=V\Lambda V^{-1}$ is diagonalizable,

$$e^{At}=Ve^{\Lambda t}V^{-1},$$

so each eigenvalue contributes a modal factor $e^{\lambda_i t}$. Its real part sets exponential growth or decay; its imaginary part sets oscillation frequency. Repeated or defective eigenvalues add polynomial factors such as $t e^{\lambda t}$ through Jordan blocks.

For continuous LTI dynamics, asymptotic stability requires every eigenvalue of $A$ to have strictly negative real part. Eigenvalues on the imaginary axis need a finer test: a repeated or defective marginal mode can grow polynomially even though its real part is zero.

Diagonalization is an explanatory tool, not the default numerical algorithm. Nearly parallel eigenvectors make $V$ ill-conditioned, and forming $V e^{\Lambda t}V^{-1}$ can lose accuracy. Production matrix-exponential routines use scaling, squaring, Padé, or Schur-based methods.

Exact sampled propagation #

With a zero-order-held command $u(t)=u_k$ over $[kT_s,(k+1)T_s)$,

$$x_{k+1}=A_dx_k+B_du_k,$$

$$A_d=e^{AT_s},\qquad B_d=\int_0^{T_s}e^{A\tau}B\,d\tau.$$

The block exponential evaluates both without requiring $A^{-1}$:

$$\exp\left( \begin{bmatrix}A&B\\0&0\end{bmatrix}T_s \right) = \begin{bmatrix}A_d&B_d\\0&I\end{bmatrix}.$$

This matters when $A$ is singular, as in the double integrator. The common shortcut $B_d=A^{-1}(A_d-I)B$ is then unavailable even though the integral is perfectly well-defined.

Time variation changes the rule #

For $\dot x=A(t)x$, a transition matrix still exists, but in general

$$\Phi(t,t_0)\ne \exp\left(\int_{t_0}^{t}A(\tau)\,d\tau\right)$$

unless the matrices $A(t_1)$ and $A(t_2)$ commute for all relevant times. The order of infinitesimal transformations matters. Numerically integrate the matrix differential equation

$$\dot\Phi(t,t_0)=A(t)\Phi(t,t_0),\qquad \Phi(t_0,t_0)=I,$$

or use an algorithm designed for time-ordered propagation.

The transition matrix is therefore more than a closed-form trick: it is the operator that carries every initial condition, control action, and disturbance through the dynamics.