Does the controller meet the required tracking accuracy?
e(t) = r(t) − y(t)
A stable system can still have a tracking error.
The long-term difference between target and output.
Compare it with the allowed error.
For a given input: zero, finite or growing error?
How large is the error?
Did it reach the right position?
Did it complete the task?
e(t) = r(t) − y(t)
Selecting K: tracking accuracy and stability
To find K at the stability boundary, first find the closed-loop pole equation.
Y(s)R(s) = KG(s)1 + KG(s)
We have the pole equation. Now find the pole location at the stability boundary.
s = σ + iω
σ is the real part. Watch the oscillations.
| σ < 0 | ? | |
|---|---|---|
| σ = 0 | ? | |
| σ > 0 | ? |
Use both facts: the pole equation, and the boundary location s = iω.
1 + KG(s) = 0
s = iω
Bode plots give the magnitude and phase of G(iω).
G(s) = 1s(s + 1)2
Unity negative feedback · K > 0
Start with the closed-loop pole equation: 1 + KG(s) = 0.
1 + KG(s) = 0
G(s) = 1s(s + 1)2
Write ω₀, m and K. Compare with your partner.
ω₀ = ?
m = |G(iω₀)| = ?
K = ?
A class challenge for practice — no grade. We will review the answers together.
Round 1 of 6
Find the boundary gain
G(s) = 1s(s + 1)(s + 1)
Plots of G(s), before multiplying by K.
Use one frequency in both plots.
ωp = ?
m = |G(iωp)| = ?
K = ?
How far is the current gain K from neutral stability?
G(s) = 1s(s + 1)2
Together · Is K = 0.2 stable?
Start with the known boundary K = 2. Then test K = 0.2.
Phase = −180°
Magnitude = 1
Both at the same frequency.
Think → show A/B/C together → explain → reveal.
G(s) = 1/[s(s + 1)²]
Question 1 of 3
How many times can we multiply the current gain to reach neutral stability?
Current K = 0.2
Neutral boundary: K = 2
0.2 × ? = 2
At ωp = 1: phase = −180°
Current |KG| = 0.1; boundary = 1
0.1 × ? = 1
GM measures how far the current gain is from neutral stability.
Start with the magnitude plot. Find where |KG(iω)| reaches 1.
Self-study · all steps are shown below
PM measures the phase distance to neutral stability.
1 · Find ωg: |KG(iωg)| = 1.
2 · Read the phase at that same ωg.
PM represents how much the phase can be reduced without breaking stability.
PM = ∠G(iωg) − (−180°)
= ∠G(iωg) + 180°
The controlled system is stable if
PM > 0
Under the assumptions from before. PM is in degrees.
Example: phase = −140°
PM = −140° + 180° = 40°.
Self-study · all steps are shown below
For the prototype second-order closed-loop system,
T(s) = ωn2s2 + 2ζωns + ωn2
PM = tan−1(2ζ√(√(1 + 4ζ4) − 2ζ2))
PM in degrees.
≈ 100ζ
Rough approximation; PM in degrees.
Therefore,
ζ ≈ PM100
Stable system: r(t) = R sin(ωt) ⇒ yss(t) = Y sin(ωt + φ)