A state vector written in physical coordinates usually consists of variables with direct engineering interpretation, such as displacement, velocity, current, voltage, pressure, or temperature. A state vector written in modal coordinates consists of amplitudes of independent dynamical modes when the state matrix is diagonalizable.

Consider the continuous-time LTI system \( \dot{\mathbf{x} }=\mathbf{A}\mathbf{x}+\mathbf{B}\mathbf{u} \), \( \mathbf{y}=\mathbf{C}\mathbf{x}+\mathbf{D}\mathbf{u} \), where \( \mathbf{x}\in\mathbb{R}^{n} \). A nonsingular coordinate transformation is defined by

\[ \mathbf{x}=\mathbf{T}\mathbf{z},\qquad \mathbf{z}=\mathbf{T}^{-1}\mathbf{x},\qquad \det(\mathbf{T})\neq 0. \]

Here \( \mathbf{x} \) is the physical-coordinate state, \( \mathbf{z} \) is the transformed state, and \( \mathbf{T} \) is the change-of-basis matrix. In modal coordinates, \( \mathbf{T} \) is usually chosen as the eigenvector matrix of \( \mathbf{A} \).

flowchart TD
  X["Physical model: x_dot = A x + B u, y = C x + D u"] --> T["Choose nonsingular T"]
  T --> Z["Coordinate map: x = T z"]
  Z --> M["Modal model: z_dot = Tinv A T z + Tinv B u"]
  M --> Y["Output map: y = C T z + D u"]
  Y --> I["Input-output behavior unchanged"]
        

Substitute \( \mathbf{x}=\mathbf{T}\mathbf{z} \) into the physical-coordinate model. Since \( \mathbf{T} \) is constant,

\[ \dot{\mathbf{x} }=\mathbf{T}\dot{\mathbf{z} }. \]

Therefore,

\[ \mathbf{T}\dot{\mathbf{z} } = \mathbf{A}\mathbf{T}\mathbf{z}+\mathbf{B}\mathbf{u}. \]

Premultiplying by \( \mathbf{T}^{-1} \) gives the transformed state equation:

\[ \dot{\mathbf{z} } = \underbrace{\mathbf{T}^{-1}\mathbf{A}\mathbf{T} }_{\mathbf{A}_z} \mathbf{z} + \underbrace{\mathbf{T}^{-1}\mathbf{B} }_{\mathbf{B}_z} \mathbf{u}. \]

The output equation becomes

\[ \mathbf{y} = \mathbf{C}\mathbf{T}\mathbf{z}+\mathbf{D}\mathbf{u} = \underbrace{\mathbf{C}\mathbf{T} }_{\mathbf{C}_z}\mathbf{z} + \mathbf{D}\mathbf{u}. \]

Hence the transformed realization is

\[ (\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D}) \quad \sim \quad (\mathbf{A}_z,\mathbf{B}_z,\mathbf{C}_z,\mathbf{D}) = \left( \mathbf{T}^{-1}\mathbf{A}\mathbf{T}, \mathbf{T}^{-1}\mathbf{B}, \mathbf{C}\mathbf{T}, \mathbf{D} \right). \]

The matrix \( \mathbf{A}_z \) is similar to \( \mathbf{A} \). Therefore, the eigenvalues of the state matrix are preserved:

\[ \det(s\mathbf{I}-\mathbf{A}_z) = \det\!\left(s\mathbf{I}-\mathbf{T}^{-1}\mathbf{A}\mathbf{T}\right) = \det\!\left(\mathbf{T}^{-1}(s\mathbf{I}-\mathbf{A})\mathbf{T}\right) = \det(s\mathbf{I}-\mathbf{A}). \]

Suppose \( \mathbf{A}\in\mathbb{R}^{n\times n} \) has \( n \) linearly independent eigenvectors \( \mathbf{v}_1,\dots,\mathbf{v}_n \) with eigenvalues \( \lambda_1,\dots,\lambda_n \). Define

\[ \mathbf{T} = \begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_n \end{bmatrix}. \]

