Worked examples — Pole placement — Ackermann's formula

3.5.34 · D3 · Physics › Guidance, Navigation & Control (GNC) › Pole placement — Ackermann's formula

Recall karo woh machine jise hum feed kar rahe hain: $K=\left[\begin{array}{cccc}0& \cdots & 0& 1\end{array}\right]{\mathcal{C}}^{-1}\alpha (A),\mathcal{C}=\left[\begin{array}{cccc}B& AB& \cdots & A^{n-1}B\end{array}\right].$ Yahan $A$ hai $n\times n$ system matrix, $B$ hai input column, $\alpha (s)$ hai desired characteristic polynomial (un poles se bana jo tum CHAHTE ho), aur $\mathcal{C}$ hai controllability matrix. Har letter parent note mein earn kiya gaya tha — ab hum sirf USE karte hain.

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Scenario matrix

Har Ackermann problem in cells mein se kisi ek mein land karta hai. Neeche ke examples cell ke naam se label hain taaki tum dekh sako ki poora space cover ho gaya hai.

Cell	Kya special hai	Covered by
A. Stable → faster	open loop already stable, bas speed chahiye	Ex 1
B. Unstable → stabilise	$\text{Re}>0$ wala pole jise left kheenchna hai	Ex 2
C. Complex target	desired poles hain $\sigma \pm j\omega$ (oscillation shaping)	Ex 3
D. Degenerate: uncontrollable	$\det \mathcal{C}=0$, formula BREAKS	Ex 4
E. Zero desired coefficient	ek pole exactly $s=0$ par (marginal)	Ex 5
F. Higher order $n=3$	selector row matter karta hai, hand-guessing fail ho jaati hai	Ex 6
G. Real-world word problem	cart / satellite English mein describe kiya gaya	Ex 7
H. Exam trap	galat polynomial / galat sign ka temptation	Ex 8

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Cell A — Stable system, bas faster chahiye

Forecast: kyunki system already stable hai, kya tumhe ek bada gain expect hai ya chhota? $k_1,k_2$ ka sign guess karo.

1. Desired polynomial. $\alpha (s)=(s+4)(s+5)=s^2+9s+20$, toh $a_1=9,a_0=20$. Kyun? Coefficients $9,20$ literally target dynamics HAIN — dekho Characteristic polynomial.
2. Controllability matrix. $AB=\left[\begin{array}{c}-1\\ -3\end{array}\right]$, toh $\mathcal{C}=\left[\begin{array}{cc}1& -1\\ 1& -3\end{array}\right]$, $\det =-3+1=-2\ne 0$. Kyun? Non-zero determinant ⟹ controllable ⟹ Ackermann legal hai.
3. Invert. ${\mathcal{C}}^{-1}=\frac{1}{-2}\left[\begin{array}{cc}-3& 1\\ -1& 1\end{array}\right]=\left[\begin{array}{cc}1.5& -0.5\\ 0.5& -0.5\end{array}\right]$. Kyun? Yeh "return ticket" hai canonical coordinates se wapas hamari coordinates mein.
4. $\alpha (A)$. $A$ diagonal hai, toh $\alpha (A)=\text{diag}(\alpha (-1),\alpha (-3))$. Compute: $\alpha (-1)=1-9+20=12$, $\alpha (-3)=9-27+20=2$. Toh $\alpha (A)=\left[\begin{array}{cc}12& 0\\ 0& 2\end{array}\right]$. Kyun? Diagonal matrix ke liye polynomial har eigenvalue par alag kaam karta hai — yeh dekhne ka sabse clean tarika ki $\alpha (A)\ne 0$.
5. Assemble. ${\mathcal{C}}^{-1}\alpha (A)=\left[\begin{array}{cc}1.5& -0.5\\ 0.5& -0.5\end{array}\right]\left[\begin{array}{cc}12& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}18& -1\\ 6& -1\end{array}\right]$. Phir $K=[01]\left[\begin{array}{cc}18& -1\\ 6& -1\end{array}\right]=[6-1]$. Kyun? Dono pieces multiply karna formula complete karta hai, aur selector row $[01]$ bottom row pick karta hai — kyunki canonical coordinates mein sirf last state actuated hota hai.

