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Mechatronics · Module 10 of 10

System Integration and Design

This is where the pieces become a machine. Integration closes the sense, decide, actuate loop end to end, and its performance, the steady-state error and the resolution, follows from every block working together.

01

Readiness check

This closing module brings the course together. Tick only what you can do closed-notes.

  • Evaluate 1/(1 + K) for a given K.
  • Divide one resolution by a gear ratio.
  • Recall the sense, decide, actuate loop.
  • Recall that feedback reduces steady-state error.
  • Recall that a system is only as good as its weakest block.
0 or 1 weak itemsContinue with this module.
2 weak itemsRevisit steady-state error in Control Systems, Module 5.
3 or more weak itemsRevisit the sense, decide, actuate loop in Module 1.
02

The core idea

Integration joins the sensing, decision, and actuation blocks into one closed loop, and the system's performance emerges from all of them. A proportional loop leaves a steady-state error of 1/(1 + K) for a step, and the end-to-end resolution is the sensor's resolution scaled by the mechanical reduction.

ess = 1/(1 + K)output resolution = sensor resolution / reductionsystem ≈ its weakest block

Every module so far built one block; integration is the discipline of making them work as one machine. It closes the loop physically and in software: the sensor feeds the conditioning, which feeds the converter and the microcontroller, which drives the actuator through the mechanism, whose motion the sensor reads again. Two things decide whether the result is good. The first is closed-loop performance. With proportional control of gain K and unity feedback, a step command leaves a steady-state error of 1/(1 + K): higher gain shrinks the error but risks instability, the trade the control course develops. The second is the resolution budget. The finest step the system can command or measure is set by the whole chain, so an encoder resolving 0.18 degrees on a motor shaft, seen through a 10:1 reduction, resolves 0.018 degrees at the output, ten times finer. Integration also means matching interfaces, voltage levels, grounds, and timing, and it never ends at first assembly: you test the integrated system, find faults, add safety interlocks, and iterate. A mechatronic system is only ever as good as its weakest block, so integration is where the design is truly proven.

The skill works when: you predict ess = 1/(1 + K) and scale resolution through the mechanism.
The skill breaks down when: blocks are tested only in isolation, or interface and grounding mismatches are ignored.
The concept. Integration closes the full loop. Its steady-state error and resolution come from every block at once, not from any one in isolation.
03

The skills, taught in order

Five skills turn a set of working blocks into a proven machine.

10.1 The integration mindset

Integration is mostly about interfaces: matching voltage levels and logic, sharing a clean ground, and meeting timing so data is fresh. Most integration faults live at the joints between blocks, not inside them.

10.2 Closing the loop

A closed loop compares output to command and drives the error toward zero. Under proportional control the residual steady-state error to a step is 1/(1 + K): raising the loop gain K reduces it, but too much gain brings oscillation, the stability trade of the control course.

Loop gain Kess = 1/(1 + K)As percent
91/1010%
491/502%
991/1001%

Higher loop gain shrinks the steady-state error, at the cost of stability margin.

10.3 The resolution budget

The finest step the whole system resolves comes from the sensor and the mechanics together. A gear reduction that slows the output improves its angular resolution by the same ratio, so output resolution is the sensor resolution divided by the reduction.

10.4 Testing, faults, and safety

Test the integrated system, not just the parts: inject known inputs and check the closed-loop response. Add interlocks and an emergency stop, and use systematic fault-finding to isolate a problem to a block or an interface.

10.5 Iteration and documentation

First assembly is rarely the last. Measure, adjust gains and interfaces, and document the wiring, signals, and control constants so the machine can be maintained and improved. The design cycle closes here and starts again.

Engineering connection: a pick-and-place robot combines every module of this course, encoders, conditioning, converters, a microcontroller, and motors, and only integration testing proves the loop meets its accuracy.

04

Worked example 1: steady-state error

A position system uses proportional control with a loop gain of K = 49 and unity feedback. Find the steady-state error to a unit step command.

