Proportional plus Integral (PI) Controller


 Ferdinand Horton
 8 months ago
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1 Proportional plus Integral (PI) Controller 1. A pole is placed at the origin 2. This causes the system type to increase by 1 and as a result the error is reduced to zero. 3. Originally a point A is on the root locus given that the angular contribution of the open loop poles to A is (2k+1) When a pole is added to the origin, the shape of the root locus has changed and A is no longer on the root locus. 5. To put A back on the root locus, a zero is added near to the pole. This zero will compensate the angle contributed by the pole at the origin, resulting point A to be on the root locus again. 6. In doing so, the system type is increased by 1, contributed by the pole at the origin. The gain at point A is similar to that before compensation since the ratio of lengths of the compensator pole and compensator zero is approximately unity. 7. In this way, the steady state error is reduced to zero without appreciably affecting the transient response. The aim of the compensator is to reduce the steady state error. But as a result, the dominant pole (point A) has been moved to a different location which consequently changes the value of damping ratio ζ and natural frequency n. This change will affect transient response performance specifications (peak time (Tp) and settling time (Ts)) since they are depend on these parameters. The introduction of compensator pole and zero bring back point A to its original position (and so ζ and n ), thus preserve the transient response of the system.
2 Example: Given the system of Figure 1.17 is operating with a damping ratio of Show that the addition of the ideal integral compensator, shown in Figure 2 reduces the steadystate error to zero for a step input without appreciably affecting transient response. Solution First we need to find the dominant pole represent: 1) Damping ratio (ζ) since the dominant pole 2) Natural frequency ( n ) where its real part is which will affect the transient response in terms of settling time (Ts) and peak time (Tp) through the following equations: All these performance specifications belong to second order system. However, the second order approximation is valid for any higher order system if the rest of the poles are at least 5 times to the left of real part of the dominant pole. If let say the real part of dominant pole is 0.9 then if the next most right pole is at least 4.5, then the approximation is valid.
3 Dominant pole calculation The dominant pole is the intersection of a line of the root locus (an acute angle of θ to real axis) with Method 1: Using calculator If only ζ or %OS is given and Ts or Tp is not given Using the system in Figure 1, ζ =0.174, therefore Using trigonometric identity, Since the dominant pole is a closedloop pole therefore all angle contributed to this pole from all open loop poles and zeros must equals to an odd multiplication of 180 0, therefore Replace Therefore
4 Method 2: Using trialerror graphical method Need to find breakaway point and jaxis crossing Breakaway point using differentiation method. Since s= lies within the locus between 2 and 1, therefore this is the breakaway point. Then look for the jaxis crossing using j method Using the characteristic equation: Let s= j Im: (No need to solve for the real part since K that will be used in calculating K p is the value of K at dominant pole and not at the jaxis)
5 d = n (1ζ 2 ) n σ d = ζ n θ The dominant pole is at s = σ d ±j d θ d σ d σ
6 From Pd, the gain K for steady state error calculation can be obtained. To find K, insert the dominant pole into characteristic equation of the system. Using calculator, The imaginary part should be 0 where in this case 0.08 is acceptable. Therefore the value of K is From here we can calculate the steady state error and transient response performance specifications: Since the system is type zero and assume the input is a unit step, therefore: This is the end of an analysis of the uncompensated system.
7 Now, let s have a look into the compensated system With the same value of ζ, the following part is equals to the uncompensated system. The dominant pole need to be recalculated since now it involve the contribution of angle from compensator pole and zero Replace Using calculator, Therefore
8 From Pd, the gain K for steady state error calculation can be obtained. To find K, insert the dominant pole into characteristic equation of the system. Using calculator, Imaginary: Real: From here we can calculate the steady state error and transient response performance specifications. Since the system is type one and assume the input is a unit step, therefore: As compared to the uncompensated system, we managed to eliminate steady state error and at the same time produced a very small change in settling time Tp (2.4% increase) and Ts (2.4% increase)