Optimizing the Controller Design to Guide the Motion of a Maglev Train

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Optimizing the Controller Design to Guide

the Motion of a Maglev Train

The challenge: To design a robust feedback controller and optimize the control

parameters to guide the motion of a Maglev train along the guideway

© Maplesoft, a division of Waterloo Maple Inc., 2008

Introduction

Problem Definition

1. Model Development1.1 Linear Model of the Magnet

1.2 Model for the Linearized Magnet Force1.3 Model of Magnet Dynamics

2. Control System Design2.1 PI Controller2.2 Control Specifications and Measurements2.3 Control System Transfer Function2.4 Controller Gains & Closed Loop System 2.5 The Final Design

3. Vehicle Model3.1 Equations of Motion3.2 Response Plots

Results


Introduction

Magnetically Levitated (Maglev) trains differ from conventional trains in that they are levitated, guided and

propelled along a guideway by a changing magnetic field rather than by steam, diesel or electric engine. The

absence of direct contact between the train and the rail allows the Maglev to reach record ground

transportation speeds, which are on par to that of commercial airplanes.

Figure 1: (A) Cross sectional view of cabin & (B) Expanded view of magnet support structure

An Electromagnetic Suspension (EMS) Maglev train uses

the attractive forces of magnets (& electro-magnets),

positioned below a guideway, to levitate a train and guide

it along the guideway (fig 1).

Three phase AC propulsion coils mounted along the steel

rails of the guideway provide a moving magnetic field that

interacts with the periodic magnetic field created by the

lifting magnets to propel the train forward. There are 24

lifting magnet pairs (48 magnets all together) mounted

along the length of the train. The lifting magnets, which

are canted at an 37° angle towards the steel rails,

consists of high temperature superconducting (HTS)

magnets and conventional (non-superconducting) coils

(fig 2). A periodic magnetic field is generate by aligning

the lifting magnet pairs so that their north and south poles

alternate.

There are many advantages and disadvantages

associated with Maglev trains in comparison to

conventional steel wheel on steel rail trains . The table

(table 1) below highlights some of these.

Figure 2: Expanded view of lifting magnets


Advantages Disadvantages

Record breaking ground transportation speeds reaching 500 km/h (300 mph)

High infrastructure costs

Less expensive to operate and maintain than traditional high-speed trains or even planes, andas a consequence lower cost per passenger per mile

Not compatible with conventional wheeled tracks making it limited to only those places where maglev lines run.

Silent at low speeds Increased noise pollution at high speeds

Significant energy savings because the vehicle does not have to carry around the weight of its propulsion system

No associated maintenance cost because there are no parts that are in contact with ground

Improved ride comfort

Table 1: Advantages and Disadvantages of Maglev trains over conventional trains

Problem Definition

EMS Maglev trains require robust controllers to guide the train along the guideway. The controller prevents the

train from colliding into the guideway by maintaining a constant air gap between the train and guideway

regardless of inconsistencies in the guideway, change in direction or angle of the track, and environmental

forces such as wind.

Control of the vehicle is achieved by a combination of passive (i.e. through the orientation of the magnets

along the vehicle) and active (i.e. with an active control system) techniques. Each magnetic coil has its own

independent control system. This independence allows the pitching (forward and aft rotations) to be controlled

by the magnets at the front and rear of the vehicle acting independently to maintain the air gap at 2 inches.

Because the magnets are canted at 37°, the lateral and yaw motions (motions that turn the vehicle left or right

relative to the direction of travel) are also controlled automatically when the air gaps are maintained at the

nominal 2 inches. For example, if a wind gust were to shift the vehicle to the right, the gap on the left side

would become smaller and the one on the right side would become bigger. Each of the magnets would then

correct for the motion by returning the magnets back to equilibrium.

Lift (up and down motion) is controlled when the left magnet and right magnet move in unison. For example,

when the rail rises it causes the gap on both the right and left side of the vehicle to get smaller. The magnet

control systems respond to this change by adjusting the current flowing through the magnets to return the gap

to the desired 2 inch setting.

