Chapter 6: The Stability of Linear Feedback Systems

Exam-Focused Study Guide

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1. Why Stability Matters

• Stability is an important topic in control system design.

• An unstable closed-loop system has no practical value.

• A stable system is a dynamic system with a bounded response to a bounded input.

• Stability of a feedback system is directly related to the location of the roots of the characteristic equation of the system transfer function.

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2. The Concept of Stability (Cone Analogy)

The response to a displacement (or initial condition) leads to:

Case	Response	Interpretation
(a)	Decreasing	Stable
(b)	Neutral	Marginal-type
(c)	Increasing	Unstable

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3. Stability Defined by Pole Locations

A system is:

• Stable if and only if all poles of the transfer function have negative real parts (all poles in the left half-plane, LHP).
• Unstable if any pole is in the right half-plane (RHP).
• Marginally stable if the characteristic equation has simple roots on the imaginary axis $j\omega$ and all the other roots are in the LHP.

Important: Marginally stable systems are not considered stable.

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4. Quick Classification Examples

For each $q(s)=0$, classify the system:

$q(s)$	Roots	Result
$(s+5)(s+3)$	$-5,\ -3$	Stable
$(s+5+j)(s+5-j)$	$-5 \pm j$	Stable
$(s+5)(s+3)(s-1)$	$-5,\ -3,\ +1$	Unstable
$(s+10)(s^2+16)$	$-10,\ \pm 4j$	Marginally stable

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5. High-Order $q(s)$: Need a Method Without Computing Roots

For a high-order

$$q(s) = a_n s^n + a_{n-1} s^{n-1} + \cdots + a_1 s + a_0 = 0$$

finding roots manually is hard. We need a method that determines stability without computing the roots.

Expanding $q(s)=a_n(s-r_1)(s-r_2)\cdots(s-r_n)$:

$$\begin{aligned} q(s) =\ & a_n s^n - a_n(\text{sum of all roots})\,s^{n-1} \\ & + a_n(\text{sum of products of roots taken two at a time})\,s^{n-2} \\ & - a_n(\text{sum of products of roots taken three at a time})\,s^{n-3} + \cdots \\ & + a_n(-1)^n (r_1 r_2 \cdots r_n) = 0 \end{aligned}$$

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6. Necessary Conditions (NC) for Stability

From the expansion above:

• NC1: All coefficients of $q(s)$ must have the same sign. Reason: e.g., $a_n(-1)^n r_1 r_2 \cdots r_n$: if a root has a different sign, the product changes sign.
• NC2: All coefficients of $q(s)$ must be non-zero. Reason: if $r_1 r_2 \cdots r_n = 0$, one root is zero → system not stable.

These are necessary but NOT sufficient.

Example — NC1, NC2 Satisfied but Still Unstable

$$q(s) = s^3 + s^2 + 2s + 8$$

• NC1 ✓, NC2 ✓
• Factor: $q(s) = (s+2)(s^2 - s + 4)$
• Roots: $s_1 = -2,\quad s_{2,3} = \tfrac{1}{2} \pm \tfrac{\sqrt{15}}{2} j$
• Roots in RHP → unstable.

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7. The Routh–Hurwitz Stability Criterion

This is a necessary and sufficient condition for stability.

7.1 Build the Routh Array

For $q(s) = a_n s^n + a_{n-1}s^{n-1} + \cdots + a_1 s + a_0 = 0$:

Row	Col 1	Col 2	Col 3	…
$s^n$	$a_n$	$a_{n-2}$	$a_{n-4}$	…
$s^{n-1}$	$a_{n-1}$	$a_{n-3}$	$a_{n-5}$	…
$s^{n-2}$	$b_1$	$b_2$	$b_3$	…
$s^{n-3}$	$c_1$	$c_2$	$c_3$	…
⋮	⋮	⋮	⋮
$s^0$

