Routh Stability Range for a Tape-Drive Servo: 22-Mec-A3 System Analysis and Control, December 2018, Q2

Question 2: Routh Stability Range for a Tape-Drive Servo (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Open-loop transfer function

$$G(s)=\frac{k(s+4)}{s(s+0.5)(s+1)(s^{2}+0.4s+4)}$$

with time measured in milliseconds, closed under unity feedback so that the characteristic equation is $1+G(s)=0$. The open-loop poles are $s=0$, $s=-0.5$, $s=-1$ and the lightly damped pair $s=-0.2\pm j1.99$ (from $s^{2}+0.4s+4$, i.e. $\omega_n=2$, $\zeta=0.1$), with a single open-loop zero at $s=-4$.

Find. The complete range of the gain $k$ for which all roots of the characteristic equation lie in the open left half plane.

Approach. Clear the denominator to obtain a fifth-order characteristic polynomial in $s$ with $k$ appearing linearly in the last two coefficients, build the Routh array symbolically in $k$, and impose positivity on the whole first column; the binding constraint turns out to be the $s^{1}$ row.

Form the characteristic polynomial. Multiplying $1+G(s)=0$ through by the open-loop denominator gives

$$s(s+0.5)(s+1)(s^{2}+0.4s+4)+k(s+4)=0$$

Expanding the product before adding the numerator term is essential; trying to read a Routh array off the factored form is not possible. Multiplying out step by step, $s(s+0.5)=s^{2}+0.5s$ and $(s+1)(s^{2}+0.4s+4)=s^{3}+1.4s^{2}+4.4s+4$, whose product is $s^{5}+1.9s^{4}+5.1s^{3}+6.2s^{2}+2s$. Therefore

$$\boxed{s^{5}+1.9s^{4}+5.1s^{3}+6.2s^{2}+(2+k)s+4k=0}$$

Note that $k$ enters only the $s^{1}$ and $s^{0}$ coefficients, which is typical when the gain multiplies a low-order numerator.

Apply the necessary condition first. Every coefficient must be positive for stability. Here $1$, $1.9$, $5.1$ and $6.2$ are fixed and positive, while $2+k>0$ and $4k>0$ together require $k>0$. So negative gains are excluded immediately, and the remaining work is to find the upper limit.
Build the Routh array in terms of $k$. The first two rows are the alternating coefficients; each later row is formed by the standard cross-multiplication.

Routh array for $s^{5}+1.9s^{4}+5.1s^{3}+6.2s^{2}+(2+k)s+4k$
Row	Column 1	Column 2	Column 3
$s^{5}$	$1$	$5.1$	$2+k$
$s^{4}$	$1.9$	$6.2$	$4k$
$s^{3}$	$b_{1}=1.8368$	$b_{2}=2-1.1053k$	—
$s^{2}$	$c_{1}=4.1312+1.1433k$	$4k$	—
$s^{1}$	$d_{1}$	—	—
$s^{0}$	$4k$	—	—

The individual entries come from

$$b_{1}=\frac{(1.9)(5.1)-(1)(6.2)}{1.9}=1.8368,\qquad b_{2}=\frac{(1.9)(2+k)-(1)(4k)}{1.9}=2-1.1053k$$

and then, using the $s^{4}$ and $s^{3}$ rows,

$$c_{1}=\frac{b_{1}(6.2)-(1.9)b_{2}}{b_{1}}=4.1312+1.1433k$$

Form the $s^{1}$ entry, which carries the binding constraint. Using the $s^{2}$ and $s^{3}$ rows,

$$d_{1}=\frac{c_{1}b_{2}-b_{1}(4k)}{c_{1}}=\frac{(4.1312+1.1433k)(2-1.1053k)-1.8368(4k)}{4.1312+1.1433k}$$

