State-Variable Analysis in Control Systems

Overview of State-Variable Analysis

• Objectives of State-Variable Analysis:     * Obtain the dynamic equations of control systems.     * Model electrical circuits in state space.     * Convert a transfer function to a state-space representation (Decomposition).     * Convert a state-space representation back to a transfer function.     * Perform linear transformations between similar systems using transformation matrices.     * Find the time-domain solution of state equations.

• Comparison Between Classical and Modern Control Theory:     * Classical Control Theory: Based on simple input-output descriptions, usually expressed as a transfer function. It does not utilize knowledge of the plant's interior structure and is limited to Single-Input Single-Output (SISO) systems. Feedback control provides only limited control of closed-loop behavior.     * Modern Control Theory: Uses a "richer" description of plant dynamics called the state-space description. It provides dynamics through a set of coupled first-order differential equations in terms of internal variables (state variables) and algebraic equations for physical output variables.

Fundamental Definitions in State-Space

• State: The minimum set of variables, known as state variables, that fully describe the system and its response to any given set of inputs.
• State-Determined System Model: A mathematical description in terms of a minimum set of variables $x_i(t)$, $i = 1, \dots, n$. Together with initial conditions at time $t_0$ and system inputs for $t \geq t_0$, these are sufficient to predict future states and outputs for all $t > t_0$. In linear time-invariant systems, $t_0$ is usually chosen as zero.
• State Variables: The smallest set of variables required to completely describe the dynamic behavior of the system.
• State Vector: If $n$ state variables $\{x_1(t), x_2(t), \dots, x_n(t)\}$ are needed, they are treated as components of a vector $\mathbf{x}(t)$, called the state vector.
• State Space: An $n$-dimensional space where the coordinate axes are the state variables $x_1, x_2, \dots, x_n$. Any state of the system is represented by a specific point in this space.

State, Output, and Dynamic Equations

• State Equations: Relations governing the derivatives of state variables as functions of state variables and system inputs.     * $\dot{x}_1 = a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n + b_{11}u_1 + \dots + b_{1m}u_m$     * $\dot{x}_n = a_{n1}x_1 + a_{n2}x_2 + \dots + a_{nn}x_n + b_{n1}u_1 + \dots + b_{nm}u_m$

• Matrix Form of State Equation:$\dot{\mathbf{x}}(t) = A\mathbf{x}(t) + B\mathbf{u}(t)$     * $\mathbf{x}(t)$: State vector ($n \times 1$).     * $\mathbf{u}(t)$: Input vector ($m \times 1$).     * $A$: System matrix ($n \times n$).     * $B$: Input matrix ($n \times m$).

• Output Equations: Relations governing the system outputs as functions of state variables and inputs.     * $y_1 = c_{11}x_1 + \dots + c_{1n}x_n + d_{11}u_1 + \dots + d_{1m}u_m$

• Matrix Form of Output Equation:$\mathbf{y}(t) = C\mathbf{x}(t) + D\mathbf{u}(t)$     * $\mathbf{y}(t)$: Output vector ($k \times 1$).     * $C$: System matrix/Output matrix ($k \times n$).     * $D$: Direct-link/Feed-forward matrix ($k \times m$).

• Dynamic Equation: The combination of both the state equation and the output equation.

• Linear Independence: State variables must be linearly independent, and a minimum number of variables must be chosen. The choice of variables is not unique.

Representing Electrical Networks in State-Space

• Modeling Strategy: Typically, choose capacitor voltages ($v_c$) and inductor currents ($i_L$) as state variables because they represent the energy stored in the system (differentiated quantities).

• Example 1 (RLC Circuit):     * Inputs: $u(t)$. Output: $v_o$. State variables: $x_1 = v_c$, $x_2 = i_L$.     * Equations:         * $\dot{v}_c = \frac{1}{C}u(t) - \frac{1}{C}i_L$         * $\dot{i}_L = \frac{1}{L}v_c - \frac{R}{L}i_L$         * $y = Ri_L$     * Matrix form:         $\begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix} = \begin{pmatrix} 0 & -\frac{1}{C} \\ \frac{1}{L} & -\frac{R}{L} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} \frac{1}{C} \\ 0 \end{pmatrix} u$$y = \begin{pmatrix} 0 & R \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + [0]u$

