Routh's Stability Criterion

First, arrange the Characteristic Equation into the form:

$$a_n{s}^n+a_{n-1}{s}^{n-1}+...+a_{1}s+a_0=0$$

Then create a table as such:

$$\begin{array}{cccccc}{s}^n& a_n& a_{n-2}& a_{n-4}& \cdots & \cdots \\ s^{n-1}& a_{n-1}& a_{n-3}& a_{n-5}& \cdots & \cdots \\ s^{n-2}& b_{n-1}& b_{n-3}& b_{n-5}& \cdots & \\ s^{n-3}& c_{n-1}& c_{n-3}& c_{n-5}& \cdots & \\ ⋮& ⋮& ⋮& ⋮& & \\ ⋮& & & & & \\ s^0& & & & & \end{array}$$

Where: $b_{n-1}=-\frac{1}{a_{n-1}}\left|\begin{array}{cc}{a}_n& a_{n-2}\\ a_{n-1}& a_{n-3}\end{array}\right|$, $b_{n-3}=-\frac{1}{a_{n-1}}\left|\begin{array}{cc}{a}_n& a_{n-4}\\ a_{n-1}& a_{n-5}\end{array}\right|$, $b_{n-3}=-\frac{1}{a_{n-1}}\left|\begin{array}{cc}{a}_n& a_{n-5}\\ a_{n-1}& a_{n-6}\end{array}\right|$

$c_{n-1}=-\frac{1}{b_{n-1}}\left|\begin{array}{cc}{a}_{n-1}& a_{n-3}\\ b_{n-1}& b_{n-3}\end{array}\right|$, $c_{n-1}=-\frac{1}{b_{n-1}}\left|\begin{array}{cc}{a}_{n-1}& a_{n-5}\\ b_{n-1}& b_{n-5}\end{array}\right|$

Worked Examples:

Example

i) $s^3+s^2+2s+24$

Solution

\begin{align} \begin{array}{c|ccccc} s^3 & a_n & a_{n-2} \\ s^{2} & a_{n-1} & a_{n-3} \\ s^{1} & b_{n-1} \\ s^{0} & \cdots & \\ \end{array} \end{align} Note: $b_{n-1}=-\frac{1}{a_{n-1}}\left|\begin{array}{cc}{a}_n& a_{n-2}\\ a_{n-1}& a_{n-3}\end{array}\right|=-\left|\begin{array}{cc}1& 2\\ 1& 24\end{array}\right|=-(24-2)=-22$
\begin{align} \begin{array}{c|ccccc} s^3 & 1 & 2 \\ s^{2} & 1 & 24 \\ s^{1} & -22 \\ s^{0} & \cdots & \\ \end{array} \end{align} As we can see, there is one change in sign in the first column and therefore a single positive pole, and therefore the system is unstable

