Problems

Problem 4.1 (TRIANGLE)

Respond to the following questions and imperatives with one or two sentences and, if needed, equations and/or sketch.

Why do we include a resistor in lumped-parameter motor models?
How are brushes used in brushed DC motors?
With regard to standard motor curves, why do we say the “braking power” is equivalent to the power that could be successfully transferred by the motor to the mechanical system?
In terms of electrical and mechanical processes, why does an efficiency versus torque motor curve have a peak?
As a DC motor’s bearings wear down, how will its efficiency curve be affected?

There is no physical resistor in a motor, but its rotor and (if applicable) stator windings have a small amount of resistance. This resistance is important to model both because it is the dominant factor in power losses in a motor and because it can thereby generate such heat that damage to the windings is a concern.
Brushes reverse the direction of current flow through the rotor windings and enable electrical contact between stationary and a rotating conductors.
In this context, “braking power” refers to the power dissipated by the brake in the experimental setup used to measure motor curves. Although this power is dissipated in the experiment, it is potentially delivered to a load (in steady state) in application.
The efficiency motor curve has a peak because there are two processes dissipating power: the electrical resistance and the mechanical rotational damping, which dominate separately at different ends of the operating spectrum. Specifically, at high-speed and low-torque, the mechanical damping dominates; conversely, at low-speed and high-torque, the electrical resistance dominates. In between these two ends of the spectrum, the trade-off peaks.
As a DC motor’s bearings wear down, the mechanical damping energy dissipation will increase, lowering the overall efficiency curve and shifting the peak toward higher-torque and lower-speed, due to the trade-off described in the previous part of this solution.

Problem 4.2 (SQUARE)

schematic of an electromechanical system for .

Consider the system presented in the schematic of . Let the DC motor have motor constant $K_{a}$ (units N-m/A) and let the motor be driven by an ideal current source $I_{S}$ . Assume the motor inertia has been lumped into $J_{1}$ and motor damping lumped into $B_{1}$ .

Draw a linear graph model.
Draw a normal tree.
Identify any dependent energy storage elements. If the motor was driven by an ideal voltage source instead, how would this change?

Draw a linear graph model and normal tree.

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

a normal tree on a linear graph model of the electromechanical system of .

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.

Problem 4.3 (RECTANGLE)

Consider the system presented in the schematic of . From the linear graph model and normal tree derived in , derive a state-space model in standard form. Let the outputs be $θ_{J_{1}}$ and $θ_{J_{2}}$ , the angular positions of the flywheels.

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: $\begin{array}{r} x = {[\begin{array}{c} Ω_{J_{1}} & T_{k_{1}} & Ω_{J_{2}} & T_{k_{2}} \end{array}]}^{⊤}, u = [\begin{array}{c} I_{S} \end{array}], y = {[\begin{array}{c} θ_{J_{1}} & θ_{J_{2}} \end{array}]}^{⊤} . \end{array}$ 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.

element
$J_{1}$	$d_{t} Ω_{J_{1}} = \frac{- 1}{J_{1}} (T_{k_{1}} + T_{B_{1}} + T_{2})$
$k_{1}$	$d_{t} T_{k_{1}} = k_{1} (Ω_{J_{1}} - Ω_{J_{2}})$
$J_{2}$	$d_{t} Ω_{J_{2}} = \frac{1}{J_{2}} (T_{k_{1}} - T_{B_{2}} - T_{k_{2}})$
$k_{2}$	$d_{t} T_{k_{2}} = k_{2} Ω_{J_{2}}$
$B_{1}$	$T_{B_{1}} = B_{1} Ω_{J_{1}}$
$B_{2}$	$T_{B_{2}} = B_{2} Ω_{J_{2}}$
$1$	$T_{2} = - K_{a} I_{S}$
$2$	$v_{1} = K_{a} Ω_{J_{1}}$
$L$	$v_{L} = L d_{t} I_{S}$
$R$	$v_{R} = I_{S} R$

