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Quoted from the ebook Introduction to Microcontroller by Gunther Gridling (pg. 43):

The class width of most classes corresponds to 1 lsb, with the exceptions of the first class (0.5 lsb) and the last class (1.5 lsb). This asymmetry stems from the requirement that the representative of the code word 0 should correspond to 0 V, so the first class has only half the width of the other classes, whereas the representative of the code word 2r − 1 should be Vref − 1 lsb to allow easy and compatible expansion to more bits. To avoid the asymmetry, we could for example use the lower bound of the class as its representative. But in this case, the worst case error made by digitization would be +1 lsb. If we use the midpoint, it is only ± 0.5 lsb

I understand that lowering Vref or using a higher bit ADC can increase the resolution of conversion or make each quantized step of 1 lsb smaller to associate with more digital/binary states. Each lsb or level is named a "class" (I don't understand why there are multiple terms to represent the same abstraction).

But why is there a 0.5 lsb or 1.5 lsb class? There is no binary state between 000 and 001 to associate with 0.5 lsb, and the digital output signal can only be integral value of lsb. So why is it there and what does it mean? If there is a 0.5 lsb class/level at the start and another 1.5 lsb class/level at the end, does that make "this" symmetric instead of of asymmetric?

Figure 2.17

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    \$\begingroup\$ 0.5 lsb is half the value of 1 lsb. \$\endgroup\$ – The Photon Jul 14 at 4:46
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1 LSB is the analog quantity corresponding to one step size in an ADC.

You write:

There is no binary state between 000 and 001 to associate with 0.5 lsb, and the digital output signal can only be integral value of lsb.

Correct. As per your picture, the step size is Vref/8. For instance digital output 010 is for analog input values of 1.5 to 2.5 LSB. That range is 1.0LSB wide.

The range of analog values that map to 000 is from 0V to Vref/16, and that's only 0.5 LSB wide.

The threshold to flip from 000 to 001 is at Vin = 0.5 LSB, so Vref/16.

But the next threshold is at Vref/16+Vref/8

The quote then goes on to say that asymmetry refers to a difference between the range at the very bottom (at 0.5LSB) vs the range at the very top top. More on that in a moment.

Technically you could set the first threshold at 0, and not 0.5LSB, but this has 2 problems.

Since in practice, there is no perfect 0V input (there is always some noise a minor offset etc..), this makes little sense. Even the smallest amount of input voltage would map to 001, which means you'd practically never have the bottom value, and in effect you have lost one LSB of resolution (the digital "000" practically never occurs, and the practical span is only from "001" to "111"). Although you could argue that this is fundamentally a poor design decision, ultimately it is only a very small loss.

Secondly, if you have a signed ADC (accepting and mapping negative input voltage to negative digital values), and you have a threshold at 0, then for instance an input of +1/10 LSB would map to digital 0 (0000), but -1/10 LSB would map to digital -1 (1111), and this introduces an undesired bias of 0.5 LSB on the digital side: a signal that fluctuates but has no bias on the analog side will appear to have a bias on the digital side.

Now more on the asymmetry: in my designs I am more concerned with the asymmetry around 0V, rather than the asymmetry between the step size around Vmax vs the step size around V=0. I would prefer negative values to be mapped symmetrically to positive values: you should be able to negate the input signal and get the exact same digital signal, but then negated.

That being said, in all higher resolution signal processing, and high resolution ADC (10 to 15 bits) that bias is really not something anyone worries about: there are plenty of sources for bias along the analog signal path leading to the ADC and they have to be digitally dealt with anyway, if not AC decoupled at the ADCs input.

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    \$\begingroup\$ Where you say "lost one bit resolution" I think you mean "lost one lsb resolution" or "lost one count resolution". Losing one bit of resolution would reduce the number of different codes by half. \$\endgroup\$ – The Photon Jul 14 at 5:29
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    \$\begingroup\$ Oh yes, just a LSB not a whole bit! Fixed. Thanks @ThePhoton, and that's why it is not always a big deal. \$\endgroup\$ – P2000 Jul 14 at 5:33
  • \$\begingroup\$ The choice of words still throwing me off here....So does the "asymmetry" here means (A) having a 0.5lsb threshold for positive signal but none for negative signal (i.e. a tiny fluctuation below 0V will fall to a negative 001 or 111 for 2's complement on the y-axis). Or (B) the 0.5 width for the lowest 000 binary and a wider 1.5 width for the last or top 111 binary hence the "asymmetry"? \$\endgroup\$ – KMC Jul 15 at 2:07
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    \$\begingroup\$ Well, @KMC, it is an odd choice of words, including the use of the term "class". If this bothers you, but you understand the concepts in both answers here (Elliot's included), I can tell you from many years of professional experience that the terminology used in this case does not matter. If you want to understand issues around the choice of threshold, quantization error, bias, saturation etc... please ask more. Don't worry about what is meant with asymmetry. \$\endgroup\$ – P2000 Jul 15 at 3:00
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The offset of 0.5 LSB is used to control the quantization error.

Suppose you provide an input voltage to an ideal ADC then take the digital output and use it as the input to an ideal DAC. The difference between the input voltage (which is continuous) and the output voltage (which has discrete steps) is the quantization error.

Suppose you start with an input voltage of 0V and gradually increase the voltage. Without the offset, the quantization error starts at 0V and increases to 1 LSB. Then the ADC goes to the next step and the error falls to 0V again. So, the error ranges from 0 to 1 LSB and has a mean value of 0.5 LSB.

If you have the 0.5 LSB offset then the quantization error ranges from -0.5 LSB to +0.5 LSB and has a mean value of 0. So, adding the offset gives you a mean error of 0 instead of 0.5 LSB.

Does this matter in practice? It depends entirely on what you plan to do with the digitized information.

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