The control for roll (rotations to the left and right along the direction of travel) is done passively by offsetting

each of the magnets pairs by 1 inch. This offset acts to provide a magnetic spring that naturally restores the

vehicle so it is perpendicular to the guideway whenever a disturbance occurs.

This worksheet will examine the design of a robust feedback controller to guide the motion of a EMS Maglev

train along the guideway.


(2)(2)

(3)(3)

(1)(1)

1. Model Development

1.1 Creating a Linear Model of the Magnet

The force from the magnet is a function of the gap distance z and the current in the coils Ic. The force can

be developed from first principles using Maxwell's equations, or by using the inductance of the magnets,

but because of the iron in the core, the actual force is a complex function of the geometry of the magnet,

the amount of iron in the magnet and the rail, and the inherent hysteresis of the iron. The nonlinear effects

of the magnet can be approximated to first order so that the magnet force is proportional to the square of

the current flowing in the magnet coils and inversely proportional to the square of the gap between the rail

and the magnet. Thus:

fd z, Ic /K1

z2Ic

2

z, Ic /K Ic

2

z2

If we assume that the vehicle mass is m and the gravitational acceleration is g, then a simple free body

model of the vertical motion gives the differential equations for the motion as follows.

Velocity:

ode1dd

d t v t Cg =

f z t , Ic t

mC

Fd t

m

d

dt v t Cg =

K Ic t2

z t2 m

CFd t

m

Displacement:

ode2dd

d t z t = v t

d

dt z t = v t

Suppose the weight of each car in a two car Maglev train is 135,000 pounds. Using the same units, the

acceleration of gravity is defined as 32.2 ft/s/s. For now let's assume the proportionality constant K is unity.

This then gives the parameters for the model as:

paramsd K = 1, g = 32.2, m =135000

32.2 :


(4)(4)

(5)(5)

(6)(6)

1.2 Creating a Model for the Linearized Magnet Force

The first step in obtaining an accurate model of the vehicle is to model the nonlinear magnet dynamics.

Therefore, we will create, using the DynamicSystems package, the nonlinear differential equation model,

using the coupled differential equations ode1 and ode2 defined above, with the parameters defined above.

ode1d eval ode1, params

d

dt v t C32.2 =

0.0002385185185 Ic t2

z t2

C0.0002385185185 Fd t


d

dt z t = v t

sys_de d DynamicSystems DiffEquation ode1, ode2 , inputvariable = Ic t , Fd t , outputvariable = z t , v t :

To see what the object sys_de looks like we use the PrintSystem command in the

DynamicSystems package.

DynamicSystems PrintSystem sys_de

Diff. Equation

continuous

2 output(s); 2 input(s)

inputvariable = Ic t , Fd t

outputvariable = z t , v t

de = v.t C32.2 =

0.0002385185185 Ic t2

z t2

C0.0002385185185 Fd t , z.t

= v t

We now have a differential equation system object with the inputs Ic and Fd, and outputs z and v.

The proportionality constant K in equation (1) is derived from the fact that the magnetic force is given by

the cross product of the B field with the current producing the magnetic field. Since the current produces

both the lifting field and the field it interacts with (through the closed loop around the C shaped magnet and

the rail), the lifting force is proportional to the square of the B field over the area A of the magnet's pole

face. The proportionality constant for this force is 1

2µ0

. Thus the force from the magnet is:


F =B

2A

2µ0

:

The B field is given by µ$H where µ is the magnetic permeability of the iron in the magnet and rail (due to

the hysteresis of the iron, this term is not constant) and the magnetic flux is given by the current in the coil

multiplied by the number of turns in the coil. Thus, the constant K is given by:

K =µ

2N

2A

2µ0

:

1.3 Using the Model to Develop an Understanding of the Magnet Dynamics

Based on a simple understanding of the physics, it should be clear that when there is no current in the

coils the vehicle will drop under the influence of gravity causing the undercarriage to hit the guideway.

