--- title: "K range for roots left of s = −2" subject: "Control Systems" chapter: 4 key: "semester_3/control_systems:rh-k-s10-s-s3-left" source: https://www.bctengineer.com/q/semester_3/control_systems/rh-k-s10-s-s3-left site: BCT Engineering --- # K range for roots left of s = −2 Control Systems · Chapter 4: Stability - Asked in: 2066 Jestha (old course), 6 marks - Video: [Applications of R-H Criteria (Part 1)](https://www.youtube.com/watch?v=S0_Hd5M7aJY) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/rh-k-s10-s-s3-left - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4 ## Question For a process control system shown below, find the range of K for which the roots of characteristic equation are more negative than s = −2. [Figure: unity negative feedback system with forward path K(s+10)/[s(s+3)]] ## Short answer Shift the axis by substituting s = z − 2 into the characteristic equation and apply Routh criterion. All roots are more negative than s = −2 for K > 1. ## Answer "Roots more negative than $s=-2$" means all closed-loop poles must lie to the left of the line $s=-2$. Shift the imaginary axis by putting $s = z - 2$, then apply the Routh criterion in $z$. #### Characteristic equation $$ \begin{aligned} 1 + \frac{K(s+10)}{s(s+3)} &= 0 \\ s^2 + (3+K)s + 10K &= 0 \end{aligned} $$ #### Shift the axis: $s = z - 2$ $$ \begin{aligned} (z-2)^2 + (3+K)(z-2) + 10K &= 0 \\ z^2 - 4z + 4 + (3+K)z - 6 - 2K + 10K &= 0 \\ z^2 + (K-1)z + (8K-2) &= 0 \end{aligned} $$ #### Routh array in z | Row | Col 1 | Col 2 | |---|---|---| | $z^2$ | 1 | $8K-2$ | | $z^1$ | $K-1$ | 0 | | $z^0$ | $8K-2$ | | #### Conditions - $z^1$ row: $K - 1 > 0 \Rightarrow K > 1$ - $z^0$ row: $8K - 2 > 0 \Rightarrow K > 0.25$ Both must hold, so $$ K > 1 $$ **Check:** at $K = 1$, $s^2 + 4s + 10 = 0$ gives $s = -2 \pm j2.45$, exactly on the line $s=-2$. For $K>1$ the real part of the roots (or both real roots) is more negative than $-2$. **Answer:** all roots are more negative than $s = -2$ for $K > 1$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does 'roots more negative than s = -2' require? A. All roots on s = -2 B. All roots with real part between -2 and 0 C. All roots real D. All roots to the left of the line s = -2 Answer: D. The roots must lie to the left of the vertical line at s = -2. (Key point: Meaning of the condition) 2. Which substitution is used, and why? A. s = z + 2, to move the axis right B. s = z - 2, so the line s = -2 becomes the imaginary axis of z C. s = 2z, to scale the roots D. s = 1/z, to invert the plane Answer: B. Putting s = z - 2 makes the line s=-2 the new jω axis, so Routh in z tests the condition. (Key point: Axis shift) 3. After shifting, the characteristic equation becomes A. z^2+(K+3)z+10K=0 B. z^2+(K-1)z+(8K-2)=0 C. z^2+(K-1)z+(10K-4)=0 D. z^2+(3+K)z+(8K+2)=0 Answer: B. Substituting s=z-2 into s^2+(3+K)s+10K gives z^2+(K-1)z+(8K-2). (Key point: Shifted polynomial) 4. Which range of K results? A. K > 0.25 B. 0.25 < K < 1 C. K > 1 D. K < 1 Answer: C. K>1 from the z^1 row and K>0.25 from the z^0 row; both hold only if K>1. 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