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

element
$J_{1}$	$d_{t} Ω_{J_{1}} = \frac{- 1}{J_{1}} (T_{k_{1}} + B_{1} Ω_{J_{1}} + - K_{a} I_{S})$
$k_{1}$	$d_{t} T_{k_{1}} = k_{1} (Ω_{J_{1}} - Ω_{J_{2}})$
$J_{2}$	$d_{t} Ω_{J_{2}} = \frac{1}{J_{2}} (T_{k_{1}} - B_{2} Ω_{J_{2}} - T_{k_{2}})$
$k_{2}$	$d_{t} T_{k_{2}} = k_{2} Ω_{J_{2}}$

The standard form of the state equation is $\begin{aligned} \frac{d x}{d t} & = [\begin{array}{c} - B_{1} / J_{1} & - 1 / J_{1} & 0 & 0 \\ k_{1} & 0 & - k_{1} & 0 \\ 0 & 1 / J_{1} & - B_{2} / J_{2} & - 1 / J_{2} \\ 0 & 0 & k_{2} & 0 \end{array}] x + [\begin{array}{c} K_{a} / J_{1} \\ 0 \\ 0 \\ 0 \end{array}] u . \end{aligned}$ The output angular positions can be related to the spring torques (which are state variables) as follows: $\begin{aligned} θ_{J_{1}} & = θ_{k_{1}} + θ_{k_{2}} \\ = T_{k_{1}} / k_{1} + T_{k_{2}} / k_{2} \end{aligned}$ and $\begin{aligned} θ_{J_{2}} & = θ_{k_{2}} \\ = T_{k_{2}} / k_{2} . \end{aligned}$ Therefore the output equation in standard form is $\begin{aligned} y & = [\begin{array}{c} 0 & 1 / k_{1} & 0 & 1 / k_{2} \\ 0 & 0 & 0 & 1 / k_{2} \end{array}] x + [\begin{array}{c} 0 \\ 0 \end{array}] u . \end{aligned}$

Problem 4.4 (QUADRILATERAL)

a linear graph model of the electromechanical system.

Consider the linear graph model of a motor coupled to a rotational mechanical system shown in . This is similar to the model from the , but includes the flexibility of the shaft coupler. An ideal voltage source drives the motor—modeled as an ideal transducer with armature resistance $R$ and inductance $L$ , given by the manufacturer in . The ideal transducer’s rotational mechanical side ( $2$ ) is connected to a moment of inertia $J_{m}$ modeling the rotor inertia and damping $B_{b}$ modeling the internal motor damping, both values given in the motor specifications. Take $B_{b} = B_{m}$ and $J_{f} = 0.324 \cdot 10^{- 3}$ kg-m $^{2}$ . Assume the shaft coupling has a torsional stiffness of $k = 100$ N-m/rad.

Derive a state-space model for the system with outputs $i_{1}$ and $Ω_{J_{f}}$ .
Create a Matlab model for the system and simulate its response from rest to an input voltage $V_{S} = 10$ V.
Plot the outputs through time until they reach steady state.

a linear graph model with normal tree of the electromechanical system.

The normal tree is shown in . The state variables are $Ω_{J_{m}}$ , $Ω_{J_{f}}$ , $i_{L}$ , and $T_{k}$ , so the system order is $n = 4$ . The state, input, and output vectors are (different orderings are valid) $\begin{aligned} x & = {[\begin{array}{c} Ω_{J_{m}} & Ω_{J_{f}} & i_{L} & T_{k} \end{array}]}^{⊤} \\ u & = [\begin{array}{c} V_{S} \end{array}] \\ y & = {[\begin{array}{c} i_{1} & Ω_{J_{f}} \end{array}]}^{⊤} \end{aligned}$

