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Why there is dc component in square signal

  1. Mar 22, 2015 #1
    when I send a square wave to a digital oscilloscope it detects the signal + dc signal. I wanted to know why. I was thinking about internal resistance of coaxial cable i was using.

    signal got out of an oscilloscope/wave generator and in channel A of the same oscilloscope/wg. It was +-1 V, while dc component was something like 17 mV.
  2. jcsd
  3. Mar 22, 2015 #2
    or probably it is due to internal impedance of the oscilloscope?
  4. Mar 22, 2015 #3


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    Gold Member

    This sounds like a small error in the generator. You can remove all DC component by placing a large capacitor in series with the oscilloscope, or selecting "AC input". It is possible that the square wave is not exactly "50/50".
  5. Mar 22, 2015 #4
    well the wave is for sure not exactly 50/50 because of rise time. I haven't thought about it anyway. I will meditate upon it, i mean, rise time could add a dc component.
    Do you think input and output resistance could not affect signal in such way too?
    I already removed the dc component using an offset.
    thank you for your advice!
  6. Mar 22, 2015 #5


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    Science Advisor

    17mV DC error in a signal of ±1V - that is an error of ≈ 1%. I would not be able to see that on an oscilloscope. And, as you said, the signal generator might easily give an error of that size.
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