Since \( \mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i \),

\[ \mathbf{A}\mathbf{T} = \begin{bmatrix} \mathbf{A}\mathbf{v}_1 & \mathbf{A}\mathbf{v}_2 & \cdots & \mathbf{A}\mathbf{v}_n \end{bmatrix} = \begin{bmatrix} \lambda_1\mathbf{v}_1 & \lambda_2\mathbf{v}_2 & \cdots & \lambda_n\mathbf{v}_n \end{bmatrix} = \mathbf{T}\boldsymbol{\Lambda}, \]

where

\[ \boldsymbol{\Lambda} = \operatorname{diag}(\lambda_1,\lambda_2,\dots,\lambda_n). \]

Premultiplying \( \mathbf{A}\mathbf{T}=\mathbf{T}\boldsymbol{\Lambda} \) by \( \mathbf{T}^{-1} \) gives

\[ \mathbf{T}^{-1}\mathbf{A}\mathbf{T} = \boldsymbol{\Lambda}. \]

Therefore, in modal coordinates,

\[ \dot{\mathbf{z} } = \boldsymbol{\Lambda}\mathbf{z} + \mathbf{B}_m\mathbf{u}, \qquad \mathbf{y} = \mathbf{C}_m\mathbf{z} + \mathbf{D}\mathbf{u}, \]

with

\[ \mathbf{B}_m=\mathbf{T}^{-1}\mathbf{B}, \qquad \mathbf{C}_m=\mathbf{C}\mathbf{T}. \]

If \( \mathbf{u}=\mathbf{0} \), then the modal states satisfy decoupled scalar equations:

\[ \dot{z}_i=\lambda_i z_i,\qquad z_i(t)=e^{\lambda_i t}z_i(0),\qquad i=1,\dots,n. \]

Modal coordinates separate a state trajectory into modal amplitudes. If \( \mathbf{x}(0)=\mathbf{x}_0 \), then

\[ \mathbf{z}(0)=\mathbf{T}^{-1}\mathbf{x}_0. \]

Let \( \mathbf{w}_i^T \) denote row \( i \) of \( \mathbf{T}^{-1} \). Then

\[ z_i(0)=\mathbf{w}_i^T\mathbf{x}_0. \]

Thus \( z_i(0) \) is the amplitude of the \( i \)-th mode in the initial condition. The homogeneous physical state is reconstructed by

\[ \mathbf{x}(t) = \mathbf{T}e^{\boldsymbol{\Lambda}t}\mathbf{z}(0) = \sum_{i=1}^{n} \mathbf{v}_i e^{\lambda_i t} z_i(0). \]

The output contribution of mode \( i \) is weighted by \( \mathbf{C}\mathbf{v}_i \). Therefore,

\[ \mathbf{y}(t) = \sum_{i=1}^{n} \mathbf{C}\mathbf{v}_i e^{\lambda_i t} z_i(0) \quad \text{for } \mathbf{u}=\mathbf{0},\; \mathbf{D}=\mathbf{0}. \]

This expression explains why a mode may exist internally but appear weak at the output: its output direction \( \mathbf{C}\mathbf{v}_i \) may be small. Conversely, an input excites mode \( i \) according to the corresponding row of \( \mathbf{T}^{-1}\mathbf{B} \).

The transfer matrix of the physical realization is

\[ \mathbf{G}(s) = \mathbf{C}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{B} + \mathbf{D}. \]

The transfer matrix of the transformed realization is

\[ \mathbf{G}_z(s) = \mathbf{C}_z(s\mathbf{I}-\mathbf{A}_z)^{-1}\mathbf{B}_z + \mathbf{D}. \]

