3.06 State Transition Matrix Properties and Solution Methods

Related Concepts: 3.01 Introduction to State-Space & Phase-Variable Formulations | 3.03 State-Space Modeling of Electrical Circuits | 3.04 State-Space Modeling of Block Diagrams & Signal Flow Graphs

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3.06 State Transition Matrix: Properties & Solution Methods

Core Idea

The State Transition Matrix (denoted $\varphi (t)$ or $e^{\mathbf{A}t}$) is a matrix-valued function of time that completely characterizes the unforced (zero-input) natural dynamic response of a continuous-time LTI state-space system. It maps the state vector from its initial state at $t=0$ to its state at any future time $t$ {acting as the matrix-valued equivalent of the scalar exponential decay factor in first-order systems}.

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1. Mathematical Definition of the Matrix Exponential

For a linear time-invariant continuous-time system governed by the state equation: $\dot{\mathbf{X}}(t)=\mathbf{AX}(t)+\mathbf{BU}(t)$

When the external input is deactivated ($\mathbf{U}(t)=0$), the system becomes homogeneous: $\dot{\mathbf{X}}(t)=\mathbf{AX}(t)$

In a simple first-order scalar system, $\dot{x}(t)=ax(t)$ has the solution $x(t)=e^{at}x(0)$. By mathematical extension, the solution to the vector-matrix differential equation is defined using the matrix exponential $e^{\mathbf{A}t}$: $\mathbf{X}(t)=e^{\mathbf{A}t}\mathbf{X}(0^{-})$

Thus, we define the State Transition Matrix (STM) $\varphi (t)$ as: $\varphi (t)\triangleq e^{\mathbf{A}t}$

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2. Fundamental Properties of the State Transition Matrix

The matrix exponential $\varphi (t)=e^{\mathbf{A}t}$ satisfies several critical algebraic properties that are heavily tested in ECE 2107 Section B theoretical questions.

Property	Mathematical Formulation	Physical/Analytical Meaning
1. Identity	$\varphi (0)=e^{\mathbf{A}\cdot 0}=\mathbf{I}$	At $t=0$, the state vector is mapped to itself (no transition has occurred).
2. Group / Additive	$\varphi (t_1+t_2)=\varphi (t_1)\varphi (t_2)=\varphi (t_2)\varphi (t_1)$	Transitioning from $0$ to $t_1+t_2$ is identical to transitioning to $t_2$, then $t_1$.
3. Inverse	${\varphi }^{-1}(t)=[e^{\mathbf{A}t}{]}^{-1}=e^{-\mathbf{A}t}=\varphi (-t)$	Transitioning backward in time is equivalent to running the inverse transition.
4. Transitive	$\varphi (t_1-t_2)\varphi (t_2-t_3)=\varphi (t_1-t_3)$	Chain multiplication allows intermediate state-taps to be bypassed.
5. Power	$[\varphi (t){]}^k=[e^{\mathbf{A}t}{]}^k=e^{\mathbf{A}kt}=\varphi (kt)$	Multiplying the transition $k$ times scales the transition duration by $k$.
6. Derivative	$\frac{d}{dt}\varphi (t)=\mathbf{A}\varphi (t)=\varphi (t)\mathbf{A}$	The rate of state transition matches the system dynamics matrix $\mathbf{A}$.

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3. Method 1: The Laplace Resolvent Method (Analytical)

The Laplace Resolvent Method is the most robust analytical technique used to calculate $\varphi (t)$ for systems of any order. It bypasses infinite series approximations by moving into the $s$-domain.

3.1 Rigorous Mathematical Derivation

We begin with the unforced, homogeneous system state equation: $\dot{\mathbf{X}}(t)=\mathbf{AX}(t)$

Taking the unilateral Laplace transform on both sides, accounting for non-zero initial conditions $\mathbf{X}(0^{-})$: $s\mathbf{X}(s)-\mathbf{X}(0^{-})=\mathbf{AX}(s)$

Gathering the $\mathbf{X}(s)$ terms to the left-hand side: $s\mathbf{X}(s)-\mathbf{AX}(s)=\mathbf{X}(0^{-})$

