Problems
Problem 3.1 (METROID)
Draw necessary sign coordinate arrows, a linear graph, a normal tree, and identify state variables and system order for each of the following schematics.
Electrical system, current source
Rotational mechanical system, torque source
Translational mechanical system, velocity source
The linear graph model is shown, below. Assign the coordinate arrows toward ground.
The normal tree is shown on the linear graph. The state variables are \(v_C\) and \(i_L\), so the system order is \(n = 2\).
The linear graph model is shown, below. Assign the coordinate arrows toward the right.
The normal tree is shown on the linear graph. The state variables are \(\Omega_{J_1}\), \(\Omega_{J_2}\), and \(T_k\), so the system order is \(n = 3\).
The linear graph model is shown, below. Assign the coordinate arrows toward ground.
The normal tree is shown on the linear graph. The state variables are \(v_{m_1}\), \(v_{m_2}\), \(f_{k_1}\), and \(f_{k_2}\), so the system order is \(n = 4\).
Problem 3.2 (MEGAMAN)
Draw necessary sign coordinate arrows, a linear graph, a normal tree, and identify state variables and system order for each of the following schematics.
Electrical system, voltage source
Rotational mechanical system, angular velocity source
Translational mechanical system, force source
The linear graph model is shown, below.
The normal tree is shown on the linear graph. The state variables are \(v_C\) and \(i_L\), so the system order is \(n = 2\).
The linear graph model is shown, below. Assign the coordinate arrows toward the right.
The normal tree is shown on the linear graph. The state variables are \(\Omega_{J_1}\), \(\Omega_{J_2}\), \(T_{k_1}\), and \(T_{k_2}\), so the system order is \(n = 4\).
The linear graph model is shown, below. Assign the coordinate arrows toward ground.
The normal tree is shown on the linear graph. The state variables are \(v_{m_1}\), \(v_{m_2}\), \(v_{m_3}\), \(f_{k_1}\), and \(f_{k_2}\), so the system order is \(n = 5\).
Problem 3.3 (SONIC)
Draw necessary sign coordinate arrows, a linear graph, a normal tree, and identify state variables and system order for each of the following schematics.
Electrical system, voltage source
Rotational mechanical system, torque source
Translational mechanical system, force source
The linear graph model is shown, below.
The normal tree is shown on the linear graph. The state variables are \(v_{C_1}\), \(v_{C_2}\), and \(i_L\), so the system order is \(n = 3\).
The linear graph model is shown, below. Assign the coordinate arrows toward the right.
The normal tree is shown on the linear graph. The state variables are \(\Omega_{J_0}\), \(\Omega_{J_1}\), \(\Omega_{J_2}\), \(T_{k_1}\), and \(T_{k_2}\), so the system order is \(n = 5\).
The linear graph model is shown, below. Assign the coordinate arrows toward ground.
The normal tree is shown on the linear graph. The state variables are \(v_{m_1}\), \(v_{m_2}\), \(v_{m_3}\), \(f_{k_1}\), \(f_{k_2}\), and \(f_{k_3}\), so the system order is \(n = 6\).
Problem 3.4 (NINTENDO)
Use the following linear graph for a circuit to answer the questions below, which are the steps to determining a state-space model of the circuit. Use the sign convention from the diagram. \(V_S\) is a voltage source.
Determine the normal tree, state variables, system order, state vector, input vector, and output vector for the outputs \(i_R\) and \(v_C\).
Write the required elemental, continuity, and compatibility equations.
Solve for the state equation in standard form.
Solve for the output equation in standard form.
The normal tree is shown, below.
