Electronic – Why would a sensor output an inverted signal

inverting-amplifieroutputplotsignal

I have a sensor that has 3 components: X, Y and Z and for every component there are 2 channels:

  • X+ :0 V ± 5 V voltage output
  • X- :0 V ± 5 V voltage output inverted

All I know is :

  • If the circuit uses "active high" logic, 5 volts represents a digital "1" and 0 volts represents a digital "0".
  • If the circuit uses "active low" logic, 5 volts represents a digital "0" and 0 volts represents a digital "1".

My question is: what's the purpose of the inverted output? How can I make use of it as I am plotting results.

Best Answer

If you are describing an encoder - not a sensor then your outputs are probably:

X+: 0 and 5V. (not ± 5 V) X-: X+ inverted.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A typical A-B-Z quadrature encoder output waveform with differential outputs.

The inverted output gives the ability to improve noise immunity in the presence of electrical noise.

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Figure 2. This diagram illustrates how a differential encoder is able to ignore noise. Channel Ā is the inverse of channel A, generated inside the encoder. However, when noise is introduced along the wire between the encoder and the encoder interface, the noise will be almost identical on both channels. Since the noise is not inverted on channel Ā, the differential encoder interface can perform a few simple operations to filter out the noise. Source: Phidgets.