7.2 Formulas for the Entries

$$b_1 = -\frac{\begin{vmatrix} a_n & a_{n-2} \\ a_{n-1} & a_{n-3} \end{vmatrix}}{a_{n-1}}, \qquad b_2 = -\frac{\begin{vmatrix} a_n & a_{n-4} \\ a_{n-1} & a_{n-5} \end{vmatrix}}{a_{n-1}}$$ $$c_1 = -\frac{\begin{vmatrix} a_{n-1} & a_{n-3} \\ b_1 & b_2 \end{vmatrix}}{b_1}, \qquad c_2 = -\frac{\begin{vmatrix} a_{n-1} & a_{n-5} \\ b_1 & b_3 \end{vmatrix}}{b_1}$$

7.3 The Criterion

• Number of roots in the RHP = number of sign changes in the first column of the array.
• System is stable ⇔ no sign changes in the first column.

7.4 Quick Example (Slide 21)

Row	Col 1
$s^3$	$1$
$s^2$	$1$
$s^1$	$-2$
$s^0$	$1$

Sign sequence: $+,+,-,+$ → 2 sign changes → 2 roots in RHP → unstable.

7.5 Worked Template — $n=4$

For $q(s)=a_4 s^4 + a_3 s^3 + a_2 s^2 + a_1 s + a_0$:

Row	Col 1	Col 2	Col 3
$s^4$	$a_4$	$a_2$	$a_0$
$s^3$	$a_3$	$a_1$	$0$
$s^2$	$b_1=\dfrac{-\begin{vmatrix}a_4&a_2\\a_3&a_1\end{vmatrix}}{a_3}$	$b_2=\dfrac{-\begin{vmatrix}a_4&a_0\\a_3&0\end{vmatrix}}{a_3}$	$0$
$s^1$	$c_1=\dfrac{-\begin{vmatrix}a_3&a_1\\b_1&b_2\end{vmatrix}}{b_1}$	$0$	$0$
$s^0$	$d_1=\dfrac{-\begin{vmatrix}b_1&b_2\\c_1&0\end{vmatrix}}{c_1}$	$0$	$0$

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8. The Four Possible Cases

Case 1 — No Element in the First Column is Zero

Standard Routh array. Apply the criterion directly.

Example: $q(s) = a_2 s^2 + a_1 s + a_0$

Row	Col 1	Col 2
$s^2$	$a_2$	$a_0$
$s^1$	$a_1$	$0$
$s^0$	$b_1 = \dfrac{a_1 a_0 - 0}{a_1} = a_0$	$0$

Stability conditions:

• NC1: $a_2,a_1,a_0$ all same sign
• NC2: $a_2,a_1,a_0$ all nonzero

If NC1 + NC2 are satisfied → all first-column entries have same sign → stable.

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Case 2 — Zero in First Column, Other Entries in That Row Nonzero

Method: replace the $0$ in the first column by a small positive $\epsilon$. Complete the array, then analyze the limit $\epsilon \to 0^+$.

Example: $q(s) = s^5 + 2s^4 + 2s^3 + 4s^2 + 11s + 10 = 0$

(NC1 ✓, NC2 ✓.)

Initial array:

Row	Col 1	Col 2	Col 3
$s^5$	$1$	$2$	$11$
$s^4$	$2$	$4$	$10$
$s^3$	$0 \to \epsilon$	$6$
$s^2$	$\dfrac{4\epsilon-12}{\epsilon}$	$10$
$s^1$	$d_1$	$0$
$s^0$	$10$

Compute $d_1$:

$$d_1 = \frac{6\!\left(\frac{-12}{\epsilon}\right) - 10\,\epsilon}{\frac{-12}{\epsilon}} = \frac{-72 - 10\epsilon^2}{-12} = 6 + \frac{\epsilon^2}{2} \approx 6$$

Also $\dfrac{4\epsilon-12}{\epsilon} \approx \dfrac{-12}{\epsilon}$ (large negative for small $\epsilon>0$).

Final first column:

Row	Value	Sign
$s^5$	$1$	$+$
$s^4$	$2$	$+$
$s^3$	$\epsilon$	$+$
$s^2$	$-12/\epsilon$	$-$
$s^1$	$6$	$+$
$s^0$	$10$	$+$

→ 2 sign changes → 2 poles in RHP → unstable.