The state variables are \(v_C\) and \(i_L\), so the order \(n=2\). The state, input, and output vectors are $$
\[\begin{aligned} \bm{x} = \begin{bmatrix} v_C \\ i_L \end{bmatrix}, \qquad \bm{u} = \begin{bmatrix} V_S \end{bmatrix}, \qquad \bm{x} = \begin{bmatrix} i_R \\ v_C \end{bmatrix}. \end{aligned}\]$$
The elemental equations are: $$
\[\begin{aligned} \frac{d v_C} {d t} &= \frac{1} {C} i_C \\ \frac{d i_L} {d t} &= \frac{1} {L} v_L \\ v_R &= i_R R. \end{aligned}\]\[ The continuity equations are: \]
\[\begin{aligned} i_R &= i_L \\ i_C &= i_L. \end{aligned}\]\[ The compatibility equation is: \]
\[\begin{aligned} v_L &= -v_R + V_S - v_C. \end{aligned}\]$$
Using the continuity and compatibility equations to eliminate secondary variables from the elemental equations, $$
\[\begin{aligned} \frac{d v_C} {d t} &= \frac{1} {C} i_L \\ \frac{d i_L} {d t} &= \frac{1} {L} (-v_R + V_S - v_C) \\ v_R &= i_L R. \end{aligned}\]\[ Using the $R$ elemental equation to eliminate $v_R$, \]
\[\begin{aligned} \frac{d v_C} {d t} &= \frac{1} {C} i_L \\ \frac{d i_L} {d t} &= \frac{1} {L} (-i_L R + V_S - v_C). \end{aligned}\]\[ Writing in standard form, \]
\[\begin{aligned} \frac{d \bm{x}}{d t} &= \begin{bmatrix} 0 & 1/C \\ -1/L & -R/L \end{bmatrix} \begin{bmatrix} v_C \\ i_L \end{bmatrix} + \begin{bmatrix} 0 \\ 1/L \end{bmatrix} \begin{bmatrix} V_S \end{bmatrix}. \end{aligned}\]$$
The outputs can be written in terms of state and input variables as $$
\[\begin{aligned} i_R = i_L \qquad v_C = v_C. \end{aligned}\]\[ Writing in standard form, \]
\[\begin{aligned} \bm{y} &= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} v_C \\ i_L \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \end{bmatrix} \begin{bmatrix} V_S \end{bmatrix}. \end{aligned}\]$$
Problem 3.5 (SUPERNINTENDO)
Use the following linear graph for a mechanical translational system to answer the questions below, which are the steps to determining a state-space model from the linear graph.
Use the sign convention from the diagram. \(F_S\) is a force source. Let the outputs be \(v_m\) and \(f_k\).
Determine the normal tree, state variables, system order, state vector, input vector, and output vector.
Write the required elemental, continuity, and compatibility equations.
Solve for the state equation in standard form.
Solve for the output equation in standard form.
The normal tree is shown, below.
The state variables are \(v_m\) and \(f_k\), so the order \(n=2\). The state, input, and output vectors are $$
\[\begin{aligned} \bm{x} = \begin{bmatrix} v_m \\ f_k \end{bmatrix}, \qquad \bm{u} = \begin{bmatrix} F_S \end{bmatrix}, \qquad \bm{y} = \begin{bmatrix} v_m \\ f_k \end{bmatrix}. \end{aligned}\]$$
The elemental equations are: $$
\[\begin{aligned} \frac{d v_m} {d t} &= \frac{1} {m} f_m \\ \frac{d f_k} {d t} &= k v_k. \end{aligned}\]\[ The continuity equation is: \]
\[\begin{aligned} f_m = F_S - f_k. \end{aligned}\]\[ The compatibility equation is: \]
\[\begin{aligned} v_k &= v_m. \end{aligned}\]$$
Using the continuity and compatibility equations to eliminate secondary variables from the elemental equations, $$
\[\begin{aligned} \frac{d v_m} {d t} &= \frac{1} {m} (F_S - f_k) \\ \frac{d f_k} {d t} &= k v_m. \end{aligned}\]\[ Writing in standard form, \]
\[\begin{aligned} \frac{d \bm{x}}{d t} &= \begin{bmatrix} 0 & -1/m \\ k & 0 \end{bmatrix} \begin{bmatrix} v_m \\ f_k \end{bmatrix} + \begin{bmatrix} 1/m \\ 0 \end{bmatrix} \begin{bmatrix} F_S \end{bmatrix}. \end{aligned}\]$$
The outputs can be written in terms of state and input variables, trivially, as $$
\[\begin{aligned} v_m = v_m \qquad f_k = f_k. \end{aligned}\]\[ Writing in standard form, \]
\[\begin{aligned} \bm{y} &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} v_m \\ f_k \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \end{bmatrix} \begin{bmatrix} F_S \end{bmatrix}. \end{aligned}\]$$
Problem 3.6 (GAMEBOY)
Use the following linear graph for a mechanical rotational system to answer the questions below, which are the steps to determining a state-space model from the linear graph.
