Chapter 3 - State-Space Representation of CT Systems

03 Chapter Map - State-Space Representation of CT Systems

Chapter 3 Overview & Map of Content (MOC)

Phase-variable matrix models, electrical networks, state-transition exponentials $e^{\mathbf{A}t}$, and continuous resolutions.

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📚 Study Notes Index

Read in order — each note assumes the previous one.

#	Note	What it covers
3.00	3.00 State-Space Analysis of CT Systems Compact Review	State-Space Compact Review, State-Space Formula Sheet
3.01	3.01 Introduction to State-Space and Phase-Variable Formulat	State-Space Fundamentals, Phase-Variable Method
3.02	3.02 State Space Modeling of Transfer Functions Canonic Form	Canonic Forms, State-Space TF, Control Canonic Form
3.03	3.03 State-Space Modeling of Electrical Circuits	Circuit State-Space, RLC State-Space, Electrical Network States
3.04	3.04 State-Space Modeling of Block Diagrams and Signal Flow	Block Diagram State Space, SFG State Space, Operational Block Realization
3.05	3.05 Discrete-Time State-Space Representation	Discrete State-Space, DT State Model
3.06	3.06 State Transition Matrix Properties and Solution Methods	State Transition Matrix, STM, Matrix Exponential, s-Domain Transition

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🎯 Exam Weight

ECE 2107 Exam Relevance

Master the core derivations, mathematical definitions, and problem-solving techniques. Refer to ECE 2107 - Signals and Systems for syllabus boundaries and past year questions.

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🔗 Related Resources

• Course Teaching Plan: ECE 2107 - Signals and Systems
• Previous chapter: 02 Chapter Map - LTI System Properties & Time-Domain Convolution
• Next chapter: 04 Chapter Map - The Sampling Theorem & Multi-Rate Processing

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Chapter 3: State-Space Representation of Continuous-Time Systems

This compact note summarizes all core concepts, matrix formulations, properties, and standard transformations of State-Space analysis. Designed as a high-density, pre-exam review card.

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1. Foundational Terminology & Definitions

*(Target: Theory Descriptive / 5-Mark Question)*

• State of a System: The absolute minimal set of variables $\{x_1(t),x_2(t),\dots ,x_n(t)\}$ such that knowledge of these variables at $t=t_0$ (initial conditions), together with the input $u(t)$ for $t\ge t_0$, uniquely and completely determines the system’s output and future behavior for all $t\ge t_0$.
• State Variables: The individual, independent dynamic variables that represent the state of the system (typically chosen as physical energy-storing quantities).
• State Vector: An $n\times 1$ column vector $\mathbf{X}(t)$ whose components are the state variables.
• State Space: The $n$-dimensional mathematical space whose coordinate axes are defined by the state variables.
• Canonic Realizations: Structured state-space representations derived from transfer functions, grouped into Control Canonic Form (CCF), Observer Canonic Form (OCF), and Diagonal/Jordan Canonic Form.

Table 3.1: Comparison: Classical Transfer Function vs. Modern State-Space

Feature	Classical Transfer Function Method	Modern State-Space Representation
Initial Conditions	Constrained strictly to zero initial conditions ($y(0)=0$).	Fully incorporates arbitrary non-zero initial conditions [$X(0)$].
System Constraints	Strictly applicable to Linear Time-Invariant (LTI) systems.	Extends seamlessly to Non-linear and Time-varying systems.
Terminal Scope	Best suited for Single-Input Single-Output (SISO) setups.	Natively handles Multiple-Input Multiple-Output (MIMO) setups.
Internal Visibility	Treats system as a “Black Box” (input-to-output only).	Unveils the complete internal dynamics and state transitions.
Computation	Frequency-domain algebraic roots.	Time-domain matrix algebra (highly optimized for computers).

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2. Standard LTI Matrix Formulations

*(Target: Numerical Solving / System Realization)*

For an LTI system with $n$ state variables (order $n$), $m$ inputs, and $p$ outputs:

2.1 State Equation (First-Order Differential Matrix Set)

$\dot{\mathbf{X}}(t)=\mathbf{AX}(t)+\mathbf{BU}(t)$

• $\dot{\mathbf{X}}(t)$ = Derivative of State Vector ($n\times 1$)
• $\mathbf{X}(t)$ = State Vector ($n\times 1$)
• $\mathbf{A}$ = State / System Matrix ($n\times n$) — Dictates natural system modes
• $\mathbf{B}$ = Input / Control Matrix ($n\times m$) — Bridges external input to state change
• $\mathbf{U}(t)$ = Input Vector ($m\times 1$)

2.2 Output Equation (Algebraic Measurement Model)

$\mathbf{Y}(t)=\mathbf{CX}(t)+\mathbf{DU}(t)$

• $\mathbf{Y}(t)$ = Output Vector ($p\times 1$)
• $\mathbf{C}$ = Output / Observation Matrix ($p\times n$) — Maps states to physical output
• $\mathbf{D}$ = Direct Transmission / Feedforward Matrix ($p\times m$) — Usually zero; non-zero if numerator order matches denominator order

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3. Selection Symmetries of State Variables

*(Target: Model Formulation / Circuit Analysis)*

3.1 Physical Variable Method (Electrical Networks)

• The Energy Rule: The number of state variables equals the number of independent energy-storing elements (inductors $L$ and capacitors $C$).
• Standard Selection Assignments:
  • State Variable $x_k(t)=v_C(t)$ (Voltage across capacitor) $⟹{\dot{x}}_k(t)=\frac{d{v}_C}{dt}=\frac{i_C(t)}{C}$
  • State Variable $x_j(t)=i_L(t)$ (Current through inductor) $⟹{\dot{x}}_j(t)=\frac{d{i}_L}{dt}=\frac{v_L(t)}{L}$
• A+ Score-Saver Trap: Avoid writing dependent loops or nodes. Verify that capacitor loop configurations and inductor node junctions do not violate linear independence, reducing matrix dimensions.

3.2 Phase Variable Method (Higher-Order ODEs)

• The Derivative Rule: State variables are assigned as the system output $y(t)$ and its successive derivatives: $x_1(t)=y(t),x_2(t)=\dot{y}(t),x_3(t)=\ddot{y}(t),\dots ,x_n(t)=y^{(n-1)}(t)$
• Matrix Structure (Companion Form): For an $n$-th order ODE: $y^{(n)}(t)+a_1{y}^{(n-1)}(t)+\cdots +a_{n}y(t)=b_{0}u(t)$ 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \\ -a_n & -a_{n-1} & -a_{n-2} & \dots & -a_1 \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ b_0 \end{bmatrix}, \quad \mathbf{C} = \begin{bmatrix} 1 & 0 & 0 & \dots & 0 \end{bmatrix}$$

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4. Operational Block Diagram Realizations

*(Target: Block Diagram Design / Signal Flow Graphs)*

State equations are mapped directly onto active realizations using Integrators ($1/s$), Adders, and Multipliers.

• The Integrator Boundary Rule:
  • The output of the $k$-th integrator is defined as the state variable $x_k(t)$ (or $X_k(s)$).
  • The input of the $k$-th integrator represents the derivative ${\dot{x}}_k(t)$ (or $s{X}_k(s)$).
• Canonic Realization Symmetries:
  • Direct Form I: Employs separate delay/integrator chains for feedback poles and feedforward zeros.
  • Direct Form II (Canonic Form): Minimizes memory requirement by utilizing a shared integrator column for both poles and zeros.

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5. The State Transition Matrix (STM)

*(Target: Mathematical Proof / Analytical Solving)*

The state transition matrix $\mathbf{\Phi }(t)=e^{\mathbf{A}t}$ maps the initial state $\mathbf{X}(t_0)$ to a future state $\mathbf{X}(t)$ under unforced (zero-input) conditions.

5.1 Evaluation Pathways

1. S-Domain Laplace Method (Resolvent Matrix inversion): $\mathbf{\Phi }(t)={\mathcal{L}}^{-1}\left\{(s\mathbf{I}-\mathbf{A}{)}^{-1}\right\}\text{(derivation)}$
2. Time-Domain Infinite Power Series Method: $\mathbf{\Phi }(t)=e^{\mathbf{A}t}=\mathbf{I}+\mathbf{A}t+\frac{{\mathbf{A}}^2{t}^2}{2!}+\frac{{\mathbf{A}}^3{t}^3}{3!}+\dots \text{(derivation)}$

5.2 Five (5) Golden Algebraic Properties of STM

For any time parameters $t,t_0,t_1,t_2$:

1. Identity Property: $\mathbf{\Phi }(0)=\mathbf{I}$
2. Inversion Property: ${\mathbf{\Phi }}^{-1}(t)=[e^{\mathbf{A}t}{]}^{-1}=e^{-\mathbf{A}t}=\mathbf{\Phi }(-t)$
3. Cyclic Decomposition (Semi-group Property): $\mathbf{\Phi }(t_2-t_1)\mathbf{\Phi }(t_1-t_0)=\mathbf{\Phi }(t_2-t_0)$
4. Time Addition Property: $\mathbf{\Phi }(t_1+t_2)=\mathbf{\Phi }(t_1)\mathbf{\Phi }(t_2)=\mathbf{\Phi }(t_2)\mathbf{\Phi }(t_1)$
5. Power Shifting Property: ${\left[\mathbf{\Phi }(t)\right]}^k=\mathbf{\Phi }(kt)$

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6. Analytical Solution of State Equations

*(Target: Mathematical Proof / Full Numerical Solution)*

6.1 Unilateral s-Domain Resolvent Solution

Taking the Laplace transform of the state equation yields: $\mathbf{X}(s)=(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{X}(0)+(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{BU}(s)\text{(derivation)}$

6.2 Total Time-Domain Response Formulation

Taking the inverse Laplace transform of the s-domain equation yields the unified state trajectory: $\mathbf{X}(t)=\underset{\text{Zero-Input Response (ZIR)}}{\underbrace{\mathbf{\Phi }(t)\mathbf{X}(0)}}+\underset{\text{Zero-State Response (ZSR)}}{\underbrace{{\int }_0^t\mathbf{\Phi }(t-\tau )\mathbf{BU}(\tau )d\tau }}\text{(derivation)}$

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7. State-Space to Transfer Function Conversion

*(Target: Mathematical Proof / Matrix Transformation)*

To convert a modern state-space representation back to a classical frequency-domain transfer function:

• Governing Algebraic Form (Single-Input Single-Output): $H(s)=\frac{\mathbf{Y}(s)}{\mathbf{U}(s)}=\mathbf{C}(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{B}+\mathbf{D}\text{(derivation)}$
• The Pole Identity: The poles of $H(s)$ correspond exactly to the eigenvalues of the system matrix $\mathbf{A}$, calculated by solving the characteristic equation: $\det (s\mathbf{I}-\mathbf{A})=0$

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8. Discrete-Time State-Space Modeling

