Example 2 For a type 0 system G (S+1)(S+2) (6.8) the zero should be used to cancel the pole at. We leave it as an exercise to prove that the compensator gives the desired closed-loop poles s+2 D,=2.5 (6.9) You should note that the lead compensator zero will still appear in the closed-loop transfer function 2022-2-3 11
2022-2-3 11 the zero should be used to cancel the pole at . We leave it as an exercise to prove that the compensator gives the desired closed-loop poles. ( 1)( 2) 2 2 s s G 3 2 2.5 2 s s D (6.8) (6.9) For a type 0 system Example 2 Note You should note that the lead compensator zero will still appear in the closed-loop transfer function
Method 2 The following graphical method maximizes the ratio between pole and zero for any given angle contribution. This minimizes the additional compensator gain needed to satisfy the gain criterion. The steps in the location of the lead-compensator pole and zero are as follows(see Figure 2) Locate the desired closed-loop pole S. Draw a line from the origin to s, and a horizontal line through s, to the left 2. Bisect the angle between the two lines drawn in step 1 3. Measure the angle either side of the line drawn in step 4. The intersections of these lines with the real axis locate the compensator pole po and zero z( 0 2022-2-3 12
2022-2-3 12 Method 2 The following graphical method maximizes the ratio between pole and zero for any given angle contribution. This minimizes the additional compensator gain needed to satisfy the gain criterion. The steps in the location of the lead-compensator pole and zero are as follows (see Figure 2). 1. Locate the desired closed-loop pole s1 . Draw a line from the origin to s1 and a horizontal line through s1 to the left. 2. Bisect the angle between the two lines drawn in step 1. 3. Measure the angle either side of the line drawn in step 4. The intersections of these lines with the real axis locate the compensator pole p0 and zero z0
N 米 2 0 Re P 2022-2-3 13
2022-2-3 13 z0 s1 p0 p Im Re 2 c 2 c z 2
Example 3 We return to the satellite attitude control problem with G(S) (6.10) Requiring a closed-loop pole, the geometry of the problem is illustrated in Figure 3. Note that the line drawn from the origin to the point s, subtends an angle of 135 to the Ir positive real B A 0 2022-2-3 14
2022-2-3 14 Requiring a closed-loop pole , the geometry of the problem is illustrated in Figure 3. Note that the line drawn from the origin to the point s1 subtends an angle of 135 to the positive real 2 1 ( ) s G s z0 s1 p0 p Im 2 c 2 c z 2 135 B A C E D -2 O +j2 l1 lp lz (6.10) Example 3 We return to the satellite attitude control problem with
We can use Matlab to help to work through the trigonometry The angle contribution of the plant and feedback at S, is obtained as follows G=tf(1,[1,0,0]); 1 GH G*Hi s1=-2+2j; The total contribution of the plant poles and zeros can be calculated directly using the Matlab equivalent of the angle criterion ∠G(s)H(s) ∑4(s1-=)-∑∠(s1-p) (6.11) S=SI 2022-2-3 15
2022-2-3 15 We can use Matlab to help to work through the trigonometry. The angle contribution of the plant and feedback at s1 is obtained as follows. G = tf(1,[1,0,0]); H = 1; GH = G*H; s1 = -2+2j; The total contribution of the plant poles and zeros can be calculated directly using the Matlab equivalent of the angle criterion m j n i j i s s G s H s s z s p 1 1 1 1 ( ) ( ) ( ) ( ) 1 (6.11)