Verify: $BK=\left[\begin{array}{c}1\\ 1\end{array}\right][6-1]=\left[\begin{array}{cc}6& -1\\ 6& -1\end{array}\right]$, toh $A-BK=\left[\begin{array}{cc}-7& 1\\ -6& -2\end{array}\right]$. Trace $=-9=-(4+5)$ ✓, $\det =14+6=20=4\cdot 5$ ✓. Char poly $s^2+9s+20$ — poles $-4,-5$ par. Notice karo $k_2=-1$: already-stable second mode ke against push karna hamesha "positive" nahi hota.

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Cell B — Unstable pole, stabilise karo

Forecast: unstable pole $+1.73$ par hai. $-3$ tak pahunchne ke liye use zero cross karna hoga — expect karo ek sizeable gain us state par jo instability "feed" kar rahi hai.

1. Desired polynomial. $\alpha (s)=(s+3{)}^2=s^2+6s+9$, toh $a_1=6,a_0=9$. Kyun? Yeh coefficients target encode karte hain: dono poles $-3$ par, koi oscillation nahi.
2. Controllability. $AB=\left[\begin{array}{c}1\\ 0\end{array}\right]$, $\mathcal{C}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$, $\det =-1$, ${\mathcal{C}}^{-1}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$. Kyun? $\det =-1\ne 0$ ⟹ controllable ⟹ Ackermann legal; yahan $\mathcal{C}$ apna khud ka inverse hai.
3. $\alpha (A)$. $A^2=\left[\begin{array}{cc}0& 1\\ 3& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 3& 0\end{array}\right]=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$. Toh $\alpha (A)=A^2+6A+9I=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]+\left[\begin{array}{cc}0& 6\\ 18& 0\end{array}\right]+\left[\begin{array}{cc}9& 0\\ 0& 9\end{array}\right]=\left[\begin{array}{cc}12& 6\\ 18& 12\end{array}\right]$. Kyun? DESIRED polynomial use karo — parent ki mistake #2 wala trap. Dekho Cayley–Hamilton theorem.
4. Assemble. ${\mathcal{C}}^{-1}\alpha (A)=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}12& 6\\ 18& 12\end{array}\right]=\left[\begin{array}{cc}18& 12\\ 12& 6\end{array}\right]$. $K=[01]\cdot \left[\begin{array}{cc}18& 12\\ 12& 6\end{array}\right]=[126]$. Kyun? Product ${\mathcal{C}}^{-1}\alpha (A)$ canonical answer ko wapas hamare coordinates mein transport karta hai; selector row $[01]$ phir bottom row read karta hai jahan gains hain.

Verify: $A-BK=\left[\begin{array}{cc}0& 1\\ 3& 0\end{array}\right]-\left[\begin{array}{cc}0& 0\\ 12& 6\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -9& -6\end{array}\right]$. Char poly $s^2+6s+9$ ✓, roots $-3,-3$. Bada $k_1=12$ us $+3$ ko overpower karta hai jo instability feed kar raha tha — exactly jaise forecast kiya tha. Dekho Eigenvalues and system stability.

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Cell C — Complex target poles (oscillation shape karo)

Forecast: complex poles $\sigma \pm j\omega$ aate hain $s^2+2|\sigma |s+({\sigma }^2+{\omega }^2)$ se. Compute karne se pehle dono coefficients predict karo.