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Case 2 Variant with Parameter $K$

$q(s) = s^4 + s^3 + s^2 + s + K = 0$, with NC1: $K>0$; NC2: $K \ne 0$.

Row	Col 1	Col 2	Col 3
$s^4$	$1$	$1$	$K$
$s^3$	$1$	$1$	$0$
$s^2$	$\epsilon$	$K$	$0$
$s^1$	$\dfrac{\epsilon - K}{\epsilon}$	$0$	$0$
$s^0$	$K$	$0$	$0$

If $K>0$, $\dfrac{\epsilon-K}{\epsilon}$ is large negative → 2 sign changes → unstable for all $K>0$.

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Case 3 — Entire Row of Zeros

Trigger: Occurs when factors such as $(s+\sigma)(s-\sigma)$ or $(s+j\omega)(s-j\omega)$ appear.

Method: The row preceding the zero row gives the auxiliary polynomial $U(s)$, which is a factor of the characteristic polynomial. The order of $U(s)$ is always even.

Example A — Range of $K$ for Stability

$q(s) = s^3 + 2s^2 + 4s + K = 0$. NC1: $K>0$. NC2: $K \ne 0$.

Row	Col 1	Col 2
$s^3$	$1$	$4$
$s^2$	$2$	$K$
$s^1$	$\dfrac{8-K}{2}$	$0$
$s^0$	$K$	$0$

Conditions:

• $K>0$ (NC1 / $s^0$ row)
• $\dfrac{8-K}{2} > 0 \Rightarrow K < 8$

$$\boxed{0 < K < 8 \text{ for stability}}$$

What if $K = 8$?

$q(s) = s^3 + 2s^2 + 4s + 8$

Row	Col 1	Col 2
$s^3$	$1$	$4$
$s^2$	$2$	$8$
$s^1$	$0$ ← row of zeros	$0$
$s^0$

Auxiliary polynomial (from $s^2$ row): $U(s) = 2s^2 + 8$.

Polynomial division:

$$\frac{s^3 + 2s^2 + 4s + 8}{2s^2 + 8} = \frac{s}{2} + 1$$

So:

$$q(s) = \left(\tfrac{s}{2}+1\right)(2s^2+8) = (s+2)(s+2j)(s-2j)$$

Roots: $-2,\ \pm 2j$ → simple roots on $j\omega$ → marginally stable.

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Case 4 — Repeated Roots on the Imaginary Axis

Critical pitfall: Routh–Hurwitz fails to detect this kind of instability — there will be no sign change, yet the system is unstable.

Example: $q(s) = (s+1)(s+j)^2(s-j)^2 = s^5 + s^4 + 2s^3 + 2s^2 + s + 1$

Row	Col 1	Col 2	Col 3
$s^5$	$1$	$2$	$1$
$s^4$	$1$	$2$	$1$
$s^3$	$\epsilon$	$\epsilon$	$0$
$s^2$	$1$	$1$
$s^1$	$\epsilon$	$0$
$s^0$	$1$

Auxiliary polynomial: $U(s) = s^4 + 2s^2 + 1 = (s^2+1)^2$ → repeated roots at $\pm j$.

No sign change → looks marginally stable, but it is unstable.

Why?

• A simple pair of poles on $j\omega$ → undamped sinusoid: $A\cos(\omega t + \phi)$ → marginally stable.
• A repeated pair of poles on $j\omega$ → response of the form

$$A\,t^{n}\cos(\omega t + \phi)$$

where $n=1$ for the second repeated pair, $n=2$ for the third, …

• Because $t^n$ grows without bound → unstable.

Memorize: Repeated roots on $j\omega$ ⇒ unstable. Routh–Hurwitz won’t show it via sign changes.

