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# note2 - Procedure for Skecth of root locus page 260 of...

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Unformatted text preview: Procedure for Skecth of root locus page 260 of text: Example 1: Gs = s + 1=s2 = z1 = ,1; p1 = p2 = 0 and n , m = 1. Step 1: Mark the poles and zeros on the s-plane. Step 2: Draw locus on the real axis: left of s = ,1. Step 3: Draw asymptotes: since n , m = 1, this step is included in Step 2. Step 4: Compute departure and arriving angles: ,2 1 = 180o + l  360o = 1 = ,90o; 2 = 90o: Step 5: Compute points where locus croesses j! axis: s2 : 1 s1 : K s0 : K Clearly for K K 0 0, there is no roots on j!-axis. 1 Step 6: Determine possible breaking point: s2 K = f s = ,1=G = , s + 1 df = , 2ss + 1 , s2 = 0 = ds s + 12 = s = 0; ,2: Thus so = ,2 is the breaking point and Km = ,4=,1 = 4. Step 7: Complete the sketch. -2 -1 2 b Example 2: Root locus for Gs = as = s2ss+ 1p , p = 4. s + Step 1: Mark zeros and poles on the s-plane. Step 2: Draw locus on real axis. 3: Draw asymtotes for n , m = 3 , 1 = 2: 90o; l = 0; l = n , m + l  n , m = ,90o; l = 1; p , zi = b1 , a1 = 1 , 4 = ,1:5: =i n,m n,m 2 Step 4: Departure angle at the two poles at 0: 180o 8 360o : P P 8 ,2 l = 180o + 360ol = l : 90o; ,90o; l = 0; l = 1: Step 5: To nd where to across j! axis, we note that s3 + 4s2 + Ks + K = 0: If s = j!, then the above is equivalent to ,4!2 + K = 0; ,!3 + K! = 0: The only solution is K = ! = 0. Thus no acrossing. 3 Step 6: Find multiple roots: d , as = b0sas , a0sbs = 0: ds bs b2s Since as = s3 + 4s2 and bs = s + 1, 2 3 6 4 7 5 b0sas , a0sbs = s3 + 4s2 , 3s2 + 8ss + 1 = s2s2 + 7s + 8 = 0: The possible roots are s1 = 0, and s2;3 = ,1:75  j 0:97. Because s2;3 are not on root locus it would result in complex K , no multiple roots exist except s = 0. -4 -2 4 -1 b Example 3: Root locus for Gs = as = s2ss+ 1p , p = 12. s + Step 1: Mark zeros and poles on the s-plane. Step 2: Draw locus on real axis. 3: Draw asymtotes for n , m = 3 , 1 = 2: 90o; l = 0; l = n , m + l  n , m = ,90o; l = 1; p , zi = b1 , a1 = 1 , 12 = ,5:5: =i n,m n,m 2 Step 4: Departure angle at the two poles at 0: 180o 8 360o : P P 8 ,2 l = 180o + 360ol = l : 90o; ,90o; l = 0; l = 1: Step 5: To nd where to across j! axis, we note that s3 + 12s2 + Ks + K = 0: If s = j!, then the above is equivalent to ,12!2 + K = 0; ,!3 + K! = 0: The only solution is K = ! = 0. Thus no acrossing. 5 Step 6: Find multiple roots: d , as = b0sas , a0sbs = 0: ds bs b2s Since as = s3 + 12s2 and bs = s + 1, 2 3 6 4 7 5 b0sas , a0sbs = s3 + 12s2 , 3s2 + 24ss + 1 = ,s2s2 + 15s + 24 = 0: The possible roots are s1 = 0, s2 = ,5:18, s3 = ,2:31. Both s2;3 are on root locus. Two roots merge at s2, and then break away on real axis. One ends with zero at ,1, and the other merge with the branch from p3 = ,12 at s3, and then break away. Tp compute departure angle, set a1s = as + Kmbs = s , s32s , r, b1s = bs = s + 1. Then a1s + K1b1s = 0 has a departure angle 1 , dep = 2 ,90o; l = 0; = o 8 1 , 180o + l  360o 6 : 90 ; l = 1: -12 -10 -8 -6 -4 -2 Compared with the case p = 4, we have a very distinct feature in term of its geometry shape from the case of p = 9. 