Robot Control Resolution and Memory Capacity: 22-Mec-B4 Integrated Manufacturing Systems, December 2018, Q7

Question 7: Robot Control Resolution and Memory Capacity (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. Two single-axis problems on the same relation between stroke, register width and the finest increment the controller can command.

Given data
Part	Quantity	Symbol	Value
(a)	Total range of the telescoping axis	$L$	0.7 m
(a)	Storage capacity for that axis	$B$	12 bits
(b)	Total range of the orthogonal slide	$L$	1.2 m
(b)	Required control resolution	CR	0.5 mm

Find. (a) the control resolution of the telescoping axis; (b) the number of bits the control memory must hold for the slide.

Figure 7.1 — The quantised axis. A $B$-bit register addresses $2^{B}$ distinct positions, which divide the stroke into $2^{B} - 1$ equal steps; the control resolution is the length of one step. Only 33 of the 4,096 addresses are drawn.

Approach. A $B$-bit register can hold $2^{B}$ distinct binary numbers, so it can address $2^{B}$ distinct positions along the axis. Those positions are the endpoints of $2^{B} - 1$ equal intervals spanning the stroke, and the control resolution is the length of one interval. Part (a) evaluates that expression; part (b) inverts it and rounds the bit count upward, because bits come only in whole numbers and the specification is a maximum.

Part (a) — count the addressable positions. With $B = 12$ bits allocated to this axis, $$\text{number of addressable points} = 2^{B} = 2^{12} = 4096$$ These 4,096 points include both ends of the stroke, so they divide the 0.7 m range into $4096 - 1 = 4095$ equal increments.
Part (a) — divide the stroke by the number of increments. $$\text{CR}_1 = \frac{L}{2^{B} - 1} = \frac{0.7}{4095} = 1.7094 \times 10^{-4} \text{ m}$$ $$\boxed{\;\text{CR}_1 = 0.171 \text{ mm}\;}$$ This is the control resolution — the finest increment the controller can command. The robot's spatial resolution also includes the mechanical errors of the arm, and its accuracy is conventionally taken as half the control resolution plus those errors, so about 0.085 mm here before mechanical inaccuracies are added.
Part (b) — invert the same relation. Requiring $\text{CR} \le 0.5$ mm $= 5 \times 10^{-4}$ m over $L = 1.2$ m, $$2^{B} - 1 \ \ge \ \frac{L}{\text{CR}} = \frac{1.2}{0.0005} = 2400 \quad\Longrightarrow\quad 2^{B} \ge 2401$$ $$B \ \ge \ \log_2 2401 = \frac{\ln 2401}{\ln 2} = 11.229$$
Part (b) — round up and confirm. Storage comes in whole bits and the requirement is a ceiling on the increment, so $B$ must be rounded up: $$\boxed{\;B = 12 \text{ bits}\;}$$ Checking both neighbours settles it. At 12 bits the axis is divided into $2^{12} - 1 = 4095$ steps and $\text{CR} = 1.2/4095 = 2.930 \times 10^{-4}$ m $= 0.293$ mm, comfortably inside the 0.5 mm specification. At 11 bits there are only $2^{11} - 1 = 2047$ steps and $\text{CR} = 1.2/2047 = 5.862 \times 10^{-4}$ m $= 0.586$ mm, which violates it. Twelve bits is therefore the smallest capacity that meets the specification.

Check: $2^{B}$ or $2^{B} - 1$ in the denominator. Groover's control resolution is $\text{CR} = L/(2^{B} - 1)$, counting the intervals between addressable points; some texts write $L/2^{B}$, counting the points themselves. The difference is 0.02 per cent at 12 bits and changes neither boxed answer here: on $L/2^{B}$, part (a) gives 0.1709 mm and part (b) still gives 12 bits, since $2^{11} = 2048 < 2400$. The Groover form is used throughout because it is the convention in the reference text for this subject.

Question 7 — results
Quantity	Value
(a) Addressable points, $2^{12}$	4,096
(a) Increments spanning the stroke, $2^{12}-1$	4,095
(a) Control resolution, $L/(2^{B}-1)$	$1.709 \times 10^{-4}$ m = 0.171 mm
(b) Increments required, $L/\text{CR}$	2,400
(b) Exact bit requirement, $\log_2 2401$	11.23
(b) Storage capacity required	12 bits
(b) Control resolution achieved at 12 bits	0.293 mm (specification met)
(b) Control resolution at 11 bits	0.586 mm (specification failed)

Topic. Robot precision of movement: control resolution, spatial resolution, accuracy and repeatability, and the sizing of the control register.

Key relations. $\text{CR} = L/(2^{B} - 1)$ for a linear joint, and $\text{CR} = R/(2^{B} - 1)$ in angular units for a rotary joint; $B = \lceil \log_2 (L/\text{CR} + 1) \rceil$ when the resolution is specified; spatial resolution $=$ control resolution $+$ mechanical inaccuracies; accuracy $\approx \tfrac{1}{2}(\text{spatial resolution})$; repeatability is a separate, statistically defined band (commonly $\pm 3\sigma$ of the return position).

Why this works. A digital controller cannot command a continuum. The storage register is the bottleneck: whatever the mechanical capability of the arm, the controller can only name $2^{B}$ points along the axis, and the axis is therefore quantised into $2^{B} - 1$ steps whether the servo could resolve more finely or not. Note that resolution is inversely proportional to $2^{B}$, so each additional bit halves the increment — which is why bit counts in this range make such large differences and why rounding up is never seriously expensive.

Common pitfall. Rounding the bit count down or to the nearest integer in part (b): 11.23 rounds to 11 arithmetically, but 11 bits gives 0.586 mm and fails the specification. The requirement is “at least this precision”, so the ceiling is always taken. The second pitfall is confusing control resolution with accuracy or repeatability — they are three different quantities, and only the first is a property of the control memory alone. The third is unit slips: the range is in metres and the resolution in millimetres, and the factor of 1,000 has to be handled explicitly.

Source. M. P. Groover, Automation, Production Systems, and Computer-Integrated Manufacturing, 5th ed., Ch. 8 (industrial robotics: precision of movement — control resolution, spatial resolution, accuracy and repeatability, with worked examples of exactly this form).

Question 7 of 7: Robot Control Resolution and Memory Capacity

Question 7: Robot Control Resolution and Memory Capacity (20 marks)