步进电机频率缓入曲线

Question

I am writing some software to control a five phase stepper motor. The speed of the stepper motor is controlled by the frequency of a pulse I am sending to the motor controller. My present challenge is that I want to ease-in and ease-out of movements. I am actually replicating the behaviour of some old software which I do not have the source code for. I have an understanding of the logic of the easing, and it behaves as so:

1. when the ending / sustained speed is slow, the ease-in is slower / takes longer
2. when the ending / sustained speed is fast, the ease-in is faster / is shorter

For example… when the sustained speed is 693 Hz, the ease-in is 766 milliseconds long. I have sampled this ease-in curve using a Saleae logic analyzer. Here is the curve:

The starting frequency is 97.77 Hz. Here is a link to the actual data . So I am trying to figure out how to implement the proper logic / formula for this in code. The following bit of code will spit out increments of Hz that are relatively close to the increments I need, but the thing I can't figure out is how to get it to repeat/hold the same current_freq for an increasingly long duration of time – which is what essentially creates the curve that you see in the graph. My multiplier creating the increments is also off, but it is relatively close…

** edit – I think the below in theory works as far as adding a dimension of incrementing time to hold the stepping up current_freq , but there's something wrong with my implementation... it's just doing each frequency once.

current_freq = 97.
end_freq = 1134

t = 4 
# number of times to send the current freqency

print current_freq

while current_freq < end_freq:
    i = 1
    t = t+t * .1673
    print i
    while i <= t:
        print current_freq
        i = i+1
        break
    current_freq = current_freq + current_freq * .1673

Any ideas? Is this a logarithm? Sin or cos? In case it isn't blatantly obvious I am horrible at math.

Answer 1

If you are trying to print out the new frequency multiple times, just remove the line with break .

This code produces the output that follows.

current_freq = 97.
end_freq = 1134

# number of times to send the current freqency
t = 4

while current_freq < end_freq:
    i = 1
    t = t + t * .1673
    print("t=%10f" % (t))
    while i <= t:
        print("i=", i, end=' ')
        print("  freq %.2f" % current_freq)
        i = i + 1
        #break
    current_freq = current_freq + current_freq * .1673

Output (truncated)

t=  4.669200
i= 1   freq 97.00
i= 2   freq 97.00
i= 3   freq 97.00
i= 4   freq 97.00
t=  5.450357
i= 1   freq 113.23
i= 2   freq 113.23
i= 3   freq 113.23
i= 4   freq 113.23
i= 5   freq 113.23
t=  6.362202
i= 1   freq 132.17
i= 2   freq 132.17
i= 3   freq 132.17
i= 4   freq 132.17
i= 5   freq 132.17
i= 6   freq 132.17
t=  7.426598
i= 1   freq 154.28
i= 2   freq 154.28
i= 3   freq 154.28
i= 4   freq 154.28
i= 5   freq 154.28
i= 6   freq 154.28
i= 7   freq 154.28
t=  8.669068
i= 1   freq 180.10
i= 2   freq 180.10
i= 3   freq 180.10
i= 4   freq 180.10
i= 5   freq 180.10
i= 6   freq 180.10
i= 7   freq 180.10
i= 8   freq 180.10
t= 10.119403
i= 1   freq 210.22
i= 2   freq 210.22
i= 3   freq 210.22
i= 4   freq 210.22
i= 5   freq 210.22
i= 6   freq 210.22
i= 7   freq 210.22
i= 8   freq 210.22
i= 9   freq 210.22
i= 10   freq 210.22
t= 11.812379
i= 1   freq 245.40
i= 2   freq 245.40
i= 3   freq 245.40
i= 4   freq 245.40
i= 5   freq 245.40
i= 6   freq 245.40
i= 7   freq 245.40
i= 8   freq 245.40
i= 9   freq 245.40
i= 10   freq 245.40
i= 11   freq 245.40
.
.
.