7 b Example 4: Root locus for Gs = as = s2ss+ 1p , p = 9. s + Step 1, 2: Mark zeros poles, and draw locus on real axis. 3: Draw asymtotes for n , m = 3 , 1 = 2: 90o; l = 0; l = n , m + l  n , m = ,90o; l = 1; p , zi = b1 , a1 = 1 , 9 = ,4: =i n,m n,m 2 Step 4: Departure angle at the two poles at 0: 8 360o 180o : P P 8 ,2 l = 180o + 360ol = l : 90o; ,90o; l = 0; l = 1: Step 5: To nd where to across j! axis, we note that s3 + 9s2 + Ks + K = 0: If s = j!, then the above is equivalent to ,9!2 + K = 0; ,!3 + K! = 0: The only solution is K = ! = 0. Thus no acrossing. 8 Step 6: Find multiple roots: d , as = b0sas , a0sbs = 0: ds bs b2s Since as = s3 + 9s2 and bs = s + 1, 2 3 6 4 7 5 b0sas , a0sbs = s3 + 9s2 , 3s2 + 18ss + 1 = ,s2s2 + 12s + 18 = ,2ss + 32 = 0: The possible roots are s1 = 0, s2 = s3 = ,3. Clearly ,3 is on root locus. Two roots merge at s2;3, and then break away exactly at s2;3. Tp compute departure angle, set a1s = as + Kmbs = s + 33, b1s = bs = s + 1. Then a1s + K1b1s = 0 has a departure angle 8 0o ; l = 0; 1 , 180o + l  360o = 120o; l = 1; dep = 3 ,120o; l = 2: as = 180o. The arriving angle at ,3 will have to use the concept of negative root locus. : 9 -9 -4 Question: What are the arriving angles at ,3? We need use continuation of root locus for negative K . 10 Example 5: Root locus for Gs = ss + 2 s1+ 12 + 4 . Step 1, 2: Mark zeros poles, and draw locus on real axis. Poles are 0; ,1; ,1  2j . No nite zero. 3: Draw asymtotes for n , m = 4 , 0 = 4: 45o; 135o; 180o 360o l = n , m + l  n , m = ,135o; ,45o; p , zi = ,2 , 1 , 1 = ,1: =i n,m 4 Step 4: Departure angle at the pole ,1 + 2j : 8 : P dep = = P l = 0; l = 1; l = 2; l = 3; , 1 , 2 , 4 , 180o + l  360o ,90o +  , 90o ,  , 90o , 180o = ,90o: Step 5: To nd where to across j! axis, we note that s4 + 4s3 + 9s2 + 10s + K = 0: If s = j!, then the above is equivalent to !4 , 9!2 + K = 0; ,4!3 + 10! = 0: 11 2 The 2nd equation solution gives !1 = 2:5 that if substituted into the rst equation gives K = 16:25. Hence there is a pair p of crossing j! axis at ! =  2:5. Step 6: To nd multiple roots, we note that as = s4 +4s3 + 9s2 + 10s and bs = 1 that gives b0s = 0; a0s = 4s3 + 12s2 + 18s + 10: Hence b0sas , a0sbs = ,4s3 + 12s2 + 18s + 10 = ,s + 14s2 + 8s + 10 = 0: The possible roots are s1 = ,1, s2;3 = ,1  1:22j . The departure and arriving angles are easier to gure out for this example. 12 -2 13 s + 0:12 + 16 Example 6: Root locus for Gs = s s + 0:12 + 25 . Step 1, 2: Mark zeros poles, and draw locus on real axis. Poles are 0; ,0:1  5j , and zeros are ,0:1  4j . 