Since $s$ is a scalar and $\mathbf{A}$ is an $n\times n$ matrix, we introduce the $n\times n$ identity matrix $\mathbf{I}$ {to make the matrix subtraction dimensionally valid}: $(s\mathbf{I}-\mathbf{A})\mathbf{X}(s)=\mathbf{X}(0^{-})$

We solve for the state vector $\mathbf{X}(s)$ by multiplying both sides by the inverse matrix $(s\mathbf{I}-\mathbf{A}{)}^{-1}$ (known as the Resolvent Matrix): $\mathbf{X}(s)=(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{X}(0^{-})$

To return to the time domain, we apply the inverse Laplace transform (${\mathcal{L}}^{-1}$): $\mathbf{X}(t)={\mathcal{L}}^{-1}\{(s\mathbf{I}-\mathbf{A}{)}^{-1}\}\mathbf{X}(0^{-})$

Comparing this to our initial definition of the state transition, $\mathbf{X}(t)=\varphi (t)\mathbf{X}(0^{-})$, we prove the fundamental relation: $\varphi (t)=e^{\mathbf{A}t}={\mathcal{L}}^{-1}\{(s\mathbf{I}-\mathbf{A}{)}^{-1}\}$

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4. Method 2: The Infinite Power Series Method

For certain simple matrices (especially upper/lower triangular or diagonal companion matrices), we can evaluate $\varphi (t)$ directly by computing the Taylor series expansion of the matrix exponential: $\varphi (t)=e^{\mathbf{A}t}=\mathbf{I}+\mathbf{A}t+\frac{{\mathbf{A}}^2{t}^2}{2!}+\frac{{\mathbf{A}}^3{t}^3}{3!}+\cdots ={\sum }_{n=0}^{\infty }\frac{{\mathbf{A}}^n{t}^n}{n!}$

4.1 Series Derivation Example [PYQ 2022 / 2025]

Question: Evaluate the state transition matrix using the infinite series method for: $\mathbf{A}=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

Step-by-Step Solution:

1. Compute Succeeding Powers of $\mathbf{A}$: ${\mathbf{A}}^1=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ ${\mathbf{A}}^2=\mathbf{A}\cdot \mathbf{A}=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$ ${\mathbf{A}}^3={\mathbf{A}}^2\cdot \mathbf{A}=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$ By mathematical induction, the $n$-th power of $\mathbf{A}$ is: ${\mathbf{A}}^n=\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$

2. Substitute into the Taylor Series Expansion: $\varphi (t)={\sum }_{n=0}^{\infty }\frac{{\mathbf{A}}^n{t}^n}{n!}=\left[\begin{array}{cc}{\sum }_{n=0}^{\infty }\frac{t^n}{n!}& {\sum }_{n=0}^{\infty }\frac{n{t}^n}{n!}\\ 0& {\sum }_{n=0}^{\infty }\frac{t^n}{n!}\end{array}\right]$

3. Evaluate the Scalar Series Sums:

   • The diagonal elements are the standard exponential expansion: ${\sum }_{n=0}^{\infty }\frac{t^n}{n!}=e^t$
   • The upper-right element is evaluated by simplifying the $n$ coefficient: ${\sum }_{n=0}^{\infty }\frac{n{t}^n}{n!}={\sum }_{n=1}^{\infty }\frac{n{t}^n}{n(n-1)!}=t{\sum }_{n=1}^{\infty }\frac{t^{n-1}}{(n-1)!}=t{e}^t$

4. Assemble the Final STM Matrix: $\varphi (t)=e^{\mathbf{A}t}=\left[\begin{array}{cc}{e}^t& t{e}^t\\ 0& e^t\end{array}\right]$

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5. State-Space to Transfer Function Conversion Derivation

Another heavily tested theoretical derivation in ECE 2107 is converting a state-space model back to an $s$-domain transfer function $H(s)$.