Substitute \( \mathbf{A}_z=\mathbf{T}^{-1}\mathbf{A}\mathbf{T} \), \( \mathbf{B}_z=\mathbf{T}^{-1}\mathbf{B} \), and \( \mathbf{C}_z=\mathbf{C}\mathbf{T} \):

\[ \begin{aligned} \mathbf{G}_z(s) &= \mathbf{C}\mathbf{T} \left( s\mathbf{I}-\mathbf{T}^{-1}\mathbf{A}\mathbf{T} \right)^{-1} \mathbf{T}^{-1}\mathbf{B} + \mathbf{D}. \end{aligned} \]

Since \( s\mathbf{I}-\mathbf{T}^{-1}\mathbf{A}\mathbf{T} = \mathbf{T}^{-1}(s\mathbf{I}-\mathbf{A})\mathbf{T} \),

\[ \left( s\mathbf{I}-\mathbf{T}^{-1}\mathbf{A}\mathbf{T} \right)^{-1} = \mathbf{T}^{-1}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{T}. \]

Therefore,

\[ \begin{aligned} \mathbf{G}_z(s) &= \mathbf{C}\mathbf{T} \mathbf{T}^{-1}(s\mathbf{I}-\mathbf{A})^{-1} \mathbf{T}\mathbf{T}^{-1}\mathbf{B} + \mathbf{D} \\ &= \mathbf{C}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{B} + \mathbf{D} = \mathbf{G}(s). \end{aligned} \]

Thus, a similarity transformation changes the internal coordinates but not the external input-output map.

flowchart TD
  A["Physical coordinates x"] --> B["Compute T from eigenvectors or chosen basis"]
  B --> C["z = Tinv x"]
  C --> D["Az = Tinv A T"]
  D --> E["Bz = Tinv B, Cz = C T"]
  E --> F["Analyze modes in z coordinates"]
  F --> G["Recover physical state: x = T z"]
        

The physical homogeneous solution is

\[ \mathbf{x}(t)=e^{\mathbf{A}t}\mathbf{x}(0). \]

If \( \mathbf{A}=\mathbf{T}\boldsymbol{\Lambda}\mathbf{T}^{-1} \), then

\[ e^{\mathbf{A}t} = e^{\mathbf{T}\boldsymbol{\Lambda}\mathbf{T}^{-1}t} = \mathbf{T}e^{\boldsymbol{\Lambda}t}\mathbf{T}^{-1}. \]

Because \( \boldsymbol{\Lambda} \) is diagonal,

\[ e^{\boldsymbol{\Lambda}t} = \operatorname{diag} \left( e^{\lambda_1 t}, e^{\lambda_2 t}, \dots, e^{\lambda_n t} \right). \]

Therefore, modal coordinates convert matrix-exponential computation into scalar exponential computation when \( \mathbf{A} \) is diagonalizable.

\[ \mathbf{z}(t)=e^{\boldsymbol{\Lambda}t}\mathbf{z}(0), \qquad \mathbf{x}(t)=\mathbf{T}e^{\boldsymbol{\Lambda}t}\mathbf{T}^{-1}\mathbf{x}(0). \]

Consider the second-order system

\[ \dot{\mathbf{x} } = \begin{bmatrix} 0 & 1\\ -2 & -3 \end{bmatrix} \mathbf{x} + \begin{bmatrix} 0\\ 1 \end{bmatrix}u, \qquad y= \begin{bmatrix} 1 & 0 \end{bmatrix}\mathbf{x}. \]

The characteristic polynomial is

\[ \det(\lambda\mathbf{I}-\mathbf{A}) = \lambda^2+3\lambda+2 = (\lambda+1)(\lambda+2). \]

Hence \( \lambda_1=-1 \) and \( \lambda_2=-2 \). Corresponding eigenvectors may be chosen as

\[ \mathbf{v}_1= \begin{bmatrix} 1\\ -1 \end{bmatrix}, \qquad \mathbf{v}_2= \begin{bmatrix} 1\\ -2 \end{bmatrix}. \]

Therefore,

\[ \mathbf{T} = \begin{bmatrix} 1 & 1\\ -1 & -2 \end{bmatrix}, \qquad \mathbf{T}^{-1} = \begin{bmatrix} 2 & 1\\ -1 & -1 \end{bmatrix}. \]