Figure 1. Proportional control drives the output close to the command but leaves a small gap. With K = 49 that gap is 1/(1 + 49) = 2 percent.
  1. ProblemFind the steady-state error for the loop in Figure 1.
  2. Given / findProportional gain K = 49, unit step, unity feedback. Find ess.
  3. AssumptionsStable type-0 loop, unity feedback, step input.
  4. ModelFor proportional control of a step, ess = 1/(1 + K).
  5. Equationsess = 1/(1 + 49)= 1/50
  6. Solveess = 1/50 = 0.02 = 2%.
  7. CheckRaising K to 99 would halve the error to 1 percent, consistent with 1/(1 + K).
  8. ConclusionThe loop tracks to within 2 percent; closing the remaining gap needs more gain or integral action, from the control course.
Result. Steady-state error ess = 2 percent.
05

Worked example 2: end-to-end resolution

A motor carries an encoder resolving 0.18 degrees per count. The motor drives the output through a 10:1 reduction. Find the resolution at the output.

Figure 2. The 10:1 reduction turns the output a tenth as far per motor count, so the output resolution is ten times finer than the encoder alone.
  1. ProblemFind the output resolution for the drive in Figure 2.
  2. Given / findEncoder 0.18 degrees per count, reduction 10:1. Find the output resolution.
  3. AssumptionsRigid, backlash-free reduction, encoder on the motor shaft.
  4. Modeloutput resolution = sensor resolution / reduction.
  5. Equationsrout = 0.18° / 10
  6. Solverout = 0.018° per count.
  7. CheckOne output revolution needs ten motor turns, so ten times the counts, giving ten times finer resolution, matching.
  8. ConclusionPlacing the encoder before the reduction buys resolution, a common integration choice, though backlash then limits real accuracy.
Result. Output resolution 0.018 degrees per count.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Wrong steady-state formPredicted error far off"Is it 1/(1 + K)?"Proportional step error is 1/(1 + K).
Adding resolutionsOutput resolution nonsensical"Did I scale by the reduction?"Divide the sensor resolution by the reduction.
Testing blocks onlyWorks apart, fails together"Did I test the closed loop?"Test the integrated system end to end.
Ignoring interfaces and groundNoise or logic errors at joints"Do levels and grounds match?"Match voltages, logic, and a common ground.
07

Practice ladder

Level 1 · Direct skill

A proportional loop has K = 9. Find the steady-state error to a unit step.

Show answer

ess = 1/(1 + 9) = 1/10 = 10 percent.

Level 2 · Mixed concept

What loop gain gives a 1 percent steady-state error?

Show answer

1/(1 + K) = 0.01, so 1 + K = 100, giving K = 99.

Level 3 · Independent problem

An encoder resolves 0.5 degrees per count and drives an output through a 5:1 reduction. Find the output resolution.

Show answer

Output resolution = 0.5/5 = 0.1 degrees per count.

Transfer task | Real engineering

Integrate a closed-loop temperature controller. List the sensor, conditioning, controller, and actuator, and describe one test that proves the loop works.

What good work looks like

Thermistor sensor, bridge and amplifier conditioning, microcontroller with a proportional or PID law, heater with PWM drive; a step test that commands a new setpoint and checks the measured temperature settles near it with an acceptable steady-state error.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I used 1/(1 + K) for this steady-state error."
"Give me three drivetrains; I will compute the end-to-end resolution."
"Integrate my whole system for me." Joining the blocks is the skill.
"Why does my loop oscillate?" Reason from gain and stability yourself.

Portfolio task

Take a full sense, decide, actuate system and analyse it end to end: predict its steady-state error and its resolution, then plan an integration test.

Must include: a block diagram of the closed loop, an ess estimate, a resolution budget, and one test plan.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the proportional steady-state error.

ess = 1/(1 + K) for a unit step.

2. How does a reduction change resolution?

It improves output resolution by the reduction ratio.

3. Where do most integration faults occur?

At the interfaces between blocks, not inside them.

4. Why test the integrated system?

Blocks can work apart yet fail together through interfaces and timing.

5. What limits overall system quality?

The weakest block in the chain.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive ess and a resolution budget from scratch.
+3 daysAnalyse one full system end to end.
+7 daysReturn to the course home and plan an integration project.
+30 daysReuse the loop, ess, and resolution view on any machine.
10

Textbook mapping

This module follows William Bolton, Mechatronics, 6th edition. Use these references to read further.

Topic in this moduleWhere to read more
Integrated mechatronic systemsBolton, Chapter 22, Mechatronics systems
Closed-loop performanceBolton, Chapter 1, Control systems
Fault finding and testingBolton, Chapter 21, Fault finding

Chapter numbers refer to Bolton's Mechatronics, 6th edition. Any edition with the same chapter titles is equivalent for study.