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I have an equation \begin{equation}{\text{L}\left(s\right)=\frac{K}{s(s+4)(s^2+6s+64)}} \end{equation}

and I am trying to determine it's root locus by hand.When I try plotting it with matlab, the root locus seems to cross the imaginary axis at about +/-5.06enter image description here

When I try to determine where the root locus will cross the imaginary axis by hand, I end up with two possible values for the imaginary axis crossing, either 5.06 like in the matlab plot or 3.52. Is there some way of discarding a value for the imaginary axis crossing? Why do I end up with two values for the imaginary axis crossing but matlab only ends up with one?

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    \$\begingroup\$ It would be helpful if you showed your manual calculations. \$\endgroup\$ – Glenn W9IQ May 18 '17 at 17:43
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I'm not sure about your calculation but Matlab yields the correct result. In this problem, the common approach is to use Routh-Hurwitz criterion and search for a row of zeros that yields the possibility for imaginary axis roots. For convert the system to the closed-loop transfer function, hence $$ \frac{K}{s^4 + 10s^3 + 88s^2 + 256s + K} $$

The Routh table is

$$ \begin{matrix} s^4 &&&& 1 &&&& 88 &&&& K \\ s^3 &&&& 10 &&&& 256 \\ s^2 &&&& 62.4 &&&& K \\ s^1 &&&& \frac{15974.4-10K}{62.4} \\ s^0 &&&& K \end{matrix} $$

The \$ s^1 \$ row is the only row that can yield a row of zeros. From the preceding row, we obtain

$$ \begin{align} &15974.4 - 10K = 0 \\ K &= \frac{15974.4}{10} = 1597.44 \end{align} $$

Now we take a look at the row above \$s^1\$ and construct the following polynomial, hence

$$ \begin{align} 62.4 s^2 + K &= 0 \\ 62.4 s^2 + 1597.44 &= 0 \\ s^2 &= \frac{-1597.44}{62.4} \\ s_{1,2} &= \pm j \sqrt{25.6} \\ s_{1,2} &= \pm j 5.0596 \\ \end{align} $$

The root locus crosses the imaginary axis at \$\pm j5.0596\$ at the gain \$K=1597.44\$. Consequently, the gain \$K\$ must be less than 1597.44 for the system to be stable.

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In order to determine \$K\$. We need to investigate:

\$1+KF(s)=0\implies 1+L(s)=0 \implies s(s+4)(s^2+6s+64)+K=0.\$

We know that \$s=j\omega\$. Plugging this into the equation and collecting the real and imaginary part we obtain.

\$\left(10\omega^3-256\omega \right)+j\left(\omega^4-88\omega^2+K \right)=0+j\cdot 0\$

By comparing the complex number on the right and left-hand side of the equation we obtain two equations:

\$10\omega^3-256\omega=0 \qquad \wedge \qquad \omega^4-88\omega^2+K=0.\$

Solving the first equation leads to \$\omega=0, \omega=\pm\frac{8}{5}\sqrt{10}\$. \$ \omega =0 \$, would lead to the trivial solution \$ K=0\$. As the second equation is even both \$\omega=\pm\frac{8}{5}\sqrt{10}\approx5.060\$ lead to the same result \$K=\frac{39936}{25}\approx1597.44\$, which is obvious because the root locus is symmetric with respect to the real axis.

Verifying this with MATLAB yields very similar results.

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  • \$\begingroup\$ you are reversing the real and the imaginary parts. \$\endgroup\$ – CroCo Dec 11 '18 at 22:10
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MATLAB is plotting the root locus only for positive values of \$K\$. Your analytical calculation may be considering both positive and negative values of \$K\$ and this is why you end up with two pair of points. Keep just the one that is obtained with \$K>0\$

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