The modal state matrix is

\[ \boldsymbol{\Lambda} = \mathbf{T}^{-1}\mathbf{A}\mathbf{T} = \begin{bmatrix} -1 & 0\\ 0 & -2 \end{bmatrix}. \]

Also,

\[ \mathbf{B}_m = \mathbf{T}^{-1}\mathbf{B} = \begin{bmatrix} 1\\ -1 \end{bmatrix}, \qquad \mathbf{C}_m = \mathbf{C}\mathbf{T} = \begin{bmatrix} 1 & 1 \end{bmatrix}. \]

Hence the modal equations are

\[ \begin{aligned} \dot{z}_1 &= -z_1 + u,\\ \dot{z}_2 &= -2z_2 - u,\\ y &= z_1+z_2. \end{aligned} \]

For the initial condition \( \mathbf{x}(0)=\begin{bmatrix}2 & -1\end{bmatrix}^T \),

\[ \mathbf{z}(0) = \mathbf{T}^{-1}\mathbf{x}(0) = \begin{bmatrix} 2 & 1\\ -1 & -1 \end{bmatrix} \begin{bmatrix} 2\\ -1 \end{bmatrix} = \begin{bmatrix} 3\\ -1 \end{bmatrix}. \]

With \( u=0 \), the modal solution is

\[ z_1(t)=3e^{-t}, \qquad z_2(t)=-e^{-2t}. \]

The physical state is recovered by

\[ \mathbf{x}(t) = \mathbf{T}\mathbf{z}(t) = \begin{bmatrix} 3e^{-t}-e^{-2t}\\ -3e^{-t}+2e^{-2t} \end{bmatrix}. \]

If a real matrix \( \mathbf{A} \) has a complex pair \( \lambda,\bar{\lambda}=\alpha\pm j\beta \), then the complex modal form is diagonal over \( \mathbb{C} \). However, engineering state variables are often kept real. If \( \mathbf{v}=\mathbf{p}+j\mathbf{q} \) is a complex eigenvector, one can use the real basis \( \mathbf{p},\mathbf{q} \). The corresponding real modal block is

\[ \begin{bmatrix} \dot{z}_1\\ \dot{z}_2 \end{bmatrix} = \begin{bmatrix} \alpha & \beta\\ -\beta & \alpha \end{bmatrix} \begin{bmatrix} z_1\\ z_2 \end{bmatrix}. \]

This block represents exponentially weighted oscillation. The decay or growth rate is controlled by \( \alpha \), and the oscillation frequency is controlled by \( \beta \).

\[ e^{\alpha t} \begin{bmatrix} \cos(\beta t) & \sin(\beta t)\\ -\sin(\beta t) & \cos(\beta t) \end{bmatrix} \]

The detailed treatment of non-diagonalizable repeated eigenvalues is postponed to the next chapter, where Jordan chains are introduced.

The core computational steps are: compute eigenvectors, assemble \( \mathbf{T} \), check conditioning, compute \( \mathbf{T}^{-1}\mathbf{A}\mathbf{T} \), \( \mathbf{T}^{-1}\mathbf{B} \), and \( \mathbf{C}\mathbf{T} \). Recommended libraries are:

• Python: NumPy, SciPy, python-control, slycot.
• C++: Eigen, Armadillo, Blaze, LAPACK/SLICOT bindings.
• Java: EJML, Apache Commons Math, Hipparchus, ojAlgo.
• MATLAB/Simulink: Control System Toolbox, Simulink, Symbolic Math Toolbox.
• Wolfram Mathematica: Eigensystem, MatrixExp, StateSpaceModel, TransferFunctionModel.

Numerically, a modal transformation is reliable only when \( \mathbf{T} \) is not ill-conditioned:

\[ \kappa(\mathbf{T}) = \|\mathbf{T}\|\|\mathbf{T}^{-1}\|. \]

Large \( \kappa(\mathbf{T}) \) means that small perturbations in physical coordinates may become large perturbations in modal coordinates.