Similarly, when the current in the magnet is very large, the magnet will be attracted to the rail and hit the

rail (in fact the assumed force function is infinite when the gap z is zero). Therefore, as we analyze the

nonlinear differential equation we should expect that the dynamics are unstable.

To investigate the dynamics, the force term f z, t must be linearized. The basic superconducting magnet

was designed to have a nominal value of N$I equal to 50,000 ampere turns. The control coils, which

regulate the current through the superconducting magnets, were designed to allow a maximum of 15,000

ampere turns. In addition, the nominal air gap for the system is 2 inches. These values establish the

numerical value for force and therefore for the term K. The linearization of the nonlinear term in (1), from

the Taylor series expansion around the nominal 2 inch gap and 50,000 ampere turns current in the coils, is

obtained from the partials Kz =v

v zf z, Ic : and KI =

v

v Ic f z, Ic :

v

v z f z, Ic = K

2 K Ic2

z3

and

v

v Ic f z, Ic =

2 K Ic

z2

When the nominal values of Ic and z are substituted into the expansion, the linearization becomes (in this

and the subsequent analysis, the variations in z and Ic, ∆z and ∆Ic, are written as z and Ic respectively):

f z t , Ic t = Kz$z t CKI$Ic t :

The nominal value for the vehicle mass, and the nominal value of K gives Kz as 281,000 pounds/foot and

the nominal value of KI as 3372000 pounds/kiloampere-turn.


(7)(7)

This then provides the linear dynamics as:

ode3dd

d t v t =

Kz$z t

mC

KI$Ic t

mC

Fd t

m:

paramsd K = 1, m = 50500 , g = 32.2, Kz = 281000, KI = 3372000 :

Now we evaluate the equation at the operating point derived above:


d

dt v t =

562

101 z t C

6744

101 Ic t C

1

50500 Fd t


(9)(9)

(8)(8)

2. Control System Design

The analyses in the previous section allow us to begin the design of the control system. The first step in this is

an investigation of the underlying dynamics in order to understand how best to control the vehicle. In this

section we will use the equations defined above to develop a controller which will effectively control the

behavior of our system to external perturbations, such as changes in guideway and disturbance forces (such

as wind)), according to a set of well defined control criteria.

To start, we define the differential equation system object that we will be working with using ode2 and ode3

above:

sys_de d DynamicSystems DiffEquation ode2, ode3 , inputvariable = Ic t , Fd t , outputvariable = z t , v t :


Diff. Equation

continuous


inputvariable = Ic t , Fd t

outputvariable = z t , v t

de = z.t = v t , v

.t =

562 z t

101C

6744 Ic t

101C

Fd t

50500

For the design of the control system, the transfer function poles and zeros for the linear differential equation

need to be derived. We start by using DynamicSystems to get the ZeroPoleGain version of sys_de:

sys d DynamicSystems ZeroPoleGain sys_de

Zero Pole Gain

continuous


inputvariable = Ic s , Fd s

outputvariable = z s , v s


(10)(10)

(11)(11)

Zero-Pole-Gain Analysis Zero-Pole Plot

Zeros

K2 K1 0 1 2

K1

K0.5

0.5

1

Zero-Pole Plot

Figure 3: Zero-Pole Plot

sys:-z = 0 0

Poles

sys:-p =

K1

101 56762 ,

1

101 56762 ,

K1

101 56762 ,

1

101 56762 ,

K1

101 56762 ,

1

101 56762 ,

K1

101 56762 ,

1

101 56762

Gains

sys:-k =

6744

101

1

50500

6744

101

1

50500

From the zero-pole plot we see immediately that there are two poles: one in the right half plane and one in the

left. As we deduced from first principles, the system is unstable. Just for completeness, if only the the transfer

function for the system were needed, it would be obtained from:

sys d DynamicSystems TransferFunction sys_de

Transfer Function

continuous


inputvariable = Ic s , Fd s

outputvariable = z s , v s

outd sys:-tf

6744

101 s2K562

1

50500 s2K281000

6744 s

101 s2K562

s

50500 s2K281000


1. 1.

Now we can design the control system.

