Dynamical Systems:

State Space

Table of Contents





1. State of a Dynamic System

A dynamic system is said to be state-determined if a finite set of variables, called state variables, can fully characterize its behavior over time in response to any input.


1.1. State Variables

The state of a system at time $t_0$ is a vector of variables:


$$ \vec{x}(t_0)^T = [x_1(t_0), x_2(t_0), \dots, x_n(t_0)] $$


If the system is state-determined, then knowledge of $\vec{x}(t_0)$ and the input $\vec{u}(t)$ for $t \geq t_0$ uniquely determines the output for all $t \geq t_0$.

The number of state variables $n$ equals the number of independent energy storage elements in the system.


1.2. State Equations

For general nonlinear systems, the state equations are written as:


$$ \begin{aligned} \dot{x}_1 &= f_1(x, u, t) \\ \dot{x}_2 &= f_2(x, u, t) \\ &\vdots \\ \dot{x}_n &= f_n(x, u, t) \end{aligned} $$


In this course, we restrict our attention to Linear Time-Invariant (LTI) systems, for which the state equations reduce to:


$$ \dot{x}_i = \sum_{j=1}^n a_{ij} x_j + \sum_{k=1}^p b_{ik} u_k, \quad i = 1, \cdots, n \qquad \text{or} $$


$$ \dot{x_1} = a_{11}x_1 + \cdots + a_{1n}x_n \;+\; b_{11}u_1 + \cdots + b_{1p}u_p \\ \dot{x_2} = a_{21}x_1 + \cdots + a_{2n}x_n \;+\; b_{21}u_1 + \cdots + b_{2p}u_p \\ \vdots \\ \dot{x_n} = a_{n1}x_1 + \cdots + a_{nn}x_n \;+\; b_{n1}u_1 + \cdots + b_{np}u_p $$


This is a set of $n$ first-order linear differential equations.



Matrix Form

The system equations can be written compactly as:


$$ \frac{d}{dt} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = A \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} + B \begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_p \end{bmatrix} $$


or simply:


$$ \dot{x} = A x + B u $$


where:

  • $x \in \mathbb{R}^n$: state vector
  • $u \in \mathbb{R}^p$: input vector
  • $A \in \mathbb{R}^{n \times n}$: system (state) matrix
  • $B \in \mathbb{R}^{n \times p}$: input matrix

1.3. Output Equations

The system output $y \in \mathbb{R}^q$ is typically given as a linear combination of the states and inputs:


$$ y_i = \sum_{j=1}^n c_{ij} x_j + \sum_{k=1}^p d_{ik} u_k, \quad i = 1, \dots, q \qquad \text{or} $$


$$ y_1 = c_{11}x_1 + \cdots + c_{1n}x_n \;+\; d_{11}u_1 + \cdots + d_{1p}u_p \\ y_2 = c_{21}x_1 + \cdots + c_{2n}x_n \;+\; d_{21}u_1 + \cdots + d_{2p}u_p \\ \vdots \\ y_q = c_{q1}x_1 + \cdots + c_{qn}x_n \;+\; d_{q1}u_1 + \cdots + d_{qp}u_p \\ $$



In matrix form:


$$ \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_q \end{bmatrix} = \begin{bmatrix} c_{11} & c_{12} & \cdots & c_{1n} \\ & & \vdots \\ & & \vdots \\ & & & c_{qn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \;+\; \begin{bmatrix} d_{11} & d_{12} & \cdots & d_{1p} \\ & & \vdots \\ & & \vdots \\ & & & d_{qp} \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_p \end{bmatrix} $$


or compactly:


$$ y = C x + D u $$


For many physical systems, the feedthrough matrix $D$ is zero: $D = 0$


1.4. Standard State-Space Representation

The complete state-space model of an LTI system is:


$$ \begin{aligned} \dot{x}(t) &= A x(t) + B u(t) \\ y(t) &= C x(t) + D u(t) \end{aligned} $$


This form is widely used in control systems, signal processing, and system identification.