Chapter17_Lesson4.py

"""
Chapter17_Lesson4.py
Transformations Between Physical and Modal Coordinates
"""

import numpy as np


def modal_transform(A: np.ndarray, B: np.ndarray, C: np.ndarray, D: np.ndarray):
    eigvals, T = np.linalg.eig(A)
    if abs(np.linalg.det(T)) < 1e-10:
        raise ValueError("Eigenvector matrix is nearly singular.")
    T_inv = np.linalg.inv(T)
    Lambda = T_inv @ A @ T
    Bm = T_inv @ B
    Cm = C @ T
    return eigvals, T, T_inv, Lambda, Bm, Cm, D


def transfer_value(A, B, C, D, s: complex):
    n = A.shape[0]
    return C @ np.linalg.inv(s * np.eye(n) - A) @ B + D


def main():
    A = np.array([[0.0, 1.0], [-2.0, -3.0]])
    B = np.array([[0.0], [1.0]])
    C = np.array([[1.0, 0.0]])
    D = np.array([[0.0]])

    eigvals, T, T_inv, Lambda, Bm, Cm, Dm = modal_transform(A, B, C, D)

    print("Eigenvalues:")
    print(eigvals)
    print("\nT:")
    print(T)
    print("\nT_inv:")
    print(T_inv)
    print("\nLambda = T_inv A T:")
    print(Lambda)
    print("\nBm = T_inv B:")
    print(Bm)
    print("\nCm = C T:")
    print(Cm)

    x0 = np.array([[2.0], [-1.0]])
    z0 = T_inv @ x0
    x0_recovered = T @ z0

    print("\nx0:")
    print(x0)
    print("z0 = T_inv x0:")
    print(z0)
    print("T z0:")
    print(x0_recovered)

    s = 1.0 + 0.5j
    G_phys = transfer_value(A, B, C, D, s)
    G_modal = transfer_value(Lambda, Bm, Cm, Dm, s)

    print("\nG_phys(s):", G_phys)
    print("G_modal(s):", G_modal)
    print("difference norm:", np.linalg.norm(G_phys - G_modal))


if __name__ == "__main__":
    main()
      

Chapter17_Lesson4.cpp

#include <iostream>
#include <array>
#include <iomanip>

using Mat2 = std::array<std::array<double, 2>, 2>;
using Vec2 = std::array<double, 2>;

Mat2 multiply(const Mat2& A, const Mat2& B) {
    Mat2 C{};
    for (int i = 0; i < 2; ++i)
        for (int j = 0; j < 2; ++j)
            for (int k = 0; k < 2; ++k)
                C[i][j] += A[i][k] * B[k][j];
    return C;
}

Vec2 multiply(const Mat2& A, const Vec2& x) {
    Vec2 y{};
    for (int i = 0; i < 2; ++i)
        for (int k = 0; k < 2; ++k)
            y[i] += A[i][k] * x[k];
    return y;
}

Mat2 inverse2x2(const Mat2& A) {
    double det = A[0][0] * A[1][1] - A[0][1] * A[1][0];
    if (det > -1e-12 && det < 1e-12) {
        throw std::runtime_error("Matrix is singular or nearly singular.");
    }
    return Mat2{ {
        { { A[1][1] / det, -A[0][1] / det } },
        { {-A[1][0] / det,  A[0][0] / det } }
    } };
}

int main() {
    Mat2 A{ { { {0.0, 1.0} }, { {-2.0, -3.0} } } };
    Mat2 T{ { { {1.0, 1.0} }, { {-1.0, -2.0} } } };
    Mat2 Tinv = inverse2x2(T);
    Mat2 Lambda = multiply(multiply(Tinv, A), T);

    Vec2 B{0.0, 1.0};
    Vec2 Bm = multiply(Tinv, B);
    Vec2 Cm{T[0][0], T[0][1]};

