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I'm trying to draw Root Locus by hand from the following block diagram and confirm the result I got from Routh Hurwitz's table:

enter image description here

The table of Routh Hurwitz that I got is the following:

enter image description here

And from the stability conditions I got K> 0.

I calculate the transfer function for the closed loop:

$$ G_0(s)=\cfrac{\cfrac{10}{(s+2)(s+5)}}{1+\cfrac{K}{s+K}\cfrac{10}{(s+2)(s+5)}}=\cfrac{10}{(s+2)(s+5)+\cfrac{K}{s+K}\times 10}\Leftrightarrow G_0(s)=\cfrac{10(s+K)}{(s+2)(s+5)(s+K)+10K}=\cfrac{10(s+K)}{(s^2+7s+10)(s+K)+10K}=\frac{10(s+K)}{s^3+7s^2+10s+Ks^2+7Ks+10K+10K}=\frac{10(s+K)}{s^3+(7+K)s^2+(7K+10)s+20K} $$

I have one zero in s=-K, but I'm trying to find the value of K. How can I draw Root Locus, if Root Locus will depend on that zero and I can't set that value, because I'm trying to figure out the K value?

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  • \$\begingroup\$ Iirc you need the open loop to draw the root locus. \$\endgroup\$
    – Huisman
    Jan 21, 2020 at 13:33

1 Answer 1

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Since you have the closed-loop system you need to put it into the from \$1+\frac{k n(s)}{d(s)}=0\ \$ to use the standard root locus techniques.

The roots depend on only the denominator of the closed-loop system.

$$ s^3+K s^2+7 s^2+7 K s+10 s+20 K =0$$

$$ s^3+7 s^2+10 s+K (s^2+7 s+20) =0$$

$$1+\frac{K (s^2+7 s+20)}{s^3+7 s^2+10 s}=0$$

And from the root locus it can also be seen that is system is stable for \$K>0\$.

enter image description here

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