Algorithm to determine the effective "phase difference" between two signals with different frequencies?

Answer 1

The overall signal, i.e. the signal that you get when you add the red and the blue, has a phase length of 16 times the blue and 9 times the red signal. You could measure the time difference between every 16th blue and every 9th red rising flank.

I guess that what you want to measure is the wear on the gears. I think that this measurement might be influenced (introducing noise) by the gears' tolerance, if there is no uniform traction.

Answer 2

I think the best thing to do would be to create an X-Y diagram of all teeth pair timings. You'd arbitrarily pick one tooth on each cog as T=0..

Answer 3

Hi, I think it is even simpler than that.

Every 16*9 samplings(of the big cog) the wheels are in the exact same spot they started.

So what you do is the following:

• pick any point in time with a sampling on the big cog. Measure the amount of time before you sample the small cog too. Remember this value.

• every 16*9 samplings of the big cog (why does this sound dubious?) do the same measurement again and compare it to your base value. When the timing begins to shift, you have a problem.

R

Answer 4

I am having some trouble visualizing your hardware setup. And the behavior you are trying to detect. A shaft slipping? Teeth wearing?

In any case, I would write a simulation of the situation so that I can get some, perhaps exaggerated, results with no noise to test algorithms against.

The algorithms I would test would be variations of the following:

Assign signal with lowest frequency to A

Time A’s rising edge. =>  Ta1

Time the next B rising edge . =>  Tb1

Calculate time Tb1 – Ta1    =>  dT1

Time next A’s rising edge. => Ta2

Time the next B rising edge. =>  Tb2

Calculate time Tb2 – Ta2    =>  dT2

Calculate second order difference  dT2 – dT1  => d2T1

Repeat precious steps to get another second order difference  => d2T2

If sign of d2T1 and d2T2 are different, 

repeat previous steps

else sign of d2T1 and d2T1 are same

    calculate third order difference  d2T2 – d2T2  =>  d3T1

Repeat previous steps to get another 3rd order difference  =>  d3T2

If d3T2 – d3T1 > max noise

    Raise alarm

Answer 5

Since we are talking about "phase", then it seems reasonable to measure the "beat" which occurs when the two waveforms reinforce each other.

Something like this, perhaps:

void cog_phase_monitor2( int cog, int t )
{
    static int last_a, last_b, last_beat, last_beat_delta = 0;;
    int beat = 0;
    if( cog == 1 ) {
     if( t - last_b < 1 )
      beat = 1;
     last_a = t;
    }
    if( cog == 2 ) {
     if( t - last_a < 1 )
      beat = 1;
     last_b = t;
    }
    if( beat ) {
     printf("**** delta beat %d \n",t-last_beat);
     if( last_beat_delta ) {
      if( last_beat_delta != t-last_beat ) {
       printf("!!!Warning beat just changed !!!\n");
       last_beat_delta = 0;
      }
     } else {
      last_beat_delta = t-last_beat;
     }
     last_beat = t;
    }

}

Now if we plug this into a simulation of two cogs, one of 9 teeth and one of 16 teeth, both turning at 10 revs per second

B at 6 msecs
A at 11 msecs
B at 12 msecs
B at 18 msecs
A at 22 msecs
B at 24 msecs
B at 30 msecs
A at 33 msecs
B at 36 msecs
B at 42 msecs
A at 44 msecs
B at 48 msecs
B at 54 msecs
A at 55 msecs
B at 60 msecs
A at 66 msecs
B at 66 msecs
**** delta beat 66
B at 72 msecs
A at 77 msecs
B at 78 msecs
B at 84 msecs
A at 88 msecs
B at 90 msecs
B at 96 msecs
A at 99 msecs
B at 102 msecs
B at 108 msecs
A at 110 msecs
B at 114 msecs
B at 120 msecs
A at 121 msecs
B at 126 msecs
A at 132 msecs
B at 132 msecs
**** delta beat 66
B at 138 msecs
A at 143 msecs
B at 144 msecs
B at 150 msecs
A at 154 msecs
B at 156 msecs
B at 162 msecs
A at 165 msecs
B at 168 msecs
B at 174 msecs
A at 176 msecs
B at 180 msecs
B at 186 msecs
A at 187 msecs
B at 192 msecs
A at 198 msecs
B at 198 msecs
**** delta beat 66

And now if we add a delay of 1 msec to one of the cogs:

B at 6 msecs
A at 11 msecs
B at 12 msecs
B at 18 msecs
A at 22 msecs
B at 24 msecs
B at 30 msecs
A at 33 msecs
B at 36 msecs
B at 42 msecs
A at 44 msecs
B at 48 msecs
B at 54 msecs
A at 55 msecs
B at 60 msecs
A at 66 msecs
B at 66 msecs
**** delta beat 66
B at 72 msecs
A at 77 msecs
B at 78 msecs
B at 84 msecs
A at 88 msecs
B at 90 msecs
B at 96 msecs
A at 99 msecs
B delayed at 102 msecs
B at 103 msecs
B at 109 msecs
A at 110 msecs
B at 115 msecs
A at 121 msecs
B at 121 msecs
**** delta beat 55
!!!Warning beat just changed !!!
B at 127 msecs
A at 132 msecs
B at 133 msecs
B at 139 msecs
A at 143 msecs
B at 145 msecs
B at 151 msecs
A at 154 msecs
B at 157 msecs
B at 163 msecs
A at 165 msecs
B at 169 msecs
B at 175 msecs
A at 176 msecs
B at 181 msecs
A at 187 msecs
B at 187 msecs
**** delta beat 66
B at 193 msecs
A at 198 msecs
B at 199 msecs
B at 205 msecs

This seems like a hopeful beginning :-)