NYQUIST STABILITY TEST OF THREE-TANK SYSTEM UNDER PID CONTROL. Consider the three-tank system investigated earlier, now operating with a PID controller, a perfect measuring device (h = 1), and a perfectly responding control valve (gv = 1). In this case the open-loop transfer function gL(s) will be

Question

NYQUIST STABILITY TEST OF THREE-TANK SYSTEM UNDER PID CONTROL.

Consider the three-tank system investigated earlier, now operating with a PID controller, a perfect measuring device (h = 1), and a perfectly responding control valve ([latex]g_v = 1[/latex]). In this case the open-loop transfer function [latex]g_L(s)[/latex] will be given by:

[latex]g_L(s) = \frac{6 K_c (1 + au_{I}s + au_{D} au_{I}s^{2})}{ au_{I}s (s + 2) (s + 4)(s + 6)}[/latex] (15.44)

Determine the range of stable controller gains when:

(a) [latex] au_D = 0, \frac{1}{ au_I} = 0[/latex] (P control)

(b) [latex] au_D = 0, \frac{1}{ au_I} = 0.05[/latex] (PI control)

(c) [latex] au_D = 10, \frac{1}{ au_I} = 1.33[/latex] (PID control)

Accepted Answer

The "gain neutral" Nyquist plots for [latex]K_{c} = 1[/latex] (obtained using the program CONSYD) are given in Figure 15.16. Note that for P control (Figure 15.16(a)), the system is stable for [latex]K_{c} = 1[/latex] and the Nyquist curve intersects the real axis at (- 0.60, 0), indicating that [latex]\gamma = 0.6[/latex]. Thus for all controller gains [latex]K_{c} \gt 1/0.60 = 1.667[/latex], the system is unstable because the curve will then encircle (-1,0) once; by the same token the system is stable for all [latex]K_{c} \lt 1/0.60 = 1.667[/latex].
For PI control (Figure 15.16(b)) the "gain neutral" Nyquist plot (for [latex]K_{c} = 1[/latex]) begins at - ∞ on the imaginary axis (because of the 1/s factor in [latex]g_{L}(s)[/latex]) and finally intersects the real axis at (- 0.732,0). Thus the system is unstable for all controller gains [latex]K_{c} \gt 1/0.732 = 1.37[/latex].
For the PID controller, the situation is a bit more complex. As indicated in Figure 15.16(c), the "gain neutral" Nyquist curve begins at - ∞ on the imaginary axis, but intersects the real axis at two places: (-19.8,0) and (-1.48,0) for [latex]K_{c} = 1[/latex]. Note however that the net number of encirclements is 0 because the point (-1,0) is never encircled for [latex]K_{c} = 1[/latex].
Observe now that:
1. As [latex]K_{c}[/latex] is increased beyond 1, the amplification will still not cause the resulting Nyquist curve to encircle the point (-1,0), and the process remains stable.
2. However, in decreasing [latex]K_{c}[/latex] the resulting curve shifts to the right, and the first encirclement of (-1,0) will occur when:

[latex]K_{c} \lt \frac{1}{1.48} = 0.674[/latex]

Subsequently, as we continue to decrease [latex]K_{c}[/latex], for [latex]K_{c}[/latex] values between 0.674 and (1/19.8) = 0.051, the system remains unstable. Beyond this point, for:

[latex]K_{c} \lt \frac{1}{19.8} = 0.051[/latex]

the entire curve shifts beyond the point (-1,0) and the system becomes stable again.
Thus with the PID controller, the system is "conditionally stable," i.e., for [latex]K_{c} \lt 0.051[/latex] and [latex]K_{c} \gt 0.674[/latex] the system is stable, but in the range [latex]0.051 \lt K_{c} \lt 0.674[/latex], the system is unstable.
The explanation for this type of behavior may be made more explicit by the Root Locus diagram of Figure 15.17. There it is seen that a pair of closed-loop poles cross into the RHP when [latex]K_{c} = 0.051[/latex], but tum around and cross back into the LHP when [latex]K_{c} = 0.674[/latex].

Question 15.7: NYQUIST STABILITY TEST OF THREE-TANK SYSTEM UNDER PID CONTRO...

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