AEEC432 Rectilinear Control

EXPERIMENT #2

Rectilinear Control System

Introduction:

This classical plant is readily transformed into the variety of configuration shown below.

It serves to vividly demonstrate both lumped parameter dynamics and generic control

issues. This system appears commonly in dynamics and controls text books and serves as

a benchmark for control method evaluation. The mechanism features adjustable masses,

interchangeable springs and adjustable air damping. Its dynamic properties are generally

the rectilinear equivalent of those of the Torsional Apparatus with additional parameter

adjustment capability. As with Model 205, this system provides vivid demonstrations of

elementary topics such as rigid body PID control, lead/lag compensators, phase and gain

margin, trajectory tracking, and regulation-as well as advanced high order collocated and

noncollocated system control. The two apparatuses also clearly demonstrate salient

properties of flexible systems such as mode shapes, natural frequencies, and

characteristic transient and frequency responses. An optional secondary drive may be

positioned at any output (mass carriage) to create a MIMO plant and provide for the

study of disturbance rejection.

1

Figure: 1 (Model 210 Rectilinear Plant)

The Model 210 holds three mass carriages which can be loaded with brass weights and

connected in a variety of configurations using springs of varying stiffness. The adjustable

dashpot can be used to provide damping for the system. A single drive motor provides

actuation to the system via the first mass carriage, and position measurements are taken

by quadrature encoders.

This report was made with the help of other students and with the help of my supervisor

and it is basic of rectilinear control system plant identification (model 210). This report is

related to those of (Torsional system). The equations and the procedure are the same way

but now we have a mass than disk. The purpose of this report is to identify the plant

parameters, implement a variety of control schemes, and demonstrate many important

control principles. This report includes experiments, which will be executed, analyzed

and also mathematical equations. The user to be done this experiments must be familiar

with the (model 210), how is work and to read or remember from the last years about

2

control systems to have the ability to solve the equations. The model to be work needs

I/O electronic unit connected with a computer that is able to show the data and the

waveform from each encoder. Finally the user must save the data and control

configuration files to avoid any problems with the procedure.

Experiment: A (system identification)

Procedure:

Step: 1

Clamp the second mass to put the mechanism in the configuration shown in Figure 1a

using a shim (i.e. ¼ inch nut) between the stop tab and stop bumper so as not to engage

the limit switch. Verify that the medium stiffness spring (nominally 400 N/m (2.25

1b/in.)) is connecting the first and second mass carriages.

Step: 2

Secure four 500g masses on the first and second mass carriages.

Step: 3

With the controller powered up, enter the Control Algorithm box via the Set-up menu and

set Ts = 0.00442. Enter the Command menu, go to Trajectory and select step, Set-up.

Select Open Loop Step and input a step size of 0 (zero), a duration of 3000 ms and 1

repetition. Exit to the background screen by consecutively selecting OK. This puts the

controller in a mode for acquiring 6 sec of data on command but without driving the

actuator. This procedure may be repeated and the duration adjusted to vary the data

acquisition period.

3

a) Setup to begin Plant Identification Procedure

b) Second Setup in Plant Identification Procedure

4

c) Third Setup in Plant Identification Procedure

Figure: 2 Configurations for Plant Identification

(model 210a. Four 500g weights on each active carriage.)

Step: 4

Go to Set up Data Acquisition in the Data menu and select Encoder # 1 and Encoder # 2

as data to acquire and specify data sampling every 2 (two) servo cycles (i.e. every 2 T s’s).

Select OK to exit. Select Zero Position from the Utility menu to zero the encoder

positions.

Step: 5

Select Execute from the Command menu. Prepare to manually displace the first mass

carriage approximately 2.5 cm. Exercise caution in displacing the carriage so as not to

engage the travel limit switch. With the first mass displaced approximately 2.5 cm in

either direction, select Run from the Execute box and release the mass approximately 1

second later. The mass will oscillate and attenuate while encoder data is collected to

record this response. Select OK after data is uploaded.

Step: 6

Select Set-up Plot from the Plotting menu and choose Encoder # 1 Position then select

Plot Data from the Plotting menu. You will see the first mass time response.

5

Figure: 3 (encoder # 1 & # 2 loaded)

Step: 7

Choose several consecutive cycles (say ~5) in the amplitude range between 5500 and

1000 counts (this is representative of oscillation amplitudes during later closed loop

control maneuvers. Much smaller amplitude responses become dominated by nonlinear

friction effects and do not reflect the salient system dynamics).