• Example 2 (Complex RLC):     * State variables: $x_1 = v_c$, $x_2 = i_L$. Output: $v_R$.     * Equations:         * $\dot{x}_1 = -\frac{1}{RC}x_1 + \frac{1}{C}x_2$         * $\dot{x}_2 = -\frac{1}{L}x_1 + \frac{1}{L}v(t)$         * $y = x_1$     * Matrix form:         $\begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix} = \begin{pmatrix} -\frac{1}{RC} & \frac{1}{C} \\ -\frac{1}{L} & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{1}{L} \end{pmatrix} v(t)$$y = \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + [0]u$

• Example 3 (Two Resistor Network):     * Output vector: $y = [I_{R2} \, V_{R1}]^T$. State variables: $x_1 = i_L$, $x_2 = v_c$.     * Resulting matrices utilize a constant $\Delta = \frac{R1}{R1 - 4R1R2 + R2}$.     * State Equation:         $\begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix} = \begin{pmatrix} -\frac{\Delta R2}{L} & \frac{\Delta}{L} \\ -(\frac{1}{C} - \frac{\Delta R2}{CR1}) & -\frac{\Delta}{CR1} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} \frac{\Delta R2}{L} \\ (\frac{1}{C} - \frac{\Delta R2}{CR1}) \end{pmatrix} i(t)$     * Output Equation:         $\begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix} -\Delta & \frac{\Delta - 1}{R_2} \\ -\Delta R_2 & \Delta \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} \Delta \\ \Delta R_2 \end{pmatrix} i(t)$

Converting Transfer Functions to State Space (Decomposition)

• Phase Variables: A set of state variables where each subsequent variable is the derivative of the previous one.

• Direct Decomposition (Constant Numerator):     * Case Example (Ex 4): $\frac{C(s)}{R(s)} = \frac{24}{s^3 + 9s^2 + 26s + 24}$     * Differential Equation: $\dddot{c} + 9\ddot{c} + 26\dot{c} + 24c = 24r$     * State Variables: $x_1 = c$, $x_2 = \dot{c}$, $x_3 = \ddot{c}$.     * State Matrix (Phase Variable Canonical Form):         $\begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -24 & -26 & -9 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ 24 \end{pmatrix} r(t)$     * Output: $c(t) = \begin{pmatrix} 1 & 0 & 0 \end{pmatrix} \mathbf{x} + [0]r(t)$.     * Note: The last row of matrix $A$ contains the denominator coefficients in negative and reverse order.

• Direct Decomposition (Polynomial Numerator - Ex 8):     * $\frac{C(s)}{R(s)} = \frac{s^2 + 7s + 2}{s^3 + 9s^2 + 26s + 24}$.     * Method: Divide by highest power of $s$, multiply by dummy variable $X(s)$, and draw state diagram.     * System Matrix ($A$) stays in phase variable canonical form.     * Output Vector ($C$): contains the numerator coefficients ($C = [2 \, 7 \, 1]$).

• Cascaded Decomposition:     * The transfer function is factored into first-order systems (e.g., $\frac{1}{s+1}, \frac{s+4}{s+2}, \dots$) connected in series.     * Ex 11 Results: Matrices reveal system poles along the diagonal of $A$ if factored completely.     * Example Result (Ex 11): $A = \begin{pmatrix} -3 & 2 & 1 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{pmatrix}$, $B = [0 \, 0 \, 5]^T$, pole values: $-1, -2, -3$.

• Parallel Decomposition:     * Uses Partial Fraction Expansion to break the Transfer Function into parallel terms ($\frac{A}{s+p1} + \frac{B}{s+p2} + \dots$).     * Results in a Diagonal Matrix (if roots are distinct) or Jordan Canonical Form (if roots are repeated).     * Ex 13 Result: $A = \text{diag}(-1, -2, -3)$, $B = [7.5 \, -10 \, 2.5]^T$, $C = [1 \, 1 \, 1]$.     * Repeated Roots (Ex 15): $\frac{5}{(s+4)(s+3)^3}$. Leads to a Jordan block for the triple root at $-3$.         $A = \begin{pmatrix} -3 & 1 & 0 & 0 \\ 0 & -3 & 1 & 0 \\ 0 & 0 & -3 & 0 \\ 0 & 0 & 0 & -4 \end{pmatrix}$

Converting State Space to Transfer Function

• Formula:$G(s) = \frac{Y(s)}{U(s)} = C[sI - A]^{-1}B + D$
• Example 19:     * $A = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}, B = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, C = \begin{pmatrix} 0 & 1 \end{pmatrix}, D = [0]$     * $[sI - A]^{-1} = \frac{1}{(s+1)(s+2)} \begin{pmatrix} s+3 & 1 \\ -2 & s \end{pmatrix}$     * Result: $G(s) = \frac{s}{(s+1)(s+2)}$.