This can be visualized as the following block diagram:




2. Homogeneous State Response

We begin by examining the system response in the absence of any input, i.e., for a zero-input system:


$$ u(t) = 0 $$


Then the state-space equation reduces to:


$$ \dot{x}(t) = A x(t) $$


2.1. Scalar Case

For intuition, consider the scalar differential equation:


$$ \dot{x}(t) = a x(t), \quad x(0) = x_0 $$


The solution is well-known:


$$ x(t) = e^{a t} x_0 $$


Verifying:


$$ \frac{d}{dt} x(t) = \frac{d}{dt} \left( e^{a t} x_0 \right) = a e^{a t} x_0 = a x(t) $$


2.2. Matrix Case

For vector-valued state $x(t) \in \mathbb{R}^n$, the equation


$$ \dot{x}(t) = A x(t), \quad x(0) = x_0 $$


has the general solution:


$$ x(t) = e^{A t} x_0 $$


where $e^{A t}$ is the matrix exponential, defined by the power series:


$$ e^{A t} = \sum_{k = 0}^{\infty} \frac{(A t)^k}{k!} $$


Just as in the scalar case, this function satisfies:


$$\frac{d}{dt}e^{At} = \frac{d}{dt} \sum_{k=0}^{\infty} \frac{(At)^k}{k!} = 0 + \sum_{k=1}^{\infty} \frac{kA^kt^{k-1}}{k!} = A\sum_{k=1}^{\infty} \frac{(At)^{k-1}}{(k-1)!} = A \sum_{k=0}^{\infty} \frac{(At)^k}{k!} = A e^{At}$$


so the solution indeed satisfies the differential equation:


$$ \frac{d}{dt} x(t) = A x(t) $$


2.3. State Transition Matrix

The matrix exponential $e^{A t}$ plays a central role in solving linear state-space models. It is called the state transition matrix, denoted:


$$ \Phi(t) = e^{A t} $$


Thus, the solution can be expressed compactly as:


$$ x(t) = \Phi(t) x(0) $$


or more generally, for any initial time $t_0$:


$$ x(t) = \Phi(t - t_0) x(t_0) $$


Properties of the State Transition Matrix

  • Identity at $t = 0$:

$$ \Phi(0) = I $$


  • Inverse property:

$$ \Phi(-t) = \Phi(t)^{-1} $$


  • Time-shifting:

$$ \Phi(t_1 + t_2) = \Phi(t_1) \Phi(t_2) $$


  • Consistency with initial condition:

$$ \begin{align*} x(t) &= \Phi(t) x(0) \\ &= \Phi(t) \Phi(-t_0) x(t_0) \\ &= \Phi(t - t_0) x(t_0) \end{align*} $$


This shows that state trajectories evolve linearly in time under the flow generated by $\Phi(t)$, governed solely by the system matrix $A$ in the absence of external inputs.




2.4. Example: Solving a Homogeneous Linear System

Given the system:


$$ \dot{x}(t) = A x(t), \qquad x(0) = x_0 $$


where


$$ A = \begin{bmatrix} -2 & 0 \\ 1 & -1 \end{bmatrix}, \qquad x_0 = \begin{bmatrix} 2 \\ 3 \end{bmatrix} $$


We wish to solve for $x(t)$ using the eigen-decomposition method:


Step 1: Eigenvalues and Eigenvectors

Find the eigenvalues of $A$:


$$ \det(A - \lambda I) = \begin{vmatrix} -2 - \lambda & 0 \\ 1 & -1 - \lambda \end{vmatrix} = (-2 - \lambda)(-1 - \lambda) $$


So the eigenvalues are:


$$ \lambda_1 = -2, \qquad \lambda_2 = -1 $$


Now compute the eigenvectors.