We choose,

t1 = 0.336 seconds 304 counts from TABLE 1


so the number of the cycles between t1 and t2 is n = 2 cycles

Divide the number of cycles by the time taken to complete them being sure to take

beginning and end times from the same phase of the respective cycles.

f = n/(t2-t1) f =frequency (Hz)

6

f = 2/(1.443-0.336)

f = 1.806 Hz

Convert the resulting frequency in Hz to radians/sec. this damped frequency, ωd,

approximates the natural frequency, ωn, according to:

(for small )

Equation: 1

ωnm11 = 2.π.f

ωnm11 = 2.π.1.806 => ωnm11 = 11.34 rad/sec

where the “m11” subscript denotes mass # 1 , trial # 1. (close the graph window by

clicking on the left button in the upper right hand corner of the graph. This will collapse

the graph to icon from where it may later be brought back up by double-clicking on it.)

step: 8

Remove the four masses from the first mass carriage and repeat step 5 through 7 to obtain

for the unloaded carriage. If necessary, repeat step 3 to reduce the execution (data

sampling only in this case) duration.

7

Figure: 4 ( encoder # 1 unloaded)

We choose,




Divide the number of cycles by the time taken to complete them being sure to take

beginning and end times from the same phase of the respective harmonic cycles.

f = n/(t2-t1) f = frequency (Hz)

f = 2/(0.717-0.053)

f = 3.01 Hz

From equation 1:

ωnm12 = 2.π.f

ωnm12 = 2.π.3.01 => ωnm12 = 18.91 rad/sec

8

Step: 9

Measure the initial cycle amplitude Xo and the last cycle amplitude Xn for the n cycles

measured in step 8. Using relationships associated with the logarithmic decrement:

(for small )

Equation: 2

find the damping ratio and show that for this small value the approximations of

equation’s 1 & 2 are valid.

Xo = 508 counts => for t1, from TABLE 2 Xo = (cycle amplitude)

Xn = 160 counts => for t2, from TABLE 2 Xn = (cycle amplitude)

Number of the cycles between t1 and t2 is n = 2 cycles

ζm12 = 1/2πn * ln(Xo/ Xn)

ζm12 = 1/12.56 *ln (508/160)

ζm12 = 0.091

ωnm11 = => ωnm11 =

The approximation of the Equation 1 and 2 are valid because the damping ratio ζd32 is

very small and it is not affect the result. ωnm11

Step: 10

9

Repeat step 5 through 9 for the second mass carriage (Figure 1b). Here in step 6 you will

need to remove Encoder #1 position and add Encoder #2 position to the plot set-up.

Figure: 5 (load encoder # 2)

Hence obtain ωnm21, ωnm22, ζm22.

How does this damping ratio compare with that for the first mass?

This damping ratio is approximately 13 times lower than the damping ratio that we found

in step: 9

We choose,





f = 2/(1.213-0.142)

10

f = 1.867 Hz

From equation 1:

ωnm21 = 2.π.f

ωnm21 = 2.π.1.867 => ωnm21 = 11.73 rad/sec





ζm21 = 1/12.56 * ln(423/388)

ζm21 = 0.0068

Figure: 6 (unloaded encoder # 2)

We choose,


11




f = 2/(0.602-0.044)

f = 3.58 Hz

From equation 1:

ωnm22 = 2.π.f

ωnm22 = 2.π.3.58 => ωnm22 = 22.49 rad/sec





ζm22 = 1/12.56 * ln(1626/449)

ζm22 = 0.101

step: 11

Connect the mass carriage extension bracket and dashpot to the second mass as shown in

figure 1c. Open the damping (air flow) adjustment knob 2.0 turns from the fully closed

position. Repeat step 5, 6, and 9 with four 500g masses on the second carriage and using

only amplitudes 500 counts in your damping ratio calculation. Hence obtain ζd where

the “d” subscript denotes “dashpot”.

12

Figure: 7 (encoder # 2 with dashpot & medium stiffness)

We choose,



so the number of the cycles between t1 and t2 is n = 0.5 cycles


f = 0.5/(0.425-0.177)

f = 2.016 Hz

From equation 1:

ωd = 2.π.f

ωd = 2.π.2.016 => ωd = 12.66 rad/sec

13



Number of the cycles between t1 and t2 is n = 0.5 cycles


ζm21 = 1/3.14 * ln(121/25)

ζm21 = 0.50

Step: 12

Each brass weight has a mass of 500 10g. (you may weight the pieces if a more precise

value is desired). Calling the mass of the four weights combined mw, use the following

relationships to solve for the unloaded carriage mass mc2, and spring constant k.

