Mr. IntegralTHE ANGLE CONDITION
Geometry → gainInspect a point.
Click the s-plane, or enter a point. Vectors point from each pole and zero to the test point. The fixed leading sign also contributes phase.
Choose a point to test the phase.
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NEGATIVE FEEDBACK · K ≥ 01 + K L₀(s) = 0Why do the branches take this path?
THE s-PLANE01 · The model
× Fixed pole○ Fixed zero● Closed-loop root◇ Test point
Overlay candidate response targets
ζ rays and σ = −4/Ts are second-order candidate guides, not guarantees for the full response.
Boundary
0Fine control near zero · changing gain opens the complete locus80
THE WHOLE SYSTEM
Unit stepCheck the response.
Current closed-loop poles and damping
TURN THE GEOMETRY INTO A CONTROLLER
Finite lead designSupply the missing phase.
Choose a desired conjugate pole pair and a real controller zero. Solve for a finite lead pole and positive gain, then check the complete system.
Try the supplied target −2 ± 2j with a zero at −2. Predict where the controller pole must go.
A SECOND VIEW OF THE SAME EQUATION
The landscape behind the line.
−180°−90°0°90°180°
Color shows phase; brightness shows log₁₀ magnitude. The white trace is the computed locus within the displayed gain range.
Height needs direction.
Poles make peaks; zeros make valleys. But magnitude alone cannot tell you whether a point lies on the locus.
The loop value must point along the negative real axis. Only then does K = 1 / |L₀(s)| select the gain.
Surface height uses log₁₀ |L₀|, capped at ±3 for display. Never reciprocate the logarithmic height. Phase is undefined at poles and zeros.
This view is illustrative between sampled points; the line and numeric readouts come from the polynomial and complex evaluations.
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A target is a hypothesis.
The response is the check.
For P(s) = 1/[s(s+1)], a finite lead C(s) = 20(s+2)/(s+8) puts the closed-loop poles at −5 and −2 ± 2j.
The complete unit-step response has about 16.45% overshoot and 1.98 s settling time in a 2% band. It meets a 25%, 2.5 s specification. The ideal second-order pair alone predicts only 4.32% overshoot: the extra pole and numerator zero matter.
Where do the branches meet?
For L₀(s) = 1/[s(s+2)], find the positive-gain meeting point and its gain. Study the visualization and worked example first; then explain your reasoning.
Hint 1 · use the same equation
D + KN = 0 gives K = −s(s+2).
Hint 2 · a repeated root
At a meeting point, dK/ds = −2s−2 = 0. Check that the resulting gain is positive.
Reveal the reasoning
s = −1 and K = 1. The closed-loop polynomial is s²+2s+1 = (s+1)². Two roots coincide. For larger K they leave the real axis as a conjugate pair. One correct answer is practice evidence, not proof of mastery.
Assumptions, numerical limits and references
Negative feedback, K ≥ 0, continuous-time SISO and real coefficients. K scales the complete fixed loop L₀. Characteristic roots include multiplicity; reduced input/output poles can differ after cancellation. This lab rejects nearly coincident open-loop pole–zero pairs rather than silently claiming internal stability.
Root sweeps have a finite maximum gain and window. Arrows and finite endpoints are not claims that branches end there. Polynomial roots use simultaneous complex iteration with residual checks. Step responses use bounded RK4 integration of the complete closed-loop transfer; sampled performance is approximate. A reported 2% settling time is observed within a finite horizon, not a formal tail bound. Unstable or marginal systems do not receive stable performance metrics.
References: Nise, Control Systems Engineering, 7th ed., §§8.3–8.7; Sundararajan, Control Systems: An Introduction, §7.1; de Oliveira, Fundamentals of Linear Control, §6.4. Original diagrams and implementation. Study alongside Swarthmore’s root-locus notes and the MATLAB root-locus definition.