How can we find the system type?
Rewrite F(s) into standard factors.
F(s) = gsn × 1τs + 1 × ⋯ × ωn²s² + 2ζωns + ωn²
Only gsⁿ can provide poles at s = 0.
| n | gsⁿ | Poles at s = 0 | Type |
|---|---|---|---|
| 2 | gs² | ||
| 1 | gs | ||
| 0 | g | ||
| −1 | g/s | ||
| −2 | g/s² |
Unit step · derive the error formula
Substitute R(s), then simplify.
ess = ?
Unit ramp · derive the error formula
Substitute R(s), then simplify.
ess = ?
Unit parabola · derive the error formula
Substitute R(s), then simplify.
ess = ?
Error constants · summary
Stable unity-feedback system · use the forward gain F(s).
| Unit input | Error constant · definition | Steady-state error |
|---|---|---|
| Step r(t) = 1 | Kp = lims→0 F(s) | ess = 11 + Kp |
| Ramp r(t) = t | Kv = lims→0 sF(s) | ess = 1Kv |
| Parabola r(t) = ½t² | Ka = lims→0 s²F(s) | ess = 1Ka |
How can we find the error constant?
Suppose n = −1, i.e., Type 1.
F(s) = gs−1 × 1τs + 1 × ⋯ × ωn²s² + 2ζωns + ωn²
Kv = lims→0 sF(s) = ?
Together · fill the Type 1 row
Type 1: F(s) ≈ gs as s → 0, g > 0
Step
11 + Kp
Ramp
1Kv
Parabola
1Ka
| Input | Find the constant | ess = ? |
|---|---|---|
| Step | Kp = lims→0 F(s) = ? | ? |
| Ramp | Kv = lims→0 sF(s) = ? | ? |
| Parabola | Ka = lims→0 s²F(s) = ? | ? |
Steady-state error · explore the types
Stable unity-feedback loop. Click a Type or an error cell to see the full derivation.
| Type | Unit stepr(t) = 1 | Unit rampr(t) = t | Unit parabolar(t) = ½t² |
|---|---|---|---|
Kp = limₛ→₀ F(s)
Step: 1/(1 + Kp)Kv = limₛ→₀ sF(s)
Ramp: 1/KvKa = limₛ→₀ s²F(s)
Parabola: 1/KaClass exercise · find the error constant
Try it first. Then we will solve it together.
F(s) = 12s(s + 3)
Rewrite into standard factors. Find the type, g and Kv.
F(s) = gs × 11 + τs
Kv = lims→0 sF(s)
Type = ?g = ?Kv = ?
Example 1 · find the steady-state error
Bode plot of the forward gain F(s)
Stable unit-feedback system
Low-frequency slope n = ? → Type ?
Kp = lims→0 F(s) = ?
| Input r(t) | 1 | t | ½t² |
|---|---|---|---|
| ess | ? | ? | ? |
Example 2 · read the low-frequency slope
Stable unit-feedback system
| Input r(t) | 1 | t | ½t² |
|---|---|---|---|
| ess | ? | ? | ? |
Can feedback make things worse?
NASA Armstrong · Enterprise, 1977 ↗21 seconds
Watch the wings as Enterprise approaches the runway.
Why can corrective feedback make the motion worse?
Neutral Stability · the boundary
K > 0
Objective
Determine closed-loop stability from the Bode plot of KG(s).
Example · Time-Domain Analysis
G(s) = 1s(s + 1)2
Unity negative feedback · K > 0
Find the stable range of K.
Example · Frequency-Domain Analysis
G(s) = 1s(s + 1)2
Bode plot of KG(s) when K = 2.
K = 2
Relative Stability
How far is the current gain K from neutral stability?
Assumptions
Relative stability · compare the gains
G(s) = 1s(s + 1)2
Compare K = 0.2, 2, and 20 for this system.
K = 0.2K = 2K = 20
At the phase crossover
|KG(iωp)| = K2
| K | Magnitude at ωp | For this system |
|---|---|---|
| 0.2 | ? | ? |
| 2 | ? | ? |
| 20 | ? | ? |
Gain Margin
Relative Stability in the Phase Plot
Define the gain crossover frequency ωg.
Example · gain crossover and phase
Bode plot of KG(s) when K = 0.2, K = 2, K = 20.
K = 0.2K = 2K = 20
| K | ωg rad/s | ∠G(iωg) | Closed loop |
|---|---|---|---|
| 0.2 | ? | ? | ? |
| 2 | ? | ? | ? |
| 20 | ? | ? | ? |
Phase Margin
PM measures the phase distance to neutral stability.
Phase Margin · prototype second-order system
For the prototype second-order closed-loop system,
Example · phase margin
Bode plot of KG(s) when K = 0.2, K = 2, K = 20.
K = 0.2K = 2K = 20
PM = ∠G(iωg) + 180°
| K | ∠G(iωg) | PM | Closed loop |
|---|---|---|---|
| 0.2 | ? | ? | ? |
| 2 | ? | ? | ? |
| 20 | ? | ? | ? |
Computing Gain/Phase Margin with Python
Compute gain and phase margins and their crossover frequencies.
control.margin(sysdata)Summary · Frequency Domain Analysis
Frequency response is determined by |G(iω)| and ∠G(iω).
r(t) = R sin(ωt) ⇒ yss(t) = Y sin(ωt + φ)