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Let's say I have following first order system with transport delay

$$G(s) = \frac{4563}{s + 64.77}\cdot e^{-0.000455\cdot s}$$

My goal is to design in the frequency domain a PI controller in the form

$$D(s) = K_p + \frac{K_i}{s} = K_p + \frac{K_p}{T_i\cdot s} = \frac{K_p\cdot T_i\cdot s + K_p}{T_i\cdot s} = \frac{K_p\cdot(T_i\cdot s + 1)}{T_i\cdot s} = \frac{K_i\cdot(s\cdot T_i + 1)}{s}$$

for that system so that I have \$PM\geq 70^{\circ}\$ and \$\omega_{BW}\geq 691\,rad\cdot s^{-1}\$. I have followed these steps in the design:

  1. Bode plot of the \$G(s)\$

enter image description here

  1. Crossover frequency \$\omega_c\$

To be able to achieve \$PM = 70^{\circ}\$ I would need the magnitude intersects the \$0\,dB\$ at such a frequency (\$\omega_c\$) at which the phase equals \$-110^{\circ}\$

enter image description here

Based on the Bode plot above I have \$\omega_c = 926\,rad\cdot s^{-1}\$.

  1. Integration gain \$K_i\$

Based on the Bode plot above I need to set the \$K_i = 10^{\frac{-13.6}{20}} = 0.2042\$ so that the magnitude intersects \$0\,dB\$ at \$\omega_c\$.

  1. Integration time constant \$T_i\$

Let's say the PI controller has following transfer function \$\frac{(s\cdot T_i + 1)}{s}\$ (I think it is justified because I have already found the \$K_i\$ value in the step 3.). The aforementioned transfer function reduces the phase at its breakpoint \$\omega = \frac{1}{T_i}\$ with value \$-45^{\circ}\$. Based on that fact I have decided to set \$\frac{1}{T_i}\$ a decade below the \$\omega_c\$ i.e. \$\frac{1}{T_i} = 0.1\cdot\omega_c = 0.1\cdot 926 \sim 90\,rad\cdot s^{-1}\$. Based on that \$T_i = \frac{1}{90} \sim 0.01\,s\$.

  1. Design verification

I have created Bode plot of the \$D(s)\cdot G(s)\$

enter image description here

and I have found that the crossover frequency has been substantially reduced in respect to the designed value \$\omega_c = 926\,rad\cdot s^{-1}\$. It means that I haven't fullfilled the requirement regarding the speed of response which is given by the \$\omega_{BW} = 691\,rad\cdot s^{-1}\$.

It seems to me that the problem is in the inappropriate selection of the breakpoint \$\omega = \frac{1}{T_i}\$ location. I think that I should take into account also the breakpoint of the integrator at \$\omega = 1\,rad\cdot s^{-1}\$.

Can anybody give me an advice how to proceed in the design of PI controller in my case? Thanks in advance for any suggestions.

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  • \$\begingroup\$ I just quickly skimmed but didn't read well what you wrote (in meeting -- mind elsewhere.) The first (and for now, only) thing that brought up my attention is that the delay factor in your first equation places a higher order s term in the numerator, which can be problematic. You can replace it with a few terms of Taylor's expansion in the denominator after changing the sign (of course) which may make a simplified analysis work better when it may otherwise be a bit misbehaved. If I get more time and inclination later, I may return to see if I can be of any use after reading better. \$\endgroup\$
    – jonk
    Jan 25 at 21:17
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    \$\begingroup\$ When you add an integrator-part your phase starts at -90° from the get-go. So you can't both have an I-term and a crossover frequency of wc = 926 with just a PI-controller. However, if you use a PI-lead or a PI-lag controller then I think you can meet both requirements. Are you allowed to use lead and lag compensators? \$\endgroup\$
    – Carl
    Jan 26 at 9:44
  • \$\begingroup\$ @Carl thank you for your reaction. As far as the lead-lag compensators. No, I have only the PI controller in disposal. What do you think about the design procedure I have described above (especially the steps 3. and 4.)? \$\endgroup\$
    – Steve
    Jan 26 at 11:47
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    \$\begingroup\$ @Steve Your design steps look reasonable. I have added an answer to show how I would handle this design task. Maybe you can take something from it? \$\endgroup\$
    – Carl
    Jan 26 at 12:20

2 Answers 2

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I'm not sure what your thoughts were during the design process so I can't comment on them, apart from that it looks reasonable. However, this is what I would do if I were to design a controller.