*(Target: Theory Descriptive / Discrete Difference Modeling)*

For digital networks represented by difference equations:

• Unified State and Output Formulation: \mathbf{X}[n+1] &= \mathbf{A}\mathbf{X}[n] + \mathbf{B}\mathbf{U}[n] \\ \mathbf{Y}[n] &= \mathbf{C}\mathbf{X}[n] + \mathbf{D}\mathbf{U}[n] \end{aligned}$$
• Discrete Unforced Total Solution: $\mathbf{X}[n]={\mathbf{A}}^n\mathbf{X}[0]+{\sum }_{k=0}^{n-1}{\mathbf{A}}^{n-1-k}\mathbf{BU}[k]\text{(derivation)}$

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9. Common Mistakes That Cost Marks

Critical Exam Pitfalls

1. Identity Matrix ($\mathbf{I}$) Omission: Writing $(s-\mathbf{A}{)}^{-1}$ instead of $(s\mathbf{I}-\mathbf{A}{)}^{-1}$. You cannot subtract a matrix from a scalar directly! Forgetting to scale $s$ by the Identity matrix $\mathbf{I}$ completely breaks the matrix dimensions.
2. Resolvent Scaling Factor Sign Swaps: Miscalculating the inverse of $(s\mathbf{I}-\mathbf{A})$. For a $2\times 2$ matrix, remember that the matrix inversion formula is: ${\mathbf{M}}^{-1}=\frac{1}{\det (\mathbf{M})}\text{adj}(\mathbf{M})$ Double-check the signs of your cofactors and determinant value before performing algebraic division.
3. Feedback Coefficient Sign Negation (Direct Form II): Forgetting to negate the feedback coefficient terms when mapping from a differential equation to feedback multiplier blocks.
4. Initial State Time-Shift Inversion: In discrete solutions, using ${\mathbf{A}}^{n-k}$ instead of the causal shift bound ${\mathbf{A}}^{n-1-k}$ inside the convolution summation.

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10. PYQ Bank — Verbatim Questions & Answer Plans

10.1 Phase Variable Derivative Realization

KUET 2024 Section B Q. 5a [13 Marks]

Represent the following differential equation given below in a state model: $\frac{d^{3}y(t)}{d{t}^3}+\frac{d^{2}y(t)}{d{t}^2}+6\frac{dy(t)}{dt}+7y(t)=2u(t)$

• Answer Plan:
  1. Define three successive phase-variables as state variables: $x_1(t)=y(t)$, $x_2(t)=\dot{y}(t)$, $x_3(t)=\ddot{y}(t)$.
  2. Write the derivatives of the first two variables: ${\dot{x}}_1(t)=x_2(t)$ and ${\dot{x}}_2(t)=x_3(t)$.
  3. Re-arrange the original differential equation to express ${\dot{x}}_3(t)$: ${\dot{x}}_3(t)=-7x_1(t)-6x_2(t)-x_3(t)+2u(t)$.
  4. Compile the state equation and output equation in standard matrix Companion Form.

10.2 Transfer Function Matrix Resolution

KUET 2019 Section B Q. 7b [10 Marks]

A state variable description of a system is given by the matrix equation: $\dot{\mathbf{X}}=\left[\begin{array}{cc}-1& 0\\ 1& -2\end{array}\right]\mathbf{X}+\left[\begin{array}{c}1\\ 0\end{array}\right]u,y=\left[\begin{array}{cc}1& 1\end{array}\right]\mathbf{X}$ Find: (i) The Transfer function, (ii) The State transition matrix, (iii) State diagram.

• Answer Plan:
  1. Identify matrices: $\mathbf{A}=\left[\begin{array}{cc}-1& 0\\ 1& -2\end{array}\right]$, $\mathbf{B}=\left[\begin{array}{c}1\\ 0\end{array}\right]$, $\mathbf{C}=\left[\begin{array}{cc}1& 1\end{array}\right]$, $\mathbf{D}=0$.
  2. Solve for $(s\mathbf{I}-\mathbf{A})=\left[\begin{array}{cc}s+1& 0\\ -1& s+2\end{array}\right]$ and evaluate its inverse resolvent matrix $(s\mathbf{I}-\mathbf{A}{)}^{-1}=\frac{1}{(s+1)(s+2)}\left[\begin{array}{cc}s+2& 0\\ 1& s+1\end{array}\right]$.
  3. Calculate the transfer function $H(s)=\mathbf{C}(s\mathbf{I}-\mathbf{A}{)}^{-1}\mathbf{B}=\frac{s+3}{(s+1)(s+2)}$.
  4. Determine the State Transition Matrix $\mathbf{\Phi }(t)={\mathcal{L}}^{-1}\{(s\mathbf{I}-\mathbf{A}{)}^{-1}\}=\left[\begin{array}{cc}{e}^{-t}& 0\\ e^{-t}-e^{-2t}& e^{-2t}\end{array}\right]$.
  5. Draw the state diagram illustrating feedback nodes and summing junctions based on individual state equations.

10.3 Discrete Time Delay Formulations

KUET 2022/2021 Section B Q. 8a [13 Marks]

Obtain the state-space representation of a discrete time system described by the following difference equation: $y(n-3)+2y(n-2)+3y(n-1)+4y(n)=U(n)$

• Answer Plan:
  1. Convert the difference equation into causal form by shifting indices: $y[n]+\frac{3}{4}y[n-1]+\frac{1}{2}y[n-2]+\frac{1}{4}y[n-3]=\frac{1}{4}u[n]$.
  2. Assign state variables to past delays: $x_1[n]=y[n-3]$, $x_2[n]=y[n-2]$, $x_3[n]=y[n-1]$.
  3. Formulate the state updating vector: $x_1[n+1]=x_2[n]$, $x_2[n+1]=x_3[n]$, $x_3[n+1]=-\frac{1}{4}{x}_1[n]-\frac{1}{2}{x}_2[n]-\frac{3}{4}{x}_3[n]+\frac{1}{4}u[n]$.
  4. Express the output equation: $y[n]=x_3[n+1]=-\frac{1}{4}{x}_1[n]-\frac{1}{2}{x}_2[n]-\frac{3}{4}{x}_3[n]+\frac{1}{4}u[n]$.
  5. Compile variables in discrete standard matrix form.

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11. Self-Check Before Moving On

• Can you define “State of a System” verbatim with its prerequisite mathematical boundaries?
• Can you write the complete s-domain state-space-to-transfer-function conversion equation and prove its eigenvalue-pole identity?
• Can you state the five algebraic properties of the State Transition Matrix ($\mathbf{\Phi }(t)$)?
• Do you know how to choose physical state variables for series/parallel capacitor-inductor networks using KVL/KCL?
• Can you draw a canonic Direct Form II realization layout from a third-order transfer function?

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Source: (k.Deergha Rao) signals and systems.pdf, ECE-2107_STATE_SPACE_ANALYSIS.pdf, Rabiul sir class notes, KUET Past Year Question Bank.

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Related Concepts: 3.02 State-Space Modeling of Transfer Functions (Canonic Forms) | 3.03 State-Space Modeling of Electrical Circuits | 3.06 State Transition Matrix: Properties & Solution Methods

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3.01 Introduction to State-Space & Phase-Variable Formulations

Core Idea

State-space analysis is a modern, time-domain mathematical model that reduces an $n$-th order linear differential equation into a system of $n$ coupled, first-order differential equations. This representation uses matrix algebra to track the internal energy states of a system alongside its inputs and outputs, providing a unified framework for multi-input multi-output (MIMO) systems and system stability analysis.

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1. The Philosophy of State-Space Representation

Classic time-domain analysis of continuous-time LTI systems relies on solving high-order linear constant-coefficient differential equations (LCCDEs) {equations relating inputs and outputs through derivatives} or converting them into $s$-domain transfer functions using Laplace transforms. While powerful, these classical methods have distinct drawbacks:

• They treat the system as a “black box,” focusing only on the input-output relationship and completely ignoring the internal states (such as capacitor voltages or inductor currents).
• They are highly tedious to apply to Multi-Input Multi-Output (MIMO) systems.
• They assume zero initial conditions when computing transfer functions, losing critical historical system state data.

State-space analysis solves these limitations by introducing a set of internal variables called state variables. By utilizing vector-matrix notation, we can analyze the internal physical states and external responses of any system simultaneously, even under non-zero initial conditions.

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2. The Core Matrix Mathematical Formulation

To mathematically represent a continuous-time system in state-space, we define three primary mathematical vectors:

1. State Vector ($\mathbf{X}(t)$): An $n\times 1$ column vector containing the $n$ state variables of the system: $\mathbf{X}(t)=\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\\ ⋮\\ x_n(t)\end{array}\right]$
2. Input Vector ($\mathbf{U}(t)$): An $m\times 1$ column vector containing the $m$ external input signals: $\mathbf{U}(t)=\left[\begin{array}{c}{u}_1(t)\\ u_2(t)\\ ⋮\\ u_m(t)\end{array}\right]$
3. Output Vector ($\mathbf{Y}(t)$): A $p\times 1$ column vector containing the $p$ external output signals: $\mathbf{Y}(t)=\left[\begin{array}{c}{y}_1(t)\\ y_2(t)\\ ⋮\\ y_p(t)\end{array}\right]$

These vectors are coupled through two fundamental, first-order matrix differential equations:

2.1 The State Equation

The State Equation relates the first derivative of the state vector to the current states and external inputs: $\dot{\mathbf{X}}(t)=\mathbf{AX}(t)+\mathbf{BU}(t)$

2.2 The Output Equation

The Output Equation expresses the system outputs as a linear combination of the current states and external inputs: $\mathbf{Y}(t)=\mathbf{CX}(t)+\mathbf{DU}(t)$

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2.3 Structural Matrix Dimension Map

Understanding the exact dimensions of these matrices is a heavily tested conceptual topic in ECE 2107 examinations. If the system order is $n$, the input count is $m$, and the output count is $p$, the matrices must comply with the following structural dimensions:

Matrix	Physical Name	Dimensions	Role in the System
$\mathbf{A}$	System Matrix	$n\times n$	Governs the internal dynamics and natural stability of the system.
$\mathbf{B}$	Input Matrix	$n\times m$	Determines how external inputs affect the internal state transitions.
$\mathbf{C}$	Output Matrix	$p\times n$	Governs how the internal states map to the measurable output terminals.
$\mathbf{D}$	Direct Transmission Matrix	$p\times m$	Controls direct feedforward paths bypassing the system states. (Usually $\mathbf{D}=0$ in physical networks).

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3. The Phase-Variable Formulation Method

The Phase-Variable Method is a systematic, algorithmic approach used to convert a single $n$-th order continuous-time differential equation into standard state-space matrices.