For $\lambda_1 = -2$:


$$ A - (-2)I = A + 2I = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} $$


Solve:


$$ \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \quad \Rightarrow \quad x_1 = -x_2 $$


An eigenvector is:


$$ \vec{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} $$


For $\lambda_2 = -1$:


$$ A + I = \begin{bmatrix} -1 & 0 \\ 1 & 0 \end{bmatrix} $$


Solve:


$$ \begin{bmatrix} -1 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \quad \Rightarrow \quad x_1 = 0 $$


An eigenvector is:


$$ \vec{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$


Step 2: Diagonalization of $A$

Construct the modal matrix $S$ from the eigenvectors and diagonal matrix $\Lambda$ from the eigenvalues:


$$ S = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}, \quad \Lambda = \begin{bmatrix} -2 & 0 \\ 0 & -1 \end{bmatrix} $$


Then


$$ e^{A t} = S e^{\Lambda t} S^{-1} $$


We compute:


$$ e^{\Lambda t} = \begin{bmatrix} e^{-2t} & 0 \\ 0 & e^{-t} \end{bmatrix} $$


Find $S^{-1}$:


$$ S^{-1} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} $$


Step 3: Compute the Solution

Now use:


$$ x(t) = e^{At} x_0 = S e^{\Lambda t} S^{-1} x_0 $$


Substitute all known values:


$$ x(t) = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} e^{-2t} & 0 \\ 0 & e^{-t} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix} $$


Evaluate step by step:


(1) Multiply $S^{-1}$ by $x_0$:


$$ \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \end{bmatrix} $$


(2) Multiply by $e^{\Lambda t}$:


$$ \begin{bmatrix} e^{-2t} & 0 \\ 0 & e^{-t} \end{bmatrix} \begin{bmatrix} 2 \\ 5 \end{bmatrix} = \begin{bmatrix} 2e^{-2t} \\ 5e^{-t} \end{bmatrix} $$


(3) Multiply by $S$:


$$ \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 2e^{-2t} \\ 5e^{-t} \end{bmatrix} = \begin{bmatrix} 2e^{-2t} \\ -2e^{-2t} + 5e^{-t} \end{bmatrix} = \begin{bmatrix} 2e^{-2t} \\ 5e^{-t} - 2e^{-2t} \end{bmatrix} $$


Solution:


$$ x(t) = \begin{bmatrix} 2e^{-2t} \\ 5e^{-t} - 2e^{-2t} \end{bmatrix} $$


In [ ]:
% method 2: use 'lsim'

A = [-2, 0; 1, -1];
B = [0, 0]';
C = [1, 0;
     0, 1];
D = 0;

G = ss(A,B,C,D);

x0 = [2; 3];

t = linspace(0,10,500);
u = zeros(size(t));

[y, tout] = lsim(G,u,t,x0);

plot(tout,y,tout,zeros(size(tout)),'k--')
xlabel('time')
legend('x_1', 'x_2')












No description has been provided for this image






















3. Forced State Response of LTI Systems
In addition to the homogeneous (zero-input) behavior, we now consider how a linear time-invariant (LTI) system responds to an external input.




Let the system be described by:




$$
\dot{x}(t) = A x(t) + B u(t)
$$




with initial condition $x(0)$.




3.1. First-Order Scalar Case
Consider a scalar system:




$$
\dot{x}(t) - a x(t) = b u(t)
$$




Multiply both sides by the integrating factor $e^{-at}$:




$$
e^{-at} \dot{x}(t) - a e^{-at} x(t) = e^{-at} b u(t)
$$




Recognize the left-hand side as the derivative of a product:




$$
\frac{d}{dt} \left( e^{-at} x(t) \right) = b e^{-at} u(t)
$$




Integrate both sides from $0$ to $t$:




$$
e^{-at} x(t) - x(0) = \int_0^t b e^{-a\tau} u(\tau) d\tau
$$




Solve for $x(t)$:




$$
x(t) = e^{at} x(0) + \int_0^t e^{a(t - \tau)} b u(\tau) d\tau
$$




This gives the complete solution:


