control theoryroots
I'm studying control theory and I encountered the root locus, I know that It plots the roots of the characteristic equation but i've some questions.
Wouldn't the centroid be at -3? Ill put it in MATLAB and see what I get (dont have enough rep to add comments) EDIT: This seems right to me, I noticed you only have H(s), do you know if there was any G(s)? I tried to click the link to the video and it wouldn't load, best of luck.
$$ c = \frac{-1.5 + \frac{\sqrt5}{2} -1.5 - \frac{\sqrt5}{2} - 0}{1}$$
$$ c = \frac{-3}{1} = -3$$
Your approach seems correct. Namely, near a breakaway point two real poles should "meetup" and become a pair of complex conjugate poles or vice versa, thus at a breakaway point one would have repeated (real) poles. Repeated poles also means that the derivative of the characteristic equation with respect to $s$ at the value of the pole should be equal to zero. For example define the general openloop as
$$ H_{ol}(s) = k \frac{N(s)}{D(s)}, \tag{1} $$
with $N(s)$ and $D(s)$ polynomials in $s$. Thus the characteristic equation would be
$$ D(s) + k\,N(s) = 0. \tag{2} $$
The derivative of the characteristic equation would be
$$ D'(s) + k\,N'(s) = 0, \tag{3} $$
with $X'(s)$ denoting the derivative of $X(s)$ with respect to $s$. However, inorder to assure that $s$ is a pole of $(2)$ one can indeed use
$$ k = -\frac{D(s)}{N(s)}. \tag{4} $$
It can be noted that the $k$ in $(2)$ should not change value and thus do not contribute additional terms to the derivative in $(3)$. Substituting $(4)$ into $(3)$ yields
$$ D'(s) - \frac{D(s)}{N(s)}N'(s) = \frac{D'(s)\,N(s) - D(s)\,N'(s)}{N(s)} = 0. \tag{5} $$
Since $N(s)$ is a polynomial $(5)$ is equivalent to
$$ D'(s)\,N(s) - D(s)\,N'(s) = 0. \tag{6} $$
Even though it is stated that $k$ should remain constant, it can be shown that $(6)$ is equivalent to setting the derivative of $k$ with respect to $s$ to zero. Namely, by using the quotient rule it can be shown that the derivative of $k$ is equal to
$$ k' = -\frac{D'(s)\,N(s) - D(s)\,N'(s)}{N(s)^2}. \tag{7} $$
When using again that $N(s)$ is a polynomial yields that setting $(7)$ to zero is equivalent to $(6)$.
Substituting your openloop transfer function does indeed yield
$$ 2s^3+18s^2+48s+38=0. \tag{8} $$
The only thing I could remark on is that you incorrectly rounded the corresponding roots of $(8)$. For example the first root rounded to two decimal places should be $s=-1.47$.
It can be noted that the other roots of $(8)$ are also breakaway points. Only, if you substitute those values into $(4)$ yields negative values for $k$, while normally a rootlocus plot considers only $k\in [0,\infty)$.
Best Answer
Since you are studying root locus, I hope you are well aware of what is characteristic equation. It is the equation formulated by equating the denominator polynomial, of the transfer function of any system, equal to zero. Since it is the denominator polynomial, its roots are nothing but the poles of the system. And because root locus is drawn for a closed loop system, as gain K is varied from zero to infinity, the characteristic equation, in this case, gives us the poles of the entire closed loop system.
Now, what are poles? or what is the physical significance of poles? To answer, let me ask you a question...How do you obtain the transfer function of system?.................To obtain the transfer function, we first write the differential equations that govern the dynamics of the system(generally applying simple laws of physics....remember free body diagrams, KVL, KCl, etc?). Now, in this differential equation the coefficients of the derivative terms are always some physical parameter of the system such as, friction coefficient or viscosity, damping coefficient or moment of inertia...etc. These physical parameters or their combination (i.e. product/ratio of any two physical parameters) gives you the poles of the system. So, poles reflect the actual physical parameters, which affect the behavior of a system. So, when we say "pole locations are changed", what we actually mean is that "some parameter(s) of the system has changed".
The response of system to any input/disturbance depends on both poles and zeros of the system. Poles are responsible for the steady state behavior, while zeros affect the transient response. But, stability depends only on poles of the system. So, in root locus we trace only the poles of the system. Since poles of the system are equal to the roots of the characteristic equation, thus I think you have the answer to your second question.
Branches of a root locus represent the path followed by the roots as you keep changing the gain K. Actually root locus can be drawn for any variable parameter of the system, if it affects the solution of the characteristic equation. It is similar to the act of keeping a value of K in the charac. eqn. and finding the solution(roots/poles) and locating the poles on a graph paper. If we repeat this procedure for several values of K, we will ultimately get a path(branch/line) showing the change in pole locations, as K is varied.
If this answer is unsatisfactory...feel free to ask. My mail-id is acharya.deep@gmail.com