PID Controller Design

PROBLEM: Given the system of Figure 9.31, design a PID controller so that the system can operate with a peak time that is two-thirds that of the uncompensated system at 20% overshoot and with zero steady-state error for a step input.

Question

PID Controller Design

PROBLEM: Given the system of Figure 9.31, design a PID controller so that the system can operate with a peak time that is two-thirds that of the uncompensated system at 20% overshoot and with zero steady-state error for a step input.

Accepted Answer

Note that our solution follows the eight-step procedure described earlier.

Step 1 Let us first evaluate the uncompensated system operating at 20% overshoot. Searching along the 20% overshoot line (ζ=0.456) in Figure 9.32, we find the dominant poles to be -5.415±j10.57 with a gain of 121.5. A third pole, which exists at -8.169, is found by searching the region between -8 and -10 for a gain equivalent to that at the dominant poles. The complete performance of the uncompensated system is shown in the first column of Table 9.5, where we

	Uncompensated	PD-compensated	PID-compensated
Plant and compensator	[latex]\frac{K(s+8)}{(s+3)(s+6)(s+10)}[/latex]	[latex]\frac{K(s+8)(s+55.92)}{(s+3)(s+6)(s+10)}[/latex]	[latex]\frac{K(s+8)(s+55.92)(s+0.5)}{(s+3)(s+6)(s+10)s}[/latex]
Dominant poles	-5.415±j10.57	-8.13±j15.87	-7.516±j14.67
K	121.5	5.34	4.6
ζ	0.456	0.456	0.456
[latex]\omega _{n}[/latex]	11.88	17.83	16.49
%OS	20	20	20
[latex]T_{s}[/latex]	0.739	0.492	0.532
[latex]T_{p}[/latex]	0.297	0.198	0.214
[latex]K_{p}[/latex]	5.4	13.27	∞
[latex]e(\infty )[/latex]	0.156	0.070	0
Other poles	-8.169	-8.079	-8.099, -0.468
Zeros	-8	-8, -55.92	-8, -55.92, -0.5
Comments	Second-order approx. OK	Second-order approx. OK	Zeros at -55.92 and -0.5 not canceled

compare the calculated values to those obtained through simulation (Figure 9.35). We estimate that the uncompensated system has a peak time of 0.297 second at 20% overshoot.

Step 2 To compensate the system to reduce the peak time to two-thirds of that of the uncompensated system, we must first find the compensated system’s dominant pole location. The imaginary part of the compensated dominant pole is

[latex]\omega _{d} =\frac{\pi }{T_{p}} =\frac{\pi }{(2/3)(0.297)} =15.87[/latex]

Thus, the real part of the compensated dominant pole is

[latex]\sigma =\frac{\omega _{d} }{ an 117.13°} =-8.13[/latex]

Next we design the compensator. Using the geometry shown in Figure 9.33, we calculate the compensating zero’s location. Using the root locus program, we find the sum of angles from the uncompensated system’s poles and zeros to the desired compensated dominant pole to be -198.37°. Thus, the contribution required from the compensator zero is 198.37° - 180° = 18.37°. Assume that the compensator zero is located at [latex]-z_{c}[/latex], as shown in Figure 9.33. Since

[latex]\frac{15.87}{z_{c}-8.13} = an 18.37°[/latex]

then

[latex]z_{c}=55.92[/latex]

Thus, the PD controller is

[latex]G_{PD}(s)=(s+55.92)[/latex]

The complete root locus for the PD-compensated system is sketched in Figure 9.34. Using a root locus program, the gain at the design point is 5.34. Complete specifications for ideal derivative compensation are shown in the third column of Table 9.5.

Steps 3 & 4 We simulate the PD-compensated system, as shown in Figure 9.35. We see the reduction in peak time and the improvement in steady-state error over the uncompensated system.

Step 5 After we design the PD controller, we design the ideal integral compensator to reduce the steady-state error to zero for a step input. Any ideal integral compensator zero will work, as long as the zero is placed close to the origin. Choosing the ideal integral compensator to be

[latex]G_{PI}(s)=\frac{s+0.5}{s}[/latex]

we sketch the root locus for the PID-compensated system, as shown in Figure 9.36. Searching the 0.456 damping ratio line, we find the dominant, second-order poles to be -7.516±j14.67, with an associated gain of 4.6. The remaining characteristics for the PID-compensated system are summarized in the fourth column of Table 9.5.

Step 6 Now we determine the gains, [latex]K_{1}[/latex], [latex]K_{2}[/latex], and [latex]K_{3}[/latex], in Figure 9.30. From Eqs. (9.26)

[latex]G_{PD}(s)=(s+55.92)[/latex]

and (9.27),

[latex]G_{PI}(s)=\frac{s+0.5}{s}[/latex]

the product of the gain and the PID controller is

[latex]G_{PID}(s) =\frac{K(s+55.92)(s+0.5)}{s}= \frac{4.6(s+55.92)(s+0.5)}{s}[/latex]

[latex]=\frac{4.6(s^{2}+56.42s+27.96 )}{s}[/latex]

Matching Eqs. (9.21)

[latex]G_{c}(s)=K_{1} +\frac{K_{2} }{s} +K_{3}s =\frac{K_{1}s+ K_{2}+K_{3}s^{2}}{s} =\frac{K_{3}\left(s^{2}+\frac{K_{1} }{K_{3} }s+\frac{K_{2} }{K_{3} } ight) }{s}[/latex]

and (9.28),

[latex]G_{PID}(s) =\frac{K(s+55.92)(s+0.5)}{s}= \frac{4.6(s+55.92)(s+0.5)}{s}[/latex]

[latex]=\frac{4.6(s^{2}+56.42s+27.96 )}{s}[/latex]

[latex]K_{1} =259.5[/latex], [latex]K_{2} =128.6[/latex], and [latex]K_{3} =4.6[/latex].

Steps 7 & 8 Returning to Figure 9.35, we summarize the results of our design. PD compensation improved the transient response by decreasing the time required to reach the first peak as well as yielding some improvement in the steady-state error. The complete PID controller further improved the steady-state error without appreciably changing the transient response designed with the PD controller. As we have mentioned before, the PID controller exhibits a slower response, reaching the final value of unity at approximately 3 seconds. If this is undesirable, the speed of the system must be increased by redesigning the ideal derivative compensator or moving the PI controller zero farther from the origin. Simulation plays an important role in this type of design since our derived equation for settling time is not applicable for this part of the response, where there is a slow correction of the
steady-state error.

Question 9.5: PID Controller Design PROBLEM: Given the system of Figure 9....

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