    Vec2 x0{2.0, -1.0};
    Vec2 z0 = multiply(Tinv, x0);
    Vec2 x0Recovered = multiply(T, z0);

    std::cout << std::fixed << std::setprecision(5);
    std::cout << "Lambda = [[" << Lambda[0][0] << ", "
              << Lambda[0][1] << "], ["
              << Lambda[1][0] << ", "
              << Lambda[1][1] << "]]\n";
    std::cout << "Bm = [" << Bm[0] << ", " << Bm[1] << "]^T\n";
    std::cout << "Cm = [" << Cm[0] << ", " << Cm[1] << "]\n";
    std::cout << "z0 = [" << z0[0] << ", " << z0[1] << "]^T\n";
    std::cout << "T z0 = [" << x0Recovered[0] << ", "
              << x0Recovered[1] << "]^T\n";

    return 0;
}
      

Chapter17_Lesson4.java

public class Chapter17_Lesson4 {
    static double[][] multiply(double[][] A, double[][] B) {
        double[][] C = new double[A.length][B[0].length];
        for (int i = 0; i < A.length; i++) {
            for (int j = 0; j < B[0].length; j++) {
                for (int k = 0; k < B.length; k++) {
                    C[i][j] += A[i][k] * B[k][j];
                }
            }
        }
        return C;
    }

    static double[] multiply(double[][] A, double[] x) {
        double[] y = new double[A.length];
        for (int i = 0; i < A.length; i++) {
            for (int k = 0; k < x.length; k++) {
                y[i] += A[i][k] * x[k];
            }
        }
        return y;
    }

    static double[][] inverse2x2(double[][] A) {
        double det = A[0][0] * A[1][1] - A[0][1] * A[1][0];
        if (Math.abs(det) < 1e-12) {
            throw new IllegalArgumentException("Matrix is singular.");
        }
        return new double[][] {
            { A[1][1] / det, -A[0][1] / det },
            {-A[1][0] / det,  A[0][0] / det }
        };
    }

    public static void main(String[] args) {
        double[][] A = { {0.0, 1.0}, {-2.0, -3.0} };
        double[][] T = { {1.0, 1.0}, {-1.0, -2.0} };

        double[][] Tinv = inverse2x2(T);
        double[][] Lambda = multiply(multiply(Tinv, A), T);

        double[] B = {0.0, 1.0};
        double[] Bm = multiply(Tinv, B);
        double[] Cm = {T[0][0], T[0][1]};

        double[] x0 = {2.0, -1.0};
        double[] z0 = multiply(Tinv, x0);
        double[] x0Recovered = multiply(T, z0);

        System.out.printf("Lambda = [[%.5f, %.5f], [%.5f, %.5f]]%n",
                Lambda[0][0], Lambda[0][1], Lambda[1][0], Lambda[1][1]);
        System.out.printf("Bm = [%.5f, %.5f]^T%n", Bm[0], Bm[1]);
        System.out.printf("Cm = [%.5f, %.5f]%n", Cm[0], Cm[1]);
        System.out.printf("z0 = [%.5f, %.5f]^T%n", z0[0], z0[1]);
        System.out.printf("T z0 = [%.5f, %.5f]^T%n", x0Recovered[0], x0Recovered[1]);
    }
}
      

Chapter17_Lesson4.m

% Chapter17_Lesson4.m
% Transformations Between Physical and Modal Coordinates

clear; clc;

A = [0 1; -2 -3];
B = [0; 1];
C = [1 0];
D = 0;