Figure — complex $s$-plane mein target poles. Do peele crosses hamare desired poles $-1\pm 2j$ hain. Blue arrow real part measure karta hai ($-1$, decay rate) aur red arrow imaginary part measure karta hai ($+2$, wobble frequency). Green half-plane woh hai jahan poles stability ke liye hone chahiye. Yeh picture dekhne se polynomial coefficients kisi bhi algebra se pehle pata chal jaate hain: decay pieces ka sum $s^1$ coefficient deta hai, origin se squared distance constant deta hai.

1. Desired polynomial. $\alpha (s)=(s+1{)}^2+2^2=s^2+2s+5$, toh $a_1=2,a_0=5$. Kyun? Complex pair $\sigma \pm j\omega$ hamesha $s^2-2\sigma s+({\sigma }^2+{\omega }^2)$ mein factor hota hai; $\sigma =-1,\omega =2$ ke saath yeh hai $s^2+2s+5$. Picture ke arrows exactly $\sigma$ aur $\omega$ hain.
2. Controllability. $AB=\left[\begin{array}{c}1\\ 0\end{array}\right]$, $\mathcal{C}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$, ${\mathcal{C}}^{-1}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$. Kyun? $\det \mathcal{C}=-1\ne 0$ ⟹ controllable, toh Ackermann mein inverse exist karta hai; yeh permutation matrix phir se self-inverse hai.
3. $\alpha (A)$. $A^2=0$ (double integrator nilpotent hai), toh $\alpha (A)=0+2A+5I=\left[\begin{array}{cc}5& 2\\ 0& 5\end{array}\right]$. Kyun? Matrix $A$ ko DESIRED polynomial mein substitute karo; kyunki $A^2$ vanish ho jaata hai, sirf $2A$ aur $5I$ terms bachte hain — sabse clean possible $\alpha (A)$.
4. Assemble. ${\mathcal{C}}^{-1}\alpha (A)=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}5& 2\\ 0& 5\end{array}\right]=\left[\begin{array}{cc}0& 5\\ 5& 2\end{array}\right]$. $K=[01]\left[\begin{array}{cc}0& 5\\ 5& 2\end{array}\right]=[52]$. Kyun? Product hamare coordinates mein wapas laata hai aur selector row $[01]$ bottom row pick karta hai jahan gains hain.

Verify: $A-BK=\left[\begin{array}{cc}0& 1\\ -5& -2\end{array}\right]$. Char poly $s^2+2s+5$ ✓. Discriminant $4-20=-16<0$ ⟹ complex roots $-1\pm 2j$ jaise design kiya tha. Compare karo LQR — optimal pole placement alternative se agar tum cost function ko $\omega ,\sigma$ choose karne dena chahte ho.

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Cell D — Degenerate: uncontrollable system

Forecast: $B$ sirf pehli state ko touch karta hai. Kya state 1 ka push kabhi state 2 ko influence kar sakta hai (woh decoupled hain)? Guess karo ki Ackermann chalega bhi ya nahi.

1. Pehle Controllability. $AB=\left[\begin{array}{c}-1\\ 0\end{array}\right]$, toh $\mathcal{C}=\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right]$, $\det =0$. Pehle kyun check karo? Agar $\det \mathcal{C}=0$ toh Ackermann mein inverse exist nahi karta — RUKO.
2. Interpret. $\mathcal{C}$ ki second row saari zeros hai: $-2$ mode unreachable hai. Uska pole $-2$ par fixed hai chahe $K$ kuch bhi choose karo. Kyun? Zero row matlab koi bhi combination $B,AB$ state 2 ko excite nahi kar sakta — woh direction actuator ko dikhai nahi deta.

Verify: koi bhi $K=[k_1{k}_2]$ try karo. $A-BK=\left[\begin{array}{cc}-1-k_1& -k_2\\ 0& -2\end{array}\right]$ — upper triangular, toh eigenvalues hain $-1-k_1$ (movable) aur $-2$ (frozen). Hum literally second pole place nahi kar sakte. Yahi "It always works" trap (mistake #3) concrete form mein hai. System yahan stable hai, toh yeh stabilisable hai; agar woh frozen pole $+2$ hota toh hum instability ke saath phanse rehte. Dekho Controllability and the controllability matrix.

