Root locus method, deriving the position of the asymptote centroid

Question

In the Root Locus Method, the linear asymptotes are centered at a point on the real axis given by $$ \sigma_A=\frac{\sum_{j=1}^n(-p_j)-\sum_{i=1}^M(-z_i)}{n-M}, $$ where $p_j$ are the $j$th open-loop poles and $z_i$ are the $i$th open-loop zeros, $n$ and $M$ are the number of open-loop poles and zeros, respectively.

So I was reading about the derivation of the position of the asymptote centroid $\sigma_A$, on Modern Control Systems 20th Edition, by Robert H. Bishop, Page 452 (shown as the following screenshot).

I can't figure out how $$ \frac{s^M+b_{M-1}s^{M-1}}{s^n+a_{n-1}s^{n-1}} $$ is reduced into $$ \frac{1}{s^{n-M}+(a_{n-1}-b_{M-1})s^{n-M-1}}. $$

Can someone elaborate the reasoning behind this? Please help!

Answer 1

This is probably polynomial long-division to two terms, treating the denominator with higher degree (since the Transfer function is proper). In particular, you can verify that the denominator poly $s^n + a_{M-1} s^{n-1} + \dots + a_0$ is equal to $$ (s^{n-M}+(a_{n-1}-b_{M-1})s^{n-M-1})(s^M + b_{M-1} s^{M-1} + \dots + b_0) + O(s^{n-2}) $$ so we can substitute that into our root locus equation to yield $$ 1 + K \frac{s^M + b_{M-1} s^{M-1} + \dots + b_0}{(s^{n-M}+(a_{n-1}-b_{M-1})s^{n-M-1})(s^M + b_{M-1} s^{M-1} + \dots + b_0) + O(s^{n-2})} $$ Dividing through by the numerator gives $$ 1+ K \frac{1}{(s^{n-M}+(a_{n-1}-b_{M-1})s^{n-M-1}) + O(s^{n-M-2})} $$ With that we eliminate the latter, lower order terms to arrive at the desired expression.

Answer 2

I suspect it has something to do with an approximation, similar to the approximation of using only the first two terms. Namely, the two expressions can be related to each other using

$$ \underbrace{\frac{s^M + b_{M-1} s^{M-1}}{s^n + a_{n-1} s^{n-1}}}_{f_1(s)} = \underbrace{\frac{1}{s^{n-M} + (a_{n-1} - b_{M-1}) s^{n-M-1}}}_{f_2(s)} + x, $$

with

$$ x = \frac{b_{M-1} (a_{n-1} - b_{M-1})}{s^{n-M} (s + a_{n-1}) (s + a_{n-1} - b_{M-1})}. $$

In the limit of $K$ to infinity some of the closedloop poles go to infinity and approach those linear asymptotes. This means that for those asymptotes you only need to use an expression which approximates the initial equation, for finding the closedloop poles, well at large values for $s$. Now, it can be noted that for large $s$ the value of $f_2(s)$ goes to zero inversely proportional to $s^{n-M}$, while $x$ goes to zero inversely proportional to $s^{n-M+2}$. Thus for large $s$ the value of $f_2(s)$ dominates over $x$.

Though, there might be a more elegant way of showing this, by starting from

$$ \frac{s^M + b_{M-1} s^{M-1} + \cdots + b_0}{s^n + a_{n-1} s^{n-1} + \cdots + a_0}, $$

instead of $f_1(s)$.