The s^3 row is all zero, so we

Differentiate the auxiliary equation from the s^4 row

The first column 1, 2, 2, 8, 6, 2.67, 16 has

No sign change, so no RHP roots

The system is marginally stable because

The jw-axis roots are simple and none are in the RHP

Asked 4/27
2071 Shrawan · 6 marks

R-H root distribution of s⁶+2s⁵+8s⁴+12s³+20s²+16s+16

Short answer

Routh array shows no sign changes, so zero roots in RHP. An all-zero s³ row gives an auxiliary equation whose solution yields four jω-axis roots (±j1.414, ±j2). Factoring out leaves s²+2s+2 giving two LHP roots at −1±j1. Thus, RHP=0, LHP=2, jω axis=4.

Using R-H criteria, tell how many roots of polynomial are in right half s-plane, in left half s-plane and on jω axis. s⁶ + 2s⁵ + 8s⁴ + 12s³ + 20s² + 16s + 16 = 0

Result: RHP = 0, LHP = 2, $j\omega$ axis = 4. The system is marginally stable.

By the Routh–Hurwitz criterion, the number of sign changes in the first column equals the number of RHP roots. A zero row shows roots symmetric about the origin, which are found from the auxiliary equation.

s^6 + 2s^5 + 8s^4 + 12s^3 + 20s^2 + 16s + 16 = 0

Routh array

Row	Col 1	Col 2	Col 3	Col 4
$s^6$	1	8	20	16
$s^5$	2	12	16	0
$s^4$	$\frac{2(8)-1(12)}{2}=2$	$\frac{2(20)-1(16)}{2}=12$	16
$s^3$	$\frac{2(12)-2(12)}{2}=0$	$\frac{2(16)-2(16)}{2}=0$

The $s^3$ row is all zeros, so there are roots placed symmetrically about the origin. Auxiliary equation from the $s^4$ row:

A(s) = 2s^4 + 12s^2 + 16, \qquad \frac{dA}{ds} = 8s^3 + 24s

Row	Col 1	Col 2
$s^3$	8	24
$s^2$	$\frac{8(12)-2(24)}{8}=6$	$\frac{8(16)-0}{8}=16$
$s^1$	$\frac{6(24)-8(16)}{6}=2.67$	0
$s^0$	16

First column: $1,\ 2,\ 2,\ 8,\ 6,\ 2.67,\ 16$ . No sign change, so no root is in the RHP.

Roots of the auxiliary equation

\begin{aligned} 2s^4+12s^2+16 &= 2(s^2+2)(s^2+4) = 0 \\ s &= \pm j1.414,\ \pm j2 \end{aligned}

These four roots lie on the $j\omega$ axis and are not repeated. The other factor is

\frac{s^6+2s^5+8s^4+12s^3+20s^2+16s+16}{(s^2+2)(s^2+4)} = s^2+2s+2 \;\Rightarrow\; s = -1\pm j1

Region	Number	Roots
Right half plane	0	none
$j\omega$ axis	4	$\pm j1.414$ , $\pm j2$
Left half plane	2	$-1\pm j1$

The $j\omega$ -axis roots are simple, so the system is marginally stable: it oscillates at 1.414 rad/s and 2 rad/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

R-H root distribution of s⁵+3s⁴+5s³+4s²+s+3 · Control Systems, chapter 4 · asked 4×
R-H root distribution of s⁵+4s⁴+2s³+8s²+s+4 · Control Systems, chapter 4 · asked 4×
Relative and absolute stability; R-H of s⁶+2s⁵+8s⁴+12s³+20s²+16s+16 · Control Systems, chapter 4 · asked 2×

All Control Systems chapter 4 questions →

R-H root distribution of s⁶+2s⁵+8s⁴+12s³+20s²+16s+16

Routh array

Roots of the auxiliary equation

Related questions