---
title: "K and T for roots left of s=−a"
subject: "Control Systems"
chapter: 4
key: "semester_3/control_systems:rh-k-t-left-of-a"
source: https://www.bctengineer.com/q/semester_3/control_systems/rh-k-t-left-of-a
site: BCT Engineering
---

# K and T for roots left of s=−a

Control Systems · Chapter 4: Stability

- Asked in: 2080 Bhadra, 6 marks
- 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-t-left-of-a
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4

## Question

For a unity feedback system having open loop transfer function G(s) = k/[s(1+Ts)], determine the values of K and T if it is desired that all the roots of closed loop system should lie in the region towards the left of s = −a.

## Short answer

Shift axis with s=z−a to obtain Tz^2+(1−2aT)z+(Ta^2−a+K)=0. Routh gives conditions: 0<T<1/(2a) and K>a(1−aT). This ensures all closed-loop roots lie to the left of s=−a.

## Answer

To make all roots lie to the left of $s=-a$, shift the imaginary axis to $s=-a$ by putting $s = z - a$. Then apply the Routh criterion to the new polynomial in $z$ (relative stability).

#### Characteristic equation

$$
\begin{aligned}
1 + \frac{K}{s(1+Ts)} &= 0 \\
Ts^2 + s + K &= 0
\end{aligned}
$$

#### Shift the axis: $s = z - a$

$$
\begin{aligned}
T(z-a)^2 + (z-a) + K &= 0 \\
Tz^2 + (1-2aT)z + (Ta^2 - a + K) &= 0
\end{aligned}
$$

#### Routh array in z

| Row | Col 1 | Col 2 |
|---|---|---|
| $z^2$ | $T$ | $Ta^2-a+K$ |
| $z^1$ | $1-2aT$ | 0 |
| $z^0$ | $Ta^2-a+K$ | |

For no sign change (all roots left of $z=0$, i.e. left of $s=-a$):

1. $T > 0$
2. $1-2aT > 0 \Rightarrow T < \dfrac{1}{2a}$
3. $Ta^2 - a + K > 0 \Rightarrow K > a - a^2T = a(1-aT)$

$$
0 < T < \frac{1}{2a}, \qquad K > a(1-aT)
$$

**Physical meaning:** the sum of the closed-loop roots is $-\frac{1}{T}$, so both roots can be left of $-a$ only if $\frac{1}{T} > 2a$. The gain must be large enough that the slower real root moves past $-a$.

**Example:** for $a=1$ and $T=0.25$: $K > 1(1-0.25)=0.75$. With $K=1$, $0.25s^2+s+1=0$ gives $s = -2, -2$, both left of $-1$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. To place all roots left of s = -a, we
   A. Substitute s = z - a and apply Routh to the polynomial in z
   B. Substitute s = z + a
   C. Apply Routh directly to the original polynomial
   D. Set the s^1 row to zero
   Answer: A. Shifting the axis to s=-a makes the question a standard stability test in z. (Key point: Method)

2. T(z-a)^2 + (z-a) + K expands to
   A. Tz^2 + (1+2aT)z + (Ta^2 + a + K)
   B. Tz^2 + (1-2aT)z + (Ta^2 - a + K)
   C. Tz^2 + z + K
   D. Tz^2 + (1-2aT)z + K
   Answer: B. Expanding gives Tz^2 - 2aTz + Ta^2 + z - a + K. (Key point: Shifted polynomial)

3. Positivity of the z^1 coefficient requires
   A. T > 1/(2a)
   B. T > 2a
   C. T < 1/(2a)
   D. T < a
   Answer: C. 1-2aT > 0 gives T < 1/(2a). (Key point: Condition on T)

4. The constant term condition gives
   A. K < a(1 - aT)
   B. K > a/T
   C. K > 0 only
   D. K > a(1 - aT)
   Answer: D. Ta^2 - a + K > 0 gives K > a - a^2 T = a(1-aT). (Key point: Condition on K)

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