K/ (mw + mc2) =ωnm212 Equation: 3

K/mc2 = ωnm222 Equation: 4

other way

other way

14

other way

other way

other way

kg other way kg

we take the other way kg

rad/sec

N/m

find the damping coefficient cm2 by equating the first order terms in the equation form:

Equation: 5

from this equation,

N/Ms

repeat the above for the first mass carriage, spring and damping mc1, cm1 and k

respectively.

Calculate the damping coefficient of the dashpot, cd.

15

other way

other way

other way

other way

other way

kg other way kg

we take the other way kg

rad/sec

N/m

from equation 3,

16

N/Ms

Step: 13

Remove the carriage extension bracket and dashpot from the second mass carriage,

replace the medium stiffness spring with a high stiffness spring (800 N/m nominally), and

repeat step 5 and 6 to obtain the resulting natural frequency .

Figure: 8 (encoder # 1 loaded with high stiffness)

We choose,





f = 1/(0.283-0.097)

17

f = 5.37 Hz

From equation 1:

ωm23 = 2.π.f

ωm23 = 2.π.5.37 => ωm23 = 33.74 rad/sec

Repeat this frequency measurement using the least stiff spring (nominally 200 N/m) to

obtain .

Figure: 9 (encoder # 1 with low stiffness)

We choose,





18

f = 1/(0.390-0.097)

f = 3.41 Hz

From equation 1:

ωm24 = 2.π.f

ωm24 = 2.π.3.41 => ωm24 = 21.42 rad/sec

Calling the value of stiffness obtained in step 12 above kmed stiffness, calculate khigh stiffness and

klow stiffness from the frequency measurement of this step.

klow stiffness N/m

khigh stiffness N/m

kmed stiffness N/m

Now all dynamics parameters have been identified! Values for m1 and m2 for any

configuration of masses may be found by adding the calculated mass contribution of the

weights to that of the unloaded carriages.

The following is necessary to establish the hardware gain for control modeling purposes.

Procedure:

Step: 14

Remove the spring connecting the first and second masses and secure four 500g masses

on the first mass carriage. (you should libel this particular spring so that the identified

parameter k2 will be consistent when used in later experiments). Use the limit clamps to

secure the second mass clear from the first. Verify that the masses are secure and that the

carriage slides freely. Hook up the drive power to the mechanism. Position the first mass

approximately 3cm to the left (negative x1 position) of its center of travel.

19

Figure: 10 (encoder # 1 loaded)

Step: 15

In the Trajectory window deselect unidirectional moves (i.e. enabling bi-directional

inputs) select Step, Set-up. Choose Open Loop Step, and input 2.00 Volts, 75 ms, 2 reps.

Execute this open loop step via the execute menu. (this move may trip a software speed

limit which disables the controller indicated by “limit exceeded” in the Controller Status

box in the “desk top”. Again to reset, simply reselect Execute from the Execute menu.)

Go to Set-up in the Plot Data menu and select Encoder # 1 velocity for plotting.

20

Figure: 11 (encoder # 1 positive-sloped)

Figure: 12 (encoder # 1 negative-sloped)

21

Step: 16

Plot this data and observe four velocity profile segments with nominal shapes of : linear

increase (constant acceleration), constant (zero acceleration), linear decrease

(deceleration), and constant. Obtain the acceleration, 1e, (counts/s2) by carefully

measuring the velocity difference and dividing by the time difference (75ms) through the

positive-sloped segment. Calculate the average magnitude of the positive and negative

accelerations for use in obtaining khw below.

The acceleration,

For positive linear segment = = 34666.66 counts/s2

For negative linear segment= = 546666.66 counts/s2

Therefore =

= 290666.66 counts/s2

Step: 17

Save any files or plots of interest. Exit the executive program and power down the

system.

Conclusion:

From the experiments that we have done we identified the plants parameters and

also there are some results between the damping frequency and the damping ratio. As we

can see in step 8 #3 and step 10 #2 (for the unloaded carriage the damping frequency are

almost the same) and in step 6 #3 and step 10 #2 the (damping frequency are almost the

same). The damping frequency for the first or second mass of unloaded cases is

approximately the double in the cases of loaded mass. The mechanical connection of

former with the motor causes frictional forces to act on it and therefore, the damping ratio

of the first mass is more than the second mass. In the cases where a dashpot is connected

22

the damping frequency is decreased because of the forces that acting on the mass

carriage.