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 $\begin{aligned} \frac{d x}{d t} & = A x + B u \\ y & = C x + D u \end{aligned}$ with $\begin{aligned} A & = [\begin{array}{c} - \frac{B_{m}}{J_{m}} & 0 & \frac{K_{a}}{J_{m}} & - \frac{1}{J_{m}} \\ - \frac{B_{b}}{J_{f}} & 0 & 0 & \frac{1}{J_{f}} \\ - \frac{K_{a}}{L} & 0 & - \frac{R}{L} & 0 \\ k & - k & 0 & 0 \end{array}] \\ B & = [\begin{array}{c} 0 \\ 0 \\ \frac{1}{L} \\ 0 \end{array}] \\ C & = [\begin{array}{c} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array}] \\ D & = [\begin{array}{c} 0 \\ 0 \end{array}] . \end{aligned}$

Matlab

We now turn to Matlab.

First, we define each of the parameters. The following, symbolic approach that stores paramter values in a params structure array is a little more work, but when we start designing, it will pay off to hold off on defining parameters as long as possible, so we get in the practice, now.

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

Now we can define the state-space model standard matrices.

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];

Now we can create an LTI system model using ss. Note that Matlab doesn’t support symbolic LTI models, so we have to substitute our parameters.

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);

Now we can simulate with lsim (or even step in this case).

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

Now let’s plot the results.

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 “wiggle” of is a contribution of the flexible shaft coupler. It does damp out over time – try plotting (with more points) up to $1$ sec – but relatively slowly. The contribution is small, but potentially not negligible.

Note the two vertical axes in the plot. This is good form when we plot two outputs on the same plot because they generally have different physical units! (In our case, they are current with units A and angular velocity with units rad/s.) It is challenging to plot more than two quantities with different units on the same plot, but such an example is the motor curves of . Note that Matlab recommends using yyaxis now, but plotyy will continue to work, although it is more limited.

Problem 4.5 (MRPOTATOHEAD)

Consider the linear graph model (with normal tree) of . This is a model of a motor with constant $K_{a}$ connected to a pair of meshing gears with transformer ratio $N$ , the output over input gear ratio. An ideal voltage source drives the motor—modeled as an ideal transducer with armature resistance $R$ and inductance $L$ . The motor’s rotational mechanical side ( $2$ ) is connected to a moment of inertia $J_{2}$ modeling the rotor and drive gear combined inertia. The damping element $B_{2}$ models the internal motor damping and the drive gear bearing damping. The output side of the gear transducer ( $4$ ) is connected to a moment of inertia $J_{4}$ modeling the output gear and load combined inertia. The damping element $B_{4}$ models the internal motor damping and the drive gear bearing damping. Use the parameter values given in .

Derive a state-space model for the system with outputs $Ω_{J_{2}}$ and $Ω_{J_{4}}$ .
Create a Matlab model for the system and simulate its response from rest to an input voltage $V_{S} = 20$ V.
Plot the outputs through time until they reach steady state.

$R$	$2 Ω$
$L$	$8$ mH
$K_{a}$	$0.2$ N-m/A
$J_{2}$	$0.1 \cdot 10^{- 3}$ kg-m $^{2}$
$B_{2}$	$50 μ$ N-m/(rad/s)
$N$	$5$
$J_{4}$	$1 \cdot 10^{- 3}$ kg-m $^{2}$
$B_{4}$	$70 μ$ N-m/(rad/s)

Figure 4.1: A linear graph model with normal tree in green of an electromechanical system with a gear reduction.

The state variables are across-variables on A-type branches and through-variables on T-type links, so $Ω_{J_{4}}$ and $i_{L}$ , so the system order is $n = 2$ .¹ The state, input, and output vectors are (different orderings are valid) $\begin{aligned} x & = {[\begin{array}{c} Ω_{J_{4}} & Ω_{i_{L}} \end{array}]}^{⊤} \\ u & = [\begin{array}{c} V_{S} \end{array}] \\ y & = {[\begin{array}{c} Ω_{J_{2}} & Ω_{J_{4}} \end{array}]}^{⊤} \end{aligned}$

