3.5.40Guidance, Navigation & Control (GNC)

Root locus — Evans' method, rules for sketching

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WHAT is the root locus?

WHY this equation? The closed-loop transfer function is

T(s)=KG(s)1+KG(s)H(s).T(s) = \frac{KG(s)}{1 + KG(s)H(s)}.

Poles are where the denominator vanishes, i.e. 1+KG(s)H(s)=01 + KG(s)H(s) = 0. Everything below flows from this one line.


Deriving the two conditions from scratch

Write L(s)=KG(s)H(s)=KN(s)D(s)L(s) = KG(s)H(s) = K\dfrac{N(s)}{D(s)}. Characteristic equation:

1+L(s)=0L(s)=1.1 + L(s) = 0 \quad\Longrightarrow\quad L(s) = -1.

Now 1-1 is a complex number with magnitude 1 and angle 180°180° (or any odd multiple). Splitting L(s)=1L(s)=-1 into magnitude and angle:


The sketching rules (each DERIVED, not memorized)

Let nn = number of open-loop poles (roots of DD), mm = number of open-loop zeros (roots of NN). Assume nmn \ge m.

Rule 1 — Start and end. Rewrite 1+KND=01 + K\frac{N}{D}=0 as D(s)+KN(s)=0D(s) + KN(s) = 0.

  • K0K \to 0: D(s)=0D(s)=0 → branches start at open-loop poles.
  • KK \to \infty: divide by KK, N(s)=0N(s)=0 → branches end at open-loop zeros. There are nn branches; mm end on finite zeros, the remaining nmn-m go to infinity.

Rule 2 — Symmetry. Coefficients are real, so complex poles come in conjugate pairs → the locus is symmetric about the real axis.

Rule 3 — Real-axis segments. A real point lies on the locus if the number of real poles+zeros to its right is odd. WHY? Angle contributions from complex-conjugate pairs cancel; each real singularity to the right contributes 180°180°. Odd count → total is an odd multiple of 180°180° → angle condition met.

Rule 4 — Asymptotes. The nmn-m escaping branches approach straight lines with angles

θa=±180°(2k+1)nm\theta_a = \frac{\pm 180°(2k+1)}{n-m}

meeting the real axis at the centroid

σa=poleszerosnm.\sigma_a = \frac{\sum \text{poles} - \sum \text{zeros}}{n-m}.

WHY the centroid? For large s|s|, G(s)H(s)1(sσa)nmG(s)H(s)\approx \frac{1}{(s-\sigma_a)^{n-m}} after matching the first two terms of the polynomial division — the far-field looks like a single pole of multiplicity nmn-m at the centroid.

Rule 5 — Breakaway/break-in points. Where branches leave or rejoin the real axis, two poles coincide → KK is at a local extremum. Solve

dKds=0,K=D(s)N(s).\frac{dK}{ds} = 0, \qquad K = -\frac{D(s)}{N(s)}.

WHY? At a double root the polynomial has a repeated root, so K(s)K(s) turns around: dK/ds=0dK/ds=0.

Rule 6 — jωj\omega-axis crossing. Use Routh–Hurwitz on D(s)+KN(s)D(s)+KN(s) to find the KK that makes a whole row zero; the auxiliary equation gives the crossing frequency ω\omega. This is the critical gain for stability.

Rule 7 — Angle of departure/arrival (from complex pole/zero pp):

θd=180°+(zerosp)(other polesp).\theta_d = 180° + \sum \angle(\text{zeros}\to p) - \sum \angle(\text{other poles}\to p).

WHY? Apply the angle condition to a test point infinitesimally close to pp.

Figure — Root locus — Evans' method, rules for sketching

Worked Example 1 — G(s)H(s)=1s(s+2)G(s)H(s)=\dfrac{1}{s(s+2)}

Poles: s=0,s=2s=0,\,s=-2 (n=2n=2). Zeros: none (m=0m=0).

Step — Real axis segment. Why? Between 00 and 2-2, exactly one pole (s=0s=0) sits to the right of any interior point → odd → segment [2,0][-2,0] is on the locus. ✔

Step — Asymptotes. nm=2n-m=2 so θa=90°,270°\theta_a = 90°, 270°. Centroid σa=(0)+(2)02=1\sigma_a = \frac{(0)+(-2)-0}{2} = -1. Why centroid at 1-1? It's the average of the poles. Branches shoot straight up and down from s=1s=-1.

Step — Breakaway. K=D/N=s(s+2)=(s2+2s)K=-D/N = -s(s+2)=-(s^2+2s). Set dK/ds=(2s+2)=0s=1dK/ds = -(2s+2)=0 \Rightarrow s=-1. Why here? Symmetry demanded the breakaway between the two real poles; the algebra confirms s=1s=-1.

Step — Gain at breakaway. K=(1)(1+2)=1K = -(-1)(-1+2) = 1. So at K=1K=1 we get a double pole at 1-1; beyond that, poles become complex with real part 1-1 (constant damping).

Insight: This system is stable for all K>0K>0 — the vertical asymptote never crosses into the right half plane.


Worked Example 2 — G(s)H(s)=1s(s+1)(s+2)G(s)H(s)=\dfrac{1}{s(s+1)(s+2)}

Poles 0,1,20,-1,-2 (n=3n=3), m=0m=0.