2.1 Using the Superconductor for Integral Control for a PI Controller

Superconductor magnets do not like rapidly varying currents. We can use the superconductor as an

integrator (this comes from physics). The magnet has essentially zero resistance, so the differential

equation for the magnet is:

Lsc d

d t Icsc t = v t :

Thus, for any voltage v t applied to the superconductor, the current in the magnet is the integral of the

voltage. Making the control signal the voltage, the superconductor will integrate this control and the gain on

this signal will be the integral control gain, which we denote by Ki.

Similarly, the control coil has the model:

Lc d

d t Ic t CRcIc t = v t :

and of course in steady state the current in the control coil is simply v t

Rv multiplied by the proportional

control gain (Kp).

This analysis shows that the control system is a proportional plus integral controller with gains Kp and Ki, but we still have not addressed how to stabilize the lift magnets. Let's investigate the measurements that

will be required to stabilize the magnets.

2.2 The Control Specifications and the Measurements NeededWhen a Maglev "flies" above the guideway (and below the rails in this design), it is moving along in inertial

coordinates that are only slightly constrained by the proximity of the guideway. This means that a slight

change in the rail location relative to the vehicle - one that is caused by an irregularity in the rail for

instance - can be ignored. Thus, measurements of the gap between the magnet and the rail are not

sufficient to tell the control system what to do. In most attractive Maglev designs, therefore, a

measurement of the gap is supplemented with a measurement of the acceleration of the magnet in the

direction of the rail (remember in this design the magnets are canted at an angle so the force is not applied

purely along the vertical (z) axis). The control system is therefore required to null accelerations.

Acceleration alone can not be the only feedback. The guideway (and the rail along with it) must follow the

terrain and the desired route (turning left and right). This means that the gap between the magnets and the

rail must be measured too. Furthermore, the availability of acceleration as a measurement would allow

derivative control (i.e. the control could be full state feedback PID control) simply by integrating the

acceleration.

The control criteria is therefore:

At a fast scale (response times between 0.1 s and 1 s), zero the acceleration.


(14)(14)

The motivation for this is twofold. First, this cancellation will result in closed loop dynamics that are second

order, and secondly the bandwidth of the closed loop servo can be set to any desired value (independent

of the magnet dynamics). As a second order system, the gains for the measurements and the proportional

feedback can be selected to make the magnet response optimal in the LQ sense (i.e. we can select the

closed loop bandwidth to be 10 Hz as required in the specifications, and we can set the damping to the

optimal value of 2

2) . The cancellation of the zero with the left half plane pole of the unstable dynamics

requires that Ki2=Kz

m:

The following simplifications are done to get the closed loop transfer function into the required form:

factor algsubs Ki = sqrtKz

m, Gcl s

sCKz

m Kf KIc m Kf KIc m Ka s

3CKf KIc m Ka s

2

Kz

m

CKf KIc m Kv s2CKf KIc m Kv s

Kz

mCKf KIc m sCKf KIc m

Kz

mCs

3 m

Ks Kz

Note that the possibility of having a common factor in the numerator and denominator is not obvious, but it

is possible to select the values of Ka, Kf and Ki to make the denominator of this transfer function equal to

sCKz

ms2 C 2ζ

CL$ω

CL$s C ω

CL

2: , where the values of the desired closed loop denominator

are:

ζCL

=2

2:

ωCL= 20$π :

To simplify the algebra, let's assume that the zeros introduced by the acceleration gain and velocity gain

are both at the same location denoted by ωz. This assumption makes the acceleration and velocity gains

as follows:

Ka =1

ωz

2: and Kv =

1

ωz

:

While it might seem like a large set of simultaneous nonlinear equations need to be solved to completely

specify the closed loop dynamics, we have systematically been eliminating the non-linearities in the

solution as we constrained the gains. Therefore, the values for the gains are completely specified by the

location of the zeros, which because of our assumption is ωz. Define the open loop undamped natural

frequency as ωOL

=Kz

m:


Then ωz=

ωCL

1C 1CωOL

ωCL

$ 2$ζCL

CωOL

ωCL

2$ζCL

CωOL

ωCL

:

The values for the feedback gains Ka and Kv and the integral gain Ki are completely determined from the

above, and the value of Kf is:

Kf =ωCL

KIc$ 1CωCL

ωz

2:

2.5 The Final Design

We can now verify that the design is correct by examining the system response to disturbance forces and

guideway changes and errors. Toward that end, let's find the numerical values for the gains based on the

equations we have derived above.