    

  • The homogeneous response $e^{at} x(0)$ depends on the initial condition
    • 

  • The forced response is the convolution of the input with the system's impulse response
    • 





3.2. General Vector Case
For higher-order systems:




$$
\dot{x}(t) = A x(t) + B u(t), \quad x(0) = x_0
$$




The solution is:




$$
x(t) = e^{At} x(0) + \int_0^t e^{A(t - \tau)} B u(\tau) d\tau
$$




This is the complete state response, where:


    

  • $e^{At} x(0)$ is the zero-input response
    • 

  • $\int_0^t e^{A(t - \tau)} B u(\tau) d\tau$ is the zero-state response
    • 





The output is given by:




$$
y(t) = C x(t) + D u(t)
$$




Substitute $x(t)$:




$$
y(t) = C e^{At} x(0) + C \int_0^t e^{A(t - \tau)} B u(\tau) d\tau + D u(t)
$$




Interpretation:


    

  • $C e^{At} x(0)$: response due to initial conditions
    • 

  • $C e^{At} B$: impulse response of the system
    • 

  • Convolution integral: response to input $u(t)$
    • 

  • $D u(t)$: feedthrough term (direct influence of input on output)
    • 







3.3. Alternative Notation Using State Transition Matrix
Let $\Phi(t) = e^{At}$ denote the state transition matrix.




Then the complete state response is:




$$
x(t) = \Phi(t) x(0) + \int_0^t \Phi(t - \tau) B u(\tau) d\tau
$$




Verifying this solution:




At $t = 0$:




$$
x(0) = \Phi(0)x(0) + \int_0^0 \Phi(0 - \tau) B u(\tau) d\tau = x(0)
$$




Differentiating using Leibniz's rule:




$$
\frac{d}{dt} \int_{0}^{t} \Phi(t - \tau) B u(\tau) d\tau
= \Phi(0) B u(t) + \int_{0}^{t} \frac{\partial}{\partial t} \Phi(t - \tau) B u(\tau) d\tau
$$




Since $\frac{\partial}{\partial t} \Phi(t - \tau) = A \Phi(t - \tau)$:




$$
\frac{d}{dt} x(t)
= A \Phi(t) x(0) + A \int_0^t \Phi(t - \tau) B u(\tau) d\tau + B u(t)
= A x(t) + B u(t)
$$




Thus, the expression satisfies the original system.




















3.4. Example: Forced Response of an LTI System
Consider the system:




$$
\begin{bmatrix}
\dot{x}_1 \\ \dot{x}_2
\end{bmatrix}
=
\underbrace{\begin{bmatrix}
-2 & 0 \\
1 & -1
\end{bmatrix}}_{A}
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
+
\underbrace{\begin{bmatrix}
1 \\ 0
\end{bmatrix}}_{B}
\cdot u(t), \qquad
u(t) = 2 \cdot u(t), \quad
x(0) =
\begin{bmatrix}
0 \\ 0
\end{bmatrix}
$$






Step 1: Use the State Transition Matrix


Since $x(0) = 0$, the state response is:




$$
x(t) = \int_0^t \Phi(t - \tau) B u(\tau) \, d\tau
$$




Given $u(\tau) = 2$ for $\tau \geq 0$, we get:




$$
x(t) = 2 \int_0^t \Phi(t - \tau) B \, d\tau
$$






Step 2: Given $\Phi(t)$


The state transition matrix is:




$$
\Phi(t) = e^{At} =
\begin{bmatrix}
e^{-2t} & 0 \\
e^{-t} - e^{-2t} & e^{-t}
\end{bmatrix}
$$




Then:




$$
x(t) = \Phi(t) \int_0^t \Phi(-\tau) B \cdot 2 \, d\tau
$$






Step 3: Evaluate the Integral


Compute:




$$
\Phi(-\tau)B =
\begin{bmatrix}
e^{2\tau} \\ e^{\tau} - e^{2\tau}
\end{bmatrix}
$$