Use the sign convention from the diagram. \(T_S\) is a torque source. Let the outputs be \(\Omega_J\) and \(T_B\).
Determine the normal tree, state variables, system order, state vector, input vector, and output vector.
Write the required elemental, continuity, and compatibility equations.
Solve for the state equation in standard form.
Solve for the output equation in standard form.
The normal tree is shown, below.
The state variables are \(\Omega_J\) and \(T_K\), so the order \(n=2\). The state, input, and output vectors are $$
\[\begin{aligned} \bm{x} = \begin{bmatrix} \Omega_J \\ T_K \end{bmatrix}, \qquad \bm{u} = \begin{bmatrix} T_S \end{bmatrix}, \qquad \bm{y} = \begin{bmatrix} \Omega_J \\ T_B \end{bmatrix}. \end{aligned}\]$$
The elemental equations are: $$
\[\begin{aligned} \frac{d \Omega_J} {d t} &= \frac{1} {J} \Omega_J \\ \frac{d T_k} {d t} &= k \Omega_k \\ T_B &= B \Omega_B. \end{aligned}\]\[ The continuity equation is: \]
\[\begin{aligned} T_J = T_S - T_k - T_B. \end{aligned}\]\[ The compatibility equations are: \]
\[\begin{aligned} \Omega_k &= \Omega_J \\ \Omega_B &= \Omega_J. \end{aligned}\]$$
Using the continuity and compatibility equations to eliminate secondary variables from the elemental equations, $$
\[\begin{aligned} \frac{d \Omega_J} {d t} &= \frac{1} {J} (T_S - T_k - T_B) \\ \frac{d T_k} {d t} &= k \Omega_J \\ T_B &= B \Omega_J. \label{eq:ss_from_lg_3_TB} \end{aligned}\]\[ Substituting into the first of these the last, we get the scalar state equations, \]
\[\begin{aligned} \frac{d \Omega_J} {d t} &= \frac{1} {J} (T_S - T_k - B \Omega_J) \\ \frac{d T_k} {d t} &= k \Omega_J \end{aligned}\]\[ Writing in standard form, \]
\[\begin{aligned} \frac{d \bm{x}}{d t} &= \begin{bmatrix} -B/J & -1/J \\ k & 0 \end{bmatrix} \begin{bmatrix} \Omega_J \\ T_k \end{bmatrix} + \begin{bmatrix} 1/J \\ 0 \end{bmatrix} \begin{bmatrix} T_S \end{bmatrix}. \end{aligned}\]$$
The outputs can be written in terms of state and input variables as $$
\[\begin{aligned} \Omega_J &= \Omega_J\\ T_B &= B \Omega_J. \tag{from \cref{eq:ss_from_lg_3_TB}} \end{aligned}\]\[ Writing in standard form, \]
\[\begin{aligned} \bm{y} &= \begin{bmatrix} 1 & 0 \\ B & 0 \end{bmatrix} \begin{bmatrix} \Omega_J \\ T_k \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \end{bmatrix} \begin{bmatrix} T_S \end{bmatrix}. \end{aligned}\]$$
Problem 3.7 (BLOWHARD)
Use the following linear graph for a mechanical rotational system to answer the questions below, which are the steps to determining a state-space model from the linear graph.
