Problems

The linear graph (a) and normal tree (b) are shown in .

We had no choice but to include the inductor \(L\) in the normal tree because we cannot include \(I_S\), needed the tree to reach the node between \(R\) and \(I_S\) (therefore \(R\) must be in the tree), needed the tree to be connected (therefore \(L\) must be in the tree), and also needed the tree to reach the electrical ground (therefore \(1\) must be in the tree).

Since \(L\) is in the normal tree and it is a T-type energy storage element, it is dependent.

This would change if the source was instead a voltage source \(V_S\), which can and should be included in the normal tree, which now can and must exclude \(L\) (so as not to form a loop). Therefore, \(L\) would be an independent energy storage element in this case.

We can use the results of the solution to , including its linear graph and normal tree of .

We begin by identifying the state, input, and output vectors: \[\begin{aligned} \bm{x} = \begin{bmatrix} \Omega_{J_1} & T_{k_1} & \Omega_{J_2} & T_{k_2} \end{bmatrix}^\top, \quad \bm{u} = \begin{bmatrix} I_S \end{bmatrix}, \quad \bm{y} = \begin{bmatrix} \theta_{J_1} & \theta_{J_2} \end{bmatrix}^\top. \end{aligned}\] Note that \(i_L\) is not a state variable because \(L\) is not an independent energy storage element.

Now we write the elemental equation and continuity or compatibility equation for each element in the following table.

Now we can eliminate all secondary variables via substitution of the continuity or compatibility equation corresponding to each element into its elemental equation, which results in the following table.

The first four equations yield the state equations when we substitute some of the last six.

The standard form of the state equation is \[\begin{aligned} \frac{\diff \bm{x}}{\diff t} &= \begin{bmatrix} -B_1/J_1 & -1/J_1 & 0 & 0 \\ k_1 & 0 & -k_1 & 0 \\ 0 & 1/J_1 & -B_2/J_2 & -1/J_2 \\ 0 & 0 & k_2 & 0 \end{bmatrix} \bm{x} + \begin{bmatrix} K_a/J_1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \bm{u}. \end{aligned}\] The output angular positions can be related to the spring torques (which are state variables) as follows: \[\begin{aligned} \theta_{J_1} &= \theta_{k_1} + \theta_{k_2}\\ &= T_{k_1}/k_1 + T_{k_2}/k_2 \end{aligned}\] and \[\begin{aligned} \theta_{J_2} &= \theta_{k_2}\\ &= T_{k_2}/k_2. \end{aligned}\] Therefore the output equation in standard form is \[\begin{aligned} \bm{y} &= \begin{bmatrix} 0 & 1/k_1 & 0 & 1/k_2 \\ 0 & 0 & 0 & 1/k_2 \end{bmatrix} \bm{x} + \begin{bmatrix} 0 \\ 0 \end{bmatrix} \bm{u}. \end{aligned}\]

The normal tree is shown in . The state variables are \(\Omega_{J_m}\), \(\Omega_{J_f}\), \(i_L\), and \(T_k\), so the system order is \(n = 4\). The state, input, and output vectors are (different orderings are valid) \[\begin{aligned} \bm{x} &= \begin{bmatrix} \Omega_{J_m} & \Omega_{J_f} & i_L & T_k \end{bmatrix}^\top \\ \bm{u} &= \begin{bmatrix} V_S \end{bmatrix}\\ \bm{y} &= \begin{bmatrix} i_1 & \Omega_{J_f} \end{bmatrix}^\top \end{aligned}\]

The following are the elementary, continuity, and compatibility equations.