3.1 General Workflow

Consider a general $n$-th order differential equation without input derivatives: $\frac{d^{n}y(t)}{d{t}^n}+a_{n-1}\frac{d^{n-1}y(t)}{d{t}^{n-1}}+\cdots +a_1\frac{dy(t)}{dt}+a_{0}y(t)=b_{0}u(t)$

Step 1: Assign State Variables

We assign the system output $y(t)$ and its successive derivatives as our phase variables: $x_1(t)=y(t)$ $x_2(t)=\dot{y}(t)={\dot{x}}_1(t)$ $x_3(t)=\ddot{y}(t)={\dot{x}}_2(t)$ $⋮$ $x_n(t)=y^{(n-1)}(t)={\dot{x}}_{n-1}(t)$

Step 2: Formulate First-Order State Derivatives

Express the derivative of each state variable in terms of the other states: ${\dot{x}}_1(t)=x_2(t)$ ${\dot{x}}_2(t)=x_3(t)$ $⋮$ ${\dot{x}}_{n-1}(t)=x_n(t)$

To find the final derivative ${\dot{x}}_n(t)$, we substitute our state variable definitions back into the original $n$-th order differential equation: ${\dot{x}}_n(t)=y^{(n)}(t)=-a_0{x}_1(t)-a_1{x}_2(t)-\cdots -a_{n-1}{x}_n(t)+b_{0}u(t)$

Step 3: Construct the Companion Matrix Form

Now, we gather these first-order differential equations directly into the canonical matrix format: $\left[\begin{array}{c}{\dot{x}}_1(t)\\ {\dot{x}}_2(t)\\ ⋮\\ {\dot{x}}_{n-1}(t)\\ {\dot{x}}_n(t)\end{array}\right]=\left[\begin{array}{ccccc}0& 1& 0& \dots & 0\\ 0& 0& 1& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \dots & 1\\ -a_0& -a_1& -a_2& \dots & -a_{n-1}\end{array}\right]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\\ ⋮\\ x_{n-1}(t)\\ x_n(t)\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ ⋮\\ 0\\ b_0\end{array}\right]u(t)$

Since our output is $y(t)=x_1(t)$, the corresponding Output Equation is: $y(t)=\left[\begin{array}{ccccc}1& 0& 0& \dots & 0\end{array}\right]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\\ ⋮\\ x_n(t)\end{array}\right]+\left[\begin{array}{c}0\end{array}\right]u(t)$

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4. Comprehensive Worked Examples (The Exam Killers)

4.1 Example 1: Third-Order System Formulation [PYQ 2024]

Question: Represent the following differential equation in a state model: $\frac{d^{3}y(t)}{d{t}^3}+\frac{d^{2}y(t)}{d{t}^2}+6\frac{dy(t)}{dt}+7y(t)=2u(t)$

Step-by-Step Solution:

1. Define Phase Variables ($n=3$): Let: $x_1(t)=y(t)$ $x_2(t)=\dot{y}(t)={\dot{x}}_1(t)$ $x_3(t)=\ddot{y}(t)={\dot{x}}_2(t)$

2. Formulate State Derivatives: From our phase variable definitions: ${\dot{x}}_1(t)=x_2(t)$ ${\dot{x}}_2(t)=x_3(t)$

   To find ${\dot{x}}_3(t)$, isolate the highest-order derivative ($\stackrel{...}{y}(t)$) in the original differential equation: $\stackrel{...}{y}(t)=-\ddot{y}(t)-6\dot{y}(t)-7y(t)+2u(t)$

   Substitute our defined state variables into this equation: ${\dot{x}}_3(t)=-7x_1(t)-6x_2(t)-x_3(t)+2u(t)$

3. Construct the State Equation: Gather the equations into vector-matrix form $\dot{\mathbf{X}}(t)=\mathbf{AX}(t)+\mathbf{BU}(t)$: $\left[\begin{array}{c}{\dot{x}}_1(t)\\ {\dot{x}}_2(t)\\ {\dot{x}}_3(t)\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -7& -6& -1\end{array}\right]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\\ x_3(t)\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 2\end{array}\right]u(t)$

4. Construct the Output Equation: Since $y(t)=x_1(t)$, map it to standard format $\mathbf{Y}(t)=\mathbf{CX}(t)+\mathbf{DU}(t)$: $y(t)=\left[\begin{array}{ccc}1& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\\ x_3(t)\end{array}\right]+\left[\begin{array}{c}0\end{array}\right]u(t)$

   Thus, the final state-space matrices are: $\mathbf{A}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -7& -6& -1\end{array}\right],\mathbf{B}=\left[\begin{array}{c}0\\ 0\\ 2\end{array}\right],\mathbf{C}=\left[\begin{array}{ccc}1& 0& 0\end{array}\right],\mathbf{D}=\left[\begin{array}{c}0\end{array}\right]$

5. State-Space Block Diagram Realization: To ensure maximum exam marks, you must be ready to sketch the physical realization of your phase variable system.

   [DIAGRAM: Continuous 3-integrator block diagram (Direct Form II style) realizing the state-space equations. Left side: input u(t) multiplied by 2 feeding the final adder. Center: a cascade of three 1/s integrator blocks whose outputs represent states x3(t), x2(t), and x1(t) from left to right. Bottom: feedback lines with multipliers -1, -6, and -7 routing from x3, x2, and x1 back into the input adder. Right side: Output y(t) tapped directly from state x1(t) - source: Textbook Sec 3.1]

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4.2 Example 2: Classroom Classical Realization

Question: Convert the following third-order system into state matrices and sketch its integrator block diagram: $\stackrel{...}{y}(t)+6\ddot{y}(t)+11\dot{y}(t)+6y(t)=u(t)$

Step-by-Step Solution:

1. Define Phase Variables: Let: $x_1(t)=y(t)$ $x_2(t)=\dot{y}(t)={\dot{x}}_1(t)$ $x_3(t)=\ddot{y}(t)={\dot{x}}_2(t)$

2. State Derivatives: ${\dot{x}}_1(t)=x_2(t)$ ${\dot{x}}_2(t)=x_3(t)$ ${\dot{x}}_3(t)=-6x_1(t)-11x_2(t)-6x_3(t)+u(t)$

3. Matrix Realization: $\left[\begin{array}{c}{\dot{x}}_1(t)\\ {\dot{x}}_2(t)\\ {\dot{x}}_3(t)\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -6& -11& -6\end{array}\right]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\\ x_3(t)\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u(t)$ $y(t)=\left[\begin{array}{ccc}1& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\\ x_3(t)\end{array}\right]+\left[\begin{array}{c}0\end{array}\right]u(t)$

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5. Common Mistakes That Cost Marks

The Zero-Coefficient Trap

If your differential equation skips a derivative term (for example, missing the $\dot{y}(t)$ term), do not skip the state variable! Keep the variable and write its coefficient as zero in your matrices. Skipping a state variable collapses the matrix order and results in a zero-mark evaluation.

The Non-Unity Coefficient Danger

If your differential equation starts with a non-unity coefficient on the highest derivative (e.g., $3\stackrel{...}{y}(t)+\ddot{y}(t)+\cdots =u(t)$), you must divide the entire equation by 3 before starting the phase variable assignments. Neglecting this division step scales all of your feedback and input matrix coefficients incorrectly.

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6. PYQ Bank — Verbatim Questions & Answer Plans

6.1 PYQ 2024 Question 2c [13 Marks]

Question: Represent the following differential equation given below in a state model: $\frac{d^{3}y(t)}{d{t}^3}+\frac{d^{2}y(t)}{d{t}^2}+6\frac{dy(t)}{dt}+7y(t)=2u(t)$

• Answer Plan: Follow the exact step-by-step mathematical formulation shown in Section 4.1. Define phase states $x_1,x_2,x_3$, calculate the derivatives, write the coupled matrices, state the dimensions, and draw the matching 3-integrator block diagram to lock in all 13 marks.

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7. Self-Check Before Moving On

• Can you define State, State Vector, State Equation, and Output Equation mathematically? [9.01, 9.02]
• Do you know how to determine the exact matrix dimensions of $\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D}$ for any $n$-th order MIMO system? [9.02]
• Can you convert any continuous-time differential equation with constant coefficients into companion matrix form using phase variables? [9.03]
• Have you memorized the warning to divide by the highest-order coefficient before building state matrices? [9.05]

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

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3.01 Introduction to State-Space & Phase-Variable Formulations | 3.03 State-Space Modeling of Electrical Circuits

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3.02 State-Space Modeling of Transfer Functions (Canonic Forms)

Core Idea

For systems represented by rational s-domain transfer functions with $s$-terms in the numerator, traditional phase-variable definitions based on output derivatives fail because they introduce derivatives of the input signal. To bypass this, we decompose the transfer function into two cascaded subsystems: an Input-to-State subsystem (with a constant numerator) and a State-to-Output subsystem (with s-terms in the numerator) using an intermediate state variable $z(t)$.

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1. The Algebra of Numerator Dynamics

1.1 The “Input Derivative” Problem

If we have a transfer function with $s$-terms in the numerator: $H(s)=\frac{Y(s)}{U(s)}=\frac{b_2{s}^2+b_{1}s+b_0}{s^3+a_2{s}^2+a_{1}s+a_0}$

Its time-domain differential equation is: $\stackrel{...}{y}(t)+a_2\ddot{y}(t)+a_1\dot{y}(t)+a_{0}y(t)=b_2\ddot{u}(t)+b_1\dot{u}(t)+b_{0}u(t)$

If we attempt to define states as derivatives of the output (e.g., $x_1=y$, $x_2=\dot{y}$, $x_3=\ddot{y}$), the derivative of the highest state ${\dot{x}}_3=\stackrel{...}{y}$ will directly depend on $\dot{u}(t)$ and $\ddot{u}(t)$. This violates the standard state equation format: $\dot{\mathbf{X}}(t)=\mathbf{AX}(t)+\mathbf{BU}(t)$ which permits only the raw input $u(t)$, not its derivatives!

