System Dynamics
System Dynamics
Home

Normal tree to state-space

At long last, we consider an algorithm to generate a state-space model from a linear graph model. In the following, we will consider a connected graph with \(E\) edges, of which \(S\) are sources (split between through-variable sources \(S_T\) and across \(S_A\)). There are \(2 E - S\) unknown across- and through-variables, so that’s how many equations we need. We have \(E-S\) elemental equations and for the rest we will write continuity and compatibility equations. \(N\) is the number of nodes.

  1. Derive \(2 E-S\) independent differential and algebraic equations from elemental, continuity, and compatibility equations.

    1. Draw a normal tree.

    2. Identify primary and secondary variables.

    3. Select the state variables to be
      across-variables on A-type branches and
      through-variables on T-type links.

    4. Define the state vector \(\bm{x}\), input vector \(\bm{u}\), and output vector \(\bm{y}\).

    5. Write an elemental equation for each passive element.1

    6. Write a continuity equation for each passive branch by drawing a contour intersecting that and no other branch. Solve each for the secondary through-variable associated with that branch.2

    7. Write a compatibility equation for each passive link by temporarily “including” it in the tree and finding the compatibility equation for the resulting loop. Solve each for the secondary across-variable associated with that link.3

  2. Eliminate variables that are not state or input variables and their derivatives. The following procedure is recommended.

    1. Eliminate all secondary variables by substitution into the elemental equations of the continuity and compatibility equations.

    2. Reduce the resulting set of equations to \(n\) (system order) in state and input variables, only. If not elimination, use linear algebra.

    3. Write the result in standard form ( or ).

    4. Express the output variables in terms of state and input variables, using any of the elemental, continuity, or compatibility equations.

    5. Write the result in standard form ( or ).

Example 3.2

For the electronic system shown, find a state-space model with outputs iL, Is, and vR2.

0. Linear graph model.
Coordinate arrows are given.

1a. The normal tree is shown on the linear graph.

1b.

Primary variables: VS, vC, iL, iR1, vR2.
Secondary variables: IS, iC, vL, vR1, iR2.

1c. State variables: vC, iL (system order n = 2).

1d. State, input, and output vectors: $$\begin{aligned} \bm{x} = \begin{bmatrix} v_C \\ i_L \end{bmatrix} \qquad \bm{u} = \begin{bmatrix} V_s \end{bmatrix} \qquad \bm{y} = \begin{bmatrix} i_L \\ I_S \\ v_{R_2} \end{bmatrix}. \end{aligned}$$

1e. Elemental equations.

C $\frac{d v_C} {d t} = \frac{1} {C} i_{C}$
L $\frac{d i_L} {d t} = \frac{1} {L} v_{L}$
R1 iR1 = vR1/R1
R2 vR2 = iR2R2

1f. Continuity equations.

Branch Equation
C (con1) iC = iR1 − iL
R2 (con2) iR2 = iL

1g. Compatibility equations.

Link Equation
L vL =  − vR2 + vC
R1 vR1 = VS − vC

2a. Eliminate secondary variables from the elemental equations.

C $\frac{d v_C} {d t} = \frac{1} {C} \left(i_{R_1} - i_L\right)$
L $\frac{d i_L} {d t} = \frac{1} {L} \left(-v_{R_2} + v_C\right)$
R1 iR1 = (VS−vC)/R1
R2 vR2 = iLR2

2b. Eliminate non-state and non-input variables from the elemental equations. In this case, iR1 and vR2 are the only variables remaining to eliminate, and they are already expressed as state- and input-variables! (This doesn’t always happen, but it does pretty frequently for relatively simple systems.)

C $\frac{d v_C} {d t} = \frac{1} {R_1 C} V_S - \frac{1} {R_1 C} v_C - \frac{1} {C} i_L$
L $\frac{d i_L} {d t} = -\frac{R_2} {L} i_L + \frac{1} {L} v_C$

2c. Write the state equation in standard form $$\begin{aligned} \frac{d\bm{x}} {d t} &= \begin{bmatrix} \frac{-1} {R_1 C} & \frac{-1} {C} \\ 1/L & -R_2/L \end{bmatrix} \begin{bmatrix} v_C \\ i_L \end{bmatrix} + \begin{bmatrix} \frac{1} {R_1 C} \\ 0 \end{bmatrix} \begin{bmatrix} V_S \end{bmatrix}. \end{aligned}$$

2d. Outputs in terms of states and inputs. $$\begin{aligned} i_L = i_L && I_S = i_{R_1} = (V_S - v_C)/R_1 && v_{R_2} = R_2 i_L. \end{aligned}$$ Note the sign for IS. This is because, for an across-variable source, we assume the through-variable flows in the opposite direction!

2e. Output in standard form. $$\begin{aligned} \bm{y} &= \begin{bmatrix} 0 & 1 \\ -1/R_1 & 0 \\ 0 & R_2 \end{bmatrix} \begin{bmatrix} v_C \\ i_L \end{bmatrix} + \begin{bmatrix} 0 \\ 1/R_1 \\ 0 \end{bmatrix} \begin{bmatrix} V_S \end{bmatrix}. \end{aligned}$$

  1. There will be \(E-S\) elemental equations.↩︎

  2. There will be \(N - 1 - S_A\) independent continuity equations.↩︎

  3. There will be \(E-N+1-S_T\) independent compatibility equations.↩︎

Online Resources for Section 3.4

No online resources.