Then:




$$
\int_0^t 2
\begin{bmatrix}
e^{2\tau} \\
e^{\tau} - e^{2\tau}
\end{bmatrix} d\tau
=
\begin{bmatrix}
e^{2t} - 1 \\
2e^t - e^{2t} - 1
\end{bmatrix}
$$






Step 4: Multiply by $\Phi(t)$


Now:




$$
x(t) = \Phi(t)
\begin{bmatrix}
e^{2t} - 1 \\
2e^t - e^{2t} - 1
\end{bmatrix}
$$




Compute:




$$
x_1(t) = e^{-2t}(e^{2t} - 1) = 1 - e^{-2t}
$$




$$
x_2(t) = (e^{-t} - e^{-2t})(e^{2t} - 1) + e^{-t}(2e^t - e^{2t} - 1) = 1 + e^{-2t}
$$






Solution




$$
x(t) =
\begin{bmatrix}
1 - e^{-2t} \\
1 + e^{-2t}
\end{bmatrix}
$$




















Appendix: Leibniz Integral Rule
Let $\mathbf{F}(t)$ be defined as an integral with time-varying limits:




$$
\mathbf{F}(t) = \int_{a(t)}^{b(t)} f(t, \tau) \, d \tau
$$




Then the total derivative of $\mathbf{F}(t)$ is given by:




$$
\frac{d \mathbf{F}(t)}{d t} = \int_{a(t)}^{b(t)} \frac{\partial f(t, \tau)}{\partial t} \, d \tau + f(t, b(t)) \frac{d b(t)}{d t} - f(t, a(t)) \frac{d a(t)}{d t}
$$






Proof Sketch




We proceed in stages to build intuition.




Case 1: Univariate Function with Time in Upper or Lower Limit


Let $f(\tau)$ be a scalar function of one variable.


(a) If:




$$
F(t) = \int_{t_0}^{t} f(\tau) \, d\tau
$$




Then by the Fundamental Theorem of Calculus:




$$
\frac{dF(t)}{dt} = f(t)
$$




(b) If:




$$
F(t) = \int_{t}^{t_0} f(\tau) \, d\tau
$$




Then:




$$
\frac{dF(t)}{dt} = -f(t)
$$




Case 2: Two-Variable Function with Fixed Limits


Let $f(t, \tau)$ be a function of two variables and define:




$$
F(t) = \int_{a}^{b} f(t, \tau) \, d\tau \quad \text{(where $a, b$ are constant)}
$$




Then the derivative is:




$$
\frac{dF(t)}{dt} = \int_{a}^{b} \frac{\partial f(t, \tau)}{\partial t} \, d\tau
$$




Case 3: Two-Variable Function with Time-Dependent Limits


Let $a(t)$ and $b(t)$ be differentiable functions of $t$, and define:




$$
\mathbf{F}(t) = \int_{a(t)}^{b(t)} f(t, \tau) \, d \tau
$$




Then by the general form of Leibniz’s Rule:




$$
\frac{d \mathbf{F}(t)}{d t}
= \int_{a(t)}^{b(t)} \frac{\partial f(t, \tau)}{\partial t} \, d \tau
+ f(t, b(t)) \frac{d b(t)}{d t}
- f(t, a(t)) \frac{d a(t)}{d t}
$$




Special Case


If $a(t) = t_0$ (constant), $b(t) = t$:




$$
\mathbf{F}(t) = \int_{t_0}^{t} f(t, \tau) \, d\tau
$$




Then:




$$
\frac{d \mathbf{F}(t)}{dt}
= \int_{t_0}^{t} \frac{\partial f(t, \tau)}{\partial t} \, d\tau + f(t, t)
$$




This version is especially useful when evaluating solutions to time-domain convolution integrals in LTI systems.
