Characteristic Equation and Eigenvalues

• Characteristic Equation:     * From Transfer Function: Denominator polynomial set to zero.     * From State Space: $\det(sI - A) = 0$.
• Eigenvalues (\lambda): are the roots of the characteristic equation and determine system stability and time behavior.
• Eigenvectors (P): For each eigenvalue $\lambda_i$, the eigenvector $P_i$ is found by solving:     $[\lambda_i I - A]P_i = \mathbf{0}$

Linear Transformations

• Transformation Matrix (P): Relates old state vector $\mathbf{x}(t)$ to a new state vector $\mathbf{z}(t)$.     $\mathbf{x}(t) = P\mathbf{z}(t) \quad \text{or} \quad \mathbf{z}(t) = P^{-1}\mathbf{x}(t)$

• Transformed System Matrices:     * $\bar{A} = P^{-1}AP$     * $\bar{B} = P^{-1}B$     * $\bar{C} = CP$

• Diagonalization:     * Case 1: $A$ is in Phase Variable Canonical Form with distinct eigenvalues. $P$ is a Vandermonde Matrix:         $P = \begin{pmatrix} 1 & 1 & \dots & 1 \\ \lambda_1 & \lambda_2 & \dots & \lambda_n \\ \lambda_1^2 & \lambda_2^2 & \dots & \lambda_n^2 \\ \vdots & \vdots & \ddots & \vdots \end{pmatrix}$     * Case 2: $A$ is not in phase variable form. $P$ is composed of the eigenvalues' corresponding eigenvectors: $P = [P_1 \, P_2 \, \dots \, P_n]$.

• Conversion to Phase Variable Canonical Form:     * Only possible if system is controllable (Matrix $S = [B \, AB \, A^2B \, \dots \, A^{n-1}B]$ is non-singular).     * Transformation matrix $Q$ is used.

Time Domain Solution of State Equations

• Homogeneous Equation (Zero-Input Response):     * $\dot{\mathbf{x}}(t) = A\mathbf{x}(t)$     * Solution: $\mathbf{x}(t) = e^{At}\mathbf{x}(0)$

• State Transition Matrix (\Phi(t)):     * $\Phi(t) = e^{At} = \mathcal{L}^{-1}\{[sI - A]^{-1}\}$     * Properties: $\Phi(0) = I$, $\Phi^{-1}(t) = \Phi(-t)$, $\Phi(t_1 + t_2) = \Phi(t_1)\Phi(t_2)$.

• Full Solution (Homogeneous + Forced/Unit Step):     * $\mathbf{x}(t) = \Phi(t)\mathbf{x}(0) + \int_{0}^{t} \Phi(t-\tau)B\mathbf{u}(\tau) d\tau$     * In Laplace domain: $\mathbf{X}(s) = [sI - A]^{-1}\mathbf{x}(0) + [sI - A]^{-1}B\mathbf{U}(s)$

• Complex MIMO System (Ex 24):     * Input: $u(t) = [1 \, 2]^T$. Initial state: $x(0) = [1 \, 0 \, 0]^T$.     * Calculates transfer matrix components individually (e.g., $C_1R_1, C_1R_2, C_2R_1, C_2R_2$).     * Final state component solution for $x_2(t)$:         $x_2(t) = 2.2 + 0.2725e^{2.3532t} - 2.47256e^{0.8234t}\cos(1.202835t) + 2.82217e^{0.8234t}\sin(1.202835t)$

Numerical Examples Summary

• Example 19 (SS to TF): $G(s) = \frac{s}{(s+2)(s+1)}$.
• Example 20 (Transfer Matrix Calculation): Uses $\text{adj}(sI - A)$ and $\Delta = s^3 + 9s^2 + 35s + 3$ to find four individual transfer functions.
• Example 23 (Unity-feedback TF to SS): System $\frac{25}{s^2 + 6s + 25}$. After $2\,ms$, $x_1(t) = 1.00199$.
• Example 25 (Direct Decomposition of 2nd order DE): $x_1(t)$ and $x_2(t)$ at $2\,ms$ results in $1.001972$ and $0.9722$ respectively.