Example

$s^5+5s^4+10s^3+10s^2+329s+325$

Solution

\begin{align} \begin{array}{c|ccccc} s^5 & a_5 & a_3 & a_1 \\ s^{4} & a_4 & a_2 & a_0 \\ s^{3} & b_4 & b_2 & b_0\\ s^{2} & c_4 & c_2 \\ s^1 & d_4 & d_2 \\ s^0 & e_4 \end{array} \end{align} Deriving B…* $b_4=-\frac{1}{a_4}\left|\begin{array}{cc}{a}_5& a_3\\ a_4& a_2\end{array}\right|=-\frac{1}{5}\left|\begin{array}{cc}1& 10\\ 5& 10\end{array}\right|=\frac{1}{5}\ast 40=8$ $b_2=-\frac{1}{a_4}\left|\begin{array}{cc}{a}_5& a_1\\ a_4& a_0\end{array}\right|=-\frac{1}{5}\left|\begin{array}{cc}1& 329\\ 5& 325\end{array}\right|=\frac{1}{5}\ast 1320=264$ $b_0=-\frac{1}{a_4}\left|\begin{array}{cc}{a}_0& 0\\ a_4& 0\end{array}\right|=-\frac{1}{5}\left|\begin{array}{cc}1& 0\\ 5& 0\end{array}\right|=0$ Deriving C… $c_4=-\frac{1}{b_4}\left|\begin{array}{cc}{a}_4& a_2\\ b_4& b_2\end{array}\right|=-\frac{1}{8}\left|\begin{array}{cc}5& 10\\ 8& 264\end{array}\right|=-\frac{1}{8}\ast -1240=155$ $c_2=-\frac{1}{b_4}\left|\begin{array}{cc}{a}_4& a_0\\ b_4& b_0\end{array}\right|=-\frac{1}{8}\left|\begin{array}{cc}5& 325\\ 8& 0\end{array}\right|=-\frac{1}{8}\ast -2600=325$ Deriving D… $d_4=-\frac{1}{c_4}\left|\begin{array}{cc}{a}_4& a_2\\ c_4& c_2\end{array}\right|=-\frac{1}{155}\left|\begin{array}{cc}5& 10\\ 155& 325\end{array}\right|=-\frac{1}{155}\ast 75=-0.48$ We have determined this system is unstable $d_2=-\frac{1}{c_4}\left|\begin{array}{cc}{a}_4& a_0\\ c_4& c_0\end{array}\right|=-\frac{1}{155}\left|\begin{array}{cc}5& 325\\ 155& 0\end{array}\right|=325$ Deriving E… $e_4=-\frac{1}{d_4}\left|\begin{array}{cc}{a}_4& a_2\\ d_4& d_2\end{array}\right|=\frac{1}{0.48}\left|\begin{array}{cc}5& 10\\ -0.48& 325\end{array}\right|=\frac{1}{0.48}\ast 1,629.8=3,395.42$ $e_2=-\frac{1}{d_4}\left|\begin{array}{cc}{a}_4& a_0\\ d_4& d_0\end{array}\right|=\frac{1}{0.48}\left|\begin{array}{cc}5& 325\\ -0.48& 0\end{array}\right|=\frac{1}{0.48}\ast 156=325$ Therefore we can fill out our table: \begin{align} \begin{array}{c|ccccc} s^5 & 1 & 10 & 329 \\ s^{4} & 5 & 10 & 325 \\ s^{3} & 8 & 264 & 0\\ s^{2} & 155 & 325 \\ s^1 & -0.48 & 325 \\ s^0 & -3395.42 & 325 \end{array} \end{align} There is one sign change present in this table This means this system is unstable, facing exponential growth.

Example

Determine the values of K at which the system is stable using the Routh’s Stability Criterion

Solution

As we can see here we have a feedback loop, which can be reduced into $\frac{G(s)}{1+G(s)H(s)}$ Where our block is $G(s)$ and $H(s)$ is 1. This gives us a transfer of: $\frac{\frac{K}{s(s^2+s+1)(s+2)}}{1+\frac{K}{s(s^2+s+1)(s+2)}}=\frac{K}{K+s(s^2+s+1)(s+2)}$ $=\frac{K}{s^4+s^3+2s^2+2s^3+2s^2+2s+K}=\frac{K}{s^4+3s^3+4s^2+2s+K}$ This means that our analysis will be on: $s^4+3s^3+4s^2+2s+K$ \begin{align} \begin{array}{c|ccccc} s^4 & a_4 & a_2 & a_0 \\ s^{3} & a_3 & a_1 \\ s^{2} & b_3 & b_1 \\ s^{1} & c_3 & c_1 \\ s^0 & d_3 \\ \end{array} \end{align} Deriving B… $b_3=-\frac{1}{a_3}\left|\begin{array}{cc}{a}_4& a_2\\ a_3& a_1\end{array}\right|=-\frac{1}{3}\left|\begin{array}{cc}1& 4\\ 3& 2\end{array}\right|=-\frac{1}{3}\ast -10=3.333$ $b_1=-\frac{1}{a_3}\left|\begin{array}{cc}{a}_4& a_0\\ a_3& 0\end{array}\right|=-\frac{1}{3}\left|\begin{array}{cc}1& K\\ 3& 0\end{array}\right|=-\frac{1}{3}\ast -3K=K$ Deriving C… $c_3=-\frac{1}{b_3}\left|\begin{array}{cc}{a}_3& a_1\\ b_3& b_1\end{array}\right|=-\frac{1}{3.333}\left|\begin{array}{cc}3& 2\\ 3.333& K\end{array}\right|=-\frac{1}{3.333}\ast -3K-6.666=0.9K+2$ $c_1=-\frac{1}{b_3}\left|\begin{array}{cc}{a}_3& 0\\ b_3& 0\end{array}\right|=0$ Deriving D… $d_3=-\frac{1}{c_3}\left|\begin{array}{cc}{a}_3& a_1\\ c_3& c_1\end{array}\right|=-\frac{1}{0.9K+2}\left|\begin{array}{cc}3& 2\\ 0.9K+2& 0\end{array}\right|=2$ Which gives us the following Routh Table: \begin{align} \begin{array}{c|ccccc} s^4 & 1 & 4 & K \\ s^{3} & 3 & 2 \\ s^{2} & 3.333 & K \\ s^{1} & 0.9K+2 & 0 \\ s^0 & 2 \\ \end{array} \end{align} $\therefore 0.9K+2\ge 0$ so: $K\ge -2.222$