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Cell E — Desired pole exactly origin par

Forecast: $s=0$ par pole matlab $\alpha$ ka constant term $a_0$ hoga... kaun sa value? Step 1 se pehle predict karo.

1. Desired polynomial. $\alpha (s)=(s-0)(s+5)=s^2+5s+0$, toh $a_1=5,a_0=0$. Kyun? Zero constant term hi is cell ka poora point hai — $s=0$ par root hone se $a_0=0$ force hota hai.
2. Controllability. $AB=\left[\begin{array}{c}1\\ -3\end{array}\right]$, $\mathcal{C}=\left[\begin{array}{cc}0& 1\\ 1& -3\end{array}\right]$, $\det =-1$, ${\mathcal{C}}^{-1}=\left[\begin{array}{cc}3& 1\\ 1& 0\end{array}\right]$. Kyun? $\det =-1\ne 0$ ⟹ controllable ⟹ inverse exist karta hai ⟹ Ackermann legal hai.
3. $\alpha (A)$. $A^2=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]=\left[\begin{array}{cc}-2& -3\\ 6& 7\end{array}\right]$. Phir $\alpha (A)=A^2+5A+0\cdot I=\left[\begin{array}{cc}-2& -3\\ 6& 7\end{array}\right]+\left[\begin{array}{cc}0& 5\\ -10& -15\end{array}\right]=\left[\begin{array}{cc}-2& 2\\ -4& -8\end{array}\right]$. $0\cdot I$ kyun include kiya? Yeh hammer karne ke liye ki $a_0=0$ identity term drop kar deta hai — ek common slip hai.
4. Assemble. ${\mathcal{C}}^{-1}\alpha (A)=\left[\begin{array}{cc}3& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}-2& 2\\ -4& -8\end{array}\right]=\left[\begin{array}{cc}-10& -2\\ -2& 2\end{array}\right]$. $K=[01]\left[\begin{array}{cc}-10& -2\\ -2& 2\end{array}\right]=[-22]$. Kyun? Product hamare coordinates mein wapas transport karta hai; selector row bottom-row gains read karta hai.

Verify: $A-BK=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]-\left[\begin{array}{cc}0& 0\\ -2& 2\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 0& -5\end{array}\right]$. Char poly $s(s+5)=s^2+5s$ ✓. Determinant $=0$ correctly reflect karta hai pole at origin. $0$ par pole matlab ek direction na badhta hai na ghatta — usable hai par marginal hai; flight hardware mein avoid karo.

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Cell F — Higher order ($n=3$), jahan guessing fail ho jaati hai

Forecast: CCF mein gains bottom row ko directly overwrite karte hain (parent Step 1). Toh $K$ desired coefficients ke equal hona chahiye. Unhe $(s+2{)}^3$ se guess karo.

1. Desired polynomial. $\alpha (s)=(s+2{)}^3=s^3+6s^2+12s+8$, toh $a_2=6,a_1=12,a_0=8$. Kyun? Teen equal poles $-2$ par; cube expand karne se yeh teen coefficients seedha milte hain.
2. Controllability. $AB=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$, $A^{2}B=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$. $\mathcal{C}=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]$ (anti-diagonal), $\det =-1$, aur ${\mathcal{C}}^{-1}=\mathcal{C}$ (self-inverse permutation). Kyun? $\det \ne 0$ ⟹ controllable; yeh anti-diagonal permutation row order reverse karta hai aur apna khud ka inverse hai.
3. $\alpha (A)$. Is $A$ ke saath: $A^2=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$, $A^3=0$. Toh $\alpha (A)=A^3+6A^2+12A+8I=\left[\begin{array}{ccc}8& 12& 6\\ 0& 8& 12\\ 0& 0& 8\end{array}\right]$. Kyun? Nilpotent shift matrix ki powers jaldi vanish ho jaati hain — polynomial upper-triangular band fill karta hai.
4. Assemble. ${\mathcal{C}}^{-1}\alpha (A)=\mathcal{C}\alpha (A)$. Kyunki $\mathcal{C}$ row order reverse karta hai, $\mathcal{C}\alpha (A)=\left[\begin{array}{ccc}0& 0& 8\\ 0& 8& 12\\ 8& 12& 6\end{array}\right]$. $K=[001]\cdot \text{(this)}=[8126]$. Kyun? Selector row $[001]$ last row pick karta hai — jo CCF mein exactly wahan hai jahan gains land karte hain.