Substituting ζCL

, ωCL

, and ωOL from the specification completely determines the gains that, along with the

other parameters, can be used to create the system model that we want to examine.

Thus the parameters are:

paramsd m = 4192.5 , Kz = 3372000 , KIc = 17984, Ka = 0.00015866 , Kv = 0.0252, Ki= 28.3599, Kf = 2463.2 :

The state space model for the complete linear system can be written from the block diagram in Figure 3, in

the following manner:

We start at the right side of the diagram, moving from integrator to integrator for each of the states. The

first integrator (for the velocity) gives:

ode1dd

d t x t = v t :

The next integration (of the acceleration) is a little complex because there is an algebraic loop involved.

Thus the differential equation for the acceleration has a term on the right that involves the derivative of the

velocity (i.e. the acceleration) that must be removed by bringing the term to the right and dividing through

by the resulting coefficient (which is 1CKf$Ka$KIc

m), so:

ode2dd

d t v t =

KzKKf$KIc

m

1CKf$Ka$KIc

m

$x t K

Kf$Kv$KIc

m

1CKf$Ka$KIc

m

$v t


(16)(16)

(18)(18)

(17)(17)

(15)(15)

C

Kf$KIc

m

1CKf$Ka$KIc

m

$x3 t C

Kf$KIc

m

1CKf$Ka$KIc

m

$g t

CFd t

m 1CKf$Ka$KIc

m

d

dt v t =

KzKKf KIc x t

m 1CKf Ka KIc

m

KKf Kv KIc v t

m 1CKf Ka KIc

m

CKf KIc x3 t

m 1CKf Ka KIc

m

CKf KIc g t

m 1CKf Ka KIc

m

CFd t

m 1CKf Ka KIc

m

The last equation comes from the integrator in the integral compensation, and this also is complex

because the right hand side also has the acceleration in it. We let Maple do the work of solving for this and

eliminating it as follows:

ode3d d

d t x3 t = KKi$Ka$

d

d t v t KKv$Ki$v t KKi$x t CKi$g t :

We now substitute all of the parameter values into these equations:


d

dt x t = v t


d

dt v t =K3647.333890 x t K99.48571587 v t C3947.845868 x3 t

C3947.845868 g t C0.0002385211688 Fd t


d

dt x3 t =K0.004499581734

d

dt v t K0.71466948 v t K28.3599 x t

C28.3599 g t

sys_de d DynamicSystems StateSpace ode1, ode2, ode3 , inputvariable = Fd t , g t , outputvariable = x t , v t :



(19)(19)

State Space

continuous

2 output(s); 2 input(s); 3 state(s)

inputvariable = Fd t , g t

outputvariable = x t , v t

statevariable = x1 t , x2 t , x3 t

a = K99.48571587, K3647.333890, 3947.845868 ,

1, 0, 0 , K0.2670253700774340814, K11.94842305075683474, K17.76365515630017511

b =

0.0002385211688 3947.845868

0 0

K1.073245494304810699 10K6 10.59624484369982489

c = 0 1 0

1 0 0

d = 0 0

0 0

We can verify the behavior of our controller by examining how it responds to disturbances and guideway

changes and/or irregularities. If we want to see what happens when there is a change in the force applied

to the vehicle (as for, example, a vertical gust), the Fd(t) input can be set to a step input, and if we want to

see what happens when there is an downward slope, the guideway input g t can be set to be a ramp.