Worked Example 1, Page 41

Question

A plant with transfer function $G(s)$ is controlled by a controller of variable proportional gain $K_p$ and unity negative feedback. Given: $G(s)=\frac{1}{s(s^2+1.2s+1)}$ show that the value of the proportional gain $K_p$ has no influence on the steady state value of the response of the plant to a unit step input. What is the effect on the stability of the system of a negative value of $K_p$? Investigate the stability of the system for the values: i) $K_p=1$ ii) $K_p=1.5$

Solution

\begin{align} \frac{G(s)}{R(s)} &= \frac{\frac{K_p}{s(s^2+1.2s+1)}}{1+ \frac{K_p}{s(s^2+1.2s+1)}} = \frac{K_p}{s(s^2+1.2s+1) + K_p} \\ \text{as }r(t) &= 1,\text{ it follows that: }R(s) = \frac{1}{s} \\ \therefore G(s) &= \frac{K_p}{s^2(s^2+1.2s+1) + K_ps} \\ c_{ss} &= \lim_{s\rightarrow0}\{sC(s)\} = \lim_{s\rightarrow0}\{\frac{K_p}{s(s^2+1.2s+1) + K_p}\} = \frac{K_p}{K_p+1} \end{align} \begin{align} \frac{G(s)}{R(s)}&= \frac{K_p}{s(s^2+1.2s+1) + K_p} \\ Q(s) &= s^3+1.2s^2+s^1+K_p(s^0) \end{align} We can see that if $K_p$ (in this case also $a_0$ was 0, we would have to have a positive pole. This would mean that the system would be unstable, therefore we are unable to have a stable system with a negative value of $K_p$. \begin{align} \begin{array}{c|ccccc} s^3 & a_3 & a_1 \\ s^{2} & a_2 & a_0 \\ s^{1} & b_2 & b_0 \\ s^{0} & c_2 \\ \end{array} \end{align} $b_2=-\frac{1}{a_2}\left|\begin{array}{cc}{a}_3& a_1\\ a_2& a_0\end{array}\right|=-\frac{1}{1.2}\left|\begin{array}{cc}1& 1\\ 1.2& K_p\end{array}\right|=-\frac{1}{1.2}\ast (K_p-1.2)=1-0.83K_p$ $b_0=0$ $c_2=-\frac{1}{b_2}\left|\begin{array}{cc}{a}_2& a_0\\ b_2& b_0\end{array}\right|=-\frac{1}{1-0.83K_p}\left|\begin{array}{cc}1.2& K_p\\ 1-0.83K_p& 0\end{array}\right|=K_p$ \begin{align} \begin{array}{c|ccccc} s^3 & 1 & 1 \\ s^{2} & 1.2 & K_p \\ s^{1} & 1-0.83Kp & 0 \\ s^{0} & K_p \\ \end{array} \end{align} The system will remain stable for all $0\le K_p\le 1.2$. Therefore: i) $K_p=1$ would be stable as it satisfies the previously mentioned condition ii) $K_p=1.5$ wouldn’t be stable as it would force a sign change in the s^1 row, therefore indicating positive roots which cause an unstable system

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