I have a question about a probably easy task, but I can't find the right K.

I have the loop with an adjustable gain element given:

$$Q(z) = \frac{H(z)}{1+ K\cdot G(z)H(z)} $$

I have given:

$$H(z) = \frac{z}{z+0.5}$$ and $$G(z) = \frac{1}{z(z-0.5)}$$

enter image description here

$$H(z) \cdot G(z) = \frac{1}{(z+0.5)(z-0.5)}$$ which give the poles 0.5 and -0.5.

I'm not sure how to go from here. Do I have to count the zeros at +/- infinity and how can I get the argument?

The solution is that the system is stable for -0.75 < K < 1.25.


The root locus plot for positive K is enter image description here

You can see that intersections the unit circle at z=±1j. The Closed-loop pole is $$\frac{-1}{G(z)H(z)} = -(z+0.5)(z-0.5)$$ For z=±1j we have \$K=-(j+0.5)(j-0.5)=-(-1-.25)=1.25\$

For negative K the root locus is enter image description here This intersects the unit circle at z=±1. For that, we have \$K=-(1+0.5)(1-0.5) = -0.75\$

So, you're region of convergence is from \$-0.75 < K < 1.25\$

  • \$\begingroup\$ Thanks, I didn't know it splits up on the imaginary axis. Now I see. \$\endgroup\$ – JavaForStarters Jan 22 '16 at 20:34

The root locus plot:

enter image description here

The stability boundaries are where it crosses the unit circle. These are the points \$\pm i\$, \$ \pm1\$.

The value of K at these points can be computed using the following:

$$ K= \frac{-1}{G(z) H(z)} = -(z-0.5) (z+0.5) $$

When \$ z=\pm i \$, \$ K = 1.25 \$ and when \$ z=\pm 1 \$, \$ K = -0.75 \$.

In between these two values the loci is inside the unit circle and the system is stable.

  • \$\begingroup\$ Shouldn't it be - 0,75 and +1.25 like cimarron said? \$\endgroup\$ – JavaForStarters Jan 22 '16 at 20:33
  • 1
    \$\begingroup\$ Yes, I have fixed the sign already. \$\endgroup\$ – Suba Thomas Jan 22 '16 at 20:39
  • \$\begingroup\$ Thanks. Is there a reason why the branches in your picture are colored like they are? \$\endgroup\$ – JavaForStarters Jan 22 '16 at 20:53
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    \$\begingroup\$ Yes, the two colors are for the two loci. When K=-0.75, the two roots are at \$ \pm 1\$. As K increases the two roots move along these two loci, meet at the origin, and exit at \$ \pm i \$ for K =1.25. (It really does not matter if the orange one ends up down and the blue up. It is arbitrary which you choose to go up or down.) \$\endgroup\$ – Suba Thomas Jan 22 '16 at 21:01

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