# Root locus of discrete time system

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 $$G(z) = \frac{1}{z(z-0.5)}$$ $$H(z) \cdot G(z) = \frac{1}{(z+0.5)(z-0.5)}$$ 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 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 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 $-0.75 < K < 1.25$

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

The root locus plot: 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.

• Shouldn't it be - 0,75 and +1.25 like cimarron said? – JavaForStarters Jan 22 '16 at 20:33
• Yes, I have fixed the sign already. – Suba Thomas Jan 22 '16 at 20:39
• Thanks. Is there a reason why the branches in your picture are colored like they are? – JavaForStarters Jan 22 '16 at 20:53
• 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.) – Suba Thomas Jan 22 '16 at 21:01