Why there is dc component in square signal

[giulioo]

hi!
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.

 

[giulioo]

or probably it is due to internal impedance of the oscilloscope?

 

[tech99]

hi!
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.

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".

 

[giulioo]

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!

 

[Svein]

It was +-1 V, while dc component was something like 17 mV.

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.