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9. Worked Example — Mars Rover

System (block diagram):

$$G_1(s) = \frac{10}{s+10},\qquad G_2(s) = \frac{1}{s^2},\qquad H(s) = Ks + 1$$

Negative feedback. Goal:

1. Find $K$ for stability.
2. Find $K$ such that one root of the characteristic equation is at $s = -5$.

(a) Stability via Routh–Hurwitz

$$T(s) = \frac{Y(s)}{R(s)} = \frac{10}{s^2(s+10) + 10(Ks+1)} = \frac{10}{s^3 + 10s^2 + 10Ks + 10}$$

Characteristic equation:

$$\Delta(s) = s^3 + 10s^2 + 10Ks + 10$$

Routh array:

Row	Col 1	Col 2
$s^3$	$1$	$10K$
$s^2$	$10$	$10$
$s^1$	$b_1$	$b_2$
$s^0$

$$b_1 = \frac{100K - 10}{10} > 0 \;\Rightarrow\; \boxed{K > 0.1}$$

(b) One root at $s = -5$

When $\Delta(s)$ is divided by $(s+5)$, the remainder must be zero.

$$\frac{s^3 + 10s^2 + 10Ks + 10}{s+5} = s^2 + 5s + (10K - 25)\quad \text{(quotient)}$$

Long division:

$$\begin{aligned} &s^3 + 10s^2 + 10Ks + 10 \\ &\underline{-(s^3 + 5s^2)} \\ &\quad 5s^2 + 10Ks \\ &\quad \underline{-(5s^2 + 25s)} \\ &\qquad (10K-25)s + 10 \\ &\qquad \underline{-((10K-25)s + 5(10K-25))} \\ &\qquad\qquad 0\quad \text{(must be zero)} \end{aligned}$$

Remainder $= 10 - 5(10K - 25) = 0$

$$10 - 50K + 125 = 0 \Rightarrow K = \frac{27}{10} = \boxed{2.7}$$

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10. Worked Example — Tracked Vehicle Turning Control

System with unity negative feedback, where:

$$G_c(s) = \frac{s+a}{s+1},\qquad G(s) = \frac{K}{s(s+2)(s+5)}$$

Goal: Select $K$ and $a$ such that:

1. The closed-loop system is stable.
2. Steady-state error for a ramp input is $\le 24\%$ of the magnitude of the command.

Step 1 — Closed-Loop Transfer Function and Characteristic Equation

$$T(s) = \frac{Y(s)}{R(s)} = \frac{G_c(s)G(s)}{1 + G_c(s)G(s)}$$ $$q(s) = s(s+1)(s+2)(s+5) + K(s+a) = s^4 + 8s^3 + 17s^2 + (K+10)s + Ka$$

Step 2 — Routh–Hurwitz

Row	Col 1	Col 2	Col 3
$s^4$	$1$	$17$	$Ka$
$s^3$	$8$	$K+10$	$0$
$s^2$	$b_1$	$Ka$
$s^1$	$c_1$	$0$
$s^0$	$Ka$

$$b_1 = \frac{8 \cdot 17 - (K+10)}{8} = \frac{126 - K}{8} > 0$$ $$c_1 = \frac{b_1(K+10) - 8Ka}{b_1} > 0$$

Stability conditions:

• (i) $K < 126$
• (ii) $(K+10)(126-K) - 64Ka > 0$
• (iii) $Ka > 0$

Step 3 — Steady-State Error for a Ramp

Ramp input: $R(s) = \dfrac{A}{s^2}$.

$$E(s) = R(s)\,[1 - T(s)] = \frac{A}{s^2}\cdot\frac{1}{1 + G_c(s)G(s)}$$ $$e_{ss} = \lim_{s \to 0} sE(s) = \frac{10A}{Ka}$$

Requirement:

$$e_{ss} = \frac{10A}{Ka} \le 0.24A \;\Rightarrow\; Ka \ge 41.6\overline{6}$$

Take $Ka = 42$.

