Calculating DC Offset for 19 High Bits Followed by Low Bit

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To calculate the DC offset for a sequence of 19 high bits followed by one low bit, the average value is determined by the formula [(19 - 1)*(amplitude)]/(20). This calculation reflects the DC offset, which is essentially the average value of the waveform. The discussion clarifies that DC offset relates to the symmetry of the waveform around zero, and if the low value is zero, the DC offset would be 50% of the high value. The RMS value is also mentioned in the context of sine waves, indicating a distinction between DC offset and RMS calculations. Understanding these concepts is crucial for accurate waveform analysis.
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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