Question 1 of 5: 8-bit Analog-to-Digital Conversion — Resolution, Codes and Quantisation Error

Question 1: 8-bit Analog-to-Digital Conversion — Resolution, Codes and Quantisation Error (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) the volt-per-code step size, (b) and (c) the output words produced by two specific inputs, (d) the quantisation error committed on the 6.200 V sample expressed three ways, and (e) the worst quantisation error the converter can ever make, referred to full scale.

Approach. Fix the LSB from the range and the word length, convert each sample by dividing by the LSB and taking the integer part, then read the error as the distance from the sample back to the level the code represents.

1. Part (a) — divide the span by the number of codes. An n-bit converter partitions the full-scale range into $2^{n}$ equal steps, so the resolution, or one least significant bit, is $$Q = \text{LSB} = \frac{FSR}{2^{n}} = \frac{10\ \text{V}}{2^{8}} = \frac{10}{256}$$ which evaluates to $\boxed{Q = 39.0625\ \text{mV} = 0.0390625\ \text{V}}$. This is the number a data-sheet quotes as the resolution, and every later part of this question is arithmetic on it.
2. Note the competing convention, and why it does not change the answers. Some texts divide by $2^{n}-1$ instead of $2^{n}$, on the grounds that 255 steps separate the 256 code centres; that gives $10/255 = 39.216$ mV. The difference is one part in 256, which is far too small to move either of the codes found below, so the answers are robust to the choice. The $FSR/2^{n}$ form is used throughout because it is the one implied by a converter whose transfer function runs from 0 to $FSR - 1\,\text{LSB}$.
3. Part (b) — convert 6.000 V by dividing and truncating. An ideal converter outputs the largest code whose level does not exceed the input, so $$N_{1} = \left\lfloor \frac{v_{1}}{Q} \right\rfloor = \left\lfloor \frac{6.000}{0.0390625} \right\rfloor = \lfloor 153.6 \rfloor = 153$$ and 153 in eight bits is $\boxed{N_{1} = 153_{10} = 1001\,1001_{2} = \text{99}_{16}}$. The level that word stands for is $153 \times 0.0390625 = 5.9766$ V, so this sample is already carrying an error of 23.4 mV even though it is a whole number of volts.
4. Part (c) — repeat for 6.200 V. The same division gives $$N_{2} = \left\lfloor \frac{6.200}{0.0390625} \right\rfloor = \lfloor 158.72 \rfloor = 158$$ so the converter returns $\boxed{N_{2} = 158_{10} = 1001\,1110_{2} = \text{9E}_{16}}$. Note that a 200 mV increase in the input moved the code by five counts, consistent with $200/39.0625 = 5.12$ — a useful arithmetic check before going on.
5. Part (d) — the error is the distance back to the code level. The word 158 represents the voltage $N_{2}Q = 158 \times 0.0390625 = 6.171875$ V, so the quantisation error committed on this sample is $$\varepsilon = v_{2} - N_{2}Q = 6.200 - 6.171875 = 0.028125\ \text{V}$$ that is, $\boxed{\varepsilon = 28.125\ \text{mV}}$, which is $0.028125/0.0390625 = 0.72$ of one LSB. It is positive because a truncating converter always rounds down.
6. Express the same error against the two references the question names. Referred to the reading itself, and then to the full-scale range, $$\begin{aligned} \frac{\varepsilon}{v_{2}} &= \frac{0.028125}{6.200} = 0.004536 \\[2pt] \frac{\varepsilon}{FSR} &= \frac{0.028125}{10} = 0.0028125 \end{aligned}$$ so the error is $$\boxed{0.454\ \text{percent of the input, and } 0.281\ \text{percent of full scale}}$$ The two differ because the sample sits at 62 percent of full scale; the percent-of-reading figure would grow without limit as the input approached zero, which is exactly why converter accuracy is always specified against full scale.
7. Part (e) — bound the error over the whole input range. For a truncating quantiser the input may sit anywhere on a tread, so the error runs from zero up to (but never reaching) one full LSB. The largest possible value is therefore $$\frac{\varepsilon_{max}}{FSR} = \frac{Q}{FSR} = \frac{1}{2^{n}} = \frac{1}{256}$$ giving $\boxed{\varepsilon_{max} = 1\ \text{LSB} = 0.391\ \text{ percent of full scale}}$.
8. State the rounding variant, because most real converters use it. If the comparator thresholds are offset by half a step so that the converter rounds to the nearest level rather than truncating, the error becomes symmetric on $\pm Q/2$ and the worst case halves to $0.5\,\text{LSB} = 0.195$ percent of full scale. Since the question gives no offset, the truncating figure is the answer, but a candidate who states both is showing exactly the understanding the examiner is testing.