Introduction to Experimental Robotics CSCI 1108 Lecture 16 Review - - PowerPoint PPT Presentation

Introduction to Experimental Robotics CSCI 1108 — Lecture 16 Review (PID, binary rep. constraints)

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PID Controller Review

PID Controller is not an exactly straightforward solution It was developed by Nicolas Minorsky in 1922 for automatic ship steering, by his analysis of behaviour

• f a helmsman (person steering a ship)

Previous development is mostly P-control:

◮ Christiaan Huygens 17th century regulate gap in

millstones in windmills based on speed

◮ James Watt “conical pendulum” governor for steam

engine, patented 1788

◮ etc. CSCI 1108 Lecture 16 2 / 11

PID in Hardware: Pneumatic PID controller

P, I, and D terms adjusted by the dials at the top

By Snip3r at Dutch Wikipedia, CC BY-SA 3.0, https: //commons.wikimedia.org/w/index.php?curid=1942158

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Some Control Examples

• 1. Thymio steering following a black tape
• 2. Car steering, particularly on a snowy road
• 3. Boat steering
• 4. Driving a car at a desired speed
• 5. Tap water temperature control (or room

thermostat) Other examples: moving robotic arm, etc.

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General Parameters in Controller Analysis

Process variable (PV , or p) — sensed value that we want to control SetPoint (SP, or s) — desired target value of the p Error (e) — difference between p and s; e = p − s Control variable or manipulated variable (CV , MV ,

• r c) — input into the process that we can control

We also consider the variable of time t in the analysis Describe these variables in the five given examples.

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Controllers: P, PI, PID, (PD)

What happens in each case if we use the P controller? When can I controller help P controller in PI controller? How does D controller helps i PID controller? How does PD controller operates?

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PID Controller Formula

The general formula for PID control is: c(t) = Kp · e(t) + Ki · t e(t′)dt′ + Kd · de(t) dt The controller depends on Kp, Ki and Kd parameters, in the proportional, integral, and derivative parts One way to look at the formula is:

◮ Proportional compensates for the error in the

present—instantaneous deviation from setpoint,

◮ Integral compensates for the errors in the past—the

accumulated deviation from setpoint, and

◮ Derivative compensates for the rror in the

future—treating the derivative as the “trend” in the error

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Binary Representation Constraints Review

Let us look again at the example of calculating circumference of a circle of diameter d = 80mm Formula c = π · d, and we know π ≈ 3.14159 Correct value: c ≈ 251.3274 ≈ 251mm We can use a very coarse integer approximation π ≈ 3 and get c ≈ 240 Other solution: represent π = PI /PId, where PI = 31416 and PId = 10000 However, “c = d * PI / PId” does not work well Why? How can we solve this problem?

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Binary Worked-out Example

Let us see what is happening in the calculation: c = d * PI / PId d=80 and PI=31416, which are in binary: d = 0101 0000 and PI = 0111 1010 1011 1000 When we multiply those numbers, d * PI we get: 0010 0110 0101 1001 1000 0000 which is the exact value of 2, 513, 280 = 80 × 31416, but it needs 32 bits to be represented (or at least 23) What happens is that Aseba has to ‘clip’ the result to the last 16 bits

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Binary Worked-out Example (2)

When we clip the following number (d * PI = 80 × 31416 = 2, 513, 280): 0010 0110 0101 1001 1000 0000 to the last 16 bits, we get: 0101 1001 1000 0000 = 128 + 256 + 2048 + 4096 + 16384 = 22912 which is indeed what Aseba shows if we make the calculation. This is why we get the final result of c = d * PI / PId to be 22913/10000=2 in Aseba

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Binary Worked-out Example (3)

If we use the native function call: call math.muldiv(c, d, PI, PId) the complete calculation will be done on 32-bit precision and we get the correct value: 80*31416/10000=2,513,280/10000=251 Note: If you wonder how Aseba does division in binary, you can try to divide 22913 by 10000 using binary representation: 22913 = 0101 1001 1000 0000 and 10000 = 0010 0111 0001 0000

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