[T, Lambda] = eig(A);
Tinv = inv(T);
Bm = Tinv * B;
Cm = C * T;

fprintf('T:'); disp(T);
fprintf('Lambda = inv(T)*A*T:'); disp(Tinv * A * T);
fprintf('Bm = inv(T)*B:'); disp(Bm);
fprintf('Cm = C*T:'); disp(Cm);

x0 = [2; -1];
z0 = Tinv * x0;
x0_recovered = T * z0;

disp('z0 = inv(T)*x0:'); disp(z0);
disp('T*z0:'); disp(x0_recovered);

s = 1 + 0.5i;
G_phys = C * inv(s*eye(size(A)) - A) * B + D;
G_modal = Cm * inv(s*eye(size(Lambda)) - Lambda) * Bm + D;

fprintf('G_phys(s)  = %.8f%+.8fi\n', real(G_phys), imag(G_phys));
fprintf('G_modal(s) = %.8f%+.8fi\n', real(G_modal), imag(G_modal));
fprintf('Difference norm = %.3e\n', norm(G_phys - G_modal));

% Optional Simulink model generation:
createSimulinkModel = false;
if createSimulinkModel
    model = 'Chapter17_Lesson4_Simulink';
    new_system(model); open_system(model);

    add_block('simulink/Sources/Step', [model '/Step']);
    add_block('simulink/Continuous/State-Space', [model '/Physical_State_Space']);
    add_block('simulink/Continuous/State-Space', [model '/Modal_State_Space']);

    set_param([model '/Physical_State_Space'], 'A', 'A', 'B', 'B', 'C', 'C', 'D', 'D');
    set_param([model '/Modal_State_Space'], 'A', 'Lambda', 'B', 'Bm', 'C', 'Cm', 'D', 'D');

    add_block('simulink/Sinks/Scope', [model '/Scope_Physical']);
    add_block('simulink/Sinks/Scope', [model '/Scope_Modal']);

    add_line(model, 'Step/1', 'Physical_State_Space/1');
    add_line(model, 'Step/1', 'Modal_State_Space/1');
    add_line(model, 'Physical_State_Space/1', 'Scope_Physical/1');
    add_line(model, 'Modal_State_Space/1', 'Scope_Modal/1');

    save_system(model);
end
      

Chapter17_Lesson4.nb

ClearAll["Global`*"]

A = { {0, 1}, {-2, -3} };
B = { {0}, {1} };
Cmat = { {1, 0} };
Dmat = { {0} };

{eigVals, eigVecs} = Eigensystem[A];

T = Transpose[eigVecs];
Tinv = Inverse[T];

Lambda = Simplify[Tinv . A . T];
Bm = Simplify[Tinv . B];
Cm = Simplify[Cmat . T];

MatrixForm /@ {T, Tinv, Lambda, Bm, Cm}

x0 = { {2}, {-1} };
z0 = Simplify[Tinv . x0];
x0Recovered = Simplify[T . z0];

MatrixForm /@ {z0, x0Recovered}

s = 1 + I/2;
Gphys = Simplify[Cmat . Inverse[s IdentityMatrix[2] - A] . B + Dmat];
Gmodal = Simplify[Cm . Inverse[s IdentityMatrix[2] - Lambda] . Bm + Dmat];

{Gphys, Gmodal, FullSimplify[Gphys == Gmodal]}

stateSolutionPhysical[t_] = MatrixExp[A t] . x0;
stateSolutionModal[t_] = T . MatrixExp[Lambda t] . z0;

FullSimplify[stateSolutionPhysical[t] == stateSolutionModal[t]]
      

Problem 1: Given \( \dot{\mathbf{x} }=\mathbf{A}\mathbf{x}+\mathbf{B}\mathbf{u} \) and \( \mathbf{x}=\mathbf{T}\mathbf{z} \), derive the transformed matrices \( \mathbf{A}_z \) and \( \mathbf{B}_z \).

Solution:

\[ \mathbf{T}\dot{\mathbf{z} } = \mathbf{A}\mathbf{T}\mathbf{z}+\mathbf{B}\mathbf{u}. \]

\[ \dot{\mathbf{z} } = \mathbf{T}^{-1}\mathbf{A}\mathbf{T}\mathbf{z} + \mathbf{T}^{-1}\mathbf{B}\mathbf{u}. \]

Therefore, \( \mathbf{A}_z=\mathbf{T}^{-1}\mathbf{A}\mathbf{T} \) and \( \mathbf{B}_z=\mathbf{T}^{-1}\mathbf{B} \).