1.2 The Auxiliary Variable $Z(s)$ Solution

To decouple the input derivatives, we insert an auxiliary variable $Z(s)$ and split $H(s)$ into two cascaded blocks:

1. Block 1 (Input-to-State): $\frac{Z(s)}{U(s)}=\frac{1}{s^3+a_2{s}^2+a_{1}s+a_0}$
2. Block 2 (State-to-Output): $\frac{Y(s)}{Z(s)}=b_2{s}^2+b_{1}s+b_0$

Multiplying these blocks yields the original transfer function: $H(s)=\left(\frac{Z(s)}{U(s)}\right)\cdot \left(\frac{Y(s)}{Z(s)}\right)=\frac{b_2{s}^2+b_{1}s+b_0}{s^3+a_2{s}^2+a_{1}s+a_0}$

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2. Phase-Variable Formulation (Control Canonic Form)

2.1 Step-by-Step Derivation Workflow

Step 1: Solve Block 1 in the Time Domain

Multiply out the s-domain terms of the Input-to-State block: $[s^3+a_2{s}^2+a_{1}s+a_0]Z(s)=U(s)$ Applying the inverse Laplace transform assuming zero initial conditions: $\stackrel{...}{z}(t)+a_2\ddot{z}(t)+a_1\dot{z}(t)+a_{0}z(t)=u(t)$

Step 2: Define the State Vector

Since this auxiliary equation has a constant numerator (1), we can safely define our state variables as successive derivatives of the auxiliary variable $z(t)$: $x_1(t)=z(t)$ $x_2(t)=\dot{z}(t)$ $x_3(t)=\ddot{z}(t)$

Step 3: Formulate State Equations

Write down the derivatives of our state variables: ${\dot{x}}_1(t)=x_2(t)$ ${\dot{x}}_2(t)=x_3(t)$ ${\dot{x}}_3(t)=\stackrel{...}{z}(t)=-a_{0}z(t)-a_1\dot{z}(t)-a_2\ddot{z}(t)+u(t)$ ${\dot{x}}_3(t)=-a_0{x}_1(t)-a_1{x}_2(t)-a_2{x}_3(t)+u(t)$

Step 4: Solve Block 2 in the Time Domain

Multiply out the s-domain terms of the State-to-Output block: $Y(s)=[b_2{s}^2+b_{1}s+b_0]Z(s)=b_2{s}^{2}Z(s)+b_{1}sZ(s)+b_{0}Z(s)$ Taking the inverse Laplace transform: $y(t)=b_2\ddot{z}(t)+b_1\dot{z}(t)+b_{0}z(t)$

Step 5: Substitute States into the Output Equation

Replace the derivatives of $z(t)$ with our defined state variables: $y(t)=b_2{x}_3(t)+b_1{x}_2(t)+b_0{x}_1(t)$ $y(t)=b_0{x}_1(t)+b_1{x}_2(t)+b_2{x}_3(t)$

Step 6: Assemble Matrix Form

Combining the state and output equations: $\dot{\mathbf{X}}(t)=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -a_0& -a_1& -a_2\end{array}\right]\mathbf{X}(t)+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u(t)$ $\mathbf{Y}(t)=\left[\begin{array}{ccc}{b}_0& b_1& b_2\end{array}\right]\mathbf{X}(t)+[0]u(t)$

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3. High-Yield Worked Classroom Problems

3.01 Constant Numerator Realization

Problem 1: Companion Matrix Mapping

Obtain the state-space model $(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})$ for the transfer function: $H(s)=\frac{Y(s)}{U(s)}=\frac{24}{s^3+9s^2+26s+24}$

• Step 1: Setup intermediate block $\frac{Z(s)}{U(s)}=\frac{1}{s^3+9s^2+26s+24}⟹\stackrel{...}{z}(t)+9\ddot{z}(t)+26\dot{z}(t)+24z(t)=u(t)$
• Step 2: State Assignments $x_1=z,x_2=\dot{z},x_3=\ddot{z}$ ${\dot{x}}_1=x_2,{\dot{x}}_2=x_3,{\dot{x}}_3=-24x_1-26x_2-9x_3+u(t)$
• Step 3: Output Mapping $Y(s)=24Z(s)⟹y(t)=24z(t)⟹y(t)=24x_1(t)$
• Step 4: Matrix Compilation $\mathbf{A}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -24& -26& -9\end{array}\right],\mathbf{B}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right],\mathbf{C}=\left[\begin{array}{ccc}24& 0& 0\end{array}\right],\mathbf{D}=[0]$

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3.02 Complete Numerator Polynomial Realization

Problem 2: Phase-Variable $s$-Numerator Realization

Obtain the state-space model $(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})$ for the transfer function: $H(s)=\frac{Y(s)}{U(s)}=\frac{s^2+7s+2}{s^3+9s^2+26s+24}$

• Step 1: Subsystem Cascading $\frac{Z(s)}{U(s)}=\frac{1}{s^3+9s^2+26s+24}\text{and}\frac{Y(s)}{Z(s)}=s^2+7s+2$
• Step 2: State Variables Mapping Since the denominator polynomial is identical to Problem 1, the system matrix $\mathbf{A}$ and input matrix $\mathbf{B}$ remain completely unchanged: $\mathbf{A}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -24& -26& -9\end{array}\right],\mathbf{B}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$
• Step 3: Output Equation Synthesis $Y(s)=[s^2+7s+2]Z(s)=s^{2}Z(s)+7sZ(s)+2Z(s)$ Converting to the time domain: $y(t)=\ddot{z}(t)+7\dot{z}(t)+2z(t)$ Substituting the states $x_1=z$, $x_2=\dot{z}$, and $x_3=\ddot{z}$: $y(t)=x_3(t)+7x_2(t)+2x_1(t)=2x_1(t)+7x_2(t)+x_3(t)$
• Step 4: Matrix Compilation $\dot{\mathbf{X}}(t)=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -24& -26& -9\end{array}\right]\mathbf{X}(t)+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u(t)$ $\mathbf{Y}(t)=\left[\begin{array}{ccc}2& 7& 1\end{array}\right]\mathbf{X}(t)+[0]u(t)$

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4. Hardware Realization Diagram

The Control Canonic Form mapped above directly dictates how an analog computer can be wired using integrators, summers, and potentiometers:

               +-----------------------------------------------------+
               |                                                     |
               v   -24                                               |
u(t) ---> ( + ) ---> [ 1/s ] ---> x_3 ---> [ 1/s ] ---> x_2 ---> [ 1/s ] ---> x_1
            ^          |                     |                      |
            |          v                     v                      v
            |         (x) 1                 (x) 7                  (x) 2
            |          |                     |                      |
            |          +------------------>( + )<--------------------+
            |                                |
            |                                v
            |                              y(t)
            |                                
            +---(x) -9 <---------------------+
            |                                |
            +---(x) -26 <--------------------+

[DIAGRAM: 3-Integrator canonic flow-graph realization of H(s) = (s^2 + 7s + 2)/(s^3 + 9s^2 + 26s + 24) showing nested feedback and feedforward summing nodes - source: Textbook Ch 2]

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5. Common Mistakes That Cost Marks

Exam Pitfall: The Sign-Flip omission & Non-Monic Coefficients

1. Denominator Sign-Flip Trap: When writing the last row of the state system matrix $\mathbf{A}$, students often copy the coefficients $a_i$ directly. Remember, they are shifted to the RHS of the state equation and must be negated: $[-a_0,-a_1,\dots ,-a_{n-1}]$.
2. Numerator Sign-Flip Mistake: Students sometimes also negate the coefficients of the output matrix $\mathbf{C}$. Do not do this! The numerator coefficients $[b_0,b_1,\dots ,b_m]$ map directly to the $\mathbf{C}$ matrix without any sign changes.
3. Non-Monic Denominator Danger: If the coefficient of the highest power of $s$ in the denominator is not unity (e.g., $3s^3+9s^2+12s+6$), you must divide the entire numerator and denominator by that factor (3) to make the polynomial monic before mapping the coefficients!

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6. Potential / Grounded Exam Practice Bank

Verbatim Class & Homework Numericals

1. Find the state variable description corresponding to the continuous system transfer function: $H(s)=\frac{s^2+3s+4}{s^3+2s^2+7s+5}$ Answer Plan: Cascade two blocks using $Z(s)$ buffer. State equations will yield $\mathbf{A}$ with last row $[-5,-7,-2]$, $\mathbf{B}=[0,0,1{]}^T$, and output $\mathbf{C}=[4,3,1]$.
2. Formulate the control canonic state-space representation for: $\frac{Y(s)}{U(s)}=\frac{5s+10}{s^2+3s+2}$ Answer Plan: System order $n=2$. Matrix $\mathbf{A}=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]$, $\mathbf{B}=\left[\begin{array}{c}0\\ 1\end{array}\right]$, $\mathbf{C}=\left[\begin{array}{cc}10& 5\end{array}\right]$.

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7. Self-Check Before Moving On

• Can you explain why numerator $s$-dynamics prevent defining state variables directly as output derivatives?
• Do you understand the mechanical step-by-step derivation of the intermediate variable $z(t)$ split?
• Can you write down the state matrices $(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})$ of any monic rational transfer function by inspection within 10 seconds?

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Source: (k.Deergha Rao) signals and systems.pdf, Section 2.8; lecture slides on State-Space representations.

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Related Concepts: 3.02 State-Space Modeling of Transfer Functions (Canonic Forms) | 3.04 State-Space Modeling of Block Diagrams & Signal Flow Graphs | 3.06 State Transition Matrix: Properties & Solution Methods

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3.03 State-Space Modeling of Electrical Circuits

Core Idea

Unlike mathematical systems where state variables are abstract phase variables, physical electrical circuits possess a natural, physically intuitive set of state variables: capacitor voltages ($v_C$) and inductor currents ($i_L$). These variables represent the physical energy storage {the capacity of capacitors to store electrical charge in electric fields, and inductors to store magnetic flux in magnetic fields} and directly determine the “history” or “memory” of the network.

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1. The Physical Philosophy of State Selection

When translating an electrical circuit into a state-space model, we do not use phase variables {derivatives of a single output}. Instead, we assign state variables to the energy-storing elements of the circuit.

Physically, the state of a system must represent the minimum amount of information at time $t_0$ needed to predict the future states for all $t>t_0$ (given the input). In RLC networks, the physical energy is stored in:

• Capacitors: Energy stored is $E_C=\frac{1}{2}C{v}_C^2(t)$. The voltage across a capacitor cannot change instantaneously {due to the conservation of electric charge}. Thus, capacitor voltage $v_C(t)$ is selected as a state variable.
• Inductors: Energy stored is $E_L=\frac{1}{2}L{i}_L^2(t)$. The current through an inductor cannot change instantaneously {due to the conservation of magnetic flux linkage}. Thus, inductor current $i_L(t)$ is selected as a state variable.

1.1 State Selection Dictionary

For any physical RLC circuit with $n_C$ capacitors and $n_L$ inductors, the total order of the state-space model is $n=n_C+n_L$. We define the state vector $\mathbf{X}(t)$ using the following variables:

Physical Element	State Variable Definition	Governing Physical Equation	Derivative Relation
Capacitor ($C$)	Voltage across the capacitor: $x_i(t)=v_{C_i}(t)$	$i_C(t)=C\frac{d{v}_C(t)}{dt}$	${\dot{x}}_i(t)=\frac{1}{C}{i}_C(t)$
Inductor ($L$)	Current flowing through the inductor: $x_j(t)=i_{L_j}(t)$	$v_L(t)=L\frac{d{i}_L(t)}{dt}$	${\dot{x}}_j(t)=\frac{1}{L}{v}_L(t)$

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2. The 4-Step Circuit Modeling Algorithm

To systematically formulate the state equation $\dot{\mathbf{X}}=\mathbf{AX}+\mathbf{BU}$ and output equation $\mathbf{Y}=\mathbf{CX}+\mathbf{DU}$ for any passive electrical network, apply this rigid 4-step workflow:

Step 1: State and Input Assignment

Identify all inductors and capacitors. Assign state variables: $x_1(t),x_2(t),\dots ,x_n(t)$ to capacitor voltages and inductor currents. Label all other branch voltages and branch currents in terms of these states and the inputs $u(t)$.

Step 2: Write Fundamental Network Equations

Apply Kirchhoff’s Voltage Law (KVL) to the loops containing inductors to find inductor voltages, and Kirchhoff’s Current Law (KCL) to the nodes containing capacitors to find capacitor currents.

• Ensure that you do not write dependent equations {e.g., redundant loops or nodes that do not contain energy storage components}.

Step 3: Isolate the Derivatives

Algebraically manipulate your equations so that the capacitor currents ($i_C$) and inductor voltages ($v_L$) are expressed only in terms of state variables (capacitor voltages and inductor currents) and external input sources.

• Once isolated, substitute the derivative relations: $i_C(t)=C{\dot{v}}_C(t)⟹{\dot{v}}_C(t)=\frac{1}{C}{i}_C(t)$ $v_L(t)=L{\dot{i}}_L(t)⟹{\dot{i}}_L(t)=\frac{1}{L}{v}_L(t)$

Step 4: Assemble the Matrix Equations

Collect the isolated first-order differential equations and write them in vector-matrix format. Express the measurable output terminal variable $y(t)$ as a linear combination of states and inputs to form the output matrix.

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3. High-Yield Worked Examples (The Exam Classics)

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3.1 Example 1: The Second-Order Series RLC Network [PYQ 2022, 2020]

Question: Derive the complete state-space model for the series RLC network shown below. Find the state and output matrices if the output is:

1. The loop current $i(t)$.
2. The capacitor voltage $v_C(t)$.

                  Inductor current i_L(t) --->
               +------------[  R  ]------------[  L  ]------------+
               |                                                  |
              (+)                                                [C] v_C(t)
             v_s(t)                                               |
              |                                                   |
               +--------------------------------------------------+

Step-by-Step Solution:

1. Assign State Variables: The circuit has one capacitor ($C$) and one inductor ($L$). The system order is $n=1+1=2$. Let our states be: $x_1(t)=v_C(t)\text{voltage across capacitor}$ $x_2(t)=i_L(t)\text{current through inductor}$ The input is the voltage source: $u(t)=v_s(t)$

2. Apply KVL and KCL:

   • Loop KVL: Summing voltages around the series loop: $v_s(t)-R{i}_L(t)-v_L(t)-v_C(t)=0⟹v_L(t)=-v_C(t)-R{i}_L(t)+v_s(t)$
   • Node KCL: The capacitor current is exactly equal to the inductor loop current: $i_C(t)=i_L(t)$

3. Isolate Derivatives:

   • For the inductor: $L{\dot{i}}_L(t)=-v_C(t)-R{i}_L(t)+v_s(t)⟹{\dot{x}}_2(t)=-\frac{1}{L}{x}_1(t)-\frac{R}{L}{x}_2(t)+\frac{1}{L}u(t)$
   • For the capacitor: $C{\dot{v}}_C(t)=i_L(t)⟹{\dot{x}}_1(t)=\frac{1}{C}{x}_2(t)$

4. Assemble Matrices: Combining these equations into matrix form $\dot{\mathbf{X}}=\mathbf{AX}+\mathbf{BU}$: \begin{bmatrix} \\dot{x}_1(t) \\\\ \\dot{x}_2(t) \\end{bmatrix} = \\begin{bmatrix} 0 & \\frac{1}{C} \\\\ -\\frac{1}{L} & -\\frac{R}{L} \\end{bmatrix} \\begin{bmatrix} x_1(t) \\\\ x_2(t) \\end{bmatrix} + \\begin{bmatrix} 0 \\\\ \\frac{1}{L} \\end{bmatrix} u(t)

   • Case 1: Output is Loop Current ($y(t)=i_L(t)=x_2(t)$): y(t) = \\begin{bmatrix} 0 & 1 \\end{bmatrix} \\begin{bmatrix} x_1(t) \\\\ x_2(t) \\end{bmatrix} + \\begin{bmatrix} 0 \\end{bmatrix} u(t) \\mathbf{C} = \\begin{bmatrix} 0 & 1 \\end{bmatrix}, \\quad \\mathbf{D} = \\begin{bmatrix} 0 \\end{bmatrix}
   • Case 2: Output is Capacitor Voltage ($y(t)=v_C(t)=x_1(t)$): y(t) = \\begin{bmatrix} 1 & 0 \\end{bmatrix} \\begin{bmatrix} x_1(t) \\\\ x_2(t) \\end{bmatrix} + \\begin{bmatrix} 0 \\end{bmatrix} u(t) \\mathbf{C} = \\begin{bmatrix} 1 & 0 \\end{bmatrix}, \\quad \\mathbf{D} = \\begin{bmatrix} 0 \\end{bmatrix}

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3.2 Example 2: The Two-Stage Cascaded RC Network [PYQ 2025, 2019, 2015]

Question: Obtain the state-space model $(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})$ of the passive cascaded RC circuit shown below. The input is $v_i(t)$ and the output is the voltage across the second capacitor $v_{C_2}(t)$.

                     R1                              R2
          o-------[  R1  ]-------+---------------[  R2  ]-------+-------o
          |                      |                             |
         (+)                    [C1] v_C1(t)                  [C2] v_C2(t) (y)
        v_i(t)                   |                             |
          |                      |                             |
          o----------------------+-----------------------------+-------o

Step-by-Step Solution:

1. Assign State Variables: The circuit has two capacitors ($C_1$ and $C_2$) and zero inductors. The system order is $n=2+0=2$. Let our states be: $x_1(t)=v_{C_1}(t)\text{voltage at node 1}$ $x_2(t)=v_{C_2}(t)\text{voltage at node 2}$ The input is the voltage source: $u(t)=v_i(t)$ The output is: $y(t)=v_{C_2}(t)=x_2(t)$

2. Apply KCL at Nodes:

   • KCL at Node 1 (Capacitor $C_1$): Sum of currents entering equals sum of currents leaving: $\frac{v_i(t)-v_{C_1}(t)}{R_1}=i_{C_1}(t)+\frac{v_{C_1}(t)-v_{C_2}(t)}{R_2}$
   • KCL at Node 2 (Capacitor $C_2$): $\frac{v_{C_1}(t)-v_{C_2}(t)}{R_2}=i_{C_2}(t)$

3. Isolate Derivatives:

   • For the first capacitor, substitute $i_{C_1}(t)=C_1{\dot{v}}_{C_1}(t)=C_1{\dot{x}}_1(t)$: $C_1{\dot{x}}_1(t)=\frac{u(t)-x_1(t)}{R_1}-\frac{x_1(t)-x_2(t)}{R_2}$ $C_1{\dot{x}}_1(t)=-\left(\frac{1}{R_1}+\frac{1}{R_2}\right)x_1(t)+\frac{1}{R_2}{x}_2(t)+\frac{1}{R_1}u(t)$ ${\dot{x}}_1(t)=-\left(\frac{1}{R_1{C}_1}+\frac{1}{R_2{C}_1}\right)x_1(t)+\frac{1}{R_2{C}_1}{x}_2(t)+\frac{1}{R_1{C}_1}u(t)$
   • For the second capacitor, substitute $i_{C_2}(t)=C_2{\dot{v}}_{C_2}(t)=C_2{\dot{x}}_2(t)$: $C_2{\dot{x}}_2(t)=\frac{x_1(t)-x_2(t)}{R_2}$ ${\dot{x}}_2(t)=\frac{1}{R_2{C}_2}{x}_1(t)-\frac{1}{R_2{C}_2}{x}_2(t)$

4. Assemble Matrices: Write in vector-matrix format: \begin{bmatrix} \\dot{x}_1(t) \\\\ \\dot{x}_2(t) \\end{bmatrix} = \\begin{bmatrix} -\\left(\\frac{1}{R_1 C_1} + \\frac{1}{R_2 C_1}\\right) & \\frac{1}{R_2 C_1} \\\\ \\frac{1}{R_2 C_2} & -\\frac{1}{R_2 C_2} \\end{bmatrix} \\begin{bmatrix} x_1(t) \\\\ x_2(t) \\end{bmatrix} + \\begin{bmatrix} \\frac{1}{R_1 C_1} \\\\ 0 \\end{bmatrix} u(t) y(t) = \\begin{bmatrix} 0 & 1 \\end{bmatrix} \\begin{bmatrix} x_1(t) \\\\ x_2(t) \\end{bmatrix} + \\begin{bmatrix} 0 \\end{bmatrix} u(t)

   Thus, the state-space matrices are: \\mathbf{A} = \\begin{bmatrix} -\\frac{R_1 + R_2}{R_1 R_2 C_1} & \\frac{1}{R_2 C_1} \\\\ \\frac{1}{R_2 C_2} & -\\frac{1}{R_2 C_2} \\end{bmatrix}, \\quad \\mathbf{B} = \\begin{bmatrix} \\frac{1}{R_1 C_1} \\\\ 0 \\end{bmatrix}, \\quad \\mathbf{C} = \\begin{bmatrix} 0 & 1 \\end{bmatrix}, \\quad \\mathbf{D} = \\begin{bmatrix} 0 \\end{bmatrix}

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3.3 Example 3: Three-Storage Multi-Loop Network [PYQ 2024 Question 3b - Verbatim 13 Marks]

Question: Represent the following electrical circuit by a state model. The output $y(t)$ is the voltage across resistor $R_2$.

                     L1 (i_L1)                         L2 (i_L2)
          o-------[  L1  ]-------+------------------[  L2  ]-------+-------o
          |                      |                                 |
         (+)                    [C] v_C(t)                        [R2] y(t)
        V(t)                     |                                 |
          |                      |                                 |
          o------[  R1  ]--------+---------------------------------+-------o

Note: This KUET 2024 circuit is an exceptional multi-loop network because resistor $R_1$ is placed in the bottom return wire, which couples both left-hand and right-hand loop currents.

