3.02 State Space Modeling of Transfer Functions Canonic Form

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.