enter image description here This one is pretty straightforward. We set K1 = 4 and want to bring out K2 as a common factor so we can plot the root locus for the system. The correct answer is:

$$\small G_{new}(s)=4K_2\frac{(s+6)}{(s-2)(s+6)+20}$$

I can't think of any algebraic trick to get this result.


1 Answer 1


To simplify calculations start with substituting \$a=s+6\$ and \$b=s-2\$.

The closed-loop transfer function is \$\frac{g}{1+g}\$, where \$g=k1(\frac{5}{a}+k2)\frac{1}{b}=\frac{k1 (5+a\ k2)}{a\ b}\$.

Thus the closed-loop transfer function \$\frac{g}{1+g}\$ becomes \$\frac{k1 (5+a\ k2)}{a\ b+k1 (5+a\ k2)}\$.

The closed-loop poles are obtained from the denominator of the closed-loop transfer function. That is,

$$a\ b+k1 (5+a\ k2)=0$$

$$a\ b+k1 \ 5+k1\ a\ k2=0$$

$$1+\frac{k1\ a\ k2}{a\ b+k1 \ 5}=0$$

Now substitute \$k1=4\$, \$a=s+6\$, and \$b=s-2\$, to get

$$1+4\ k2 \frac{s+6}{(s+6)(s-2)+20}=0$$

  • \$\begingroup\$ Amazing. Thank you so much! \$\endgroup\$
    – Or Perez
    Jan 25, 2023 at 20:54

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