Calculating DC Offset for 19 High Bits Followed by Low Bit

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SUMMARY

The discussion focuses on calculating the DC offset for a sequence of 19 high bits followed by a low bit. The correct formula for determining the DC offset is derived as [(19 - 1) * (the amplitude)] / 20, which represents the average value of the waveform. Participants clarify that the DC offset relates to the symmetry of the waveform around zero, and if the low value is zero, the DC offset equals 50% of the high value. This understanding is crucial for accurate waveform analysis.

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Chrispp
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If I have a string of 19 continuous bits 'high' followed by a bit 'low', how do I calculate the DC Offset if the pattern is repeated? Is it just [(19 - 1)*(the amplitude)]/(20)?

Thanks for the help.
Chris
 
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Yes, that looks correct. You are just calculating the average value -- that would be the DC offset.
 
DC offset??
Don't you mean equivilent DC voltage value?
This would be the RMS value in a sine wave.

AFAIK DC offset has to do with symmetry of the waveform around 0.
It would be 50% of the high value if the low value is 0.
 

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