5.1 Rigorous Mathematical Derivation

Given the complete continuous-time system model: $\dot{\mathbf{X}}(t)=\mathbf{AX}(t)+\mathbf{BU}(t)$ $\mathbf{Y}(t)=\mathbf{CX}(t)+\mathbf{DU}(t)$

1. Apply Laplace Transform assuming Zero Initial Conditions ($\mathbf{X}(0^{-})=0$): $s\mathbf{X}(s)=\mathbf{AX}(s)+\mathbf{BU}(s)$ $(s\mathbf{I}-\mathbf{A})\mathbf{X}(s)=\mathbf{BU}(s)$

2. Isolate the State Vector $\mathbf{X}(s)$: $\mathbf{X}(s)=(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{BU}(s)$

3. Substitute $\mathbf{X}(s)$ into the Output Equation: $\mathbf{Y}(s)=\mathbf{C}\left[(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{BU}(s)\right]+\mathbf{DU}(s)$ $\mathbf{Y}(s)=\left[\mathbf{C}(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{B}+\mathbf{D}\right]\mathbf{U}(s)$

4. Extract the Transfer Function $H(s)$: $H(s)=\frac{\mathbf{Y}(s)}{\mathbf{U}(s)}=\mathbf{C}(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{B}+\mathbf{D}$

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6. Comprehensive Worked Example (The Exam Killer)

6.1 Solving the Standard Transition and Transfer Function Matrix [PYQ 2024 / 2022]

Question: Given the state-space matrices of a continuous-time system: $\mathbf{A}=\left[\begin{array}{cc}-1& 0\\ 1& -2\end{array}\right],\mathbf{B}=\left[\begin{array}{c}1\\ 0\end{array}\right],\mathbf{C}=\left[\begin{array}{cc}1& 1\end{array}\right],\mathbf{D}=[0]$ Find:

1. The s-domain transfer function $H(s)$.
2. The state transition matrix $\varphi (t)=e^{\mathbf{A}t}$.

Part 1 Solution: Transfer Function $H(s)$

1. Formulate $(s\mathbf{I}-\mathbf{A})$: $s\mathbf{I}-\mathbf{A}=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}-1& 0\\ 1& -2\end{array}\right]=\left[\begin{array}{cc}s+1& 0\\ -1& s+2\end{array}\right]$

2. Compute the Resolvent Matrix Inverse $(s\mathbf{I}-\mathbf{A}{)}^{-1}$:

   • Determinant: $\text{Det}(s\mathbf{I}-\mathbf{A})=(s+1)(s+2)-(0)(-1)=(s+1)(s+2)$
   • Adjoint: $\text{Adj}(s\mathbf{I}-\mathbf{A})=\left[\begin{array}{cc}s+2& 0\\ 1& s+1\end{array}\right]$
   • Inverse: $(s\mathbf{I}-\mathbf{A}{)}^{-1}=\frac{1}{(s+1)(s+2)}\left[\begin{array}{cc}s+2& 0\\ 1& s+1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{s+1}& 0\\ \frac{1}{(s+1)(s+2)}& \frac{1}{s+2}\end{array}\right]$

3. Perform Matrix Multiplication for $H(s)=\mathbf{C}(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{B}$: $H(s)=\left[\begin{array}{cc}1& 1\end{array}\right]\left[\begin{array}{cc}\frac{1}{s+1}& 0\\ \frac{1}{(s+1)(s+2)}& \frac{1}{s+2}\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]$ Multiply $\mathbf{C}$ with $(s\mathbf{I}-\mathbf{A}{)}^{-1}$ first: $\mathbf{C}(s\mathbf{I}-\mathbf{A}{)}^{-1}=\left[\begin{array}{cc}\frac{1}{s+1}+\frac{1}{(s+1)(s+2)}& \frac{1}{s+2}\end{array}\right]=\left[\begin{array}{cc}\frac{(s+2)+1}{(s+1)(s+2)}& \frac{1}{s+2}\end{array}\right]=\left[\begin{array}{cc}\frac{s+3}{(s+1)(s+2)}& \frac{1}{s+2}\end{array}\right]$ Now multiply by $\mathbf{B}$: $H(s)=\left[\begin{array}{cc}\frac{s+3}{(s+1)(s+2)}& \frac{1}{s+2}\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\frac{\mathbf{s}+3}{(\mathbf{s}+1)(\mathbf{s}+2)}$

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Part 2 Solution: State Transition Matrix $\varphi (t)$

We calculate $\varphi (t)={\mathcal{L}}^{-1}\{(s\mathbf{I}-\mathbf{A}{)}^{-1}\}$ by performing partial fraction expansion on the individual terms of the resolvent matrix.

