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I have been measuring signals in an oscilloscope and I have stumbled upon a practical exercise of calculating the DC offset of a signal. Basically what I did was turn on the DC coupling, take the maximum voltage (eg. 2.36 volts) then turn on the AC coupling , take the maximum voltage again (e.g., 560mV) and finally DC offset= 2.360-0.560 Volts=1.8 Volts (or so I thought) Though the exercise continues and states that the DC coupling shows us peak-peak 1.04V and the AC coupling 890mV peak-peak. So my question is the following: If the DC offset is responsible for "lifting the signal" , how is it that the peak-peak is different for the same signal in the DC and AC coupling?

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An oscilloscope's AC coupling feature removes the DC part of the signal by placing a capacitor in series with the signal. This is essentially a High Pass Filter. So some attenuation is expected.

Generic High Pass Filter:

Generic High Pass Filter

In general, AC coupling is a convenient way to keep the scope trace on the screen when observing small signals "riding" on top of larger signals that are not of interest. But, as you pointed out, to make an accurate amplitude measurement, DC coupling should be used along with the time it takes to recenter and adjust the oscilloscope's amplitude to make an accurate assessment of the smaller signal.

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  • \$\begingroup\$ @stf2000 so you are saying that the difference from 1.04V peak-peak on DC coupling from 890 mV peak-peak on AC coupling can be attributed to the attenuation and not something else? \$\endgroup\$
    – user170589
    Jan 4, 2018 at 18:06
  • \$\begingroup\$ This is likely true. You can reinforce your observations by changing the frequency of the signal you are observing. Since this is a High Pass Filter, the lower the frequency the more attenuation you should see when AC coupled. Try lowering the frequency and repeating your test. \$\endgroup\$
    – st2000
    Jan 5, 2018 at 1:02

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