We now use StateMint: statemint.stmartin.edu The .rnd file is at the following url: ricopic.one/assets/quadrilateral.rnd

StateMint gives a state-space model that has standard form \[\begin{aligned} \frac{\diff\bm{x}}{\diff t} &= A \bm{x} + B \bm{u} \\ \bm{y} &= C \bm{x} + D \bm{u} \end{aligned}\] with \[\begin{aligned} A &= \left[\begin{matrix}- \frac{B_m} {J_m} & 0 & \frac{K_a} {J_m} & - \frac{1} {J_m}\\- \frac{B_b} {J_f} & 0 & 0 & \frac{1} {J_f}\\- \frac{K_a} {L} & 0 & - \frac{R} {L} & 0\\k & - k & 0 & 0\end{matrix}\right] \\ B &= \left[\begin{matrix}0\\0\\\frac{1} {L}\\0\end{matrix}\right] \\ C &= \left[\begin{matrix}0 & 0 & 1 & 0\\0 & 1 & 0 & 0\end{matrix}\right] \\ D &= \left[\begin{matrix}0\\0\end{matrix}\right]. \end{aligned}\]

syms R L Ka k Bm Jm Bb Jf

Kv_spec = 10.2; % V/krpm ... voltage constant spec 
rads_krpm = 1e3*2*pi/60; % (rad/s)/krpm
Kv_si = Kv_spec/rads_krpm; % V/(rad/s)
params.Ka = Kv_si; % N-m/A ... TF/motor constant
m_in = 0.0254; % m/in
d = 2.5*m_in; % m ... flywheel diameter
thick = 1*m_in; % m ... flywheel thickness
vol = pi*(d/2)^2*thick; % flywheel volume
rho = 8000; % kg/m^3 ... flywheel density (304) 
m = rho*vol; % kg ... flywheel mass
params.Jf = 1/2*m*(d/2)^2; % kg-m^2 ... flywheel in.
params.Jm = 56.5e-6; % kg-m^2 ... inertia of rotor
params.Bm = 16.9e-6; % N-m/s^2 ... motor damping coef
params.Bb = params.Bm; % N-m/s^2 ... brng damp coef
params.k = 100; % N-m/rad ... torsional stiffness
params.R = 1.6; % Ohm ... armature resistance
params.L = 4.1e-3; % H ... armature inductance

A = [...
  -Bm./Jm 0 Ka./Jm -1./Jm; ...
  -Bb./Jf 0 0 1./Jf; ...
  -Ka./L 0 -R./L 0; ...
  k -k 0 0 ...
];
B = [0; 0; 1./L; 0];
C = [0 0 1 0; 0 1 0 0];
D = [0; 0];

A_s = double(subs(A,params)); % sym2num
B_s = double(subs(B,params)); % sym2num
C_s = double(subs(C,params)); % sym2num
D_s = double(subs(D,params)); % sym2num
sys = ss(A_s,B_s,C_s,D_s);

t = linspace(0,.3,400);
u = 10*ones(size(t)); % V ... step input
y = lsim(sys,u,t);

figure
[ax,l1,l2] = plotyy(t,y(:,1),t,y(:,2));
set(l1,'linewidth',1.5); set(l2,'linewidth',1.5);
ylabel(ax(1),'i_1 (A)')
ylabel(ax(2),'\Omega_{J_f} (rad/s)')
xlabel('time (s)')

The state variables are across-variables on A-type branches and through-variables on T-type links, so \(\Omega_{J_4}\) and \(i_L\), so the system order is \(n = 2\).¹ The state, input, and output vectors are (different orderings are valid) \[\begin{aligned} \bm{x} &= \begin{bmatrix} \Omega_{J_4} & \Omega_{i_L} \end{bmatrix}^\top \\ \bm{u} &= \begin{bmatrix} V_S \end{bmatrix}\\ \bm{y} &= \begin{bmatrix} \Omega_{J_2} & \Omega_{J_4} \end{bmatrix}^\top \end{aligned}\]

The following are the elementary, continuity, and compatibility equations.