Step 4 — Combine Conditions (using $Ka=42$)

Plug $Ka = 42$ into condition (ii):

$$(K+10)(126-K) - 64 \cdot 42 > 0$$ $$\Rightarrow K^2 - 116K + 1428 < 0$$

Solve quadratic for boundaries:

$$K = \frac{116 \pm \sqrt{116^2 - 4 \cdot 1428}}{2} = \frac{116 \pm \sqrt{7744}}{2} = \frac{116 \pm 88}{2}$$ $$K_1 = 14,\qquad K_2 = 102$$ $$\boxed{14 \le K \le 102 \quad \text{(with } Ka = 42\text{ )}}$$

Also need $K < 126$ (already implied) and $Ka > 0$ (✓ since $Ka=42$).

Acceptable Designs (Examples)

• $(K, a) = (70,\ 0.6)$
• $(K, a) = (50,\ 0.84)$
• … (any $(K,a)$ in the stable region with $Ka \ge 41.6$)

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11. Practical Tip — Finding Roots of a Polynomial Manually

When you must find roots by hand (e.g., to verify or to factor a low-order $q(s)$):

Example: $q(s) = s^3 + 4s^2 + 6s + 4$

Step 1 — Try integer candidates.

• $s = -1$: $q(-1) = -1 + 4 - 6 + 4 = 1 \ne 0$ → not a root.
• $s = -3$: $q(-3) = -27 + 36 - 18 + 4 = -5 \ne 0$ → not a root.
• Sign change between $-1$ and $-3$ → a root lies between them.
• $s = -2$: $q(-2) = -8 + 16 - 12 + 4 = 0$ ✓ → $s = -2$ is a root.

Step 2 — Polynomial division to get the remaining quadratic.

$$\frac{s^3 + 4s^2 + 6s + 4}{s+2} = s^2 + 2s + 2$$

Long division:

$$\begin{aligned} &s^3 + 4s^2 + 6s + 4 \\ &\underline{-(s^3 + 2s^2)} \\ &\quad 2s^2 + 6s + 4 \\ &\quad \underline{-(2s^2 + 4s)} \\ &\qquad 2s + 4 \\ &\qquad \underline{-(2s + 4)} \\ &\qquad\qquad 0 \end{aligned}$$

Step 3 — Solve the quadratic.

$$s^2 + 2s + 2 = 0 \Rightarrow s_{2,3} = -1 \pm j$$

Final roots: $s_1 = -2,\ s_{2,3} = -1 \pm j$.

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12. Quick-Reference Summary

Concept	Statement
Stability ↔︎ poles	All poles in LHP
Marginal stability	Simple roots on $j\omega$, all others in LHP — not stable
NC1	All coefficients of $q(s)$ have the same sign
NC2	All coefficients of $q(s)$ are nonzero
NC1 + NC2	Necessary, not sufficient
Routh–Hurwitz	Necessary and sufficient
#RHP roots	= #sign changes in first column of Routh array
Case 2 (zero in first column)	Replace 0 by $\epsilon \to 0^+$
Case 3 (row of zeros)	Form auxiliary polynomial $U(s)$ from preceding row; $U(s)$ is a factor of $q(s)$; order of $U(s)$ is even
Case 4 (repeated $j\omega$ roots)	Routh–Hurwitz fails to detect; system is unstable

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13. Exam Strategy Checklist

1. Check NC1 & NC2 first — fast reject of unstable systems.
2. Build Routh array; track signs in the first column.
3. If you see a single zero in column 1 → Case 2 ($\epsilon$method).
4. If you see a whole row of zeros → Case 3 (auxiliary polynomial). Then test whether roots are simple or repeated on $j\omega$.
5. For parametric stability (find range of $K$, or $K$ and $a$):
   • Force every first-column entry $> 0$.
   • Combine inequalities into a range / region.
6. For error specs combined with stability (Tracked Vehicle pattern):
   • Compute $e_{ss}$ from $E(s)=R(s)/(1+G_cG)$.
   • Convert spec into an inequality on parameters.
   • Intersect with Routh–Hurwitz inequalities.
7. Watch out for repeated $j\omega$ roots — Routh–Hurwitz cannot catch them.