Control Response to Disturbance Forces and Guideway Changes

Generate Response PlotThe response obtained below was calculated for a

10,000 pound aerodynamic gust of wind, and for a

guideway slope that descends 1 ft over 10 seconds

of travel.

time [sec]2 4 6 8 10

Slope [inches]

K9

K7

K5

K3

0Slope

Figure 6A: Input Signal

time [sec]0 2 4 6 8 10

Gap Error [inches]

0

0.005

0.010

0.015

0.020

0.025

Gap Error

Figure 6B:

System Response Plot

A quick look at the control response plots shows that the controller is able to correct for air gap changes

due to disturbance forces, such as a 10,000 pound aerodynamic gust of wind, in about 0.1 seconds.

Moreover, the air gap due to such a strong wind is only 0.004 inches. In addition, we can see that the

controller is able to maintain the air gap even when the guiderail path is perturbed. These results show that

the controller meets the design criteria specified in Section 2.2.

3. Vehicle Model

By maintaining the air gap between the guideway and train we expect the control system to be able control the

motion of the vehicle in terms of heave, sway, pitch and yaw. In this section we will use the magnet model and

control system equations determined previously to derive the equations of motion for the entire Maglev

system. We will then examine the Maglev train system response to disturbance forces and guideway changes.

3.1 Equations of Motion

Figure 7 shows the magnets as they are grouped along the sides of the vehicle. Twelve groups of four

magnets are mounted along the vehicle (4 magnets are grouped as a single assembly and the force they

apply is along the axis of the magnets through a compliant joint). Because of the cant angle the vertical

forces (along the z-axis) are given by f i cos β and the horizontal forces along the y axis are given by


f i sin β , where β is the magnet cant angle.

Figure 7. Maglev train geometry

Location of the 12 magnet modules, their produced forces f1 through f

12, and the three lever

arms for the moments (magnets are placed symmetrically so these 3 define all 6 lever arms).

The moments and forces created by the magnets depend, of course, on the geometric location of the

magnets. All of the magnets are canted at an angle of 35°. Thus the attractive force of the magnet (which

is what has been modeled so far) must be resolved into vertical and sideways (sway) forces. Once this has

been done, the moments created by the magnets can be defined. We create a vector of forces for the

magnets and then also create a vector of lever arms for each of the rotational degrees of freedom. When

defining the equations we assume that all of the rotational angles are small so the gyroscopic forces and

Coriolis forces can be neglected. The equations of motion then become:


(24)(24)

Equations of Motion:

Iθ $

d

d t

d

d t θ t = L

ΘΘΘΘf$$$$cos β

Iθ

d2

dt2

θ t = LΘ f cos β (20)

Iψ$

d

d t

d

d t ψ t = L

ψf$$$$sin β

Iψ

d2

dt2

ψ t = Lψ f sin β (21)

m$d

d t

d

d t y t = Lyf$$$$cos β

m d

2

dt2

y t = Ly f cos β (22)

m$d

d t

d

d t z t = Lzf$$$$sin β

m d

2

dt2

z t = Lz f sin β (23)

The row vectors LΘΘΘΘ, L

ψ, Ly, and L

z are each an array of distances (or +1's and -1's in the case of the

forces) that multiply the forces to give the forces or moments (summed together at the center of gravity)

for the 12 magnet modules. They are given by:

Row Vectors LΘ

, Lψ, Ly, and L

z:

LΘ

=

1 .. 12 Vectorrow

Data Type: anything

Storage: rectangular

Order : Fortran_order

: Lψ

=

1 .. 12 Vectorrow

Data Type: anything



:

Ly=

1 .. 12 Vectorrow

Data Type: anything

Storage: empty


: Lz =

1 .. 12 Vectorrow

Data Type: anything



:

The equations of motion defined above are combined with the magnet module equations to obtain the

closed-loop system equations that describe the behavior of the Maglev train.