Note: Leibniz Integral Rule


If $\mathbf{F}(t) = \int_{a(t)}^{b(t)} f(t, \tau) d \tau$, then


$$\frac{d \mathbf{F}}{d t} = \int_{a(t)}^{b(t)} \frac{\partial f(t, \tau)}{\partial t} d \tau + f(t, b(t)) \frac{d b}{d t} - f(t, a(t)) \frac{d a}{d t}$$




Proof.


Suppose $f$ is a univariate function



$$
\begin{align*}
F(t) = \int_{t_0}^{t} f(\tau) d \tau &\quad \implies \quad
\frac{d F (t)}{d t} = f(t) \\\\
F(t) = \int_{t}^{t_0} f(\tau) d \tau &\quad \implies \quad
\frac{d F (t)}{d t} = -f(t)
\end{align*}
$$



Now suppose $f$ is a function of two variables ($a$ and $b$ are constant)



$$
\begin{align*}
F(t) = \int_{a}^{b} f(t, \tau) d \tau \quad \implies \quad
\frac{d F(t)}{d t} = \int_{a}^{b} \frac{\partial f(t, \tau)}{\partial t} d \tau
\end{align*}
$$



Now suppose the limits of integration $a$ and $b$ themselves depend on $t$,



$$\mathbf{F}(t) = F(t, a(t), b(t)) = \int_{a(t)}^{b(t)} f(t, \tau) d \tau$$



$$
\begin{align*}
\frac{d \mathbf{F}(t)}{d t} &= \frac{\partial F}{\partial t} + \frac{\partial F}{\partial a} \frac{d a}{d t} + \frac{\partial F}{\partial b}\frac{d b}{d t} \\\\
&= \int_{a(t)}^{b(t)} \frac{\partial f(t, \tau)}{\partial t} d \tau + f(t, b(t))\frac{d b}{d t} - f(t,a(t))\frac{d a}{d t}\\\\
\frac{d}{dt} \int_{t_0}^{t} f(t,\tau) d \tau &=  \int_{t_0}^{t} \frac{\partial f(t,\tau)}{\partial t}  d \tau + f(t,t)
\end{align*}
$$


















3.5. Response of LTI Systems to Singularity Inputs
The general state-space solution to an LTI system is:




$$
x(t) = e^{At}x(0) + \int_0^t e^{A(t - \tau)} Bu(\tau)\, d\tau
$$




This expression shows the total response as a combination of the homogeneous solution and the forced response due to input $u(t)$.




3.5.1. Impulse Input
Suppose the input is an impulse:




$$
u(t) = K\delta(t), \quad
K =
\begin{bmatrix}
k_1 \\
k_2 \\
\vdots \\
k_p
\end{bmatrix}
$$




Then the state becomes:




$$
\begin{align*}
x(t) &= e^{At}x(0) + \int_0^t e^{A(t - \tau)} B K \delta(\tau) \, d\tau \\
&= e^{At}x(0) + e^{At} B K \\
&= e^{At} \left(x(0) + BK\right)
\end{align*}
$$




Interpretation: The effect of an impulse at $t = 0$ is equivalent to a jump in the initial state:




$$
x(0^+) = x(0^-) + BK
$$




This is particularly useful in modeling discontinuities in physical systems such as impulsive forces.




$$
u(t) = Ku_{\text{step}} =
\begin{bmatrix}
k_1 \\
k_2 \\
\vdots \\
k_p
\end{bmatrix} u_{\text{step}}(t)
$$




3.5.2. Step Input
Now consider a step input:




$$
u(t) = K \cdot u_{\text{step}}(t)  =
\begin{bmatrix}
k_1 \\
k_2 \\
\vdots \\
k_p
\end{bmatrix} u_{\text{step}}(t)
$$




