Reducing noise in strain gauge using an RC circuit

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SUMMARY

This discussion focuses on designing an RC low pass filter to reduce noise from a motor operating at 40Hz, affecting the signal from a strain gauge monitoring periodic load changes at 1/minute. A 90% reduction in noise corresponds to a -20dB attenuation, which can be achieved by selecting appropriate resistor (R) and capacitor (C) values. The suggested approach involves starting with a capacitor value of 100 μF and calculating the corresponding resistor value using the corner frequency formula. Additionally, participants emphasize the importance of verifying signal magnitudes through the transfer function of the filter.

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liam1992
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Vibrations of a motor (40Hz) cause noise on the signal of a strain gauge that monitors load on a cantilever that measures periodic load changes at the rate of 1/minute. Design a filter that reduces the signal caused by the vibrations by 90%.

I thought about a low pass filter that has a cut off frequency of 40Hz but I'm not sure about the reduction by 90% bit. Is there an equation linking frequency to the gain of a circuit because a 90% reduction would be -20dB according to gain=20log(Vout/Vin) or -10dB if you used gain=10log(Pout/Pin). I am unsure how to do this and any help would be greatly appreciated.
 
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liam1992 said:
Vibrations of a motor (40Hz) cause noise on the signal of a strain gauge that monitors load on a cantilever that measures periodic load changes at the rate of 1/minute. Design a filter that reduces the signal caused by the vibrations by 90%.

I thought about a low pass filter that has a cut off frequency of 40Hz but I'm not sure about the reduction by 90% bit. Is there an equation linking frequency to the gain of a circuit because a 90% reduction would be -20dB according to gain=20log(Vout/Vin) or -10dB if you used gain=10log(Pout/Pin). I am unsure how to do this and any help would be greatly appreciated.

I suppose it makes a difference if the question wants exactly 90% reduction in the motor noise signal (is that voltage or power?), or whether they mean at least 90%.

If the true signal from the sensor has a frequency of 1/min ~ 0.017 Hz and the motor noise is at 40Hz then there's more than two decades of frequency to play with; You can afford to put your sensor signal closer to the center of the pass-band of your filter rather than have it degraded by the "turn" at the cutoff corner. So, for example, you might stick your corner frequency at 40Hz/100.

For an RC filter a simple approach is to choose a 'likely' value for the capacitor; for low frequencies like this start with a large but manageable capacitor value -- say 100 μF. Then use the formula for the corner frequency to find a corresponding R value. If you don't like the values you get, pick another C and try again.

Write the transfer function for the filter and verify your signal magnitudes.

You could also use the transfer function itself to choose an appropriate co-component (the R for a given C, or C for a given R) by setting the transfer function magnitude expression equal to the desired gain (attenuation) for the signal to be suppressed.
 

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