Verify: kyunki hum CCF mein the, $K=[a_0{a}_1{a}_2]=[8126]$ ✓ — selector row ne sirf bottom row read ki, exactly parent ka Step 1 claim. Closed-loop bottom row $[-8-12-6]$ ban jaati hai, char poly deta hai $s^3+6s^2+12s+8=(s+2{)}^3$ ✓. Ise aankhon se guess karne ki koshish karo — nahi kar sakte; isliye hmare paas formula hai.

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Cell G — Real-world word problem

Forecast: "koi oscillation nahi" ⟹ real poles. "Time constant $0.5$ s" ⟹ pole $-1/0.5=-2$ par. Target poles padhne se pehle guess karo.

1. Spec ko poles mein translate karo. Real, repeated, $s=-2$ par (critically damped, $\tau =0.5\text{ s}$). $\alpha (s)=(s+2{)}^2=s^2+4s+4$. Kyun? Decay $e^{\lambda t}$ ka time constant hai $1/|\lambda |$; equal real poles = koi wobble nahi.
2. Yeh exactly parent Example 1 hai, toh $\mathcal{C}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$, $\alpha (A)=\left[\begin{array}{cc}4& 4\\ 0& 4\end{array}\right]$. Kyun? Same plant, same steps — controllability aur $\alpha (A)$ scratch se dobara karne ki zaroorat nahi.
3. Assemble. $K=[01]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}4& 4\\ 0& 4\end{array}\right]=[44]$. Kyun? Selector row transported $\alpha (A)$ ki bottom row pick karta hai — yahi gains hain.

Verify (units ke saath!): control law hai $u=-4p-4v$. Units: $[k_1]={\text{s}}^{-2}$ (force-per-position), $[k_2]={\text{s}}^{-1}$ (force-per-velocity) — dimensionally ek spring ($-4p$) plus ek damper ($-4v$): humne ek ideal spring-damper banaya. Closed loop $s^2+4s+4$, poles $-2,-2$, settling time $\approx 4\tau =2\text{ s}$ ✓, koi imaginary part nahi ⟹ koi oscillation nahi ✓. Poora plan bas yahi hai: poles choose karo → Ackermann chalao.

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Cell H — Exam trap (galat polynomial / galat sign)

Forecast: Cayley–Hamilton theorem se, $\chi (A)=0$. Isse kaun sa gain milta hai? Aur galat sign ek correct gain ke saath kya karta hai?

