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Reducing noise in strain gauge using an RC circuit

  1. Apr 9, 2012 #1
    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.
     
  2. jcsd
  3. Apr 9, 2012 #2

    gneill

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    Staff: Mentor

    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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