Since $c_{1}>0$ for every $k>0$, the sign of $d_{1}$ is the sign of its numerator. Expanding that numerator gives a downward parabola in $k$:

$$N(k)=-1.2636k^{2}-9.6267k+8.2623$$

Solve $N(k)=0$ for the stability boundary. Writing it as $1.2636k^{2}+9.6267k-8.2623=0$ and taking the positive root,

$$k=\frac{-9.6267+\sqrt{(9.6267)^{2}+4(1.2636)(8.2623)}}{2(1.2636)}=\frac{-9.6267+11.5946}{2.5273}=0.7787$$

Because $N(k)$ opens downward and is positive at $k=0$, it stays positive on $0<k<0.7787$ and turns negative beyond. The remaining first-column entries are not binding: $b_{2}=2-1.1053k>0$ only demands $k<1.8095$, a weaker requirement, and $c_{1}$ and $4k$ are positive throughout. Collecting the two active conditions,

$$\boxed{0<k<0.7787}$$

Locate the oscillation that appears at the limit. At $k=0.7787$ the $s^{1}$ row vanishes, the classical signature of a pair of roots on the imaginary axis. The auxiliary polynomial is formed from the row above:

$$A(s)=c_{1}s^{2}+4k=5.0215\,s^{2}+3.1147=0\;\Longrightarrow\;s=\pm j\sqrt{0.6203}=\boxed{\pm j0.7876}$$

So the closed loop does not merely lose stability at $k=0.7787$; it breaks into a sustained oscillation at $0.7876$ rad per millisecond. Because the question states that time is in milliseconds, that is $787.6$ rad/s, or about $125$ Hz, with a period near $8.0$ ms. Quoting the answer in real time like this is the useful engineering statement: it identifies the mechanical resonance ($\omega_n=2$ rad/ms, $\zeta=0.1$) as the culprit, since the crossing frequency sits well below it only because the lightly damped pair contributes phase so aggressively.

Question 2 — final results
Quantity	Value
Characteristic polynomial	$s^{5}+1.9s^{4}+5.1s^{3}+6.2s^{2}+(2+k)s+4k$
Lower limit (from $4k>0$)	$k>0$
Upper limit (from the $s^{1}$ row)	$k<0.7787$
Stability range	$\mathbf{0<k<0.7787}$
Frequency at the upper limit	$\omega=0.7876$ rad/ms $=787.6$ rad/s ($\approx125$ Hz)
Non-binding condition	$b_{2}>0\Rightarrow k<1.8095$

Concept & Theory

Topic. Routh–Hurwitz stability testing of a parameterised characteristic polynomial.

Key relations. Characteristic equation $1+G(s)=0$; Routh entries $b_i=\dfrac{a_1a_{2i}-a_0a_{2i+1}}{a_1}$ and similarly down the array; stability $\Leftrightarrow$ no sign change in the first column; a vanishing row means imaginary-axis roots, found from the auxiliary polynomial built on the row above.

Why this works. The Routh test counts right-half-plane roots without ever computing them, which is what makes a symbolic gain tractable: $k$ propagates through the array as a polynomial, and each first-column entry becomes an inequality in $k$. Because the array is built from the top down, the highest-order rows here are free of $k$ and only the lower rows constrain it. The $s^{1}$ row is quadratic in $k$ while the others are linear, so it is the row most likely to bind — and the auxiliary polynomial hands over the oscillation frequency at the boundary for no extra work, which is information a simple “stable or not” answer discards.

Common pitfall. Adding $k(s+4)$ to the factored denominator instead of expanding first — the array cannot be built from a factored form. A second frequent error is stopping at the necessary condition (all coefficients positive) and concluding stability: here every coefficient is positive for any $k>0$, yet the system is unstable for $k>0.7787$. Positive coefficients are necessary, never sufficient, beyond second order.

Source. Ogata, Modern Control Engineering, 5th ed., §5.6 (Routh's stability criterion, including the vanishing-row case); Franklin, Powell & Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed., §3.6; Nise, 8th ed., §6.2–6.4.

Question 2 of 7: Routh Stability Range for a Tape-Drive Servo

Question 2: Routh Stability Range for a Tape-Drive Servo (25 marks)