1. Trap 1 — open-loop polynomial. $\chi (A)=A^2+A-2I$. Compute karo $A^2=\left[\begin{array}{cc}2& -1\\ -2& 3\end{array}\right]$, phir $A^2+A-2I=\left[\begin{array}{cc}2& -1\\ -2& 3\end{array}\right]+\left[\begin{array}{cc}0& 1\\ 2& -1\end{array}\right]+\left[\begin{array}{cc}-2& 0\\ 0& -2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$. Kyun? Cayley–Hamilton guarantee karta hai ki matrix apni KHUD ki char poly ko zero kar deta hai. Toh $K=[01]{\mathcal{C}}^{-1}\cdot 0=[00]$ — bilkul koi feedback nahi, poles $1,-2$ par unchanged (abhi bhi unstable!). Yahi parent ki mistake #2 hai, quantify ki gayi.
2. Correct computation — Step 2a: desired polynomial. $\alpha (s)=(s+1{)}^2+1=s^2+2s+2$, toh $a_1=2,a_0=2$. Kyun? Poles $-1\pm j$ dete hain $\sigma =-1,\omega =1$, toh $s^2-2\sigma s+({\sigma }^2+{\omega }^2)=s^2+2s+2$ — DESIRED polynomial, open-loop wala nahi.
3. Step 2b: controllability. $AB=\left[\begin{array}{c}1\\ -1\end{array}\right]$, $\mathcal{C}=\left[\begin{array}{cc}0& 1\\ 1& -1\end{array}\right]$, $\det =-1$, ${\mathcal{C}}^{-1}=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$. Kyun? $\det \ne 0$ ⟹ controllable ⟹ Ackermann apply hota hai.
4. Step 2c: $\alpha (A)$. $A^2=\left[\begin{array}{cc}2& -1\\ -2& 3\end{array}\right]$, toh $\alpha (A)=A^2+2A+2I=\left[\begin{array}{cc}2& -1\\ -2& 3\end{array}\right]+\left[\begin{array}{cc}0& 2\\ 4& -2\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right]$. Kyun? $A$ ko DESIRED polynomial mein substitute karo — same computation jaisi parent Example 2 mein thi.
5. Step 2d: assemble. ${\mathcal{C}}^{-1}\alpha (A)=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right]=\left[\begin{array}{cc}6& 4\\ 4& 1\end{array}\right]$. $K=[01]\left[\begin{array}{cc}6& 4\\ 4& 1\end{array}\right]=[41]$. Kyun? Product hamare coordinates mein wapas laata hai; selector row bottom-row gains $[41]$ read karta hai.
6. Trap 2 — galat sign. Agar dost $u=+Kx$ use kare par $K=[41]$ rakhe, closed loop hai $A+BK=\left[\begin{array}{cc}0& 1\\ 6& 0\end{array}\right]$, char poly $s^2-6$, poles $\pm \sqrt{6}$ — shuruat se zyada unstable. Kyun? $u=+Kx$ ke saath closed loop hai $A+BK$, $A-BK$ nahi; fix yeh hai ki $K=-[41]=[-4-1]$ use karo.

Verify: correct case $A-BK=\left[\begin{array}{cc}0& 1\\ 2& -1\end{array}\right]-\left[\begin{array}{cc}0& 0\\ 4& 1\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -2& -2\end{array}\right]$, char poly $s^2+2s+2$, roots $-1\pm j$ ✓. Wrong-polynomial case: $\chi (A)=0$ ✓ toh $K=0$, poles $1,-2$ par rahte hain. Wrong-sign case: $A+B[41]$ ka char poly $s^2-6$ ✓ (unstable) — dono traps numerically confirm hote hain.

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Recall Self-test across all cells

Kis cell mein $\det \mathcal{C}=0$ hai, aur kya hota hai? ::: Cell D — Ackermann breaks; unreachable pole frozen ho jaata hai. CCF ($n=3$) mein, $K$ directly kiske barabar hota hai? ::: Desired polynomial coefficients $[a_0{a}_1{a}_2]$ (Ex 6, $=[8126]$). "Time constant $\tau$, koi oscillation nahi" spec ka matlab kaun se poles hain? ::: Repeated real poles $-1/\tau$ par (Ex 7). OPEN-loop char poly plug karne se kaun sa gain milta hai? ::: $K=0$ — kyunki $\chi (A)=0$ Cayley–Hamilton se (Ex 8). $s=0$ par desired pole hone se kaun sa coefficient zero force hota hai? ::: Constant term $a_0=0$ (Ex 5).

Dekho bhi Observer design and duality (Ackermann for observers) — wahi formula, transposed, observer poles place karta hai.