1
\$\begingroup\$

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.

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

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

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

\$\endgroup\$
1
  • \$\begingroup\$ Amazing. Thank you so much! \$\endgroup\$
    – Or Perez
    Jan 25, 2023 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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