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How to plot root loci of loop gain G(z)= k/(z-0.5)(z-0.5)(z-0.5) in z plane?

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  • \$\begingroup\$ Just like you would in the s domain. The root locus is really just playing with polynomial ratios; it doesn't care about z or Laplace. Moreover, it's all about the relative locations of the poles -- so three poles each on top of the other has the same root locus no matter where those poles may be. The only real difference is what the pole locations mean in terms of physical behavior. \$\endgroup\$ – TimWescott Dec 30 '18 at 20:15
  • \$\begingroup\$ There is no z-plane for root loci, likewise with the s-domain. \$\endgroup\$ – KingDuken Dec 30 '18 at 20:22
  • \$\begingroup\$ the root locus design process in the z-plane is used to to obtain compensation \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 30 '18 at 21:07
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As @TimWescott points out, it doesn't matter if it's s-plane, z-plane, or any other-plane, the problem is one of finding the roots of a polynomial. In this particular case, however,the problem is easily solved without recourse to classical root locus analysis.

Generally, we can express the case of three repeated roots as solving the characteristic equation:

$$\small (z-\alpha)^3+K=0$$

where \$\small K\ge 0\$ is the loop gain.

This is conveniently achieved by solving for the roots of:

$$\small (z-\alpha)=(-K)^{1/3}=a+jb$$

that leads to the root locus: $$\small z=\alpha+(-K)^{1/3}=\alpha +a+jb$$

Thus:

$$\small (a+jb)^3=-K$$ Expanding: $$\small(a^3-3ab^2)+j(3a^2b-b^3)=-K $$

Equating the imaginary term to zero: $$\small b(3a^2-b^2)=0$$

which reveals the three solutions: $$\small b=0; \:\:b=\pm ja\sqrt 3$$ Equating the real term to \$\small -K\$ and setting \$\small b=0\$: $$\small a=(-K)^{1/3} $$

Thus, \$\small a\$ is negative.

The three loci are, therefore:

(1) \$\small z=\alpha +a \$, which starts at \$\small z=\alpha\$ and travels to \$\small -\infty\$ along the real axis

(2) \$\small z=\alpha +a +ja\sqrt{3}\$, which starts at \$\small z=\alpha\$ and travels to \$\small \infty\$ along a line at \$\small -60^o\$ to the real axis

(3) \$\small z=\alpha +a -ja\sqrt{3}\$, which starts at \$\small z=\alpha\$ and travels to \$\small \infty\$ along a line at \$\small 60^o\$ to the real axis

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