1
$\begingroup$

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?

|improve this question
$\endgroup$
  • $\begingroup$ Is $s10$ a typo, or is that "s times 10"? $\endgroup$ – Jason R Apr 23 '15 at 15:56
  • $\begingroup$ it is an s/10 sir. $\endgroup$ – minimal risk Apr 23 '15 at 15:58
  • $\begingroup$ @minimalrisk do you know how to get the closed loop transfer function and how to get its poles? $\endgroup$ – fibonatic Apr 23 '15 at 17:07
  • $\begingroup$ yes, its $$ 1+K(s)*G(s)$$ and the poles are -2, -10, and the two zeroes are in -1 $\endgroup$ – minimal risk Apr 23 '15 at 17:17
2
$\begingroup$

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,

                                  Closed loop block diagram

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.

|improve this answer
$\endgroup$
  • $\begingroup$ Nice answer; welcome to the site. $\endgroup$ – Jason R Apr 24 '15 at 0:34
  • $\begingroup$ Thanks, but I think it missed my question. Lets have a look at the denominator We got 2 poles,2 zeros, thus we conclude that each pole has its zero buddy that it tends to as K goes to negative infinity. We have no asymptotes, so no pole is going to infinity. Here comes the But, for the negative K, root locus diagram, by rule number 5, the portion of the real axis of the right of the -1 zeroes and the portion of the real axis of the left of the -10 poles belongs to the $$RL$$, plus we have no breach/escape points beyond these points, so a loop is out of consideration. $\endgroup$ – minimal risk Apr 24 '15 at 6:05
  • $\begingroup$ @minimalrisk Could you maybe elaborate what you mean by rule number 5, because not all literature uses the same (order of) rules for drawing root locus. And these rules are mainly useful to estimate it for higher order systems. For a second order system you can calculate it analytically. $\endgroup$ – fibonatic Apr 24 '15 at 9:41
  • $\begingroup$ Rule number 5 states that the portion of the real axis is part of the RL if and only if the number of open loop poles and zeroes to its right is an even number. $\endgroup$ – minimal risk Apr 24 '15 at 10:38
  • $\begingroup$ @minimalrisk I added an expression for the poles of the closed loop system, from which it will become obvious why it tends $\pm\infty$. For simple second order systems like this it can be worth it to finding the analytical solution for the poles. $\endgroup$ – fibonatic Apr 24 '15 at 11:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.