Transfer function calculation:

The so-called hardware gain, Khw, of the system is comprised of the product:

Khw = kckaktkmpkekepks

Where,

Kc = the DAC gain, = 10V / 32.768 DAC counts

Ka = the Servo Amp gain, = approx. 2 (amp/V)

Kt = the Servo Motor Torque constant = approx. 0.1 (N-m/amp)

Kmp = the Motor Pinion pitch radius inverse = 26.25 m-1

Ke = the Encoder gain = 16.000 pulses / 2 radians

Kep= the Encoder Pinion pitch radius inverse = 89 m-1

Ks = the Controller Software gain = 32 (controller counts / encoder or ref input counts)

khw = 11.625.103

In step 15, we obtained the acceleration (counts/s2) of a known inertia, m1 = mw +

mc1 with a known voltage applied at the DAC. This relates to the applied force during the

acceleration according to:

Applied Force =

And we have a direct measurement of the five-term product kaktkmpkekep. i.e:

2.00V kaktkmp = Applied Force in step 15

23

Experiment: B (Rigid Body PD & PID Control)

The experiments demonstrates some key concepts associated with proportional plus

derivative (PD) control and subsequently the effects of adding integral action (PID). The

system is using two steel shafts of different stiffness each one, as springs. This control

scheme it is used in such diverse areas as machine tools, automobiles and spacecraft,

because is acting on plants modeled as rigid bodies. The diagram for forward path PID

control of a rigid body is shown in Figure: 13a. Figure: 13b shows the case when the

derivatives term is in the return path. The experiment: B includes 10 steps.

The close loop transfer functions for the respective cases are:

equation: 6

equation: 7

24

Figure: 13 (Rigid body PID control-control block diagram)

For the first portion of this exercise we shall consider PD control only (k i = 0). For the

case of kd in the return path the transfer function reduces to:

Equation: 8

25

by defining:

we may express:

Equation: 9

the effect of kp and kd on the roots of the denominator (damped second order oscillator) of

c(s) is studied in the work that follows.

Procedure: (Proportional & Derivative Control Actions)

Step: 1

Using the result of experiment A construct a model of the plant with four 500g mass

pieces on the first mass carriage with no springs or damper attached. You may neglect

friction.

Step: 2

Set-up the plant in the configuration described in Step 1. There should be no springs or

damper connected to the first carriage and the other carriages should be secured away

from the range of motion of the first carriage.

26

Step: 3

From Eq determine the value of kp (kd = 0) so that the system behaves like a Hz

spring-mass oscillator.

Step:4

Set-up to collect Encoder #1 and Commanded Position information via the Set-up Data

Acquisition box in the Data menu. Set-up a closed-loop step of 0 (zero) counts, dwell

time = 3000ms, and 1 (one) rep (Trajectory in the Command menu).

Step:5

Enter the Control Algorithm box under Set-up and set Ts = 0.00442 s and select

Continuous Time Control. Select PID and set-up Algorithm. Enter the kp value

determined above for Hz oscillation (kd & ki = 0, do not input values greater than

kp = 0.08) and select OK.

Place the first mass carriage at approximate the-0.5 cm (negative is toward the motor)

mark.

In this and all future work, be sure to stay clear of the mechanism before doing the next

step. Selecting implement Algorithm immediately implements the specified controller; if

there is an instability or large control signal, the plant may react violently. If the system

appears stable after implementing the controller, first displace the disk with a light, non

sharp object (e.g. a plastic ruler) to verify stability prior to touching plant

Select Implement Algorithm, then OK.

27

Step:6

Select Execute under Command. Prepare to manually displace the mass carriage roughly

2 cm. Select Run, displace the mass approximately 3 cm and release it. Do not hold the

mass position for longer than about 1 second as this may cause the motor drive thermal

protection to open the control loop.

step:7

Plot encoder #1 output. Determine the frequency of oscillation. What will happen when

proportional gain, kp, is doubled? Repeat Steps 5 & 6 and verify your prediction. (Again,

for system stability, do not input values greater than kp = 0.06).

Figure: 14 (encoder # 1 kp value)

28

Figure: 15 (encoder # 1 kp double value)

step:8

Determine the value of the derivative gain, kd, to achieve kdkhw = 50N/(m/s). Repeat step

5, except input the above value for kd and set kp & ki = 0. (do not input values greater

than kd = 0.04)

29

Figure: 16 (encoder # 1 kd value)

step:9

After checking the system for stability by displacing it with a ruler, manually move the

mass back and forth to feel the effect of viscous damping provided by kd. Do not

excessively coerce the mass as this will again cause the motor drive thermal protection to

open the control loop.

