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Differential Nonlinearity: What happens when DNL = -1LSB
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[QUOTE="Peter Alexander, post: 6125308, member: 635102"] [h2]Homework Statement [/h2] Question is simple: what happens when ##DNL = -1 LSB## where DNL signifies differential nonlinearity and LSB stands for Least Significant Bit. It is also required to try and sketch such condition. [h2]Homework Equations[/h2] Equation for differential nonlinearity: $$DNL(i) = \frac{V_{out}(i) - V_{out}(i - 1)}{\text{ideal LSB step width}} - 1$$ [h2]The Attempt at a Solution[/h2] I understand what DNL is, is represent a deviation between two analog values corresponding to adjacent input digital values (from Wikipedia). Ideally, two sequential digital codes should belong to analog values that are 1LSB apart, so the deviation from such step is called DNL. At ##DNL \leq -1LSB##, missing codes appear in the transfer function. Those are binary representations that have no such analog signal to cause them. I am having troubles visualizing such case. What are the conditions that have to be met in order for this to occur? Is such case an example of bad analog-digital converter, or is it something that can happen to properly functional instruments as well? [/QUOTE]
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Differential Nonlinearity: What happens when DNL = -1LSB
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