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 $\begin{aligned} \frac{d x}{d t} & = A x + B u \\ y & = C x + D u \end{aligned}$ with $\begin{aligned} A & = [\begin{array}{c} - \frac{B_{2} N^{2} + B_{4}}{J_{2} N^{2} + J_{4}} & \frac{K_{a} N}{J_{2} N^{2} + J_{4}} \\ - \frac{K_{a} N}{L} & - \frac{R}{L} \end{array}] \\ B & = [\begin{array}{c} 0 \\ \frac{1}{L} \end{array}] \\ C & = [\begin{array}{c} N & 0 \\ 1 & 0 \end{array}] \\ D & = [\begin{array}{c} 0 \\ 0 \end{array}] . \end{aligned}$

Matlab

We now turn to Matlab.

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

Now we can define the state-space model standard matrices.

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];

Now we can create an LTI system model using ss. Note that Matlab doesn’t support symbolic LTI models, so we have to substitute our parameters.

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);

Now we can simulate with lsim (or even step in this case).

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

Now let’s plot the results.

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)')

Problem 4.6 (CLUNKER)

Draw a linear graph, a normal tree, identify state variables, identify system order, and denote any dependent energy storage elements for each of the following schematics.

The electronic system of figure 4.21, voltage and current sources, and transformer with transformer ratio $N$ .
The electromechanical system of figure 4.22 with motor model parameters shown, coordinate arrow in green. Model the propeller as a moment of inertia $J_{2}$ and damping $B_{2}$ .
The translational mechanical system of figure 4.23, force source, coordinate arrow in green.

Figure 4.22: Sketch of a motor coupled to a fan.

Figure 4.23: Schematic of a mechanical system.

problem 4.6 part (a) linear graphs and normal trees in green – four possible choices! {#clunker4 .figure h=“clunker4” caption_plain=” part (a) linear graphs and normal trees in green – four possible choices!“}

Figure : problem 4.6 part (b) linear graphs and normal tree in green

Figure : problem 4.6 part (c) linear graphs and normal tree in green

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.
The electromechanical system of figure 4.22 has linear graph and normal trees of . The state variables are $Ω_{J_{m}}$ , $Ω_{J_{2}}$ , $i_{L}$ , and $T_{k_{1}}$ . The system order is $n = 4$ . There are no dependent energy storage elements.
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.

Problem 4.7 (CURVY)

Consider the DC motor curves of figure 4.13, reproduced in figure 4.24.

At peak efficiency, what is the steady-state motor speed?
At peak efficiency, what is the steady-state motor torque?
You are to use this motor to drive a load at a constant angular speed of $100$ rad/s with at least $1$ N-m of torque. You wisely choose to use a gear reduction between the motor and load. What should the gear ratio be to meet the above requirements and optimize efficiency? Justify your answer in terms of the motor curves of figure 4.24.

Figure 4.24: The motor curve figure 4.13.

From the motor curves, peak efficiency corresponds to about $580$ rad/s.
From the motor curves, $0.2$ N-m.
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.

Problem 4.8 (CHAIR)

Consider the opamp circuit of figure 4.25, which will be used to drive a PMDC motor. The input can supply a variable $V_{S} \in [0, 10]$ V, the motor has constant $K_{a} = 0.05$ V/(rad/s) and coil resistance $R_{m} = 1 Ω$ , and the opamp has differential supplies $\pm 24$ V. Assume the maximum torque magnitude required from the motor at top speed is $| T_{2} | = 0.1$ N-m and ignore any voltage drop in the motor due to the coil inductance.² Select $R_{1}$ and $R_{2}$ to demonstrably meet the following design requirements:

drivable motor speeds of at least $[0, 400]$ rad/s,
no saturation of the opamp (i.e. $| v_{o} | < 24$ V), and
a maximum combined power dissipation by $R_{1}$ and $R_{2}$ less than $300$ mW.