1 Bode Plot of \$G_{ol}\$

enter image description here

This is a nice Bode Plot, but ideally we want the magnitude response to be a linearly decreasing curve. For this we can use an I-term.

Placing the break frequency of the I-term at \$\omega_i=100 \text{rad/s}\$ will do the trick. \$\tau_i = \frac{1}{\omega_i} = 0.01\text{s}\$

$$C_i = \frac{\tau_is+1}{\tau_is} = \frac{0.01s+1}{0.01s} $$

2 Bode Plot of \$G_{ol}C_i \$

enter image description here

A nice strictly decreasing magnitude response. To get a 70° phase margin we need a gain of \$K_p = 20^{\frac{-17}{20}} = 0.0738\$

3 Bode Plot of controlled system and step response

enter image description here

We see a wide bandwidth and large phase margin from the Bode Plot. We also see a very nice step response.


From the comments

Please can you tell me how did you find the breakpoint frequency \$ \frac{1}{\tau_i} = 100\text{rad/s}\$?

Sure. I chose \$\omega_i = 100\text{rad/s}\$ because I want a strictly linear decreasing magnitude response (gain response). An I-term raises the gain of the low frequencies. \$\omega_i\$ is the frequency where you want your I-term to "stop" increasing the gain. If this happens at around \$\omega = 100\text{rad/s}\$ then we get a nice magnitude response for your system with the controller implemented.

Have a look at the Bode Plot for the \$C_i\$ compensator alone. See it has high gain for low frequencies, and at \$100 \text{rad/s}\$ the gain is around \$0\text{dB}\$: -

enter image description here

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  • \$\begingroup\$ Thank you very much for your answer. It is very helpful for me. If I compare your approach with the approach used by me, I can see that you placed the breakpoint \$\frac{1}{T_i}\$ at first and then you set the gain \$K_p\$. So you have let's say fixed phase characteristic before gain selection. Please can you tell me how did you find the breakpoint frequency \$\frac{1}{T_i} = 100\,rad\cdot s^{-1}\$? \$\endgroup\$
    – Steve
    Jan 26 at 14:13
  • \$\begingroup\$ @Steve I have added an explanation to my answer. The reason is because we want a strictly decreasing magnitude response. \$\endgroup\$
    – Carl
    Jan 26 at 14:23
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Designers are usually not comfortable dealing with \$k_i\$ and \$k_p\$ (or \$k_d\$) coefficients when it comes down to stabilizing a system. The vast majority of them - and me in particular : ) deal with poles and zeroes placement, specifically positioned to force crossover at the desired frequency with the proper phase margin. I have rewritten the original expression to plot it in Mathcad:

enter image description here

Then I determine the magnitude and phase lag of that transfer function at the considered crossover frequency:

enter image description here

From these results, I will know how the compensator must be tailored to force crossover and bring adequate phase boost to meet the phase margin criterion. In your case, a PI compensator relates to a type 2a compensator featuring a pole at the origin for high dc gain and a zero. However, if you exactly want a 70° PM and not more, you will need an extra pole bringing down the phase boost created by the zero: the compensator becomes a type 2. The calculation steps are here:

enter image description here

And if you now multiply this transfer function with that of your process, we have this:

enter image description here

Crossover and phase margin goals are met.

If I now convert these elements (except the pole) in \$\tau_i\$ and \$k_p\$ values following calculations from my blue book, you have:

enter image description here

and you see how the two compensators slightly differ in terms of responses: the type 2 properly rolls off the gain at high frequency and will ensure a good noise immunity with improved gain margin. It also exactly tailors the phase boost to what is needed with the extra pole what the PI can't do since it does not bring a pole. As a result, the phase margin is not exactly meeting the target of 70° once the second compensator is involved to compensate the process:

enter image description here

Nevertheless, this is meeting the target. A similar methodology can be applied to a filtered PID compensator which once supplemented with an extra pole becomes a type 3 compensator.

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