Mathematical Rules of Thumb, illustrated reader · Chapter 25

25Control Theory

Performance, Robustness, and Implementation Reality

5 demonstrations follow the chapter's rules. Choose a value, watch the figure and the numbers change, and check your prediction. Every choice is precomputed from the notebook calculations.

Ask the chapter skill

“Help me use Chapter 25 for my question. Choose a rule, check its assumptions, and show how the result changes when an input changes.”

Use math-thumb-control-theory from the companion's skill package. The demonstrations below also work on their own.

Examples use constructed inputs or the book's own values, disclosed in each panel. A picture illustrates a rule; its assumptions set its scope.

1Demonstration 1 of 5

Connect a real step response with bandwidth

Why can two systems with similar bandwidth have different overshoot?

Plot the exact normalized first-order closed-loop response and compare rise-time formulas.

y(t)=1−e−2πfbt,t10−90=log⁡92πfb y(t)=1-e^{-2\pi f_b t},\quad t_{10-90}=\frac{\log9}{2\pi f_b}

First-order bandwidth (Hz). Single real pole and unit gain; the .35/f rule is not universal for higher-order loops.

Predict first. Why can two systems with similar bandwidth have different overshoot?

Choose an example

Connect a real step response with bandwidth. For this normalized first-order closed-loop model, 10–90% rise time is ln(9)/(2πf)=0.17485 s. The .35/f rule agrees closely here, but different pole patterns need separate verification.
First-order bandwidth (Hz): 2
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Bandwidth (Hz)
2
Exact 10–90% rise time (s)
0.17485
Rule .35 / bandwidth (s)
0.175

For this normalized first-order closed-loop model, 10–90% rise time is ln(9)/(2πf)=0.17485 s. The .35/f rule agrees closely here, but different pole patterns need separate verification.

Use the idea

Use rule 25.1.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Single real pole and unit gain; the .35/f rule is not universal for higher-order loops.

Check your understanding: Why can two systems with similar bandwidth have different overshoot?
Their pole and zero structures differ. The first-order model here has no overshoot.

Book source: Rule 25.1.1: Estimate rise time from closed-loop bandwidth. Demonstration C25-D01. Worked illustration.

2Demonstration 2 of 5

Budget latency at an assumed crossover

Does a positive remaining nominal margin prove the implemented loop is robust?

Plot delay phase lag across frequency and compute its cost at a stipulated crossover.

ϕdelay=360fcτ degrees \phi_{delay}=360f_c\tau\text{ degrees}

Pure delay τ (s). fc=5Hz, nominal margin 60°; fixed-crossover screen, not full Nyquist analysis or implemented-loop validation.

Predict first. Does a positive remaining nominal margin prove the implemented loop is robust?

Choose an example

Budget latency at an assumed crossover. Delay 0.01 s contributes 18° lag at 5 Hz. Subtracting it from a nominal 60° margin leaves 42°. That is below the usual 45° floor, so expect overshoot and ringing. This fixed-crossover screen is not a full stability proof.
Pure delay τ (s): 0.01
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Delay (s)
0.01
Lag at 5 Hz (degrees)
18
Nominal margin (degrees)
60
Fixed-crossover remaining margin
42

Delay 0.01 s contributes 18° lag at 5 Hz. Subtracting it from a nominal 60° margin leaves 42°. That is below the usual 45° floor, so expect overshoot and ringing. This fixed-crossover screen is not a full stability proof.

Use the idea

Use rule 25.3.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

fc=5Hz, nominal margin 60°; fixed-crossover screen, not full Nyquist analysis or implemented-loop validation.

Check your understanding: Does a positive remaining nominal margin prove the implemented loop is robust?
No. Other crossovers, model uncertainty, discretization and nonlinear actuator behavior also need checks.

Book source: Rule 25.3.1: Convert every delay into phase lag at crossover. Demonstration C25-D02. Worked illustration.

3Demonstration 3 of 5

Simulate integral windup and a remedy

Why does the old integral keep pushing after the target drops?

Drive a normalized first-order plant with an unreachable setpoint, then lower the target. Compare the stored integral and delayed recovery.

uraw=2(r−y)+I,İ=r−y,u=clip⁡(uraw,−1,1) u_{raw}=2(r-y)+I,\quad\dot I=r-y,\quad u=\operatorname{clip}(u_{raw},-1,1)

Conditional integration. PI gains Kp=2, Ki=1, step .005, actuator ±1. Conditional integration is one illustrative anti-windup design.

Predict first. Why does the old integral keep pushing after the target drops?

