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Unformatted text preview: Procedure for Skecth of root locus page 260 of text:
Example 1: Gs = s + 1=s2
= z1 = ,1; p1 = p2 = 0 and n , m = 1.
Step 1: Mark the poles and zeros on the splane.
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 = , 2ss + 1 , s2 = 0
=
ds
s + 12
= 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 Gs = as = s2ss+ 1p , p = 4.
s
+
Step 1: Mark zeros and poles on the splane.
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 , as = b0sas , a0sbs = 0:
ds bs
b2s
Since as = s3 + 4s2 and bs = s + 1,
2 3 6
4 7
5 b0sas , a0sbs = s3 + 4s2 , 3s2 + 8ss + 1
= s2s2 + 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 Gs = as = s2ss+ 1p , p = 12.
s
+
Step 1: Mark zeros and poles on the splane.
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 , as = b0sas , a0sbs = 0:
ds bs
b2s
Since as = s3 + 12s2 and bs = s + 1,
2 3 6
4 7
5 b0sas , a0sbs = s3 + 12s2 , 3s2 + 24ss + 1
= ,s2s2 + 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 a1s = as + Kmbs =
s , s32s , r, b1s = bs = s + 1. Then a1s + K1b1s = 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 Gs = as = 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 , as = b0sas , a0sbs = 0:
ds bs
b2s
Since as = s3 + 9s2 and bs = s + 1,
2 3 6
4 7
5 b0sas , a0sbs = s3 + 9s2 , 3s2 + 18ss + 1
= ,s2s2 + 12s + 18
= ,2ss + 32 = 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 a1s = as + Kmbs =
s + 33, b1s = bs = s + 1. Then a1s + K1b1s = 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 Gs = ss + 2 s1+ 12 + 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 as = s4 +4s3 +
9s2 + 10s and bs = 1 that gives b0s = 0; a0s = 4s3 + 12s2 + 18s + 10:
Hence b0sas , a0sbs = ,4s3 + 12s2 + 18s + 10
= ,s + 14s2 + 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 + 0:12 + 16
Example 6: Root locus for Gs = s s + 0:12 + 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 + 0:12 + 25
Example 7: Root locus for Gs = s s + 0:12 + 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 + 0:2s2 + 0:2K + 16:01s + 25:01K = 0:
The RouthHurwitz table is s3 1
s2 K + 0:2
s
s0 25:01K 0:2K + 16:01
25:01K
0
0 where 0:2K + 16:01K + 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 + 0:2s2 + 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 + 12
Example 8: Root locus for Gs = s2 + 13 .
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 aj! + Kbj! = 1 , !23 + K 1 + j!2 = 0;
1 , !23 + K 1 , !2 = 0; 2K! = 0
that gives ! = 0 and 1 , !23
K = , 1 , !2 = ,1
!=0
Thus there is no j!axis crossing. 0: Step 6: To nd breaking point, or multiple roots, solve a0sbs , asb0s = 0;
3ss + 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.
 Fall '10
 GU

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