DIRECT SYNTHESIS CONTROLLER TUNING FOR THE MULTIPLE-TANK PROCESS. Use the direct synthesis tuning procedure to obtain PID controller parameters for the three-tank system of Example 15.1. Recall that for this system: g(s) = 6/(2s + 1)(4s + 1)(6s + 1) (14.73) Assume that a reasonable closed-loop

Question

DIRECT SYNTHESIS CONTROLLER TUNING FOR THE MULTIPLE-TANK PROCESS.

Use the direct synthesis tuning procedure to obtain PID controller parameters for the three-tank system of Example 15.1. Recall that for this system:

[latex]g(s) = \frac{6}{(2s + 1) (4s + 1) (6s + 1)}[/latex] (14.73)

Assume that a reasonable closed-loop response given by Eq. (15.14) is desired. Repeat the design for the situation in which there are only two tanks so that:

[latex]\frac{y(s)}{y_{d}(s)} = q(s) = \frac{1}{ au_{r} ^{2} s^{2} + 2 au_{r} \zeta_{r} s + 1}[/latex] (15.14)

[latex]g(s) = \frac{6}{(4s + 1) (6s + 1)}[/latex] (15.24)

Accepted Answer

Using Eq. (15.20) for the tuning parameters, let us find a good reference trajectory for which [latex]\phi[/latex] defined by Eq. (15.17) is close to one of the three time constants in Eq. (14.73). Let us choose [latex] au_{r} = 4.0 , \zeta_{r} = 0.5[/latex] in the reference trajectory Eq. (15.14) to yield:

[latex]\phi = \frac{ au_{r}}{2 \zeta_{r}}[/latex]

[latex]\beta = 2K au_{r} \zeta_{r} = 4K(\zeta_{r})^{2} \phi[/latex] (15.17)

[latex]K_{c} = \frac{( au_{1} + au_{2})}{\beta}[/latex]

[latex] au_{I} = ( au_{1} + au_{2})[/latex] (15.20)

[latex] au_{D} = \left( \frac{ au_{1} au_{2}}{ au_{1} + au_{2}} ight)[/latex]

[latex]\frac{y(s)}{y_{d}} = q(s) = \frac{1}{16 s^{2} + 4 s + 1}[/latex] (15.25)

which produces the underdamped trajectory shown in Figure 15.5. Note that from Eq. (15.17) this yields [latex]\phi = 4 , \beta = 24[/latex]. In this case [latex] au_{3} = 4.0[/latex] is canceled exactly, and the controller tuning parameters from Eq. (15.20) become [latex]K_{c} = 0.333 , 1/ au_{I} = 0.125 , au_{D} = 1.50[/latex] for this PID controller. The desired reference trajectory given in Eq. (15.25) is achieved exactly as illustrated in Figure 15.5.

If we apply the same type of design to the two-tank system with the model given by Eq. (15.24), we see that we may choose [latex] au_{r} = 4.0 , \zeta_{r} = 0.5[/latex] again yielding the reference trajectory in Eq. (15.25) with [latex]\phi = 4 , \beta = 24[/latex]. The resulting PI controller parameters are now given by [latex]K_{c} = 0.250, 1/ au_{I} = 0.1667[/latex]. With this controller, the desired reference trajectory is achieved exactly - this time by a PI controller for the two-tank system. Note that because we chose .the same reference trajectory for each problem, the closed-loop responses for both problems are identical and shown in Figure 15.5.

Question 15.3: DIRECT SYNTHESIS CONTROLLER TUNING FOR THE MULTIPLE-TANK PRO...

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