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I know this subject tends more to control theory but I am certain its part of the global knowledge basis for engineers, thus I believe I can find the answer here. I'm trying to draw Root Locus for $K<0$, but can't understand how the system behaves.
I have this open loop system:
$$ \frac{20(1+s)^2}{(1+\frac{s}{10})(1+\frac{s}{2})} $$
I know it has no asymptotes, but yet by one of the drawing rules for $k<0$, it tends to $\pm \infty$ in the real axis. How is it possible?
I will denote the given system as,
$$ G(s) = \frac{20(1+s)^2}{(1+\frac{s}{2})(1+\frac{s}{10})} $$
The input or reference signal to the closed loop will be denoted by $R(s)$ and the output by $Y(s)$. The controller will initially be denoted by $H(s)$, while in this case it is equal to $K$. The block diagram of the closed loop system can then be drawn as,
With its transfer function from input to output equal to,
$$ \frac{Y(s)}{R(s)} = \frac{G(s)}{1 + H(s) G(s)}. $$
For root locus analysis you want to find the poles of this transfer function as function of $H(s)$. The poles can be found by solving when the denominator is equal to zero. However the denominator (and numerator) are polynomials of $s$, so can't be fractions themselves. In order to ensure this I will slit $G(s)$ up in to two parts, its numerator, $G_n(s)$, and denominator, $G_d(s)$. The same can be done for $H(s)$. The resulting transfer function can then be written as,
$$ \frac{Y(s)}{R(s)} = \frac{\frac{G_n(s)}{G_d(s)}}{1 + \frac{H_n(s)}{H_d(s)} \frac{G_n(s)}{G_d(s)}} = \frac{H_d(s) G_n(s)}{H_d(s) G_d(s) + H_n(s) G_n(s)}. $$
As stated before, in this case $H(s)$ is equal to $K$, thus $H_d(s)$ is equal to one. By substituting in $K$ for $H_n(s)$ and the two polynomials, from this first equation, for $G_n(s)$ and $G_d(s)$, then the following expression can be obtained,
$$ \frac{Y(s)}{R(s)} = \frac{20(1+s)^2}{(1+\frac{s}{2})(1+\frac{s}{10}) + 20K(1+s)^2}. $$
The resulting denominator is a second order polynomial, so solving for the poles yields two solutions, which can be simplified to,
$$ p=-\frac{400 K \pm 4 \sqrt{1 - 225 K} + 6}{400 K + 1}. $$
For $K=-\frac{1}{400}$ the limits of both $p$ are $\pm\infty$ depending if you approach that value from the left or the right. Thus near $K=-\frac{1}{400}$ there will be values for the poles of the closed loop transfer function which will have a positive real values.
