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).