Use the sign convention from the diagram. \(\Omega_S\) is an angular velocity source. Let the outputs be the angular velocity \(\Omega_J\) of the inertia and the angular displacement \(\theta_k\) across the spring.
Determine the normal tree, state variables, system order, state vector, input vector, and output vector.
Write the required elemental, continuity, and compatibility equations.
Solve for the state equation in standard form.
Solve for the output equation in standard form.
The normal tree is shown, below.
The state variables are \(\Omega_J\) and \(T_k\), so the order \(n=2\). The state, input, and output vectors are $$
\[\begin{aligned} \bm{x} = \begin{bmatrix} \Omega_J \\ T_k \end{bmatrix}, \qquad \bm{u} = \begin{bmatrix} \Omega_S \end{bmatrix}, \qquad \bm{y} = \begin{bmatrix} \Omega_J \\ \theta_k \end{bmatrix}. \end{aligned}\]$$
The elemental equations are: $$
\[\begin{aligned} \frac{d \Omega_J} {d t} &= \frac{1} {J} T_J \\ \frac{d T_k} {d t} &= k \Omega_k \\ T_B &= \Omega_B B. \end{aligned}\]\[ The continuity equations are: \]
\[\begin{aligned} T_B &= T_k \\ T_J &= T_k. \end{aligned}\]\[ The compatibility equation is: \]
\[\begin{aligned} \Omega_k &= -\Omega_B + \Omega_S - \Omega_J. \end{aligned}\]$$
Using the continuity and compatibility equations to eliminate secondary variables from the elemental equations, $$
\[\begin{aligned} \frac{d \Omega_J} {d t} &= \frac{1} {J} T_k \\ \frac{d T_k} {d t} &= k (-\Omega_B + \Omega_S - \Omega_J) \\ \Omega_B &= \frac{1} {B} T_k. \end{aligned}\]\[ Using the $B$ elemental equation to eliminate $\Omega_B$, \]
\[\begin{aligned} \frac{d \Omega_J} {d t} &= \frac{1} {J} T_k \\ \frac{d T_k} {d t} &= k \left(-\frac{1} {B} T_k + \Omega_S - \Omega_J\right). \end{aligned}\]\[ Writing in standard form, \]
\[\begin{aligned} \frac{d \bm{x}}{d t} &= \begin{bmatrix} 0 & 1/J \\ -k & -k/B \end{bmatrix} \begin{bmatrix} \Omega_J \\ T_k \end{bmatrix} + \begin{bmatrix} 0 \\ k \end{bmatrix} \begin{bmatrix} \Omega_S \end{bmatrix}. \end{aligned}\]$$ Note that, if we had chosen (arbitrarily) to switch the two state variables in \(\bm{x}\), the \(A\)-matrix would be flipped vertically and horizontally and the \(B\)-matrix would be flipped vertically.
The outputs can be written in terms of state and input variables as $$
\[\begin{aligned} \Omega_J = \Omega_J \end{aligned}\]\[ and, by Hooke's law, \]
\[\begin{aligned} \theta_k = \frac{1} {k} T_k. \end{aligned}\]\[ Writing in standard form, \]
\[\begin{aligned} \bm{y} &= \begin{bmatrix} 1 & 0 \\ 0 & 1/k \end{bmatrix} \begin{bmatrix} \Omega_J \\ T_k \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \end{bmatrix} \begin{bmatrix} \Omega_S \end{bmatrix}. \end{aligned}\]$$
Problem 3.8 (BLINKEN)
Use the following linear graph for an electrical system to answer the questions below, which are the steps to determining a state-space model from the linear graph.
Use the sign assignments from the diagram. \(I_S\) is a current source. Let the outputs be the voltage across the capacitor \(v_C\) and the current through the resistor \(i_R\).
Determine the normal tree, state variables, system order, state vector, input vector, and output vector.
Write the required elemental, continuity, and compatibility equations.
Solve for the state equation in standard form.
