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My transfer function is \$\frac{4s + 2}{s(s-1)(s+2)} \$

Given that there are 3 poles and 1 zeroes, the amount of loci going to infinity is 3-1 = 2 loci = r. I calculated where the centroid of the asymptotes should be begin. The equation is the (sum of poles - sum of zeroes)/r = location of centroid on real axis. The centroid location I got was -.25, so I can't understand why MATLAB plotted the root locus otherwise. Given that there are 3 poles, there are to be 3 asymptotes. (1 on the real axis and 2 going off the opposite direction.) I illustrated the supposed asymptotes in black in the following image. enter image description here My problem is that the calculated asymptotes do not match up with my expectations. My calculated asymptotes are clearly not what MATLAB has calculated. Why and how so? I also did an angle computation using the formula (2q + 1)/r.

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  • \$\begingroup\$ "do not match up with my expectations" and those expectations are.....? \$\endgroup\$ – Andy aka Nov 14 '16 at 8:36
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The angle the asymptotes make with the real-axes is \$\frac{(2 k+1)\pi}{2} \$. The denominator is the number of finite poles minus the number of finite zeros (\$3-1\$). This turns out as \$\frac{\pi}{2}\$ and \$\frac{3 \pi}{2}\$.

Because of one finite zero, one of the loci ends up at that zero. Only the other two go off to \$\infty\$.

enter image description here

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