Step-by-Step Solution:

1. Assign State Variables: The circuit has three energy-storing elements: capacitor $C$, inductor $L_1$, and inductor $L_2$. The system order is $n=1+2=3$. Let our states be: $x_1(t)=v_C(t)\text{voltage across capacitor C}$ x_2(t) = i_{L_1}(t) \quad \text{{current through inductor } L_1 \text{ (left-to-right)}} x_3(t) = i_{L_2}(t) \quad \text{{current through inductor } L_2 \text{ (left-to-right)}} The input is: $u(t)=V(t)$ The output $y(t)$ is the voltage across the load resistor $R_2$. Since the current through $R_2$ is exactly $i_{L_2}(t)$: $y(t)=R_2{i}_{L_2}(t)=R_2{x}_3(t)$

2. Write KVL/KCL and Isolate Derivatives:

   • Capacitor State Equation (${\dot{x}}_1$): Apply KCL at the top central node (above capacitor $C$). The current entering from the left is $i_{L_1}(t)=x_2(t)$. The currents leaving are the capacitor current $i_C(t)$ and the right inductor current $i_{L_2}(t)=x_3(t)$. $x_2(t)=i_C(t)+x_3(t)⟹i_C(t)=x_2(t)-x_3(t)$ Since $i_C(t)=C{\dot{v}}_C(t)=C{\dot{x}}_1(t)$: $C{\dot{x}}_1(t)=x_2(t)-x_3(t)⟹{\dot{\mathbf{x}}}_1(\mathbf{t})=\frac{1}{\mathbf{C}}{\mathbf{x}}_2(\mathbf{t})-\frac{1}{\mathbf{C}}{\mathbf{x}}_3(\mathbf{t})$

   • Left Inductor State Equation (${\dot{x}}_2$): Write KVL around the left-hand loop containing $V(t)$, $L_1$, $C$, and $R_1$. Warning: Note that $R_1$ is in the bottom path, which carries only the left-hand loop current $i_{L_1}(t)=x_2(t)$: $V(t)-v_{L_1}(t)-v_C(t)-R_1{i}_{L_1}(t)=0$ Since $v_{L_1}(t)=L_1{\dot{i}}_{L_1}(t)=L_1{\dot{x}}_2(t)$: $L_1{\dot{x}}_2(t)=-v_C(t)-R_1{i}_{L_1}(t)+V(t)$ $L_1{\dot{x}}_2(t)=-x_1(t)-R_1{x}_2(t)+u(t)⟹{\dot{\mathbf{x}}}_2(\mathbf{t})=-\frac{1}{{\mathbf{L}}_1}{\mathbf{x}}_1(\mathbf{t})-\frac{{\mathbf{R}}_1}{{\mathbf{L}}_1}{\mathbf{x}}_2(\mathbf{t})+\frac{1}{{\mathbf{L}}_1}\mathbf{u}(\mathbf{t})$

   • Right Inductor State Equation (${\dot{x}}_3$): Write KVL around the right-hand loop containing $C$, $L_2$, and $R_2$: $v_C(t)-v_{L_2}(t)-R_2{i}_{L_2}(t)=0$ Since $v_{L_2}(t)=L_2{\dot{i}}_{L_2}(t)=L_2{\dot{x}}_3(t)$: $L_2{\dot{x}}_3(t)=v_C(t)-R_2{i}_{L_2}(t)$ $L_2{\dot{x}}_3(t)=x_1(t)-R_2{x}_3(t)⟹{\dot{\mathbf{x}}}_3(\mathbf{t})=\frac{1}{{\mathbf{L}}_2}{\mathbf{x}}_1(\mathbf{t})-\frac{{\mathbf{R}}_2}{{\mathbf{L}}_2}{\mathbf{x}}_3(\mathbf{t})$

3. Matrix Assembly: Gather the equations to formulate the state and output matrices: $\left[\begin{array}{c}{\dot{x}}_1(t)\\ {\dot{x}}_2(t)\\ {\dot{x}}_3(t)\end{array}\right]=\left[\begin{array}{ccc}0& \frac{1}{C}& -\frac{1}{C}\\ -\frac{1}{L_1}& -\frac{R_1}{L_1}& 0\\ \frac{1}{L_2}& 0& -\frac{R_2}{L_2}\end{array}\right]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\\ x_3(t)\end{array}\right]+\left[\begin{array}{c}0\\ \frac{1}{L_1}\\ 0\end{array}\right]u(t)$ $y(t)=\left[\begin{array}{ccc}0& 0& R_2\end{array}\right]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\\ x_3(t)\end{array}\right]+\left[\begin{array}{c}0\end{array}\right]u(t)$

   The final matrices are: $\mathbf{A}=\left[\begin{array}{ccc}0& \frac{1}{C}& -\frac{1}{C}\\ -\frac{1}{L_1}& -\frac{R_1}{L_1}& 0\\ \frac{1}{L_2}& 0& -\frac{R_2}{L_2}\end{array}\right],\mathbf{B}=\left[\begin{array}{c}0\\ \frac{1}{L_1}\\ 0\end{array}\right],\mathbf{C}=\left[\begin{array}{ccc}0& 0& R_2\end{array}\right],\mathbf{D}=\left[\begin{array}{c}0\end{array}\right]$

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3.4 Example 4: Series RLC with Parallel Admittance (Numerical Practice)

Question: Derive the numerical state matrices $(\mathbf{A},\mathbf{B})$ for the electrical circuit shown below. Given component values: $R_s=2\Omega$, $L=1\text{ H}$, $C=0.5\text{ F}$.

               Rs = 2 ohms
          o----[  Rs  ]------+-----------------------+----o
          |                  |                       |
         (+)                [C] C = 0.5 F           [L] L = 1 H
        v_i(t)               |                       |
          |                  |                       |
          o------------------+-----------------------+----o

Step-by-Step Solution:

1. Assign State Variables:

   • State variables: $x_1(t)=v_C(t)\text{voltage across capacitor}$ $x_2(t)=i_L(t)\text{current through inductor}$
   • Input: $u(t)=v_i(t)$

2. Write KCL and KVL:

   • KCL at top right node: The current entering through resistor $R_s$ equals the sum of capacitor and inductor currents: $\frac{v_i(t)-v_C(t)}{R_s}=i_C(t)+i_L(t)⟹i_C(t)=-\frac{1}{R_s}{v}_C(t)-i_L(t)+\frac{1}{R_s}{v}_i(t)$
   • KVL for Inductor branch: The inductor is in parallel with the capacitor, so the voltage across it is exactly $v_C(t)$: $v_L(t)=v_C(t)$

3. Isolate Derivatives:

   • For capacitor: $C{\dot{x}}_1(t)=-\frac{1}{R_s}{x}_1(t)-x_2(t)+\frac{1}{R_s}u(t)⟹{\dot{x}}_1(t)=-\frac{1}{R_{s}C}{x}_1(t)-\frac{1}{C}{x}_2(t)+\frac{1}{R_{s}C}u(t)$
   • For inductor: $L{\dot{x}}_2(t)=x_1(t)⟹{\dot{x}}_2(t)=\frac{1}{L}{x}_1(t)$

4. Evaluate Numerically: Substitute $R_s=2$, $L=1$, $C=0.5$: ${\dot{x}}_1(t)=-\frac{1}{2\times 0.5}{x}_1(t)-\frac{1}{0.5}{x}_2(t)+\frac{1}{2\times 0.5}u(t)⟹{\dot{x}}_1(t)=-x_1(t)-2x_2(t)+u(t)$ ${\dot{x}}_2(t)=\frac{1}{1}{x}_1(t)⟹{\dot{x}}_2(t)=x_1(t)$

   In matrix form: $\left[\begin{array}{c}{\dot{x}}_1(t)\\ {\dot{x}}_2(t)\end{array}\right]=\left[\begin{array}{cc}-1& -2\\ 1& 0\end{array}\right]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\end{array}\right]+\left[\begin{array}{c}1\\ 0\end{array}\right]u(t)$

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4. Common Mistakes That Cost Marks

The Inductor Current Polarities in KVL

Always write the inductor voltage as $+L\frac{d{i}_L}{dt}$ in the direction of the defined current loop. If you write loop KVL equations and flip the polarity of $v_L(t)$, you will generate negative signs in your $\mathbf{A}$ matrix, indicating a system with unstable poles. Stable physical circuits containing only passive resistors, inductors, and capacitors must have state eigenvalues with negative real parts!

Dependent State Redundancy

Only write states for independent energy-storing elements. If a circuit has a loop of only capacitors, or a cut-set of only inductors, the states are dependent {since the voltages or currents are constrained by KVL/KCL and cannot change independently}. In such cases, the total system order $n$ is reduced, and writing standard independent matrices will result in singular, non-invertible state equations.

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5. PYQ Bank — Verbatim Questions & Answer Plans

5.1 PYQ 2024 Question 3b [13 Marks]

Question: Represent the following electrical circuit by a state model: [CIRCUIT: Voltage source V(t) in series with R1 on bottom wire, inductor L1 on top-left, capacitor C in middle shunt, inductor L2 on top-right, and output resistor R2 on right-hand shunt - source: KUET 2024 Q3b]

• Answer Plan: Define the three energy storage states $x_1(t)=v_C(t)$, $x_2(t)=i_{L_1}(t)$, and $x_3(t)=i_{L_2}(t)$. Follow the exact algebraic formulation steps detailed in Section 3.3 to derive the isolated derivatives and present the correct $3\times 3$ system matrices to secure all 13 marks.

5.2 PYQ 2022 Question 4b [15 Marks - Verbatim]

Question: Find the Z & Y parameters of two port network shown in Fig. 4(b) (which can be derived using the state modeling conversions we cover in Note 3.06).

• Answer Plan: This problem bridges network theory and state space, solved by calculating s-domain equivalent matrices as outlined in Note 3.06.

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6. Self-Check Before Moving On

• Can you explain why capacitor voltage ($v_C$) and inductor current ($i_L$) are selected as natural state variables? [1.1]
• Do you know how to write independent KVL and KCL equations and completely isolate first-order derivatives (${\dot{v}}_C,{\dot{i}}_L$)? [2.0]
• Can you solve a multi-loop circuit containing three energy storage elements and compile them into $3\times 3$ state matrices? [3.3]
• Have you mastered the polarity convention for inductors and capacitors in KVL/KCL loops? [4.0]

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Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3 & 4), Rabiul sir -2309008.pdf, Azmat Sir-2309008.pdf (KUET Class Notes & Lecture Slides).

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Related Concepts: 3.01 Introduction to State-Space & Phase-Variable Formulations | 3.02 State-Space Modeling of Transfer Functions (Canonic Forms) | 3.03 State-Space Modeling of Electrical Circuits

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3.04 State-Space Modeling of Block Diagrams & Signal Flow Graphs

Core Idea

Operational block diagrams and signal flow graphs (SFGs) are graphical representations of a system’s mathematical relationships. By assigning state variables to the outputs of integrators (in continuous-time systems) or unit delays (in discrete-time systems), we can systematically extract the coupled first-order matrix state equations ($\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D}$) directly from the graphical signal paths.

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1. State-Variable Assignment Rules for Graphical Systems

To systematically formulate state equations from any block diagram or signal flow graph, we apply two rigid assignment rules depending on the system domain:

Rule 1: Continuous-Time (CT) Systems (Integrators)

The state variables represent the output signals of the integrator blocks {integrators accumulate signals over time}.

