--- title: "R-H stability of s⁵+2s⁴+2s³+4s²+s+2" subject: "Control Systems" chapter: 4 key: "semester_3/control_systems:rh-s5-2-2-4-1" source: https://www.bctengineer.com/q/semester_3/control_systems/rh-s5-2-2-4-1 site: BCT Engineering --- # R-H stability of s⁵+2s⁴+2s³+4s²+s+2 Control Systems · Chapter 4: Stability - Asked in: 2080 Baisakh, 6 marks - Video: [Special Case of Routh Array II : Numerical 2](https://www.youtube.com/watch?v=3E8MevPWNRA) — Learning Electronics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/rh-s5-2-2-4-1 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4 ## Question Check for stability of the system using R-H criterion whose characteristic equation is given by q(s) = s⁵ + 2s⁴ + 2s³ + 4s² + s + 2 = 0. Also determine the number of poles in RH plane and on imaginary axis. ## Short answer There are no right half-plane roots, but four repeated imaginary roots at s = ±j1. The system is unstable because repeated jω-axis poles cause unbounded response. ## Answer **Result:** RHP = 0, $j\omega$ axis = 4 ($\pm j1$ repeated), LHP = 1 ($s=-2$). The system is **unstable**. The Routh array is formed. A row of zeros is handled with the auxiliary polynomial, whose roots are the roots symmetric about the origin. $$ q(s) = s^5 + 2s^4 + 2s^3 + 4s^2 + s + 2 = 0 $$ #### Routh array | Row | Col 1 | Col 2 | Col 3 | |---|---|---|---| | $s^5$ | 1 | 2 | 1 | | $s^4$ | 2 | 4 | 2 | | $s^3$ | $\frac{2(2)-1(4)}{2}=0$ | $\frac{2(1)-1(2)}{2}=0$ | | Zero row. Auxiliary polynomial from the $s^4$ row: $$ A(s) = 2s^4 + 4s^2 + 2, \qquad \frac{dA}{ds} = 8s^3 + 8s $$ | Row | Col 1 | Col 2 | |---|---|---| | $s^3$ | 8 | 8 | | $s^2$ | $\frac{8(4)-2(8)}{8}=2$ | $\frac{8(2)-0}{8}=2$ | | $s^1$ | 0 (zero row again) | | New auxiliary polynomial $A_1(s) = 2s^2+2$, $\frac{dA_1}{ds} = 4s$: | Row | Col 1 | |---|---| | $s^1$ | 4 | | $s^0$ | 2 | First column: $1,\ 2,\ 8,\ 2,\ 4,\ 2$. There is **no sign change**, so no pole is in the RHP. #### Poles on the imaginary axis $$ 2s^4 + 4s^2 + 2 = 2(s^2+1)^2 = 0 \Rightarrow s = \pm j1,\ \pm j1 $$ The remaining factor is $\frac{q(s)}{(s^2+1)^2} = s + 2$, giving $s = -2$. | Region | Number of poles | |---|---| | Right half plane | 0 | | On $j\omega$ axis | 4 ($\pm j1$, repeated) | | Left half plane | 1 ($s=-2$) | #### Stability The poles on the $j\omega$ axis are **repeated**, so the response contains $t\sin t$ terms that grow with time. The system is **unstable**, even though no pole is in the RHP. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the s^3 row become all zeros for s^5+2s^4+2s^3+4s^2+s+2? A. The polynomial has a zero coefficient B. K is chosen so C. There is a pole at the origin D. The s^4 row is proportional to the s^5 row (2,4,2 = 2x(1,2,1)), so the cross-products vanish Answer: D. Since the s^4 row is a multiple of the s^5 row, every cross-product is zero, indicating roots symmetric about the origin. (Key point: Zero row) 2. How is the zero row replaced? A. By the derivative of the auxiliary polynomial from the row above (s^4 row) B. By epsilon C. By the s^5 row D. By ones Answer: A. The auxiliary polynomial A(s)=2s^4+4s^2+2 is differentiated and its coefficients (8, 8) fill the zero row. (Key point: Auxiliary polynomial) 3. What is the root distribution? A. RHP 0, jω axis 4 (±j1 repeated), LHP 1 (s = -2) B. RHP 2, LHP 3 C. RHP 0, jω 2, LHP 3 D. RHP 1, jω 2, LHP 2 Answer: A. A(s)=2(s^2+1)^2 gives ±j1 twice; the remaining factor is s+2. (Key point: Root distribution) 4. Why is the system unstable though no root is in the RHP? A. The jω roots are repeated, giving growing t sin t terms B. The roots at -2 are too slow C. A coefficient is negative D. The first column has a sign change Answer: A. Repeated imaginary-axis poles produce terms like t sin t that grow without bound. 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