Problem 2: Prove that \( \mathbf{A} \) and \( \mathbf{A}_z=\mathbf{T}^{-1}\mathbf{A}\mathbf{T} \) have the same characteristic polynomial.

Solution:

\[ \begin{aligned} \det(\lambda\mathbf{I}-\mathbf{A}_z) &= \det(\lambda\mathbf{I}-\mathbf{T}^{-1}\mathbf{A}\mathbf{T})\\ &= \det(\mathbf{T}^{-1}(\lambda\mathbf{I}-\mathbf{A})\mathbf{T})\\ &= \det(\mathbf{T}^{-1}) \det(\lambda\mathbf{I}-\mathbf{A}) \det(\mathbf{T})\\ &= \det(\lambda\mathbf{I}-\mathbf{A}). \end{aligned} \]

Problem 3: For \( \mathbf{A}=\begin{bmatrix}0 & 1\\ -6 & -5\end{bmatrix} \), find a modal transformation.

Solution: The characteristic polynomial is

\[ \lambda^2+5\lambda+6=(\lambda+2)(\lambda+3). \]

For \( \lambda_1=-2 \), choose \( \mathbf{v}_1=\begin{bmatrix}1\\-2\end{bmatrix} \). For \( \lambda_2=-3 \), choose \( \mathbf{v}_2=\begin{bmatrix}1\\-3\end{bmatrix} \). Hence

\[ \mathbf{T} = \begin{bmatrix} 1 & 1\\ -2 & -3 \end{bmatrix}, \qquad \mathbf{T}^{-1}\mathbf{A}\mathbf{T} = \begin{bmatrix} -2 & 0\\ 0 & -3 \end{bmatrix}. \]

Problem 4: Let \( \mathbf{x}(0)=\mathbf{T}\mathbf{z}(0) \). Show that the homogeneous physical-coordinate solution can be reconstructed from modal amplitudes.

Solution:

\[ \mathbf{z}(t)=e^{\boldsymbol{\Lambda}t}\mathbf{z}(0). \]

Since \( \mathbf{x}(t)=\mathbf{T}\mathbf{z}(t) \),

\[ \mathbf{x}(t) = \mathbf{T}e^{\boldsymbol{\Lambda}t}\mathbf{z}(0) = \sum_{i=1}^{n}\mathbf{v}_i e^{\lambda_i t}z_i(0). \]

Problem 5: Explain why a similarity transformation changes state variables but does not change the transfer function.

Solution: A similarity transformation is only a change of basis in the internal state space. The physical input \( \mathbf{u} \) and output \( \mathbf{y} \) remain the same. Algebraically,

\[ \mathbf{C}_z(s\mathbf{I}-\mathbf{A}_z)^{-1}\mathbf{B}_z+\mathbf{D} = \mathbf{C}(s\mathbf{I}-\mathbf{A})^{-1}\mathbf{B}+\mathbf{D}. \]

Therefore, the internal representation changes, but the input-output behavior is invariant.

A physical-coordinate model represents states in variables connected to the original engineering system, while a modal-coordinate model represents the same state trajectory as a combination of independent modes. The transformation \( \mathbf{x}=\mathbf{T}\mathbf{z} \) produces \( \mathbf{A}_z=\mathbf{T}^{-1}\mathbf{A}\mathbf{T} \), \( \mathbf{B}_z=\mathbf{T}^{-1}\mathbf{B} \), and \( \mathbf{C}_z=\mathbf{C}\mathbf{T} \). When \( \mathbf{T} \) is the eigenvector matrix, the state matrix becomes diagonal for distinct eigenvalues, allowing each modal coordinate to describe one independent exponential component. The eigenvalues, characteristic polynomial, transfer function, and input-output behavior are invariant under the transformation.