Measuring rotational speed for Tachometer vs Oscilloscope

In summary, the conversation discusses an issue with the oscilloscope showing double the value of rotational speed compared to the tachometer. Possible explanations are discussed, such as the hall effect sensor only sensing one polarity and the motor type affecting the frequency. It is suggested to change the gain of the oscilloscope to fix the problem and to measure the period of two cycles of the signal to accurately calculate the frequency.
  • #1
Special One
32
1
Homework Statement
f = 64.3 [Hz]
w = 1930 [RPM]
Relevant Equations
w=2*pi*f
Question: Why does the oscilloscope double almost the exact value of rotational speed measured by Tachometer?

Rotational speed from Tachometer = 1930 [RPM]
Frequency of 1 period = 64.3 [Hz] which means 3857.91 [RPM]

The output waveform of hall-effect sensor is attached.
Can you have any explanation please?
 

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  • #2
We need a link to the experimental setup.

A magnet has two poles, but it appears the oscilloscope is seeing only one polarity, so that should be OK. What is the hall effect sensor sensing? How is the magnet mounted?

What type of tachometer are you using?
Maybe it is locking to a harmonic of the rotation.

What type of motor is it?
At 65 Hz, it is probably not an AC induction motor.
The distributor of a 4 stroke motor, where there might be a Hall effect sensor, rotates at half the speed of the crankshaft.
 
  • #3
Baluncore said:
We need a link to the experimental setup.

A magnet has two poles, but it appears the oscilloscope is seeing only one polarity, so that should be OK. What is the hall effect sensor sensing? How is the magnet mounted?

What type of tachometer are you using?
Maybe it is locking to a harmonic of the rotation.

What type of motor is it?
At 65 Hz, it is probably not an AC induction motor.
The distributor of a 4 stroke motor, where there might be a Hall effect sensor, rotates at half the speed of the crankshaft.
The answer is because it puts out 2 pulses for every rotation.
So, in order fix this problem we should change the gain in oscilloscope from
+60.000,000 to +30.000,000
Now we will be getting the exact value as it is in the tachometer.
 
  • #4
Special One said:
The answer is because it puts out 2 pulses for every rotation.
What is "it", and why does it "put out" two pulses per rotation ?

Special One said:
So, in order fix this problem we should change the gain in oscilloscope from
+60.000,000 to +30.000,000
The gain of an oscilloscope sets the vertical scale, not the sweep rate.

I would not assume that the two pulses per rev are equally spaced. So you should measure the period of two cycles of the signal with the oscilloscope, then take the reciprocal to convert that to frequency.
 

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