Then the response is:




$$
\begin{align*}
x(t) &= e^{At}x(0) + \int^t_0 e^{A(t-\tau)}Bu(\tau)d\tau \\
&= e^{At}x(0) + \int^t_0 e^{A(t-\tau)}BKd\tau \\
&= e^{At}x(0) + e^{At}\int^t_0 e^{-A\tau}d\tau BK \\
&= e^{At}x(0) + e^{At}[-A^{-1}e^{-A\tau} \big\rvert^t_0] BK \\
&= e^{At}x(0) + e^{At}[-A^{-1}e^{-At} + A^{-1}] BK \\
&= e^{At}x(0) + e^{At}A^{-1}[I - e^{-At}] BK \\
&= e^{At}x(0) + A^{-1}[e^{At} - I] BK \qquad (Ae^{At} = e^{At}A)
\end{align*}
$$




Steady-State Behavior


If the system is stable (i.e., all eigenvalues of $A$ have negative real parts), then:




$$
\lim_{t \to \infty} e^{At} = 0 \quad \Rightarrow \quad
\lim_{t \to \infty} x(t) = -A^{-1} BK
$$




This gives the final state in response to a constant step input.




















Example: Step Response of a Mass-Spring-Damper System


Consider the second-order differential equation of a damped mass-spring system under a step input force:




$$
\ddot{y} + \frac{c}{m} \dot{y} + \frac{k}{m} y = g \cdot u_{\text{step}}(t)
$$




Step 1: State-Space Representation


Let us define state variables:




$$
x_1 = y, \quad x_2 = \dot{y}
$$




Then the system can be written in state-space form:




$$
\dot{x} =
\begin{bmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{bmatrix}
=
\underbrace{\begin{bmatrix}
0 & 1 \\
-\frac{k}{m} & -\frac{c}{m}
\end{bmatrix}}_{A}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
+
\underbrace{\begin{bmatrix}
0 \\
g
\end{bmatrix}}_{B}
u(t)
$$




with output equation:




$$
y(t) =
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
= x_1(t)
$$




Step 2: Steady-State Response to Step Input


For a unit step input, $u(t) = 1 \cdot u_{\text{step}}(t)$, we use the steady-state result for LTI systems:




$$
x(\infty) = -A^{-1} B \cdot 1
$$




Compute the inverse of the matrix $A$:




$$
A = \begin{bmatrix}
0 & 1 \\
-\frac{k}{m} & -\frac{c}{m}
\end{bmatrix}, \quad
\det(A) = 0 \cdot \left(-\frac{c}{m}\right) - 1 \cdot \left(-\frac{k}{m}\right) = \frac{k}{m}
$$




Using the $2 \times 2$ inverse formula:




$$
A^{-1} =
\frac{1}{\frac{k}{m}}
\begin{bmatrix}
-\frac{c}{m} & -1 \\
\frac{k}{m} & 0
\end{bmatrix}
=
\frac{m}{k}
\begin{bmatrix}
-\frac{c}{m} & -1 \\
\frac{k}{m} & 0
\end{bmatrix}
=
\begin{bmatrix}
-\frac{c}{k} & -\frac{m}{k} \\
1 & 0
\end{bmatrix}
$$




Then:




$$
x(\infty) = -A^{-1} B K =
- \begin{bmatrix}
-\frac{c}{k} & -\frac{m}{k} \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
0 \\
g
\end{bmatrix} \cdot 1
=
- \begin{bmatrix}
-\frac{m}{k} g \\
0
\end{bmatrix}
=
\begin{bmatrix}
\frac{m g}{k} \\
0
\end{bmatrix}
$$




Interpretation


    

  • The steady-state displacement of the mass is:
    • 





$$
y(\infty) = x_1(\infty) = \frac{mg}{k}
$$




    

  • The velocity goes to zero as $t \to \infty$, i.e., $x_2(\infty) = 0$
    • 





This result is physically intuitive: the final displacement under a constant force $mg$ balances the spring force:




$$
k y = mg \quad \Rightarrow \quad y = \frac{mg}{k}
$$






























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