Step:10

Repeat step 8 & 9 for a value of kd fives times as large (Again ). Can you feel

the increased damping?

khw = 11625.103 , kb = 0

30

By multiplying the kd 5 times we can see that the damping increased 5 times because in

this case we have the kb=0.

So ,

Figure: 17 (encoder # 1 kd value five time larger)

Conclusion:

As we can see in this experiment the PI & PID Control are found commonly in

application and also the effect of kp and kd on the roots of the denominator. Because we

use the PI & PID we have to do with the gain of controls. The equations of closed loop

transfer function that we found in these two cases, the proportional and derivative gains

played important rule for resulting value and scheme. In the procedure of the experiment

31

and specific in step: 7 we can see that if we doubled the value of proportional gain do not

affect the result as well as the case of multiplying the derivative gain five times as large

in step: 10. Finally in this experiment we can see the behavior of the second order-

system.

32

EXERIMENT: A

TABLE: 1 (DATA FROM STEP: 6)

Sample time encoder # 1 pos

38 0.336 304

39 0.345 637

40 0.354 957

41 0.363 1263

42 0.372 1553

43 0.381 1821

44 0.390 2069

45 0.398 2293

46 0.407 2489

47 0.416 2656

. . .

. . .

. . .

. . .

. . .

163 1.443 2

164 1.452 38

165 1.461 72

166 1.470 103

167 1.479 131

168 1.488 155

169 1.496 176

170 1.505 193

171 1.514 206

172 1.523 215

33

EXERIMENT: A



6 0.053 508

7 0.062 972

8 0.071 1388

9 0.080 1744

10 0.089 2027

11 0.097 2230

12 0.106 2344

13 0.115 2370

14 0.124 2308

15 0.133 2168

. . .

. . .

. . .

. . .

. . .

76 0.673 125

77 0.682 146

78 0.691 159

79 0.699 165

80 0.708 166

81 0.717 160

82 0.726 148

83 0.735 133

84 0.744 116

85 0.753 98

34

EXERIMENT: A



7 0.062 4882

8 0.071 4520

9 0.080 4112

10 0.089 3663

11 0.097 3179

12 0.106 2664

13 0.115 2125

14 0.124 1567

15 0.133 998

16 0.142 423

. . .

. . .

. . .

. . .

. . .

129 1.142 3008

130 1.151 2752

131 1.160 2469

132 1.169 2162

133 1.178 1834

134 1.186 1489

135 1.195 1130

136 1.204 762

137 1.213 388

138 1.222 13

35

EXERIMENT: A

TABLE: 4 (DATA FROM STEP: 10unload)


5 0.044 449

6 0.053 1228

7 0.062 1933

8 0.071 2534

9 0.080 3006

10 0.089 3328

11 0.097 3490

12 0.106 3490

13 0.115 3330

14 0.124 3023

. . .

. . .

. . .

. . .

. . .

63 0.558 57

64 0.567 463

65 0.576 839

66 0.584 1167

67 0.593 1433

68 0.602 1626

69 0.611 1738

70 0.620 1767

71 0.629 1715

72 0.638 1584

36

EXERIMENT: A



11 0.097 1688

12 0.106 1525

13 0.115 1355

14 0.124 1179

15 0.133 999

16 0.142 818

17 0.151 637

18 0.159 459

19 0.168 287

20 0.177 121

. . .

. . .

. . .

. . .

. . .

48 0.425 25

49 0.434 101

50 0.443 174

51 0.452 244

52 0.460 310

53 0.469 372

54 0.478 429

55 0.487 481

56 0.496 527

57 0.505 567

37

EXERIMENT: A

TABLE: 6 (DATA FROM STEP: 13hi)


2 0.018 746

3 0.027 1024

4 0.035 1202

5 0.044 1290

6 0.053 1278

7 0.062 1160

8 0.071 956

9 0.080 712

10 0.089 431

11 0.097 130

. . .

. . .

. . .

. . .

. . .