• If an integrator block has output $x_k(t)$, then its input must be the first derivative of that state: $\text{Output of Integrator }k=x_k(t)⟹\text{Input of Integrator }k={\dot{x}}_k(t)$
• For a system of order $n$, there will be exactly $n$ cascaded or parallel integrator blocks ($1/s$ in the s-domain), yielding $n$ state variables.

Rule 2: Discrete-Time (DT) Systems (Unit Delays)

The state variables represent the output signals of the unit delay blocks {delays hold discrete samples for one interval}.

• If a unit delay block ($z^{-1}$) has output $x_k[n]$, then its input must be the next advanced sample state: $\text{Output of Delay Block }k=x_k[n]⟹\text{Input of Delay Block }k=x_k[n+1]$
• For a discrete system of order $n$, there will be exactly $n$ delay elements, yielding $n$ discrete state variables.

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2. Rigorous 5-Step Graphical-to-Matrix Algorithm

Follow this mechanical checklist to convert any operational flowchart into state-space matrix format:

graph TD
    A[Step 1: Label Outputs of Integrators/Delays as x_i] --> B[Step 2: Label Inputs of Integrators/Delays as dx_i or x_i_plus_1]
    B --> C[Step 3: Write Equations for Adder Nodes in terms of x_i and u]
    C --> D[Step 4: Express System Output y in terms of x_i and u]
    D --> E[Step 5: Assemble Vector-Matrix Equations A, B, C, D]

1. Step 1: Identify all integrator blocks ($1/s$) or delay blocks ($z^{-1}$). Label their outputs as state variables $x_1,x_2,\dots ,x_n$.
2. Step 2: Label the inputs to these blocks as ${\dot{x}}_1,{\dot{x}}_2,\dots ,{\dot{x}}_n$ (for CT) or $x_1[n+1],x_2[n+1],\dots ,x_n[n+1]$ (for DT).
3. Step 3: Systematically write algebraic equations for each adder node {summation points} in the diagram. Express the derivative/advanced states (${\dot{x}}_i$ or $x_i[n+1]$) strictly as a linear combination of the state variables $x_j$ and the external inputs $u$.
4. Step 4: Write the output equation by expressing $y(t)$ or $y[n]$ strictly in terms of the state variables $x_j$ and input $u$.
5. Step 5: Group coefficients and assemble into the canonical matrix format: \text{State Equation:} & \quad \mathbf{\dot{X}} = \mathbf{A}\mathbf{X} + \mathbf{B}\mathbf{U} \\ \text{Output Equation:} & \quad \mathbf{Y} = \mathbf{C}\mathbf{X} + \mathbf{D}\mathbf{U} \end{aligned}$$

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3. Continuous-Time (CT) Block Diagram Modeling

Consider a continuous-time system containing three integrators. We want to extract the state-space representation directly from its hardware block wiring.

3.1 Worked Numerical: Multi-Feedback Integrator Diagram

Question: Obtain the state model $(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})$ of the continuous-time block diagram represented below:

[DIAGRAM: Continuous 3-integrator block diagram. Input u(t) enters an adder on the left. The output of this adder goes into Integrator 1. The output of Integrator 1 is state x3(t), which feeds into Integrator 2. The output of Integrator 2 is state x2(t), which feeds into Integrator 3. The output of Integrator 3 is state x1(t). Feedback gains of -5, -4, and -2 are taken from state outputs x1(t), x2(t), and x3(t) respectively, and routed back to the main input adder. The output y(t) is a linear sum of x1(t) and x2(t) - source: Azmat Sir Slide 15]

Step-by-Step Solution:

• Step 1: Assign States Let the outputs of the three integrators be assigned as: $x_1(t),x_2(t),x_3(t)$

• Step 2: Assign State Derivatives The inputs to the integrators are: ${\dot{x}}_1(t),{\dot{x}}_2(t),{\dot{x}}_3(t)$

• Step 3: Establish Integrator Connections From the cascading connections:

  \dot{x}_1(t) &= x_2(t) \\ \dot{x}_2(t) &= x_3(t) \end{aligned}$$

• Step 4: Write Main Adder Equation The input to the first integrator is ${\dot{x}}_3(t)$, which is the output of the main feedback adder node. Let us trace the signals entering this adder: ${\dot{x}}_3(t)=-5x_1(t)-4x_2(t)-2x_3(t)+u(t)$

• Step 5: Write Output Equation The system output $y(t)$ is the sum of $x_1(t)$ and $x_2(t)$: $y(t)=x_1(t)+x_2(t)$

• Step 6: Assemble State-Space Matrices $\left[\begin{array}{c}{\dot{x}}_1(t)\\ {\dot{x}}_2(t)\\ {\dot{x}}_3(t)\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -5& -4& -2\end{array}\right]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\\ x_3(t)\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u(t)$ $y(t)=\left[\begin{array}{ccc}1& 1& 0\end{array}\right]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\\ x_3(t)\end{array}\right]+\left[\begin{array}{c}0\end{array}\right]u(t)$

  Thus, the continuous-time state matrices are: $\mathbf{A}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -5& -4& -2\end{array}\right],\mathbf{B}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right],\mathbf{C}=\left[\begin{array}{ccc}1& 1& 0\end{array}\right],\mathbf{D}=\left[\begin{array}{c}0\end{array}\right]$

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4. Discrete-Time (DT) Block Diagram Modeling

Discrete-time diagrams utilize unit delays ($z^{-1}$). The mathematical extraction requires careful tracking of step indices.

4.1 Worked Numerical: Unit Delay Network [PYQ 2016]

Question: Find the state-space description $(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})$ corresponding to the system depicted below by choosing the state variables to be the outputs of the unit delays:

[DIAGRAM: Discrete-time realization block diagram containing two delay blocks (z^-1). Node labels: Input u[n] enters on the left, feeding into a top-path adder and a bottom-path adder. Delay Block 1 has output state x1[n], and Delay Block 2 has output state x2[n]. Tracing feedback path loops: the input to Delay 1 (x1[n+1]) is fed by u[n] + 1/3 x2[n]. The input to Delay 2 (x2[n+1]) is fed by the output of a multiplier (4/3) from x2[n] + x1[n] + u[n]. The output y[n] is tapped from the final right-side adder node - source: KUET 2016 Question 4a]

Step-by-Step Solution:

1. Assign State Variables: Let the outputs of the delay blocks be the states: $\mathbf{X}[n]=\left[\begin{array}{c}{x}_1[n]\\ x_2[n]\end{array}\right]$

2. Establish Advanced State Equations: Tracing the signals at the inputs of the delays ($x_i[n+1]$): $x_1[n+1]=\frac{1}{3}{x}_2[n]+u[n]$ $x_2[n+1]=x_1[n]+\frac{4}{3}{x}_2[n]+u[n]$

3. Establish Output Equation: Tracing the output path $y[n]$ from the final right-side adder node: $y[n]=x_1[n]+\frac{4}{3}{x}_2[n]+u[n]$

4. Formulate Coupled Matrices: $\left[\begin{array}{c}{x}_1[n+1]\\ x_2[n+1]\end{array}\right]=\left[\begin{array}{cc}0& \frac{1}{3}\\ 1& \frac{4}{3}\end{array}\right]\left[\begin{array}{c}{x}_1[n]\\ x_2[n]\end{array}\right]+\left[\begin{array}{c}1\\ 1\end{array}\right]u[n]$ $y[n]=\left[\begin{array}{cc}1& \frac{4}{3}\end{array}\right]\left[\begin{array}{c}{x}_1[n]\\ x_2[n]\end{array}\right]+\left[\begin{array}{c}1\end{array}\right]u[n]$

   Thus, the final discrete state-space matrices are: $\mathbf{A}=\left[\begin{array}{cc}0& \frac{1}{3}\\ 1& \frac{4}{3}\end{array}\right],\mathbf{B}=\left[\begin{array}{c}1\\ 1\end{array}\right],\mathbf{C}=\left[\begin{array}{cc}1& \frac{4}{3}\end{array}\right],\mathbf{D}=\left[\begin{array}{c}1\end{array}\right]$

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5. Signal Flow Graph (SFG) State-Space Formulations

A Signal Flow Graph (SFG) is a simplified directed graph where nodes represent variables and branches represent gains. It is mathematically equivalent to block diagrams.

5.1 The SFG State Extraction Rule

• Identify all branch transmittance terms that contain the integrator factor $s^{-1}$ (or delay factor $z^{-1}$).
• The State Node Rule: The output node of every branch with transmittance $s^{-1}$ is designated as a state variable $x_i(t)$. The input node to that same branch represents the derivative of that state, ${\dot{x}}_i(t)$.

               \dot{x}_i(t)     1/s      x_i(t)
              o-------------( Integrator )------------->o

• Write nodal equations for all derivative nodes (${\dot{x}}_i$) by summing all incoming branch signals entering that specific node.

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6. Common Mistakes That Cost Marks

The Integrator Output-Input Swapping Trap

Never swap the input and output variables of the integrator! The output of the integrator block is the state $x_i(t)$, and the input is ${\dot{x}}_i(t)$. If you mistakenly write the input as $x_i(t)$ and output as ${\dot{x}}_i(t)$, your derivative equations will involve unwanted integral calculus, completely violating the algebraic first-order state format.

The Direct Feedforward Path ($\mathbf{D}$) Oversight

Always verify if there is a direct feedforward path bypassing the delay/integrator chains directly from the input terminal $u(t)$ to the output terminal $y(t)$. If such a path exists (for example, a gain branch running from start to finish), your direct transmission matrix $\mathbf{D}$ is non-zero ($\mathbf{D}\ne 0$). Forgetting to include this feedforward term in your output equation yields a major structural grading deduction.

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7. PYQ Bank — Verbatim Questions & Answer Plans

7.1 PYQ 2016 Question 4a [12 Marks]

Question: Find the state variable description corresponding to the system depicted in Fig. 4(a) by choosing the state variables to be the outputs of the unit delays.

• Answer Plan: Follow the step-by-step discrete mathematical derivation provided in Section 4.1. Clearly label the state variables $x_1[n]$ and $x_2[n]$ at the delay outputs, trace the inputs $x_1[n+1]$ and $x_2[n+1]$, formulate the node state equations, write the output equation, and present the final matrices ($\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D}$) with their respective dimensions to secure all 12 marks.

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8. Self-Check Before Moving On

• Can you define state-variable assignment rules for continuous integrators and discrete unit delays?
• Do you know how to write algebraic equations for adder nodes by summing incoming branch signals?
• Have you mastered the 5-step graphical-to-matrix mapping algorithm?
• Can you identify non-zero $\mathbf{D}$ matrices caused by direct input-output feedforward branches?

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

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3.04 State-Space Modeling of Block Diagrams & SFGs | 3.06 State Transition Matrix: Properties & Solution Methods

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3.05 Discrete-Time State-Space Representation

Core Idea

Just as continuous-time (CT) systems can be represented in state-space using derivatives, discrete-time (DT) systems described by linear constant-coefficient difference equations (LCCDEs) {equations relating discrete input and output sequences through delays} can be modeled using first-order shift-operator difference matrix equations. Instead of continuous integrators ($1/s$), DT systems utilize unit delays ($z^{-1}$) as their physical memory storage elements, where the state variables represent delayed values of the output or internal signals.

