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Unformatted text preview: Zerodegree Loci for Negative K
We consider sm + b1sm,1 + + bm :
KGs = ,jK j sn + a sn,1 + + a
1
n By convention, we have 1 + KGs = 0; K = 0 ! ,1:
The angle of Gs is 0o + 360o l
Change: 180o + 360o is replaced by 0o + 360o l.
On real axis: the number of poles and zeros on RHS of a
testing point so = even number 0 is counted even.
Centroid is the same but the angle of asymptotes are l = l 360o=n , m; l = 0; 1; :::; n , m , 1:
This is rotated by 180o=n , m from 180o loci. j !axis crossing: locus crosses for K 0. 1 The departure and arriving angles:
1X
X
= i , i , l 360o ;
dep
q
1X
X arr = i , i + l 360o q where l = 0; :::; q , 1 also rotated by 180o=n , m from
180o loci. Example: Find root locus of +
KGs = K s s2 1 for both negative and positive K . 1 Solid line for positive K , and dashed line for negative K 2 Example: Find zerodegree root locus for K+
K +
KGs = s2s +s9s1 3 = K s4 + s s3 1 27 :
+
12 + Step 1, 2: Mark poles zeros and raw locus on real axis.
Step 3: Find centroid of asymptotes:
P
pi , P zi = 0 + 0 , 9 , 3 + 1 = ,3:667;
=
n,m
4,1
and angles for asymptotes:
8 0o
360o l = l n , m = 120o
: ,120o l = 0;
l = 1;
l = 2: Step 4: Departure angle: not necessary as they are clear from
locus on real axis. The same is true for arriving angle.
Step 5: j!axis crossing: aj! + Kbj! = !4 , 12j!3 , 27!2 + K j! + 1 = 0
K! , 12!3 = 0; !4 , 27!2 + K = 0:
The solutions are ! = 0 and K = 12!2 that are not crossing
as K 0 for zerodegree root locus. 3 Step 6: Breaking points, or multiple roots:
0 = b0sas , a0sbs = s4 + 12s3 + 27s2
,4s3 + 36s2 + 54ss + 1
= ,s3s3 + 28s2 + 63s + 54 = 0:
The possible roots are 0; ,6:545; ,1:395 j 0:8980. The only
possible breaking point is ,6:545.
Step 7: Complete sketch. 9 3 4 1 Selecting Gain from Loci: K = 0 ! 1
K = , G1s K = jG1sj :
Hence if so is a desired closedloop pole, then K= Q
jso , pij
Q
jso , zij L L :::L
= 12 n S1S2:::Sm
where Li: length of vector from so to pi and Si: from so to zi. Example: Determine K such that = 0:5 where 1
Gs = s s + 42 + 16 ; p1 = 0; p2;3 = ,4 4j: First sketch root locus: Since n , m = 3 , 0 = 3, there
are three braches: one on real axis, and two approaching
asymptotes:
8 60o;
l = 0;
8
= , ; l = 180o; l = 1;
3
: ,60o; l = 2:
We need also determine departure angle and j!axis crossing:
dep = , 1 , 3 + 180
j! j! + 42 + 16 = 0
o = !3 , 32! = 0; ,135o , 90o + 180o = ,45o;
K = 8!2 p that gives !c = 32 = 5:66, and Kc = 8 32 = 256.
5 Since the desired closedloop poles are stable, K Kc = 256. = 0:5 implies that the angle between the vector to the
desired pole and j!axis is = sin,10:5 = 30o:
To achieve the desired pole location, we need
K = jG1s j = jsojjso , p2jjso , p3j:
o
Since jsoj 4, jso , p2j 2:1, and jso , p3j 7:7, we obtain
desired K as K = 65.
ζ=0.5
 s o p 2  4j
so p 1  so p 3
−4 6 Dynamic Compensators
Often static gain may not achieve the desired performance, or
stability. In this case, dynamic compensators need be employed.
2
Example: Gs = 1=s2 + !o . Root locus shows that the
closedloop is unstable no matter what K we choose. ωo j −ωo j 7 However, if we replace K by PD controller Ds = K s + z; z 0:
Then we have stable closedloop system for any K 0. Since
such Ds is not realizable, we may use lead compensator
s
Ds = K + + z ; p z:
1 s=p p z Root locus with lead compensator. 8 Lead Compensator:
We consider compensators of the form KDs where
Ds = s + z :
s+p
If p z 0, then we call it lead compensator. If z p
then we call it lag compensator. 0, The argument to the desired pole location generated by lead
compensator is given by
m by + m + 180 = , o , = 180o. φm φ ψ 9 Design of Lead Compensator:
The purpose is to improve transient response, through assigning
the dominant poles at the desired location, while all other poles
far away from j!axis.
Step 1: Sketch root locus for KDsGs.
Step 2: Translate speci cations into desired pole location.
Step 3: Compute the argument to be compensated for by
using the formula:
m= X
, X i + 180o:
i
i
i Make sure that the rest of the poles are at least 3 5 times
away from j!axis than the dominant poles.
Step 3: Compute z and p.
Step 4: Compute required gain via gain condition:
K = , Ds 1Gs 0:
o
o
In general, each lead compensator can have at most 90o compensation. If the required m 90o, more than one lead compensator need be employed.
10 1
Example: Gs =
performance requirement ss + 1 with p
and !n 7rad=sec. Note !d = 1 , 2!n = 6:0622. 0:5 The root locus is sketched rst which has three branches. = 0:5 implies gives the dashed ray with = 30o, and
!n = 7 gives a dashed circle of radius 7. The intersection is
the desired pole location.
Compute the argument
m m as = 1 + 2 + 180o = 30o + 20:92o 50:92o: where 1 = 120o and 2 = 90o + tan,12:5=6:022. φ ψ 11 φ2 φ1 The desired poles are so = ,3:5 6:06j by = 30o. Since
= 0 is the lower bound, 51o. A general rule is that = o
m+3 5o = 55o; = 3o 5o = 4o: p The distance from so to the real axis is 72 , 3:52 = 6:0622.
See the gure below. Thus z , 3:5 = 6:0622 ctan55o = 4:2448;
p , 3:5 = 6:0622 ctan4o = 86:6932:
that gives z = 7:7448, and p = 90:1932.
Compute K as
s
K = ss +s1z + p
= 618:0427:
+
s=,3:5+j 6:06 7 55 o 4o
p 3.5 z 12 Veri cation of the Design:
After the design, we need verify if the design satis es the performance requirement.
Find the poles of the closedloop system: cs = ss + 1s + p + K s + z
= s3 + p + 1s2 + p + K s + Kz
= s + 83:4s + 3:9 + 6:5j s + 3:9 , 6:5j :
The dominant poles are not exactly ,3:5 7j , but the corresponding damping ratio and natural frequency are
p2
3:9 = = 0:5148; !n = 3:9 + 6:52 = 7:5763 !n that satisfy the performance requirement. