Root locus — Evans' method, rules for sketching: Exercises

3.5.40 · D4 · Physics › Guidance, Navigation & Control (GNC) › Root locus — Evans' method, rules for sketching

Poore notes mein, hum wohi vocabulary use karte hain jo parent note mein hai:

$G (s) H (s) = K \frac{N ( s )}{D ( s )}$ open-loop transfer function hai; $K \geq 0$ gain ka knob hai.
$n$ = poles ki sankhya (bottom polynomial $D$ ki roots), $m$ = zeros ki sankhya (top polynomial $N$ ki roots).
Pole woh value of $s$ hai jahan transfer function blow up hoti hai; zero woh hai jahan woh vanish ho jaata hai. Dono complex $s$ -plane mein marked dots hain (horizontal axis = real part $σ$ , vertical axis = imaginary part $j ω$ ).

Figure s01 dekho. Cyan crosses poles hain (X), white circle ek zero hai (O), aur amber vertical line imaginary axis hai — woh "danger line" jise ek pole stable rehne ke liye cross nahi kar sakta. Neeche ke har exercise isi plane pe khele jaate hain.

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

Level 1 — Recognition

Goal: $n$ , $m$ , poles, zeros padho, aur branches count karo. Abhi paths ki koi computation nahi.

Exercise 1.1

$G (s) H (s) = \frac{s + 3}{s ( s + 2 ) ( s + 5 )}$ ke liye open-loop poles aur zeros list karo, $n$ aur $m$ do, aur batao kitne branches infinity tak jaate hain.

Recall Solution 1.1

Poles bottom ki roots hain: $s = 0, - 2, - 5$ . Toh $n = 3$ . Zeros top ki roots hain: $s = - 3$ . Toh $m = 1$ . Branches ki sankhya $= n = 3$ (degree-3 polynomial ke closed-loop root ke per ek branch — upar WHY box dekho). Finite zeros pe khatam hone waale branches $= m = 1$ . Infinity ki taraf jaane waale branches $= n - m = 3 - 1 === 2 ==$ .

Exercise 1.2

$G (s) H (s) = \frac{K}{( s + 1 ) ^{2} + 4}$ ke liye poles identify karo. Kya woh real hain ya complex?

Recall Solution 1.2

Denominator ko zero set karo: $(s + 1)^{2} + 4 = 0 \Rightarrow (s + 1)^{2} = - 4 \Rightarrow s + 1 = \pm 2 j$ . Toh poles hain $s = - 1 + 2 j$ aur $s = - 1 - 2 j$ — ek complex-conjugate pair (real axis ke across mirror images, bilkul jaisa Rule 2 symmetry promise karta hai). $n = 2$ , $m = 0$ .

Level 2 — Application

Goal: ek rule cleanly apply karo.

Exercise 2.1 (Real-axis rule)

$s = 0, - 4$ par poles aur $s = - 2$ par ek zero ke liye, real axis ke kaun se hisse locus se belong karte hain?

Recall Solution 2.1

Rule yeh hai: ek real point locus par tab hai jab us point ke strictly daayein real poles aur zeros ki count odd ho.

Odd-count criterion kyun? (geometry — Figure s02 dekho.) Ek point locus par tab hota hai jab woh angle condition satisfy kare: $G (s) H (s)$ ka total angle odd multiple of $180°$ hona chahiye. $G (s) H (s)$ ka angle hai (har zero se test point tak ke angles ka sum) minus (har pole se test point tak ke angles ka sum). Ab socho ek test point real axis par baitha hai:

Poles/zeros ka koi bhi complex-conjugate pair do angles contribute karta hai jo exact mirror images hain (ek $+ ϕ$ , ek $- ϕ$ ). Woh cancel ho jaate hain — toh complex singularities real-axis test point ko kabhi affect nahi karti.
Test point ke left wali ek real singularity positive real direction mein point karti hai → angle $0°$ contribute karti hai.
Test point ke right wali ek real singularity negative real direction mein point karti hai → exactly $180°$ contribute karti hai.

Toh total angle $180° \times (number of real singularities to the right)$ hai. Us total ke odd multiple of $180°$ (angle condition) hone ke liye, daayein count odd hona chahiye. Yahi rule ki poori origin hai.

Apply karo. Axis ko right se left mark karo: pole $0$ , phir zero $- 2$ , phir pole $- 4$ .

$s = - 1$ par point ( $0$ aur $- 2$ ke beech): daayein ek singularity ( $s = 0$ ) → count $= 1$ , odd → ON.
$s = - 3$ par point ( $- 2$ aur $- 4$ ke beech): daayein do ( $0, - 2$ ) → count $= 2$ , even → OFF.
$s = - 5$ par point ( $- 4$ ke left): daayein teen → odd → ON. Locus par: segment $[- 2, 0]$ aur ray $(- \infty, - 4]$ — Figure s02 mein amber segments.

Exercise 2.2 (Centroid & asymptote angles)

$G (s) H (s) = \frac{1}{s ( s + 2 ) ( s + 4 )}$ ke liye asymptote centroid $σ_{a}$ aur asymptote angles nikalo.

Recall Solution 2.2

$n = 3$ poles ${0, - 2, - 4}$ , $m = 0$ zeros, toh $n - m = 3$ branches escape karte hain.

Ye angles $180°/ (n - m)$ ke odd multiples kyun hain. Saare poles aur zeros se bahut door, har ek roughly ek hi jagah baitha lagta hai, toh $G (s) H (s) \approx \frac{1}{s ^{n - m}}$ (yahan $n - m$ poles minus zeros denominator mein bache hue hain). Angle condition tab padhti hai $∠ \frac{1}{s ^{n - m}} = - (n - m) \cdot ∠ s = \pm 180° (2 k + 1)$ . $s$ ke angle ke liye solve karne par milta hai $∠ s = \frac{\pm 180° ( 2 k + 1 )}{n - m}$ . Odd multiples $(2 k + 1)$ directly " $- 1$ ka angle $180°, 540°, \dots$ hota hai" fact se aate hain — isliye numerator hamesha $180°$ ka odd multiple hota hai, aur $(n - m)$ se divide karne par $(n - m)$ equally-spaced escape directions fan out hoti hain.

Centroid kyun average hai. Polynomials divide karo: large $s$ ke liye $\frac{N}{D}$ behave karta hai $\frac{1}{( s - σ _{a} ) ^{n - m}}$ ki tarah jahan pehle do leading coefficients match karne se $σ_{a} = \frac{\sum poles - \sum zeros}{n - m}$ milta hai. Physically: far-field poles ko alag nahi pehchaan sakta, toh woh sabko ek single equivalent pole of multiplicity $n - m$ ki tarah replace karta hai jo unke balance point (unke average) par baitha hai — woh balance point wahan hai jahan $(n - m)$ ramps mili hain.

Compute karo: Centroid: $σ_{a} = \frac{( 0 - 2 - 4 ) - 0}{3} = \frac{- 6}{3} === - 2 ==$ . Angles: $θ_{a} = \frac{180° ( 2 k + 1 )}{3}$ for $k = 0, 1, 2 \Rightarrow θ_{a} = 60°, 180°, 300°$ . Teen straight ramps point $s = - 2$ se un angles par nikalti hain.

Level 3 — Analysis

Goal: kai rules combine karo — breakaway, gain, crossing.

Exercise 3.1 (Breakaway point + gain wahan)

$G (s) H (s) = \frac{1}{s ( s + 4 )}$ ke liye breakaway point aur us par gain $K$ nikalo.

Recall Solution 3.1

Yahan $D (s) = s (s + 4) = s^{2} + 4 s$ , $N (s) = 1$ . $d K / d s = 0$ kyun? Real segment par do branches approach karti hain, ek double root mein collide hoti hain, phir peel off ho jaati hain. Double root wahan hoti hai jahan gain function $K (s)$ turn around karta hai — ek local extremum — toh uska derivative zero hai. $K = - \frac{D ( s )}{N ( s )} = - (s^{2} + 4 s) .$ $\frac{d K}{d s} = - (2 s + 4) = 0 \Rightarrow s === - 2 == .$ Gain wahan: $K = - ((- 2)^{2} + 4 (- 2)) = - (4 - 8) === 4 == .$ Toh $K = 4$ par $s = - 2$ par repeated pole milta hai; us se aage poles complex ho jaate hain constant real part $- 2$ ke saath.

Exercise 3.2 ( $j ω$ -crossing via Routh)

$G (s) H (s) = \frac{1}{s ( s + 1 ) ( s + 3 )}$ ke liye woh gain $K$ nikalo jis par locus imaginary axis cross karta hai, aur crossing frequency $ω$ .

Recall Solution 3.2

Characteristic equation $D (s) + K N (s) = 0$ : $s (s + 1) (s + 3) + K = s^{3} + 4 s^{2} + 3 s + K = 0.$ Routh array banao:

s^{3} s^{2} s^{1} s^{0} 14 \frac{4 \cdot 3 - K}{4} K 3 K 0

Zero row ka matlab $j ω$ -crossing kyun hota hai. Routh test count karta hai kitni roots right half plane mein hain. Imaginary axis par exactly baitha root ( $s = \pm j ω$ , real part $0$ ) stable aur unstable ke beech ka knife-edge hai — woh moment jab ek root cross kar raha hai. Routh us exact instant ko signal karta hai poora row vanish karke, kyunki array "just stable" aur "just unstable" mein fark nahi kar paati: purely imaginary roots ka ek pair appear ho gaya hai. Toh woh $K$ jo ek row zero karta hai precisely woh $K$ hai jis par ek branch axis touch karti hai. Auxiliary polynomial frequency kyun deta hai. Jab ek row zero ho jaata hai, us ke upar wala row directly characteristic polynomial ke ek factor ki coefficient list hoti hai — auxiliary polynomial — aur woh factor exactly woh hai jo imaginary-axis roots carry karta hai. Use solve karne par $\pm j ω$ directly milta hai.

Stability ko $s^{1}$ entry $> 0$ chahiye: $\frac{12 - K}{4} > 0 \Rightarrow K < 12$ . Critical gain: $K === 12 ==$ ( $s^{1}$ row zero ho jaati hai). Crossing frequency: zero row ke upar wala row $s^{2}$ row hai, auxiliary deta hai $4 s^{2} + K = 0 \Rightarrow 4 s^{2} + 12 = 0 \Rightarrow s^{2} = - 3 \Rightarrow s = \pm j 3$ . Toh $K = 12$ par $ω === 3 \approx 1.732 ==$ .

Exercise 3.3 (Angle of departure)

Poles $s = - 1 \pm 2 j$ aur pole $s = 0$ ke liye (no zeros), pole $p = - 1 + 2 j$ se departure angle nikalo.

Recall Solution 3.3

Formula kahan se aata hai (Figure s03 dekho). Pole $p$ se infinitesimal step door ek test point $s$ rakho, unknown departure direction $θ_{d}$ mein. Locus par koi bhi point angle condition satisfy karta hai: $\sum ∠ (zeros \to s) - \sum ∠ (poles \to s) = 180°.$ Pole $p$ khud term $- ∠ (p \to s) = - θ_{d}$ contribute karta hai (us tiny test point ki taraf $p$ se vector departure direction mein point karta hai). Har doosra pole aur zero itna door hai tiny step ke comparison mein ki woh angle sirf $p$ tak ka angle hai. Substitute karo aur $θ_{d}$ isolate karo: $- θ_{d} + \sum ∠ (zeros \to p) - \sum ∠ (other poles \to p) = 180°,$ $θ_{d} = - 180° + \sum ∠ (zeros \to p) - \sum ∠ (other poles \to p)$ aur kyunki $- 180°$ aur $+ 180°$ ek hi direction name karte hain, standard form $+ 180°$ likhti hai. Woh $180°$ aur kuch nahi balki characteristic equation se " $- 1$ ka angle $180°$ " fact hai — wohi odd- $180°$ jo har doosre rule ko shape karta hai.

Apply karo. Koi zeros nahi, toh woh sum $0$ hai. Do "other poles": conjugate $- 1 - 2 j$ aur $0$ .

$- 1 - 2 j$ se $p = - 1 + 2 j$ tak vector: change $(0, + 4)$ → straight upar point karta hai → angle $= 90°$ .
$0$ se $p = - 1 + 2 j$ tak vector: change $(- 1, + 2)$ → quadrant II → $arctan (\frac{2}{- 1}) = 180° - 63.43° = 116.57°$ . $θ_{d} = 180° + 0 - (90° + 116.57°) = 180° - 206.57° = - 26.57°.$ Positively name karte hue: $θ_{d} === 333.43° == (\equiv - 26.57°)$ . Branch upper complex pole se down-and-right ki taraf nikalti hai — Figure s03 mein cyan arrow.

Level 4 — Synthesis

Goal: design — target hit karne ke liye elements chunna ya add karna.

Exercise 4.1 (Damping ratio ke liye gain chunna)

$G (s) H (s) = \frac{1}{s ( s + 2 )}$ ke liye (parent ke Example 1 se), $K > 1$ ke liye poles hain $s = - 1 \pm j K - 1$ . Woh $K$ nikalo jo damping ratio $ζ = 0.5$ deta hai.

Recall Solution 4.1

Second-order pole $s = - σ \pm j ω_{d}$ ki natural frequency $ω_{n} = σ^{2} + ω_{d}^{2}$ hoti hai aur $ζ = \frac{σ}{ω _{n}}$ (pole negative real axis ke saath jo angle banata hai uska cosine). Yahan $σ = 1$ aur $ω_{d} = K - 1$ . $ζ = \frac{σ}{σ ^{2} + ω _{d}^{2}} = \frac{1}{1 + ( K - 1 )} = \frac{1}{K} .$ $\frac{1}{K} = 0.5$ set karo $\Rightarrow K = 2 \Rightarrow K === 4 == .$ Check: $K = 4$ par, poles hain $- 1 \pm j 3$ ; negative real axis se angle mein $cos θ = 1/2$ hai, yaani $θ = 60°$ , jisse $ζ = 0.5$ milta hai. ✔

Exercise 4.2 (Stabilise karne ke liye zero add karna — PD action)

Plant $\frac{1}{s ( s + 1 ) ( s + 2 )}$ $K = 6$ par unstable ho jaata hai (parent Example 2). Ek PD controller ek zero add karta hai. $s = - 0.5$ par ek zero insert karo, toh open loop ban jaata hai $\frac{s + 0.5}{s ( s + 1 ) ( s + 2 )}$ . Centroid recompute karo, escaping branches count karo, aur naya Routh test carry out karo critical gain nikalne ke liye. Stability par comment karo.

Recall Solution 4.2

Ab $n = 3$ poles ${0, - 1, - 2}$ , $m = 1$ zero ${- 0.5}$ , toh $n - m = 2$ branches escape karti hain (3 se kam). Centroid: $σ_{a} = \frac{( 0 - 1 - 2 ) - ( - 0.5 )}{2} = \frac{- 3 + 0.5}{2} = \frac{- 2.5}{2} === - 1.25 ==$ . Asymptote angles: $\frac{180° ( 2 k + 1 )}{2} = 90°, 270°$ — escaping branches ab straight up aur down jaati hain $60°$ par right half plane mein bend hone ki jagah.

Naya Routh test (woh edge case jo hum miss nahi kar sakte). Characteristic equation $D (s) + K N (s) = 0$ : $s (s + 1) (s + 2) + K (s + 0.5) = s^{3} + 3 s^{2} + (2 + K) s + 0.5 K = 0.$ Routh array:

s^{3} s^{2} s^{1} s^{0} 13 \frac{3 ( 2 + K ) - 0.5 K}{3} 0.5 K 2 + K 0.5 K 0

$s^{1}$ entry hai $\frac{6 + 3 K - 0.5 K}{3} = \frac{6 + 2.5 K}{3}$ . $K > 0$ ke liye yeh hamesha positive hai, aur $s^{0}$ entry $0.5 K > 0$ bhi. Toh koi bhi positive $K$ koi row vanish nahi karta — locus kabhi imaginary axis cross nahi karta. Comment: $- 0.5$ par zero ne ek aisa system jo $K = 6$ par unstable tha use saare $K > 0$ ke liye stable bana diya. Yahi reason hai kyun ek PD controller (jo left-half-plane zero contribute karta hai) ek stabilising tool hai — woh zero ek branch ko leftward drag karta hai aur escaping pair ko vertical asymptotes ke upar steer karta hai right half plane ki jagah.

Level 5 — Mastery

Goal: poora hand sketch, saare rules stitched, saare cases covered.

Exercise 5.1 (Complete sketch)

$G (s) H (s) = \frac{1}{s ( s + 2 ) ( s + 4 )}$ ko fully analyse karo: real-axis segments, asymptotes (angles + centroid), breakaway point aur uska gain, aur $j ω$ -crossing gain aur frequency.

Recall Solution 5.1

Poles/zeros: $n = 3$ poles ${0, - 2, - 4}$ , $m = 0$ . Teen branches; saare $n - m = 3$ infinity tak escape karte hain.

Real-axis segments (Rule 3). Right se: pole $0$ , pole $- 2$ , pole $- 4$ .

