--- title: "R-H root distribution of s⁶+3s⁵+4s⁴+6s³+5s²+3s+2" subject: "Control Systems" chapter: 4 key: "semester_3/control_systems:rh-s6-3-4-6-5" source: https://www.bctengineer.com/q/semester_3/control_systems/rh-s6-3-4-6-5 site: BCT Engineering --- # R-H root distribution of s⁶+3s⁵+4s⁴+6s³+5s²+3s+2 Control Systems · Chapter 4: Stability - Asked in: 2079 Bhadra, 8 marks; 2074 Asoj, 8 marks; 2071 Chaitra, 4 marks; 2069 Chaitra, 6 marks; 2081 Baisakh, 6 marks (asked 5 times) - Video: [Example on Routh Array Stable System](https://www.youtube.com/watch?v=BApzhX2WBe0) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/rh-s6-3-4-6-5 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4 ## Question Construct Routh array and determine the stability of the system whose characteristic equation is s⁶ + 3s⁵ + 4s⁴ + 6s³ + 5s² + 3s + 2. Comment on the location of the roots of characteristic equation. ## Short answer The Routh array produces a row of zeros. The auxiliary equation 2(s²+1)²=0 gives repeated roots ±j1. No sign changes occur, so there are no RHP poles, but the repeated imaginary poles yield unstable response. Roots: 2 in LHP (−1, −2), 4 on the jω axis ( ±j1 twice). ## Answer **Result:** the system is **unstable**. It has 2 roots in the LHP, a repeated pair on the $j\omega$ axis at $\pm j1$, and no roots in the RHP. The Routh–Hurwitz criterion states that the number of roots in the right half of the s-plane equals the number of sign changes in the first column of the Routh array. A row of all zeros means there are roots placed symmetrically about the origin; these are found from the auxiliary equation. $$ q(s) = s^6+3s^5+4s^4+6s^3+5s^2+3s+2 = 0 $$ All coefficients are present and positive, so the necessary condition is satisfied. #### Routh array | Row | Col 1 | Col 2 | Col 3 | Col 4 | |---|---|---|---|---| | $s^6$ | 1 | 4 | 5 | 2 | | $s^5$ | 3 | 6 | 3 | 0 | | $s^4$ | $\frac{3(4)-1(6)}{3}=2$ | $\frac{3(5)-1(3)}{3}=4$ | 2 | | | $s^3$ | 0 | 0 | | | The $s^3$ row is all zeros. Auxiliary equation from the $s^4$ row: $$ A(s) = 2s^4+4s^2+2, \qquad \frac{dA}{ds} = 8s^3 + 8s $$ Replace the $s^3$ row by $8,\ 8$ and continue: | Row | Col 1 | Col 2 | Col 3 | |---|---|---|---| | $s^3$ | 8 | 8 | | | $s^2$ | $\frac{8(4)-2(8)}{8}=2$ | $\frac{8(2)-2(0)}{8}=2$ | | | $s^1$ | 0 | | | The $s^1$ row is zero again. New auxiliary equation $A_1(s) = 2s^2+2$, $\frac{dA_1}{ds} = 4s$: | Row | Col 1 | |---|---| | $s^1$ | 4 | | $s^0$ | 2 | First column: $1,\ 3,\ 2,\ 8,\ 2,\ 4,\ 2$. **No sign change**, so no root is in the right half plane. #### Roots of the auxiliary equation $$ 2s^4+4s^2+2 = 2(s^2+1)^2 = 0 \;\Rightarrow\; s = \pm j1,\ \pm j1 $$ So $\pm j1$ is a **repeated** pair on the imaginary axis. Dividing $q(s)$ by $(s^2+1)^2$ leaves $s^2+3s+2 = (s+1)(s+2)$. #### Location of roots | Region | Number | Roots | |---|---|---| | Right half plane | 0 | none | | On $j\omega$ axis | 4 | $\pm j1$ (each twice) | | Left half plane | 2 | $-1,\ -2$ | #### Comment on stability There is no RHP root, but the imaginary-axis roots are **repeated**. Repeated poles on the $j\omega$ axis give terms like $t\sin t$ in the response, which grow without limit. Therefore the system is **unstable**. If the $j\omega$ roots were non-repeated, it would only be marginally stable. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When the whole s^3 row is zero, what is done next? A. Stop; the system is unstable B. Form the auxiliary equation from the row above, differentiate it, and use its coefficients in the zero row C. Multiply the s^4 row by -1 D. Replace the row with ones Answer: B. The auxiliary equation A(s) from the s^4 row is differentiated and its coefficients replace the zero row. (Key point: Zero row handling) 2. The auxiliary equation 2s^4+4s^2+2=0 gives A. s = +-1, +-j B. s = +-j1 each repeated twice C. s = -1, -2 D. s = +-j2 Answer: B. 2(s^2+1)^2 = 0 so s = +-j1, each twice. (Key point: Auxiliary roots) 3. How many roots lie in each region (RHP, jw axis, LHP)? A. 2, 2, 2 B. 0, 2, 4 C. 0, 4, 2 D. 1, 2, 3 Answer: C. No sign changes means no RHP roots; four roots at +-j1 on the axis; the rest are -1 and -2. (Key point: Root count) 4. Why is the system unstable though no root is in the RHP? A. The first column has a zero B. The jw-axis roots are repeated, producing t sin t terms that grow C. The roots -1 and -2 are too close D. Because the order is 6 Answer: B. Repeated imaginary roots give terms like t sin(t), which grow without bound; simple ones would be only marginally stable. 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