Choose an example

Simulate integral windup and a remedy. A normalized first-order plant uses PI gains Kp=2, Ki=1 and actuator limits ±1. The setpoint is unreachable for four seconds, then drops to .5. With conditional integration on, the integral stops growing while the actuator is pinned, so the output reaches 0.486, close to the .5 target. Any controller with integral action and a limited actuator needs this guard.
Conditional integration: on
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Anti-windup
on
Actuator limits
[−1,1]
Final output
0.485664
Integral state just before t=4
0

A normalized first-order plant uses PI gains Kp=2, Ki=1 and actuator limits ±1. The setpoint is unreachable for four seconds, then drops to .5. With conditional integration on, the integral stops growing while the actuator is pinned, so the output reaches 0.486, close to the .5 target. Any controller with integral action and a limited actuator needs this guard.

Use the idea

Use rule 25.3.3 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

PI gains Kp=2, Ki=1, step .005, actuator ±1. Conditional integration is one illustrative anti-windup design.

Check your understanding: Why does the old integral keep pushing after the target drops?
During saturation it accumulated an error the actuator could not correct. Limiting integration prevents much of that stored excess.

Book source: Rule 25.3.3: Add anti-windup whenever integral control can saturate. Demonstration C25-D03. Worked illustration.

4Demonstration 4 of 5

Remove steady error with integral action

Why does Ki=0 never reach the setpoint?

Hold a setpoint of 1, then add a constant disturbance at t=5 s. Compare how the output recovers.

u=Kpe+Ki∫edt u=K_p e+K_i\int e\,dt

Integral gain Ki. Plant y′=−y+u+d, Kp=2, disturbance d=−0.5 from t=5 s, step .005 s, no actuator limits.

Predict first. Why does Ki=0 never reach the setpoint?

Choose an example

Remove steady error with integral action. Integral action keeps adding up the error until it is gone. Final error is 0.039, but the slow integral takes seconds to clean up the disturbance.
Integral gain Ki: 1
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Integral gain Ki
1
Output before disturbance (t=5 s)
0.959138
Final output (t=10 s)
0.961052
Final error
0.038948

Integral action keeps adding up the error until it is gone. Final error is 0.039, but the slow integral takes seconds to clean up the disturbance.

Use the idea

Use rule 25.1.2 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Plant y′=−y+u+d, Kp=2, disturbance d=−0.5 from t=5 s, step .005 s, no actuator limits.

Check your understanding: Why does Ki=0 never reach the setpoint?
A proportional controller only pushes when error exists, so some error must remain. The integral keeps pushing until the error is zero.

Book source: Rule 25.1.2: Add integral action for zero steady-state error to constant disturbances. Demonstration C25-D04. Worked illustration.

5Demonstration 5 of 5

See what phase margin looks like in a step response

Why do designers start near 45 to 60 degrees?

Loop L(s)=k/(s(s+a)) is tuned so it crosses unit gain at 1 rad/s with the chosen phase margin. Watch the closed-loop step response.

PM=180∘+∠L(iωc) PM=180^\circ+\angle L(i\omega_c)

Phase margin (degrees). Second-order loop, unity negative feedback, no delay or saturation. Other loops map margin to overshoot differently.

Predict first. Why do designers start near 45 to 60 degrees?

Choose an example

See what phase margin looks like in a step response. The loop L(s)=k/(s(s+a)) crosses unity gain at 1 rad/s with 45° of phase margin. Its step response overshoots by 23.3%. 45° gives a modest overshoot and quick settling, a common working compromise. Margin is a starting screen; check the actual response.
Phase margin (degrees): 45
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Phase margin (degrees)
45
Open-loop pole (rad/s)
1
Loop gain
1.41421
Overshoot (%)
23.3006
Peak output
1.23301

The loop L(s)=k/(s(s+a)) crosses unity gain at 1 rad/s with 45° of phase margin. Its step response overshoots by 23.3%. 45° gives a modest overshoot and quick settling, a common working compromise. Margin is a starting screen; check the actual response.

Use the idea

Use rule 25.2.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Second-order loop, unity negative feedback, no delay or saturation. Other loops map margin to overshoot differently.

Check your understanding: Why do designers start near 45 to 60 degrees?
At 30° this loop overshoots about 42% and rings; at 60° about 9%. Below 45° a little extra delay can push it toward instability.

Book source: Rule 25.2.1: Start loop shaping near 45 to 60 degrees phase margin. Demonstration C25-D05. Worked illustration.

Bring the idea to a question of your own

Choose the relationship that answers your question, check its conditions, and compare the result with the accuracy or decision threshold you need.

The chapter skill can adapt these calculations to your inputs. It should name the assumptions, explain what the result supports, and say what still needs evidence. The chapter workbook adds a lab and three exercises with answers.