What counter method to use for measuremen​ts on combustion engine

Question

Hello,
Here is my dilemma. I want to measure the rpm of crank/cam shaft and thus I need to determine which of the three methods to use to get accuracy +-5rpm and couple of readings during one revolution of the shaft.
The hardware: NI PCI 6120-two counters
IRC 360-3600 pulses
Lets say I want to measure the range of rpm of the crankshaft from 100rpm(1.7Hz) to 10000rpm(166.7Hz). If I use the IRC with 720pulses I get the frequency range I will be measuring 1200Hz-120kHz assuming the IRC is on the crankshaft.
So my question is which of the methods to use? I did some calculations to determine that
1) 1 counter - the calculated rpm if the real rpm was 6000 would be between 5973-6017 which is not good enough
2) 2 counters, high freq - I would have to be using the frequency of the gate around 1Hz to get desired accuracy at 100rpm which is not suitable since I also want to have more readings through one revolution - do I understand it right? That would mean for example at 6000rpm I would have to wait 100revolutions to get one rpm reading, right?
3) seems as the right choice but still the acuracy is not good enough. If I use the divider of 10 it gives me 72readings per one revolution if the IRC is set to 720 pulse. If I measure at 6000rpm the diversion is then +-3rpm. But with 10000rpm and 3600pulses per revolution of the irc the diversion is already +-40
What method is used for this kind of measurement? How to improve  accuracy?
Thanks
Message Edited by ceties on 07-09-2009 03:12 PM
LV 2011, Win7

Answer 1

Hi Ceties,
Yes, if you want to measure 600 kHz signal, the error will increase. I thought you were intend to measure upto 120 kHz.
In Method III, we do the following:
We have a known time period (1 ms), we have a frequency to measure (100 kHz), we will know how many pulses are going to be in that period ([100*e3]*[1*e-3]=100). First, we would like to divide this frequency with the first counter: we generate another known pulse train for 1 ms period with i.e. 500 pulses. In this case the rate will be 100/500 = 1/5.
Know, we are able to measure a lower frequency signal (100 kHz*1/5 = 20 kHz) with Method I. Which results a much more accurate measurement. That is the theory behind Method III.
Pulses per revolution to read will remain the same in each case: (frequency to measur)/(basis clock); in different methods we are trying to increase accuracy with measuring either a longer period (Method II) or diveded frequency (Method III).
I hope it will help moving forward with your project!
Regards,
Matyas