Solve for the output equation in standard form.
The normal tree is shown, below.
The primary variables are \(v_C\) and \(i_R\).
The secondary variables are \(i_C\) and \(v_R\).
The state variable is \(v_C\), so the order \(n=1\). The state, input, and output vectors are $$
\[\begin{aligned} \bm{x} = \begin{bmatrix} v_C \end{bmatrix}, \qquad \bm{u} = \begin{bmatrix} I_S \end{bmatrix}, \qquad \bm{y} = \begin{bmatrix} v_C \\ i_R \end{bmatrix}. \end{aligned}\]$$
The elemental equations are: $$
\[\begin{aligned} \frac{d v_C} {d t} &= \frac{1} {C} i_C \\ v_R &= i_R R. \end{aligned}\]\[ The continuity equation is: \]
\[\begin{aligned} i_C = I_S - i_R. \end{aligned}\]\[ The compatibility equation is: \]
\[\begin{aligned} v_R = v_C. \end{aligned}\]$$
Using the continuity and compatibility equations to eliminate secondary variables from the elemental equations, $$
\[\begin{aligned} \frac{d v_C} {d t} &= \frac{1} {C} (I_S - i_R) \\ v_C &= i_R R. \end{aligned}\]\[ Using the $R$ elemental equation to eliminate $i_R$, the single state equation is \]
\[\begin{aligned} \frac{d v_C} {d t} &= \frac{1} {C} (I_S - v_C/R) \end{aligned}\]\[ Writing in standard form, \]
\[\begin{aligned} \frac{d \bm{x}}{d t} &= \begin{bmatrix} -\frac{1} {R C} \end{bmatrix} \begin{bmatrix} v_C \end{bmatrix} + \begin{bmatrix} \frac{1} {C} \end{bmatrix} \begin{bmatrix} I_S \end{bmatrix}. \end{aligned}\]$$
The outputs can be written in terms of state and input variables as $$
\[\begin{aligned} v_C = v_C \end{aligned}\]\[ and, from the $R$ elemental equation, \]
\[\begin{aligned} i_R &= v_R/R \\ &= v_C/R. \tag{compatibility equation} \end{aligned}\]\[ Writing in standard form, \]
\[\begin{aligned} \bm{y} &= \begin{bmatrix} 1 \\ 1/R \end{bmatrix} \begin{bmatrix} v_C \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \end{bmatrix} \begin{bmatrix} I_S \end{bmatrix}. \end{aligned}\]$$
Problem 3.9 (CHUNKER)
Use the assigned coordinate arrows to draw a linear graph, a normal tree, and identify state variables and system order for each of the following schematics.
Electrical system, voltage and current source
Rotational mechanical system, torque source, coordinate arrow
Translational mechanical system, force sources (2)
Problem 3.10 (STEVENUNIVERSE)
Use the assigned coordinate arrows to draw a linear graph, a normal tree, and identify state variables, system order, and dependent energy storage elements for each of the following schematics.
Electrical system, voltage and current source
Rotational mechanical system, angular velocity source
Translational mechanical system, force sources (2)
Problem 3.11 (WINKEN)
Use the assigned coordinate arrows to draw a linear graph, a normal tree, and identify state variables and system order for each of the following schematics.
Electrical system, voltage source
Rotational mechanical system, torque source, coordinate arrow
Translational mechanical system, force source, coordinate arrow
Problem 3.12 (GRANADA)
Use the assigned coordinate arrows to draw a linear graph, a normal tree, and identify state variables and system order for each of the following systems.
Rotational mechanical system, two torque sources
Translational mechanical system, velocity source
For the rotational mechanical system with two torque sources, the linear graph and normal tree are shown below.
From the linear graph and normal tree the state variables are \(\Omega_{J_1}\), \(\Omega_{J_2}\), \(\Omega_{J_3}\), and \(\tau_K\). This gives a system order of \(n=4\).
For the translational mechanical system with a single velocity source, the linear graph and normal tree are shown below.