We now use StateMint: statemint.stmartin.edu The .rnd file is at the following url: ricopic.one/dynamic_systems/source/mrpotatohead.rnd

StateMint gives a state-space model that has standard form \[\begin{aligned} \frac{\diff\bm{x}}{\diff t} &= A \bm{x} + B \bm{u} \\ \bm{y} &= C \bm{x} + D \bm{u} \end{aligned}\] with \[\begin{aligned} A &= \left[\begin{matrix}- \frac{B_{2} N^{2} + B_{4}}{J_{2} N^{2} + J_{4}} & \frac{K_a N} {J_{2} N^{2} + J_{4}}\\- \frac{K_a N} {L} & - \frac{R} {L}\end{matrix}\right] \\ B &= \left[\begin{matrix}0\\\frac{1} {L}\end{matrix}\right] \\ C &= \left[\begin{matrix}N & 0\\1 & 0\end{matrix}\right] \\ D &= \left[\begin{matrix}0\\0\end{matrix}\right]. \end{aligned}\]

syms R L Ka B2 J2 N B4 J4

params.R = 2; % Ohm ... armature resistance
params.L = 8e-3; % H ... armature inductance
params.Ka = 0.2; % N-m/A ... motor constant
params.J2 = .1e-3; % kg-m^2 ... inertia
params.B2 = 50e-6; % N-m/(rad/s) ... damping
params.N = 5; % gear ratio
params.J4 = 1e-3; % kg-m^2 ... inertia
params.B4 = 70e-6; % N-m/(rad/s) ... damping

A = [...
  -(B2.*N.^2 + B4)./(J2.*N.^2 + J4), ...
  Ka.*N./(J2.*N.^2 + J4); ...
  -Ka.*N./L, ...
  -R./L ...
];
B = [0; 1./L];
C = [N 0; 1 0];
D = [0; 0];

A_s = double(subs(A,params)); % sym2num
B_s = double(subs(B,params)); % sym2num
C_s = double(subs(C,params)); % sym2num
D_s = double(subs(D,params)); % sym2num
sys = ss(A_s,B_s,C_s,D_s);

t = linspace(0,.23,300);
u = 20*ones(size(t)); % V ... step input
y = lsim(sys,u,t);

figure
[ax,l1,l2] = plotyy(t,y(:,1),t,y(:,2));
set(l1,'linewidth',1.5); 
set(l2,'linewidth',1.5,'linestyle','--');
set(ax(1),'YLim',[0,120])
set(ax(2),'YLim',[0,24])
ylabel(ax(1),'\Omega_{J_2} (rad/s)')
ylabel(ax(2),'\Omega_{J_4} (rad/s)')
xlabel('time (s)')

{#clunker4 .figure h=“clunker4” caption_plain=” part (a) linear graphs and normal trees in green – four possible choices!“}

1. The electronic system of figure 4.21 has linear graph and normal trees of . The state variables are \(v_{C_2}\) and either \(i_{L_1}\) or \(i_{L_2}\). The system order is \(n = 2\). Capacitor \(C_1\) and either \(L_1\) or \(L_2\) are dependent energy storage elements.

2. The electromechanical system of figure 4.22 has linear graph and normal trees of . The state variables are \(\Omega_{J_m}\), \(\Omega_{J_2}\), \(i_L\), and \(T_{k_1}\). The system order is \(n = 4\). There are no dependent energy storage elements.

3. The translational mechanical system of figure 4.23 has linear graph and normal trees of . The state variables are \(v_{m_1}\), \(v_{m_2}\), \(v_{m_3}\), \(f_{k_1}\), and \(f_{k_2}\). The system order is \(n = 5\). There are no dependent energy storage elements.

1. From the motor curves, peak efficiency corresponds to about \(580\) rad/s.

2. From the motor curves, \(0.2\) N-m.

3. The gear reduction allows us to trade speed and torque. Without the gear reduction, you can easily meet the \(100\) rad/s requirement with ample torque, around \(3\) N-m. However, in those operating conditions, efficiency is around \(20\) percent (due to resistive electronic dissipation, mostly). In order to optimize efficiency, we’d like to operate as closely as possible to \(0.2\) N-m and \(580\) rad/s. If we gear down the full \(580\) to \(100\) rad/s (\(N = 5.8\)), we get a torque of \(0.2\cdot 5.8 = 1.16\) N-m. This is more than enough torque to meet our \(1\) N-m torque requirement, therefore \(N \approx 5.8\) is the optimal gear ratio.

