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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)\$:

enter image description here

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?

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    \$\begingroup\$ Are you sure that \$ C(z)=K(z-\beta)(z-\alpha) \$ and not some \$ C(z)=K\frac{z-\beta}{z-\alpha} \$? \$\endgroup\$ – jDAQ Jan 25 at 19:49
  • \$\begingroup\$ @jDAQ Sorry, you're right. I already put the right formula in the post. \$\endgroup\$ – Carmen González Jan 25 at 20:29
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Looking at the root locus with the damping factor and natural frequency isolines, we can better see where the poles move.

rlocus with damping factor and natural frequency isolines

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.

rlocus with damping factor and natural frequency isolines, and some rule-of-thumb marking of places with good settling times

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%!)

step response showing non-minimum phase and settling time of one

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  • \$\begingroup\$ 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\$? \$\endgroup\$ – Carmen González Jan 25 at 22:46
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    \$\begingroup\$ 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}\$. \$\endgroup\$ – jDAQ Jan 25 at 22:52
  • \$\begingroup\$ 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. \$\endgroup\$ – Carmen González Jan 25 at 23:08
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    \$\begingroup\$ You can do that, it's easier than what I suggested. \$\endgroup\$ – jDAQ Jan 25 at 23:19
  • \$\begingroup\$ 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? \$\endgroup\$ – Carmen González Jan 26 at 0:06

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