From the linear graph and normal tree the state variables are \(v_{m_1}\), \(v_{m_2}\), and \(F_K\). This gives a system order of \(n=3\).
Problem 3.13 (VALENCIA)
Use the following linear graph for a mechanical translational system to answer the questions below, which are the steps to determining a state-space model from the linear graph.
Use the sign convention from the diagram. \(F_S\) is a force source. Let the outputs be \(v_{m_1}\) and \(v_{m_2}\).
Determine the normal tree, state variables, system order, state vector, input vector, and output vector.
Write the required elemental, continuity, and compatibility equations.
Solve for the state equation in standard form.
Solve for the output equation in standard form.
From the linear graph the following can be found to be the normal tree.
From this normal tree the state variables can be found to be \(v_{m_1}\), \(v_{m_2}\), and \(F_K\). This gives a system order \(n=3\). We can now define the state, input, and output vectors as, $$
\[\begin{aligned} x&=\left[v_{m_1}\ v_{m_2}\ F_K\right]^T\\ y&=\left[v_{m_1}\ v_{m_2}\right]^T\\ u&=\left[F_S\right]. \end{aligned}\]$$
For this system the elemental equations are, $$
\[\begin{aligned} \frac{dv_{m_1}}{dt}&=\frac{1} {m_1}F_{m_1}\\ \frac{dv_{m_2}}{dt}&=\frac{1} {m_2}F_{m_2}\\ \frac{dF_K} {dt}&=Kv_K\\ F_B&=Bv_B. \end{aligned}\]\[ By drawing loops cutting a single passive normal tree branch the following equations can be written, \]
\[\begin{aligned} F_{m_1}&=F_S-F_K\\ F_{m_2}&=F_K-F_B. \end{aligned}\]\[ By adding normal tree links to the normal tree and summing around loops the following equations are derived, \]
\[\begin{aligned} v_K&=v_{m_1}-v_{m_2}\\ v_B&=v_{m_2}. \end{aligned}\]$$
Starting with the elemental equation for \(m_1\), $$
\[\begin{aligned} \frac{dv_{m_1}}{dt}&=\frac{1} {m_1}F_{m_1}\\ &=\frac{1} {m_1}\left(F_S-F_K\right).\tag{continuity} \end{aligned}\]\[ Working on the equation for $m_2$, \]
\[\begin{aligned} \frac{dv_{m_2}}{dt}&=\frac{1} {m_2}F_{m_2}\\ &=\frac{1} {m_2}\left(F_K-F_B\right)\tag{continuity}\\ &=\frac{1} {m_2}\left(F_K-Bv_B\right)\tag{elemental}\\ &=\frac{1} {m_2}\left(F_K-Bv_{m_2}\right).\tag{compatibility} \end{aligned}\]\[ Finally, the equation for $F_K$, \]
\[\begin{aligned} \frac{dF_K} {dt}&=Kv_K\\ &=K\left(v_{m_1}-v_{m_2}\right).\tag{compatibility} \end{aligned}\]\[ These equations can now be converted into a state equation in standard form, \]=x+u$$
The output vector \(y\) is a subset of the state variable \(x\). Therefore, in standard form the output equation is, \[y=\left[ \begin{matrix} 1&0&0\\ 0&1&0 \end{matrix} \right]x+\left[ \begin{matrix} 0\\0 \end{matrix} \right]u\]
Problem 3.14 (STEVENASH)
Use the linear graph of for an electrical system to answer the questions below, which are the steps to determining a state-space model from the linear graph.
Use the sign convention from the diagram. \(V_S\) is a voltage source. Let the outputs be \(v_{C_2}\), \(v_R\), and \(i_S\) (i.e., the source current).
Determine the normal tree, state variables, system order, state vector, input vector, and output vector.
Write the required elemental, continuity, and compatibility equations.
Solve for the state equation in standard form.
Solve for the output equation in standard form.
Online Resources for Section 3.7
No online resources.