Asymptotes. θa=60°,180°,300°\theta_a=60°,180°,300°; σa=33=1\sigma_a=\frac{-3}{3}=-1. Why 3 asymptotes? nm=3n-m=3 branches all escape.

jωj\omega crossing (Routh). Char. eqn: s3+3s2+2s+K=0s^3+3s^2+2s+K=0. Routh array first column needs 32K3>0K<6\frac{3\cdot2 - K}{3}>0 \Rightarrow K<6. At K=6K=6: auxiliary 3s2+6=0s=±j23s^2+6=0 \Rightarrow s=\pm j\sqrt2. Why this matters: the locus crosses the imaginary axis at ω=2\omega=\sqrt2 when K=6K=6 — that is the maximum stable gain.


Recall Feynman: explain to a 12-year-old

Imagine two magnets (the poles) on a table and a dial. When the dial is at zero, two little beads sit exactly on the magnets. As you turn the dial up, the beads slide along fixed paths — sometimes toward "target dots" (zeros), sometimes flying off toward the edges along straight ramps. The root locus is just the drawing of those bead paths. If a bead crosses a red danger line (the imaginary axis), the machine starts shaking uncontrollably (unstable). You use the drawing to pick a dial setting that keeps all beads safely on the good side.


Flashcards

What equation defines every point of the root locus?
The characteristic equation 1+KG(s)H(s)=01+KG(s)H(s)=0, i.e. G(s)H(s)=1G(s)H(s)=-1.
Which condition determines the SHAPE of the locus (independent of K)?
The angle condition G(s)H(s)=±180°(2k+1)\angle G(s)H(s)=\pm180°(2k+1).
Which condition gives the gain K at a chosen point?
The magnitude condition K=1/G(s)H(s)K=1/|G(s)H(s)|.
Where do branches start and end as K goes 0→∞?
Start at open-loop poles (K→0), end at open-loop zeros or infinity (K→∞).
How many branches go to infinity?
nmn-m (poles minus zeros).
Real-axis rule for the locus?
A point is on the locus if the number of real poles+zeros to its RIGHT is odd.
Asymptote angles formula?
θa=±180°(2k+1)/(nm)\theta_a=\pm180°(2k+1)/(n-m).
Asymptote centroid formula?
σa=(poleszeros)/(nm)\sigma_a=(\sum\text{poles}-\sum\text{zeros})/(n-m).
How do you find breakaway points?
Solve dK/ds=0dK/ds=0 where K=D(s)/N(s)K=-D(s)/N(s).
How do you find where the locus crosses the imaginary axis?
Apply Routh–Hurwitz to find critical K, then use the auxiliary equation for ω.
Why is the locus symmetric about the real axis?
Because the polynomial has real coefficients, so complex poles occur in conjugate pairs.
For 1/[s(s+2)]1/[s(s+2)], where is the breakaway point and its gain?
s=1s=-1, K=1K=1.

Connections

  • Characteristic equation & closed-loop poles
  • Routh–Hurwitz stability criterion
  • PID / PD controller design — adding zeros to reshape the locus
  • Bode plot & frequency response — magnitude/phase alternative view
  • Nyquist criterion — encirclement view of L(s)=1L(s)=-1
  • Damping ratio and settling time — reading ζ,ωn\zeta,\omega_n off pole locations
  • GNC control loops — where gain tuning lands in autopilots

Concept Map

denominator=0

rearrange

angle part

magnitude part

shapes locus

labels gain

as K sweeps 0 to inf

D+KN=0, K to 0

n branches

real coeffs

odd count to right

n-m escaping

reveal

Closed-loop TF

Characteristic eq 1+KGH=0

L s = -1

Angle condition ±180 2k+1

Magnitude condition K=1/GH

Root Locus map

Closed-loop poles move

Rule 1 start at poles end at zeros

Rule 2 symmetry about real axis

Rule 3 real-axis segments

Rule 4 asymptotes and centroid

Stability and damping

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, root locus ka basic idea simple hai: tumhare feedback system mein ek gain knob hota hai jise hum KK bolte hain. Jaise-jaise tum KK ko 00 se \infty tak ghumate ho, closed-loop poles complex plane mein move karte hain. Root locus bas unn poles ka rasta (path) draw kar deta hai. Iska sabse bada faayda — tumhe polynomial solve karne ki zaroorat hi nahi. Sirf kuch geometric rules se path bana lo aur turant pata chal jaayega ki system stable rahega ya nahi, aur response kitna fast/damped hoga.

Sab kuch ek hi equation se nikalta hai: 1+KG(s)H(s)=01+KG(s)H(s)=0, yaani G(s)H(s)=1G(s)H(s)=-1. Ab 1-1 ka magnitude 11 hai aur angle 180°180°. Isse do conditions banti hain — angle condition jo locus ki shape deti hai (gain se independent!), aur magnitude condition jo har point pe KK ki value batati hai. Mantra yaad rakho: "Shape it, then gain it."

Rules practical hain: branches poles se start hoti hain aur zeros pe end (ya infinity par jaati hain, total nmn-m branches). Real axis par ek point locus par hai agar uske right mein poles+zeros ki count odd ho. Infinity waali branches asymptotes follow karti hain jinka centroid σa=(poleszeros)/(nm)\sigma_a=(\sum\text{poles}-\sum\text{zeros})/(n-m) hota hai. Breakaway point dK/ds=0dK/ds=0 se, aur imaginary axis crossing Routh-Hurwitz se — yahi crossing tumhara maximum stable gain deta hai.

GNC (autopilot, missile guidance) mein yeh bahut kaam ka hai kyunki gain tuning direct pole locations se juda hota hai. Agar system slow ya oscillatory hai, tum ek zero (PD controller) add karke locus ko left half plane ki taraf kheench lete ho — isse stability aur speed dono improve hoti hai. Isliye root locus sirf theory nahi, real controller design ka backbone hai.

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Connections