The motor has elemental equations \[\begin{aligned} \Omega_2 = v_1/K_a\quad\text{and}\quad \ T_2 = -K_a i_1. \end{aligned}\] The first elemental equation gives at top speed \(v_1 = K_a \Omega_2 = (0.05)(400) = 20\) V. And here we need a current that meets the torque requirement \(i_1 = -T_2/K_a = 0.1/.05 = 2\) A, which corresponds to a resistive coil voltage drop of \(v_{R_m} = (2)(1) = 2\) V. So we need \(22\) V at top \(400\) rad/s. This is within the saturation of the opamp (\(24\) V), so if \(v_o = 22\) V, we meet all requirements so far.

We recognize the non-inverting opamp circuit, which delivers an output voltage \[\begin{aligned} v_o = \frac{R_1+R_2} {R_2} V_S \end{aligned}\] as long as it doesn’t saturate (\(|v_o| < 24\) V). To produce a top \(v_o = 22\) V with max \(V_S = 10\) V, the resistance ratio must be \[\begin{aligned} \label{chair:eq1} \frac{R_1+R_2} {R_2} = 22/10. \end{aligned}\] The last requirement limits power dissipation by \(R_1\) and \(R_2\) to \(300\) mW. Since these share a current, they are effectively in series and the total voltage drop across them is $$\begin{align} v_{R_{12}} = i_{R_{12}} (R_1 + R_2). \tag{Ohm's law} \end{align}$$ From the circuit diagram, \(v_{R_{12}} = v_o\). The power dissipated by the resistors then is $$\begin{align} \mathcal{P}_{R_{12}} &= v_{R_{12}} i_{R_{12}} \tag{def} \\ &= \frac{v_{R_{12}}^2}{R_1 + R_2} \tag{Ohm's law} \\ &= \frac{v_o^2} {R_1 + R_2} \end{align}$$ which gives the constraint \[\begin{aligned} \frac{v_o^2} {R_1 + R_2} < 0.3\ \text{W} \quad\Rightarrow\\ R_1 + R_2 > \frac{22^2} {0.3} \approx 1.613\ k\Omega. \end{aligned}\] Combining this with the yields \[\begin{aligned} 2.2 R_2 > 1613 \quad\Rightarrow\quad R_2 > 733.3\ \Omega \end{aligned}\] or \[\begin{aligned} R_1 > 880.0\ \Omega \end{aligned}\] and \[\begin{aligned} R_1 = 1.2 R_2. \end{aligned}\] So, for instance, a valid solution is \[\begin{aligned} R_1, R_2 = 12, 10\ \text{k}\Omega. \end{aligned}\] So it turned out to be relatively easy to meet the power dissipation requirement by selecting large enough resistors.

1. At \(400\ \frac{\mathrm{rad}}{\mathrm{s}}\) the motor produces approximately \(1.3\ \mathrm{N-m}\) of torque.

2. At \(1.3\ \mathrm{N-m}\) the output power is approximately \(520\ \mathrm{W}\)

3. At \(1.3\ \mathrm{N-m}\) the efficiency is approximately 65%. This means the output power is 65% lower than the input power. Therefore, \(520\ \mathrm{W}=0.65\mathcal{P}_\mathrm{in}\), so \(\mathcal{P}_\mathrm{in}=\frac{520} {0.65}=800\ \mathrm{W}\).

4. The output power is lower than the input power because the resistance of the motor armature and friction of the bearings cause energy to be lost. This ends up causing the motor to heat up as it is run.