$(- 2, 0)$ : daayein ek → odd → ON.
$(- 4, - 2)$ : daayein do → even → OFF.
$(- \infty, - 4)$ : daayein teen → odd → ON.

Asymptotes (Rule 4). $σ_{a} = \frac{0 - 2 - 4}{3} = - 2$ ; angles $\frac{180° ( 2 k + 1 )}{3} = 60°, 180°, 300°$ .

Breakaway (Rule 5). $K = - D / N = - (s) (s + 2) (s + 4) = - (s^{3} + 6 s^{2} + 8 s)$ . $\frac{d K}{d s} = - (3 s^{2} + 12 s + 8) = 0 \Rightarrow 3 s^{2} + 12 s + 8 = 0.$ $s = \frac{- 12 \pm 144 - 96}{6} = \frac{- 12 \pm 48}{6} = - 2 \pm \frac{4 3}{6} = - 2 \pm 1.1547.$ Roots: $s = - 0.845$ aur $s = - 3.155$ . Sirf $s = - 0.845$ real segment par hai jo locus par ON hai ( $(- 2, 0)$ ), toh breakaway $= - 0.845$ . ( $- 3.155$ root OFF gap $(- 4, - 2)$ mein hai, toh reject kar do.) Gain at breakaway: $K = - ((- 0.845)^{3} + 6 (- 0.845)^{2} + 8 (- 0.845)) === 3.079 ==$ (3 d.p. tak).

$j ω$ -crossing (Rule 6). Char. eqn: $s^{3} + 6 s^{2} + 8 s + K = 0$ . Routh: $s^{1}$ entry hai $\frac{6 \cdot 8 - K}{6} = \frac{48 - K}{6}$ . Zero at $K === 48 ==$ . $s^{2}$ row se auxiliary $6 s^{2} + K = 0 \Rightarrow 6 s^{2} + 48 = 0 \Rightarrow s^{2} = - 8 \Rightarrow ω = 8 = 22 \approx== 2.828 ==$ .

Poori picture: do branches $0$ aur $- 2$ se nikalti hain, $- 0.845$ par milti aur breakaway karti hain, phir $60°/300°$ asymptotes ki taraf curve karti hain aur imaginary axis $\pm j 22$ par $K = 48$ pe cross karti hain; teesri branch $- 4$ se $- \infty$ tak $180°$ asymptote ke along left jaati hai. $0 < K < 48$ ke liye stable.

Exercise 5.2 (Degenerate / limiting case)

$G (s) H (s) = \frac{1}{s ^{2}}$ (origin par exactly do poles) ka locus kaisa dikhta hai? $K \to 0$ aur $K \to \infty$ limits aur stability discuss karo.

Recall Solution 5.2

$n = 2$ poles dono $s = 0$ par (ek double pole), $m = 0$ . Characteristic equation: $s^{2} + K = 0 \Rightarrow s = \pm j K$ .

$K \to 0$ : dono poles origin par hain (wahan se shuru karte hain).
Jaise $K$ badhta hai: poles straight upar aur neeche imaginary axis par $\pm j K$ ki taraf move karte hain.
$K \to \infty$ : woh $\pm j \infty$ tak axis ke along escape karte hain. Centroid check: $σ_{a} = \frac{0 + 0}{2} = 0$ , angles $90°, 270°$ — asymptotes imaginary axis HI hain, exact answer ke consistent. Stability: poles har $K > 0$ ke liye imaginary axis par hain (real part $= 0$ ), toh system saare gains ke liye marginally stable / oscillatory hai — kabhi asymptotically stable nahi. Yeh degenerate case hai jahan poora locus stability boundary par rehta hai. Ek pure double integrator sirf gain se stabilise nahi ho sakta; zero add karna padta hai (PD action).

Connections

Root locus — Evans' method, rules for sketching — parent jise yeh drill karta hai.
Characteristic equation & closed-loop poles — har exercise $D (s) + K N (s) = 0$ se shuru hoti hai.
Routh–Hurwitz stability criterion — har $j ω$ -crossing ke liye use hota hai.
PID / PD controller design — L4/L5 mein zero-adding synthesis.
Damping ratio and settling time — L4 gain-picking.
GNC control loops — jahan yeh gains ek real autopilot mein set hoti hain.
Bode plot & frequency response · Nyquist criterion — stability ke alternative views.