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1. The Discrete-Time State-Space Formulation

In the discrete-time domain, the state-space model is formulated as a set of coupled, first-order difference matrix equations that track state transitions from sample index $n$ to $n+1$ {the discrete sample counter representing sequential clock ticks}.

1.1 The Discrete State Equation

The state equation governs how the internal state vector at the next clock cycle ($\mathbf{X}[n+1]$) is computed from the current state vector ($\mathbf{X}[n]$) and the current input vector ($\mathbf{U}[n]$): $\mathbf{X}[n+1]=\mathbf{AX}[n]+\mathbf{BU}[n]$

1.2 The Discrete Output Equation

The output equation maps the current states and current inputs directly to the system’s measurable output terminals ($\mathbf{Y}[n]$): $\mathbf{Y}[n]=\mathbf{CX}[n]+\mathbf{DU}[n]$

Where:

• $\mathbf{X}[n]$ is the $n\times 1$ State Vector containing the internal memory states.
• $\mathbf{U}[n]$ is the $m\times 1$ Input Vector containing the external excitation sequences.
• $\mathbf{Y}[n]$ is the $p\times 1$ Output Vector containing the system outputs.
• $\mathbf{A}$ is the $n\times n$ System Matrix (governing natural dynamics and discrete poles).
• $\mathbf{B}$ is the $n\times m$ Input Matrix.
• $\mathbf{C}$ is the $p\times n$ Output Matrix.
• $\mathbf{D}$ is the $p\times m$ Direct Transmission Matrix.

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2. Phase-Variable Selection in Discrete Time

For an $N$-th order recursive LCCDE with output delays: $y[n]+a_{1}y[n-1]+a_{2}y[n-2]+\cdots +a_{N}y[n-N]=b_{0}u[n]$

To eliminate high-order delays and construct first-order state-space matrices, we apply a standard Delay-Chain Phase Variable Mapping.

Step-by-Step Selection Workflow

Step 1: Assign State Variables to Bounded Delays

We define our state variables as the output of the unit delay elements, starting from the longest delay ($y[n-N]$) up to the single-sample delay ($y[n-1]$): $x_1[n]=y[n-N]$ $x_2[n]=y[n-N+1]$ $x_3[n]=y[n-N+2]$ $⋮$ $x_N[n]=y[n-1]$

Step 2: Write First-Order Shift Equations

Advance each state variable by one sample index ($n\to n+1$) to find its state-transition behavior: $x_1[n+1]=y[n-N+1]=x_2[n]$ $x_2[n+1]=y[n-N+2]=x_3[n]$ $⋮$ $x_{N-1}[n+1]=y[n-1]=x_N[n]$

Step 3: Solve for the Final State Transition

The transition of the final state variable ($x_N[n+1]$) represents the current output $y[n]$. Isolate $y[n]$ in the original difference equation: $x_N[n+1]=y[n]=-a_{1}y[n-1]-a_{2}y[n-2]-\cdots -a_{N}y[n-N]+b_{0}u[n]$

Substitute the state variable definitions from Step 1 into this expression: $x_N[n+1]=-a_N{x}_1[n]-a_{N-1}{x}_2[n]-\cdots -a_1{x}_N[n]+b_{0}u[n]$

Step 4: Assemble the Canonical Companion Matrix

Gather the state transitions into the discrete-time companion matrix format: $\left[\begin{array}{c}{x}_1[n+1]\\ x_2[n+1]\\ ⋮\\ x_{N-1}[n+1]\\ x_N[n+1]\end{array}\right]=\left[\begin{array}{ccccc}0& 1& 0& \dots & 0\\ 0& 0& 1& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \dots & 1\\ -a_N& -a_{N-1}& -a_{N-2}& \dots & -a_1\end{array}\right]\left[\begin{array}{c}{x}_1[n]\\ x_2[n]\\ ⋮\\ x_{N-1}[n]\\ x_N[n]\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ ⋮\\ 0\\ b_0\end{array}\right]u[n]$

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3. Comprehensive Worked Examples (The Exam Killers)

3.1 Example 1: Third-Order Recursive Difference Equation [Passage 84]

Question: Obtain the state-space model $(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})$ in companion form for the system described by the difference equation: $4y[n]+3y[n-1]+2y[n-2]+y[n-3]=u[n]$

Step-by-Step Solution:

1. Normalize the Difference Equation (Monic Lead Coefficient): Divide the entire difference equation by the coefficient of $y[n]$ (which is $4$): $y[n]+\frac{3}{4}y[n-1]+\frac{1}{2}y[n-2]+\frac{1}{4}y[n-3]=\frac{1}{4}u[n]$

2. Define Phase-Variable States ($N=3$): Assign our state variables to the outputs of the cascade of three delay elements: $x_1[n]=y[n-3]$ $x_2[n]=y[n-2]$ $x_3[n]=y[n-1]$

3. Formulate State Shifts: Shift each state variable forward by one step: $x_1[n+1]=y[n-2]=x_2[n]$ $x_2[n+1]=y[n-1]=x_3[n]$

   Solve for the final state shift $x_3[n+1]=y[n]$ by isolating $y[n]$ in the normalized LCCDE: $y[n]=-\frac{1}{4}y[n-3]-\frac{1}{2}y[n-2]-\frac{3}{4}y[n-1]+\frac{1}{4}u[n]$ $x_3[n+1]=-\frac{1}{4}{x}_1[n]-\frac{1}{2}{x}_2[n]-\frac{3}{4}{x}_3[n]+\frac{1}{4}u[n]$

4. Assemble the State Equation Matrix: $\left[\begin{array}{c}{x}_1[n+1]\\ x_2[n+1]\\ x_3[n+1]\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -\frac{1}{4}& -\frac{1}{2}& -\frac{3}{4}\end{array}\right]\left[\begin{array}{c}{x}_1[n]\\ x_2[n]\\ x_3[n]\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ \frac{1}{4}\end{array}\right]u[n]$

5. Assemble the Output Equation Matrix: Since our output sequence is $y[n]=x_3[n+1]$ (the current updated output), we substitute the state transition of $x_3[n+1]$ directly to write our Output Equation strictly in terms of current states and current inputs: $y[n]=-\frac{1}{4}{x}_1[n]-\frac{1}{2}{x}_2[n]-\frac{3}{4}{x}_3[n]+\frac{1}{4}u[n]$ $y[n]=\left[\begin{array}{ccc}-\frac{1}{4}& -\frac{1}{2}& -\frac{3}{4}\end{array}\right]\left[\begin{array}{c}{x}_1[n]\\ x_2[n]\\ x_3[n]\end{array}\right]+\left[\begin{array}{c}\frac{1}{4}\end{array}\right]u[n]$

   Thus, the final discrete companion state-space matrices are: $\mathbf{A}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -\frac{1}{4}& -\frac{1}{2}& -\frac{3}{4}\end{array}\right],\mathbf{B}=\left[\begin{array}{c}0\\ 0\\ \frac{1}{4}\end{array}\right],\mathbf{C}=\left[\begin{array}{ccc}-\frac{1}{4}& -\frac{1}{2}& -\frac{3}{4}\end{array}\right],\mathbf{D}=\left[\begin{array}{c}\frac{1}{4}\end{array}\right]$

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4. Hardware Realization (Direct Form II Canonic Structure)

To visualize this discrete-time state-space system, we can map the matrices directly to a Canonic Direct Form II block diagram. Each state variable $x_i[n]$ corresponds to the output of a physical unit-delay element ($z^{-1}$).

               u[n] 
                │
                v [x 1/4]
                │
                +─────────> ( + ) <─────────────────────────────────┐
                             │                                      │
                             v                                      │
                         ┌───────┐                                  │
                         │ z^-1  │───────> x3[n] ───( x -3/4 )──────┤
                         └───────┘                                  │
                             │                                      │
                             v                                      │
                         ┌───────┐                                  │
                         │ z^-1  │───────> x2[n] ───( x -1/2 )──────┤
                         └───────┘                                  │
                             │                                      │
                             v                                      │
                         ┌───────┐                                  │
                         │ z^-1  │───────> x1[n] ───( x -1/4 )──────┘
                         └───────┘
                             │
                             +───────────> Output y[n] = x3[n+1]

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5. Common Mistakes That Cost Marks

The Monic Coefficient Oversight

Students frequently forget to normalize the lead coefficient of $y[n]$ to 1 before assigning phase-variable states. In the equation $4y[n]+3y[n-1]+\cdots =u[n]$, starting state assignments without dividing by 4 first yields feedback gains and input scaling coefficients that are off by a factor of 4, leading to a critical grading penalty.

The Current Output ($y[n]$) State Definition Trap

Defining $x_1[n]=y[n]$ as a state variable is a major error. The current output $y[n]$ is not a memory state; it is a dependent response that changes instantaneously with the current input. State variables must represent delayed storage elements (e.g., $y[n-1],y[n-2]$) to avoid algebraic loops during transition matrix calculations.

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6. Potential/Grounded Exam Questions

6.1 Discrete LCCDE Conversion

Question: Represent the second-order recursive system $y[n]-\frac{1}{6}y[n-1]-\frac{1}{6}y[n-2]=x[n]$ in state-space companion form.

• Answer Plan:
  1. Define $x_1[n]=y[n-2]$ and $x_2[n]=y[n-1]$.
  2. Write shifts: $x_1[n+1]=x_2[n]$, and $x_2[n+1]=y[n]=\frac{1}{6}{x}_1[n]+\frac{1}{6}{x}_2[n]+u[n]$.
  3. Assemble state-space matrices: $\mathbf{A}=\left[\begin{array}{cc}0& 1\\ \frac{1}{6}& \frac{1}{6}\end{array}\right],\mathbf{B}=\left[\begin{array}{c}0\\ 1\end{array}\right],\mathbf{C}=\left[\begin{array}{cc}\frac{1}{6}& \frac{1}{6}\end{array}\right],\mathbf{D}=\left[\begin{array}{c}1\end{array}\right]$

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7. Self-Check Before Moving On

• Can you write out the discrete-time State and Output matrix equations with correct variable dimensions?
• Do you know how to assign delay-chain state variables to an $N$-th order difference equation?
• Have you mastered the requirement to normalize the coefficient of $y[n]$ to 1 before formulating state companion matrices?
• Can you draw a Canonic Direct Form II block diagram from a discrete state-space model?

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Source: (k.Deergha Rao) signals and systems.pdf (Chapter 6 & 8), ECE-2107_STATE_SPACE_ANALYSIS.pdf, Rabiul sir -2309008.pdf (Class Lecture Notes).

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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).