# Ending points of the root locus

Let $$D(s) + KN(s) = 0 \tag{1}$$where $$D(s)$$ and $$N(s)$$ are polynomials of $$s \in \mathbb{C}$$ such that $$\text{Deg}(D) = n, \ \text{Deg}(N) = m$$ and $$n\ge m$$. The root locus method tells us how the solutions of $$(1)$$ changes as we change parameter $$K$$ from $$K=0$$ to $$K = \infty$$.

I'm trying to understand these extreme cases. Let $$K \to 0$$ and we have $$D(s) = 0$$ so in this case, the set of the solutions is $$A = \{s \in \mathbb{R} | D(s) = 0\}$$. Now let $$K \to \infty$$, if we choose $$s$$ such that $$N(s) \not = 0$$ then the answer will be infinity. So we should choose $$s$$ such that $$N(s) = 0$$. In that case, if $$(1)$$ holds, we also have $$D(s) =0$$ which means $$N(s)$$ and $$D(s)$$ have the same factor but this isn't the result that should be obtained. Curiously, if we rewrite $$(1)$$ $$\frac{N(s)}{D(s)} = -\frac{1}{K} \tag{2}$$ and let $$K\to \infty$$, one possible case that $$(2)$$ holds is $$N(s) = 0$$ which gives us $$m$$ solutions and this doesn't require $$D(s)$$ have the same factor as $$N(s)$$! Why this happens? And why the first solution is wrong?

Example: Let $$D(s) = s^2 - 4$$ and $$N(s) = s + 3$$. So $$(1)$$ becomes $$s^2 - 4 + K(s+3) = 0$$If $$K \to \infty$$ and $$s = -3$$ then $$9 - 4 = 0$$ which is clearly wrong. On the other hand, rewriting the equation $$\frac{s+3}{s^2 - 4} = -\frac{1}{K}$$ If $$K \to \infty$$ and $$s = -3$$ then $$\frac{0}{9-4} = 0$$ which is true, of course.

• These visuals may offer some insight, though not directly addressing the question. Jan 5 at 12:23
• @OverLordGoldDragon Nice visualization! Thanks. Jan 5 at 13:10

The problem with your example is that $$\infty\cdot 0$$ isn't necessarily equal to zero. The only way to judge what is happening in the limit $$K\to\infty$$ is to divide the original equation by $$K$$:

$$\frac{D(s)}{K}+N(s)=0\tag{1}$$

Now it is obvious that for $$K\to\infty$$ the actual value of $$D(s)$$ is irrelevant, as long as it is finite. Consequently, the only necessary condition for $$(1)$$ to be true as $$K$$ becomes large is that $$N(s)=0$$.

This problem is a bit similar to the problem of determining the limit

$$\lim_{x\to\infty}\frac{x}{x+c}\tag{2}$$

The value of the limit $$(2)$$ is independent of the choice of the constant $$c$$, which becomes negligible compared to $$x$$. The same is true in the first equation of the question: the actual value of $$D(s)$$ becomes irrelevant compared to $$KN(s)$$.

• Thanks. You are right that $\infty\cdot 0$ isn't necessarily equal to zero but my mean was that first choose $s$ such that $N(s)$ becomes identically zero and then $K \to \infty$. In this way we are not faced with the problem of $\infty\cdot 0$. Jan 6 at 22:25
• @S.H.W: But that's not the way limits work. In the form $D(s)+KN(s)=0$ you can't directly deal with the limit $K\to\infty$, you have to divide by $K$ to see what's happening. Jan 7 at 12:33
• I think the problem is that we shouldn't evaluate the expression before taking the limit since we are looking for $s$ such that solves $D(s) + KN(s) = 0$ for different values of $K$. So $s$ is unknown and we should first take the limit and then see what values of $s$ solve the equation. Do you think that this is the problem? Jan 7 at 15:29
• @S.H.W: Yes, but the problem is that the limit is not finite, so you can't get much information from that limit. The only way is really to divide by $K$. Jan 7 at 16:07