3: Draw asymtotes for n , m = 3 , 2 = 1: on real axis. Step 4: Departure angle at the pole ,0:1 + 5j : dep , 1 , 3 + 1 + 2 + 180o  ,90o , 90o + 90o + 90o + 180o = 180o: = Arriving angle at the zero ,0:1 + 4j : arr =  + 3 , 2 , 180o 90o , 90o + 90o , 90o , 180o = 180o: 1 + 2 5j -5j 14 s + 0:12 + 25 Example 7: Root locus for Gs = s s + 0:12 + 16 . This example is similar to the previous one except that the pair of complex poles are interchanged with the pair of complex zeros. Step 1, 2: Mark zeros poles, and draw locus on real axis. Poles are 0; ,0:1  4j , and zeros are ,0:1  5j . 3: Draw asymtotes for n , m = 3 , 2 = 1: on real axis. Step 4: Departure angle at the pole ,0:1 + 4j : dep , 1 , 3 + 1 + 2 + 180o  ,90o , 90o + 90o , 90o + 180o = 0o: = Arriving angle at the zero ,0:1 + 4j : arr =  + 2 + 3 , 2 , 180o 90o + 90o + 90o , 90o , 180o = 0o: 1 15 5: To nd where to across j! axis, we note that s3 + K + 0:2s2 + 0:2K + 16:01s + 25:01K = 0: The Routh-Hurwitz table is s3 1 s2 K + 0:2 s s0 25:01K 0:2K + 16:01 25:01K 0 0 where 0:2K + 16:01K + 0:2 , 25:01 = : K + 0:2 We now have that = 0  0:2K 2 , 8:96K + 3:2 = 0 that gives K = 44:4 and K = 0:362. Set up auxiliary polynomial s = K + 0:2s2 + 25:01K: Then s = 0 yields 25 01K s = j K :+ 0:2  ! = 4:99; 4:01: v u u u t 16 Step 6, 7: No need to compute multiple roots. 5j -5j 17 s + 12 Example 8: Root locus for Gs = s2 + 13 . Step 1, 2: Mark zeros poles, and draw locus on real axis. Poles are p1 = p2 = p3 = j , and p4 = p5 = p6 = ,j . Zeros are z1 = z2 = ,1. Real axis is not on root locus. Step 3: Draw asymtotes for n , m = 6 , 2 = 4 on real axis. Centroid: p , zi = 3j , 3j + 2 = 0:5: =i n,m 4 Angles: P P 45o; o o + l  90o = 135 ; l = 45 ,135o; ,45o; 8 : l = 0; l = 1; l = 2; l = 3: Step 4: Departure angle at the poles j : 2 , 3 , 180o , l  360o dep = 3 0; l = 0; = ,120o  l = ,120o; l = 1; 120o; l = 2: 8 : 18 Arriving angle at the zero ,1: 3 j + 3 ,j + 180o + l  360o arr = 2 o o + l  180o = 90 ; l = 0; = 90 ,90o; l = 1: where j = 180o + 45o, and ,j = 180o , 45o. 8 : 5: To nd where to across j! axis, we note that aj! + Kbj! = 1 , !23 + K 1 + j!2 = 0;  1 , !23 + K 1 , !2 = 0; 2K! = 0 that gives ! = 0 and 1 , !23 K = , 1 , !2 = ,1 !=0 Thus there is no j!-axis crossing. 0: Step 6: To nd breaking point, or multiple roots, solve a0sbs , asb0s = 0;  3ss + 1 , s2 + 1 = 2s2 + 3s , 1 = 0; p that gives possible multiple roots at s = ,3  17=14 that are real. 19 But no portion of the real axis is not on the root locus except the zeros at ,1. Hence no multiple roots or breaking point. Step 7: Complete root locus. j -1 -0.5 -j 20 ...
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## This note was uploaded on 01/23/2012 for the course EE 4580 taught by Professor Gu during the Fall '10 term at LSU.

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