Hint: start with the elemental equations of the DC motor to determine the necessary amplifier output $v_{o}$ , then constrain $R_{1}$ and $R_{2}$ to meet the gain requirements, and finally further constrain $R_{1}$ and $R_{2}$ to meet the power dissipation requirement.

The motor has elemental equations $\begin{array}{r} Ω_{2} = v_{1} / K_{a} and T_{2} = - K_{a} i_{1} . \end{array}$ The first elemental equation gives at top speed $v_{1} = K_{a} Ω_{2} = (0.05) (400) = 20$ 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 $\begin{array}{r} v_{o} = \frac{R_{1} + R_{2}}{R_{2}} V_{S} \end{array}$ 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 $\begin{array}{r} \frac{R_{1} + R_{2}}{R_{2}} = 22 / 10. \end{array}$ 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 $\tag not allowed in aligned environment$ From the circuit diagram, $v_{R_{12}} = v_{o}$ . The power dissipated by the resistors then is $\tag not allowed in aligned environment$ which gives the constraint $\begin{array}{r} \frac{v_{o}^{2}}{R_{1} + R_{2}} < 0.3 W \Rightarrow \\ R_{1} + R_{2} > \frac{22^{2}}{0.3} \approx 1.613 k Ω . \end{array}$ Combining this with the yields $\begin{array}{r} 2.2 R_{2} > 1613 \Rightarrow R_{2} > 733.3 Ω \end{array}$ or $\begin{array}{r} R_{1} > 880.0 Ω \end{array}$ and $\begin{array}{r} R_{1} = 1.2 R_{2} . \end{array}$ So, for instance, a valid solution is $\begin{array}{r} R_{1}, R_{2} = 12, 10 k Ω . \end{array}$ So it turned out to be relatively easy to meet the power dissipation requirement by selecting large enough resistors.

Problem 4.9 (ONOMATOPOEIA)

Consider the DC motor curves of figure 4.13, reproduced in figure 4.24. If this motor is running at $400 \frac{rad}{s}$ ,

How much torque is produced?
What is the output power?
What is the input power?
Why are the input and output power the same or different?

At $400 \frac{rad}{s}$ the motor produces approximately $1.3 N - m$ of torque.
At $1.3 N - m$ the output power is approximately $520 W$
At $1.3 N - m$ the efficiency is approximately 65%. This means the output power is 65% lower than the input power. Therefore, $520 W = 0.65 P_{in}$ , so $P_{in} = \frac{520}{0.65} = 800 W$ .
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.

Problem 4.10 (DEGLAZIFICATION)

Explain in your own words what lumped parameter elements should be used when modeling an electric motor and why.

Problem 4.11 (CONFUZZLED)

In the linear graph below a system is depicted consisting of a motor with its related damping and inertia driven by a voltage source and connected to a set of gears driving a second inertia. A rotary spring is attached between the two inertias.
Given this linear graph:

draw a normal tree,
determine the state variables and system order, and
list any dependent energy storage elements and explain what this implies.

Problem 4.12 (LEVITATION)

In the linear graph and normal tree below a system is depicted consisting of a motor driven by a voltage source $V_{S}$ with inertia $J$ driving a rotary damper and spring connected in series. Let the motor constant be $K a$ , and outputs of the system be the rotational velocity of the inertia, $Ω_{J}$ , and the change in rotational velocity across the rotational damper, $Ω_{B}$ .
Given this linear graph and normal tree:

determine the state variables,
define the state, input, and output vectors,
write the elemental, continuity, and compatibility equations, and
solve for the state and output equations.

Do not ignore the voltage drop across $R_{m}$ , though. Note that this amounts to an assumption of steady-state operation at top speed. By requiring a specific $T_{2}$ , we are also implicitly ignoring torque losses due to motor bearing damping.↩︎
Do not ignore the voltage drop across $R_{m}$ , though. Note that this amounts to an assumption of steady-state operation at top speed. By requiring a specific $T_{2}$ , we are also implicitly ignoring torque losses due to motor bearing damping.↩︎

Online Resources for Section 4.10

No online resources.