We rename the state variable x3(t) to aux(t):

contorlODEs d subs x3 t = aux t , ode1, ode2, ode3

d

dt x t = v t ,

d

dt v t =K3647.333890 x t K99.48571587 v t

C3947.845868 aux t C3947.845868 g t C0.0002385211688 Fd t ,d

dt aux t


(28)(28)

(27)(27)

(25)(25)

(26)(26)

=K0.004499581734 d

dt v t K0.71466948 v t K28.3599 x t

C28.3599 g t

Next, the equations are duplicated 12 times for each of the magnet modules:

listVarnames d x, v, aux, g :allcoils d seq op subs map dummy/dummy = dummy i, listVarnames , contorlODEs , i

= 1 ..12 :

Expression for the force generated per module (for n modules) is defined as:

coilForceExpr d F t =m$ diff v t , t

n

F t =

m d

dt v t

n

Parameter values are substituted:

coilForce d subs n = 12 , params, coilForceExpr

F t = 349.3750000 d

dt v t

Creating the force expressions for y and z axes:

YForceExpr d Fy t = F t $sin β

Fy t = F t sin β

ZForceExpr d Fz t = F t $cos β

Fz t = F t cos β

Substituting parameter values and generating the force for each individual module:

listVarnames d F, Fy, Fz, v :

YForce d map eval, seq subs map dummy/dummy = dummy i, listVarnames ,

subs coilForce, β = evalf convert 35,'units ','deg ','rad' , YForceExpr , i = 1 ..12 :

ZForce d map eval, seq subs map dummy/dummy = dummy i, listVarnames ,

subs coilForce, β = evalf convert 35,'units ','deg ','rad' , ZForceExpr , i = 1 ..12 :


Vehicle Dynamics

Heave

The vehicle dynamics in the z-direction is then given by summing up all of the Z forces:

odeZ d eval subs params, ZForce, m$diff z t , t$2 = add x, x = seq Fz i t , i = 1 ..12

4192.5 d

2

dt2

z t = 286.1912455 d

dt v1 t C286.1912455

d

dt v2 t

C286.1912455 d

dt v3 t C286.1912455

d

dt v4 t

C286.1912455 d

dt v5 t C286.1912455

d

dt v6 t

C286.1912455 d

dt v7 t C286.1912455

d

dt v8 t

C286.1912455 d

dt v9 t C286.1912455

d

dt v10 t

C286.1912455 d

dt v11 t C286.1912455

d

dt v12 t

(29)

Sway

The vehicle dynamics in the y-direction is then given by summing up all of the Y forces:

odeY d eval subs params, YForce, m$diff y t , t$2 = add x, x = seq K1i$ Fy i t , i

= 1 ..12

4192.5 d

2

dt2

y t =K200.3932674 d

dt v1 t C200.3932674

d

dt v2 t

K200.3932674 d

dt v3 t C200.3932674

d

dt v4 t

K200.3932674 d

dt v5 t C200.3932674

d

dt v6 t

K200.3932674 d

dt v7 t C200.3932674

d

dt v8 t

K200.3932674 d

dt v9 t C200.3932674

d

dt v10 t

K200.3932674 d

dt v11 t C200.3932674

d

dt v12 t

(30)


Pitch

The dynamic equation is a function of the level arm distance from the center of gravity for each of the

modules.

Collecting the level arm into a list:

LPitch d K56.7, K56.7, K11.52, K11.52, K5.76, K5.76 :LPitch d op LPitch , op map x/Kx, LPitch :

The equation of motion of the pitching motion, with augmented forces generated by the motion of the

modules relative to the rail, is given by:

odeTheta d evalf subs Iθ = 882500, β = convert 37,'units','deg ','rad' , params, ZForce, I

θ

$diff θ t , t$2 = add x, x = zip `* ,̀ LPitch, seq Fz i t , i = 1 ..12 K θ t

Ccot β $ψ t $add x, x =map x/x$x, LPitch

8.82500 105

d2

dt2

θ t =K16227.04362 d

dt v1 t K16227.04362

d

dt v2 t

K3296.923148 d

dt v3 t K3296.923148

d

dt v4 t

K1648.461574 d

dt v5 t K1648.461574

d

dt v6 t

C16227.04362 d

dt v7 t C16227.04362

d

dt v8 t

C3296.923148 d

dt v9 t C3296.923148

d

dt v10 t

C1648.461574 d

dt v11 t C1648.461574

d

dt v12 t K13523.1120 θ t

K17945.77574 ψ t

(31)


