So , I was reading a course book : ''Modern Control Engineering '' by Ogata and I came forward this statement that made me skeptical:
''The points of root locus only satisfiy the angle - condition. Closed - loop poles are the roots of the characteristic equation and therefore satisfiy both the magnitude and angle condition. If , we want to find the closed - loop poles for a given gain value K , then we have to ...''

I am not sure why does not every point of root locus correspond to a closed - loop pole for some K. Is it true? And if so , these points that are not closed - loop poles for any K , do they have any meaning/ physical/geometrical/control theory ?

I even have an assumption to this: Maybe he means , that if K has to be integer ( K may has to be integer , in ''real world applications''(?))? Then of course , root locus would have a discrete-represantion graph , so he means that if K - is integer , in root locus we connect the dots (to have a continoous represantantion) and thefore not every point of it is closed-loop pole and everything would make sense.
I even graphed, a root-locus in python : If you click an any point python returns the corresponding K for this point (which may be float) . So is this the reason? Let me know if I am wrong/right


1 Answer 1


Suppose we have a system $G(s) = \frac{1}{s(s+1)}$ and controller $K$ (this is purely a gain) and we close the loop: $$T(s) = \frac{KG(s)}{1+KG(s)} = \frac{K\frac{1}{s(s+1)}}{1+K\frac{1}{s(s+1)}}$$ $$ = \frac{K}{s^2+s+K} $$ As you might notice, the poles of this closed-loop equation depend on the value of $K$: $$s^2 + s + K = 0 \rightarrow s = -0.5\pm\sqrt{0.25 - K}$$ This means that one can influence the behaviour by only changing this $K$ to any arbitrary real value (imaginary might sound cool in simulation, but its kinda hard to put an imaginary voltage to a system for instance). The root locus plot represents for how the poles shift for changing values of $K$ (where $K > 0$). As you stated, if $K$ can only be an integer, you will indeed not get a continuous function, only dots at the places where $K$ exists. But as I stated earlier $K$ can by any real value, however for negative values you can quickly see the system becomes unstable.

EDIT: I have rewritten this part in a more elaborate proof. Instead of looking at a specific system, suppose an arbitrary system $$G(s) = \frac{N(s)}{D(s)}$$ The angle condition is the point at which the phase of the open loop system is an odd multiple of $-180^o$ or in other words: $$\mathcal{Im}\left\{\frac{KN(s)}{D(s)}\right\} = 0 ~~\text{ and }~~ \mathcal{Re}\left\{\frac{KN(s)}{D(s)}\right\} < 0$$ The magnitude condition is the point where the magnitude of the open loop transfer equals 1. If both the magnitude condition and the angle condition match, the denominator of the closed loop transfer function becomes 0 (which is something we tend to avoid at all cost). Now, suppose we take a value for $K$ and calculate the value for $s = s_0$ such that the open loop transfer equals $-1$: $$\frac{KN(s_0)}{D(s_0)} = -1 \rightarrow \frac{N(s_0)}{D(s_0)} = -\frac{1}{K}$$ We know this $s_0$ lies on the root locus. What you might also notice is that the sign does not change if I change $K$ to any positive, real value. Therefore, if $s$ lies on the root locus, the angle condition holds for any positive, real $K$. You could also reflect this in the bode plot, as $K$ only changes the magnitude, not the phase plot. As you might expect, this also implies the inverse holds: the angle condition holds for any value of $s$ that lies on the root locus of $G(s)$. The magnitude condition only holds for a finite set of values of $s$ on the root locus (the solution at which the characteristic equation equals 0). I hope I explained it a bit better, you can try also to just write out a couple cases and you quickly notice the same.

  • $\begingroup$ yes , but as you said every point of root locus satisfies $KG(s)=-1$ so it satifies both the angle condition and the magnitude , s can't be anything . s has for every K , a certain value in order $|G(s)K| =1$ .. doesn't it? I agree that every point of rlocus should satisfiy the angle condition , but I don't get why not the magnitude as well $\endgroup$ Dec 8, 2020 at 16:11
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    $\begingroup$ yes, in fact only a limited set of values for $s$ will set $KG(s)=-1$ for some $K$. This value is the point on the s-plane. Suppose $K=4.25$, then if and only if $s = -0.5\pm2j$ the open loop transfer function becomes -1. If we change the value of $K$, the solution of $s$ changes accordingly. However, any value $s$ which is on the root locus satisfies the angle condition, yet the magnitude condition is only satisfied if $s$ has the value that sets the characteristic equation to 0. $\endgroup$
    – Petrus1904
    Dec 8, 2020 at 16:39
  • $\begingroup$ do you mean , that he is suggesting that for a given K , not every point of the root locus , satisfies the magnitude condition (for a given K only some s , of the root locus will satisfiy the current charcteristic equation) but the angle - condition , is satisfied by all point - at any given K ? because yes that makes sense $\endgroup$ Dec 8, 2020 at 17:01
  • $\begingroup$ maybe the translation in my mother - tongue of the book is not so successful , because I felt like he implied that there are point of root locus which are not closed loop poles for any K ..and that made me wonder.. $\endgroup$ Dec 8, 2020 at 17:02
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    $\begingroup$ yes. exactly that! I have changed my answer with an elaborate proof for any system. As I stated, even only a couple points satisfy the magnitude condition. these points being the exact poles of the closed-loop system. The amount of points are equal to the order of the system. $\endgroup$
    – Petrus1904
    Dec 8, 2020 at 17:40

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