Chapter 7 Usage of Various Functions
7-
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7) pulse out operation explanation
Condition 1)
Set up as acceleration slop = 1, max. frequency = 1000, no of pulse out = 5000.
x
If as acceleration slop = 1, 1 pulse is output on the 1st step (velocity: 50pps).
Pulse velocity is 50pps, so time consuming is 20ms.
y
2 pulses are output on the 2nd step (velocity: 100pps) and time consumes 20ms
z
By calculation in the same way, the time to reach to 1000pps is
20ms x (20-1) = 380ms, and the no. of output pulses are 1+2+3...+18+19 = 190 units.
{
Decreasing velocity inclination is 1, thus 190 units of pulses are needed.
|
The no. of pulses in the uniform velocity region are 5000-190-190=4,620 units.
}
Whole spent time is 50,380ms
Time
velocity
Acceleration time:380ms
Accelerating pulses:190
Deceleration
20ms
50pps
Acceleration step : 19
Uniform velocity
Time :4,620ms
Pulses :4,620
Deceleration time :380ms
Decelerating pulses:190
1st step
2
nd
step
example: when acceleration is 1.
