servo drive systems_chapter 4_part1

7/23/2019 Servo Drive Systems_Chapter 4_part1

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Trajectory Control

Chapter 4

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4.1 Point-to-Point Control

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Point to point motion is applied for some kinds of robots suchas: point welding, packaging, In this type of motion, we

only take into account the position of the beginning, ending

points and time of motion.

Problem:

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0

We have to determine the rule of

change of the angular q between

the initial position

and the

finally position in the timeinterval 0, .Assuming that: inertial moment of motor

: torque of motor


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Note that the problem has plenty of solutions. Therefore, weinclude an optimal criterion: the least consuming energy

Mathematic description:

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=

where = subjec to: 2MT

The general solution has the following form:

= Then the motion trajectory will be the third order form:

=

(4.1)

(4.2)

(4.3)


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We need to determine 4 coefficients: , , , . We havethe following equations:

The velocity will follow the second order form:

= 3 2 And the acceleration changes w.r.t. the first order rule

= 6 2

=

= = 3 2 =

(4.4)

(4.5)

(4.6)


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The velocity gets the its maximum value:

For example: = 0 , = , = 1 ; the initial and finalvelocity= = 0From the above equations, we find the solutions:

= = 0

=3/2, when

= 1 / 2

= 3 , =2

And the acceleration gets its the maximum value:

= 6, when = 0 = 1The disadvantage of this rule is the maximum accelerations at

the initial and final point.


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0

3

0 1

1

3/2/

6

6

/

0

3/2


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The famous trajectory that is usually used in most industrial

controllers is Blended Polynomial with the trapezoidal

velocity.

Assuming that

= = 0, the time of acceleration and

deceleration is equal. We have:

= ()/2, at = /2In order to guarantee the trajectory is continuous, the velocities

at the transition points have to be smooth. That means:

= where, is the position at the time when the motoraccelerates at

(/

)

(4.7)

(4.8)


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We have:

= From Eq. (4.8) & (4.9), we got:

= 2 12

4( )

where,

(4.9)

= 0 (4.10)Given , , , , finding /2, we have:

(4.11)

(4.12)

It is obvious that if

=

,

= =

(4.13)


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Therefore, given , and , from Eq. (4.12) we calculateand choose , then calculate from Eq. (4.11). Finally, thetrajectory is determined by a piecewise function

=

12 0 2

12 (4.14)

Example: Determine the Blended Polynomial trajectory in

order to control the position of the motor in 20 rounds within

2s. The rated speed is 2000 (RPM).


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4.2 Application for Step and DC motors

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The motor step angle: ; position: ; and speed are given by Step motor

To rotate a stepper motor at a constant speed, pulses must be

generated at a steady rate

= 2

()

= () = ()

where

is the number of pulses

is the delay time

(4.15)


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To start and stop the stepper motor in a smooth way, control of

the acceleration and deceleration is needed. Figure in previous

section shows the relation between acceleration, speed and

position.

Linear speed ramp

The time delay between the stepper motor pulses controlsthe speed. These time delays must be calculated in order to

make the speed of the stepper motor follow the speed rampas

closely as possible.

Discrete steps control the stepper motor motion, and the

resolution of the time delay between these steps is given by

the frequency of the timer.


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Assuming that the time

delay is generated by a

timer with a frequency

(Hz), = (s)

0 t t t t t

We have: = =

=

(4.16)


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The position at the time is given by equation:

(4.17)

=

= 12

Assuming that is corresponding to the nthpulse, then = 12 =

= 2 And the time delay between two steps is:

= + =2 ( 1 )

(4.18)


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Finally the expression for the counter delay is found:

(4.19)

When = 0 :

= 12 ( 1 )

(4.20) =1

2

= ( 1 )In order to reduce the burden of calculations, we use the Taylor

series approximation


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(4.21)Finally,

=

( 1 )( 1)

4 14 1

= 24 1

This calculation is much faster than the double square root, but

introduces an error at n =1. A way to compensate for this erroris by multiplying with 0,676.


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DC servo motorA trajectory generation algorithm is essential for optimum

motion control. A linear piecewise velocity trajectory is

implemented in this application.

For a position move, the velocity is incremented by aconstant acceleration value until a specified maximum

velocity is reached.

The maximum velocity is maintained for a required amount

of time and then decremented by the same acceleration(deceleration) value until zero velocity is attained.

The velocity trajectory is therefore trapezoidal for a long

move and triangular for a short move where maximum

velocity was not reached.


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The SetMovesubroutine is invoked once at the beginning of

a move to calculate the trajectory limits.

velocity

limit

shortmove

long

moveSlope =

Accel.

limit

velocity

time

The DoMoveroutine is then invoked at every sample time

to calculate new desiredvelocity and position values as

follows (from Eq. (4.14))


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= . = 12 .

In position mode, the actual shape of the velocity profile

depends on the values of velocity, acceleration, and the size of

the movement.

The block diagram of position controller

Kp PI DC 1TrajectoryGenerator

Position feedback

Velocity feedback

(4.22)


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4.3 Continuous Path Motion

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In many applications, the servo systems (Robots, CNC) have

to be controlled in a predefined continuous path.

=

=

= 0 =

The problem is to determine the trajectory that goes through N

path points with the constraint conditions. There are three

cases that usually use in trajectory generator:


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4.3 Continuous Path Motion

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The velocities, , at the path points are predetermined The velocities, , at the path points are calculated by a

specific criterion.

Guarantee the continuity of acceleration, , at the pathpoints.

For example, in the case of the velocities are predetermined

Assuming that the trajectory has N path points, then the third

order polynomialis adopted to interpolate two adjacent points

(, +) . Each polynomial has to satisfy the followingconditions: = + = + = + = +

(4.23)

servo drive systems_chapter 4_part1

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