(33)(33)

Yaw

LYaw d 56.7, K56.7, 11.52, K11.52, 5.76, K5.76 :LYaw d op LYaw , seq op Ki, map x/Kx, LYaw , i = 1 .. nops LYaw :

odePsi d evalf subs Iψ = 882500, β = convert 35,'units','deg','rad' , params, YForce, I

ψ

$diff ψ t , t$2 = add x, x = zip `* ,̀ LYaw, seq Fy i t , i = 1 ..12 K ψ t

Ctan β $θ t $add x, x =map x/x$x, LYaw

8.82500 105

d2

dt2

ψ t = 11362.29826 d

dt v1 t K11362.29826

d

dt v2 t

C2308.530440 d

dt v3 t K2308.530440

d

dt v4 t

C1154.265220 d

dt v5 t K1154.265220

d

dt v6 t

C1154.265220 d

dt v7 t K1154.265220

d

dt v8 t

C2308.530440 d

dt v9 t K2308.530440

d

dt v10 t

C11362.29826 d

dt v11 t K11362.29826

d

dt v12 t K13523.1120 ψ t

K9468.984961 θ t

(32)

The closed-loop system equations defining the behavior of the entire Maglev train in terms of heave, sway,

pitch and yaw is obtained by combining the twelve magnet module equations and the four equations of

motion defined above:

sysOverall dDynamicSystems NewSystem op allcoils , odeY, odeZ, odeTheta, odePsi ,

inputvariable = Fd t , seq g i t , i = 1 ..12 , outputvariable = y t , z t , θ t , ψ t

;

Diff. Equation

continuous


inputvariable = Fd t , g1 t , g2 t , g3 t , g4 t , g5 t , g6 t , g7 t , g8 t , g9 t ,g10 t , g11 t , g12 t

outputvariable = y t , z t , θ t , ψ t

3.2 Response Plots

The heave response plots to a vertical gust of wind and a descending guideway slope of 0.1ft/sec are

shown in Figures 8 and 9, respectively. From the plots we see that the control system developed for each

magnet is able to work together to minimize the heave displacement.


Heave Response

Response to Vertical Gust of Wind

time [secs]0 2 4 6 8 10

Wind [lbs]

0

2000

4000

6000

8000

10000Input - Vertical Gust of Wind


time [secs]0 2 4 6 8 10

eave Displacement [in

0

0.001

0.002

0.003

Heave response to a gust of wind

Figure 8B: Input Signal

Response to Descending Guideway

time [secs]2 4 6 8 10

Slope [inches]

K12

K10

K8

K6

K4

K2

0Input - Descending Guideway


Time [secs]2 4 6 8 1012

ve displacement [inch

K3.#10-8

K2.#10-8

K1.#10-8

0

Heave response to descending guideway

Figure 9B: Input Signal


Results

This worksheet documented the design and development of a controller to guide a EMS Maglev train along a

guideway. In section 1, we determined a mathematical model to describe the magnet dynamics of the lifting

magnets. In section 2, we developed a PID controller with feedback acceleration to maintain an air gap

between the train and the guideway of 2 inches and thus counteract the inherent unstable nature of the lifting

magnets (i.e. when there is no current in the coils the vehicle will drop under the influence of gravity causing

the undercarriage to hit the guideway and when the current in the magnet is very large, the magnet will be

attracted to the rail and hit the rail). We showed that the behavior of our controller adhered well to a set control

criteria by investigating the system response to disturbance forces and guideway changes. In section 3, we

used the magnet model and control system equations to derive the equations of motion for the entire Maglev

system. These equations were then used to examine the Maglev train's response to disturbance forces and

guideway changes. We were able to show that the control system defined in section 2 was able to control the

motion of the train in terms of heave displacement.

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Optimizing the Controller Design to Guide the Motion of a Maglev Train

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