# Design a digital controller for minimize the settling time

I have a system of digital control with unitary feedback, with the following transfer function:

$$G(z)=2\frac{(z-0.5)(z-2)}{(z+0.2)(z+1.5)}$$

I need to use a digital controller:

$$C(z)=K\frac{z-\beta}{z-\alpha}$$

And the purpose is minimize the settling time as much as possible and calculate the settling time for this new situation.

I design the root locus for $$\G(z)\$$:

Poles: $$\z=-1.5\$$; $$\z=-0.2\$$

Zeros: $$\z=0.5\$$; $$\z=2\$$

Break points:

$$\z=-0.615\$$; $$\z=0.948\$$

What are the values for $$\\alpha\$$ and $$\\beta\$$ for the controller and what is the value for the new settling time?

• Are you sure that $C(z)=K(z-\beta)(z-\alpha)$ and not some $C(z)=K\frac{z-\beta}{z-\alpha}$? – jDAQ Jan 25 at 19:49
• @jDAQ Sorry, you're right. I already put the right formula in the post. – Carmen González Jan 25 at 20:29

## 1 Answer

Looking at the root locus with the damping factor and natural frequency isolines, we can better see where the poles move.

For continuous-time systems, we have that the settling time is given by $$T_s = \frac{\ln(0.05\sqrt{1-\zeta^2})}{\zeta \omega_0},$$

As an approximation, to reach the smallest possible settling time you would want to maximize

$$\max_{K,\alpha, \beta} \zeta \omega_0.$$

Or, as a proper minimization,

$$\min_{K,\alpha, \beta} T_s = \min_{K,\alpha, \beta} \frac{\ln(0.05\sqrt{1-\zeta^2})}{\zeta \omega_0}.$$

I have marked a ballpark point where the settling time would be minimized, and by using a system where $$\ C(z)=K \$$ the best place is where they first break off of the real line.

By using a $$\ C(z)=K\frac{z-\beta}{z-\alpha} \$$, as you suggested, it is possible to (in theory) exactly cancel the pole $$\ z_p = -0.2\$$ and the zero $$\ z_z = 0.5\$$, then the system would become just

$$\hat{G} = 2K\frac{z-2}{z+1.5},$$

And the feedback system would be

$$H(z) = \frac{\hat{G}}{1+\hat{G}} = \frac{1}{\left(2K\frac{z-2}{z+1.5}\right)^{-1}+1} = \frac{2K(z-2)}{(z+1.5)+2K(z-2)},$$ $$H(z) = \frac{2K(z-2)}{(2K+1)z+1.5-4K}.$$

We then find $$\K\$$ such that the only remaining pole is at zero, which is

$$(2K+1)z+1.5-4K = \alpha z + 0 \Leftrightarrow 1.5-4K = 0 \Leftrightarrow K = \frac{1.5}{4} = 0.375.$$

That way,

$$H(z) = \frac{2 \cdot 0.375 (z-2)}{(2 \cdot 0.375 +1) z} = \frac{3}{7}\frac{(z-2)}{z} = \frac{3}{7}(1-2z^{-1}).$$

This way the settling time will be just $$\T_s = 1\$$, but do notice that the system is non-minimum phase ("goes in the wrong direction first" and has undershoot of 100%!)

• What were the poles and zeros that you canceled? I think I have to cancel the pole furthest from the origin (z = -1.5) and the zero furthest from the origin (z = 2), right? It's just that in the graph you made, I can't see any canceled pole or zero. How did you calculate the settling time equal to 2? Did you use the formula $|\sigma|=4.5/T_s$? – Carmen González Jan 25 at 22:46
• In my graph I used the root locus of $C(z)=K$ to exemplify the best places to land the poles, I did not plot/do the final solution. I would advise against cancelling unstable poles, try cancelling $z_p = -0.2$ and positioning a pole a bit more negative than the zero $z_z = 0.5$, the exact value of the pole will depend on the the equation that calculates the break point, make sure that it breaks at the origin, then figure $K_{critical}$. – jDAQ Jan 25 at 22:52
• Why can't I cancel the zero at z = 0.5 in addition to canceling the pole at z = -0.2? So I got a straight line at Root Locus that would go from -1.5 to 2. – Carmen González Jan 25 at 23:08
• You can do that, it's easier than what I suggested. – jDAQ Jan 25 at 23:19
• If I do that, I got a straight line between -1.5 and 2. What is the settling time in this case? Is the settling time zero? – Carmen González Jan 26 at 0:06