1. Analyze Resolvent Matrix Elements: $(s\mathbf{I}-\mathbf{A}{)}^{-1}=\left[\begin{array}{cc}\frac{1}{s+1}& 0\\ \frac{1}{(s+1)(s+2)}& \frac{1}{s+2}\end{array}\right]$

2. Partial Fraction Expansion for element $(2,1)$: $\frac{1}{(s+1)(s+2)}=\frac{K_1}{s+1}+\frac{K_2}{s+2}$ $K_1={\frac{1}{s+2}|}_{s=-1}=1$ $K_2={\frac{1}{s+1}|}_{s=-2}=-1$ $⟹\frac{1}{(s+1)(s+2)}=\frac{1}{s+1}-\frac{1}{s+2}$

3. Take Inverse Laplace of Each Element:

   • ${\mathcal{L}}^{-1}\left\{\frac{1}{s+1}\right\}=e^{-t}$
   • ${\mathcal{L}}^{-1}\{0\}=0$
   • ${\mathcal{L}}^{-1}\left\{\frac{1}{s+2}\right\}=e^{-2t}$
   • ${\mathcal{L}}^{-1}\left\{\frac{1}{(s+1)(s+2)}\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s+1}-\frac{1}{s+2}\right\}=e^{-t}-e^{-2t}$

4. Reassemble the Time-Domain Matrix $\varphi (t)$: $\varphi (t)=\left[\begin{array}{cc}{e}^{-t}& 0\\ e^{-t}-e^{-2t}& e^{-2t}\end{array}\right]u(t)$

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7. Common Mistakes That Cost Marks

The Identity Matrix Subtraction Oversight

When constructing $(s\mathbf{I}-\mathbf{A})$, students frequently subtract the matrix $\mathbf{A}$ directly from the scalar variable $s$ (e.g., writing $s-a_{11}$ instead of $s\cdot 1-a_{11}$). Always map $s$ to the diagonal elements of an identity matrix of matching dimensions first to prevent mathematically invalid matrix subtractions.

The Inverse Order Multiplier Trap

In the transfer function derivation, remember that matrix multiplication is non-commutative. The transfer function is strictly defined as $\mathbf{C}(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{B}+\mathbf{D}$. Swapping the multiplication order to $\mathbf{B}(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{C}$ will trigger dimensional mismatches or mathematically incorrect companion polynomials.

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8. PYQ Bank — Verbatim Questions & Answer Plans

8.1 PYQ 2022 Question 6a / 2024 Question 6a [10 Marks]

Question: Determine the relationship between Laplace transform and Fourier transform.

• Answer Plan: Define the bilateral Laplace transform $X(s)$ with $s=\sigma +j\omega$. Show that by setting $\sigma =0$ (meaning we evaluate along the imaginary $j\omega$-axis), the Laplace transform collapses exactly into the continuous-time Fourier transform. State the absolute convergence condition (the imaginary axis must lie inside the Laplace Region of Convergence (ROC) for the Fourier transform to exist).

8.2 PYQ 2022 / 2025 Question 8d [13 Marks]

Question: Find the state transition matrix by the infinite series method for the system matrix $\mathbf{A}=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$.

• Answer Plan: Follow the complete step-by-step Taylor series expansion shown in Section 4.1. Calculate ${\mathbf{A}}^2$ and ${\mathbf{A}}^3$, state the mathematical induction pattern for ${\mathbf{A}}^n$, write out the scalar summation series, and assemble the final exponential matrix.

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9. Self-Check Before Moving On

• Can you state and explain all 6 algebraic properties of the state transition matrix?
• Do you know how to derive the Laplace Resolvent equation $\varphi (t)={\mathcal{L}}^{-1}\{(s\mathbf{I}-\mathbf{A}{)}^{-1}\}$ from the unforced homogeneous state equation?
• Can you mathematically derive the transfer function conversion formula $H(s)=\mathbf{C}(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{B}+\mathbf{D}$?
• Do you know how to calculate $\varphi (t)$ using both analytical Laplace inversions and infinite power series methods?

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Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3 / 8), ECE-2107_STATE_SPACE_ANALYSIS.pdf, Azmat Sir-2309008.pdf (Class Lecture Slides).