25 0.221 404

26 0.230 578

27 0.239 701

28 0.248 760

29 0.257 745

30 0.266 672

31 0.274 554

32 0.283 402

33 0.292 227

34 0.301 41

38

EXERIMENT: A

TABLE: 7 (DATA FROM STEP: 13le)


10 0.089 166

11 0.097 680

12 0.106 1131

13 0.115 1536

14 0.124 1879

15 0.133 2149

16 0.142 2339

17 0.151 2441

18 0.159 2456

19 0.168 2385

. . .

. . .

. . .

. . .

. . .

43 0.381 39

44 0.390 299

45 0.398 544

46 0.407 722

47 0.416 844

48 0.425 935

49 0.434 991

50 0.443 1010

51 0.452 996

52 0.460 950

39

EXERIMENT: A



1 0.009 5832

2 0.018 5615

3 0.027 5347

4 0.035 5032

5 0.044 4671

6 0.053 4269

7 0.062 3831

8 0.071 3360

9 0.080 2862

10 0.089 2343

. . .

. . .

. . .

. . .

. . .

46 0.407 265

47 0.416 700

48 0.425 1122

49 0.434 1529

50 0.443 1918

51 0.452 2283

52 0.460 2622

53 0.469 2930

54 0.478 3206

55 0.487 3445

40

EXERIMENT: A

TABLE: 9 (DATA FROM STEP: 15po)


5 0.044 229

6 0.053 1059

7 0.062 1951

8 0.071 2889

9 0.080 3858

10 0.089 4788

11 0.097 5662

12 0.106 5463

13 0.115 5035

14 0.124 4631

. . .

. . .

. . .

. . .

. . .

23 0.204 -2074

24 0.213 -3174

25 0.221 -4320

26 0.230 -5509

27 0.239 -6655

28 0.248 -7729

29 0.257 -8254

30 0.266 -7899

31 0.274 -7446

32 0.283 -6981

41

EXERIMENT: A

TABLE: 10 (DATA FROM STEP: 15ne)


0 0.000 2620

1 0.009 2297

2 0.018 2026

3 0.027 1813

4 0.035 1663

5 0.044 1578

6 0.053 1558

7 0.062 1600

8 0.071 1704

9 0.080 1861

. . .

. . .

. . .

. . .

. . .

23 0.204 360

24 0.213 -282

25 0.221 -1000

26 0.230 -1783

27 0.239 -2566

28 0.248 -3321

29 0.257 -4043

30 0.266 -4719

31 0.274 -5347

32 0.283 -5920

42

EXERIMENT: B



0 0.000 -3377

1 0.009 -3293

2 0.018 -3212

3 0.027 -3134

4 0.035 -3061

5 0.044 -2992

6 0.053 -2927

7 0.062 -2866

8 0.071 -2810

9 0.080 -2756

. . .

. . .

. . .

. . .

. . .

51 0.452 28

52 0.460 80

53 0.469 129

54 0.478 176

55 0.487 221

56 0.496 265

57 0.505 308

58 0.514 351

59 0.522 392

60 0.531 433

43

EXERIMENT: B



0 0.000 5463

1 0.009 5584

2 0.018 5702

3 0.027 5817

4 0.035 5931

5 0.044 6041

6 0.053 6150

7 0.062 6256

8 0.071 6359

9 0.080 6460

. . .

. . .

. . .

. . .

. . .

41 0.363 5962

42 0.372 5954

43 0.381 5947

44 0.390 5942

45 0.398 5937

46 0.407 5934

47 0.416 5931

48 0.425 5930

49 0.434 5929

50 0.443 5929

44

EXERIMENT: B



0 0.000 -1062

1 0.009 -992

2 0.018 -924

3 0.027 -858

4 0.035 -794

5 0.044 -732

6 0.053 -671

7 0.062 -611

8 0.071 -554

9 0.080 -498

. . .

. . .

. . .

. . .

. . .

46 0.407 299

47 0.416 296

48 0.425 292

49 0.434 289

50 0.443 286

51 0.452 284

52 0.460 282

53 0.469 281

54 0.478 280

55 0.487 280

45

EXERIMENT: B



0 0.000 1419

1 0.009 1336

2 0.018 1256

3 0.027 1179

4 0.035 1105

5 0.044 1035

6 0.053 968

7 0.062 903

8 0.071 840

9 0.080 779

. . .

. . .

. . .

. . .

. . .

45 0.398 -13

46 0.407 -13

47 0.416 -12

48 0.425 -11

49 0.434 -11

50 0.443 -10

51 0.452 -9

52 0.460 -9

53 0.469 -8

54 0.478 -8

46

Documents

AEEC432 Rectilinear Control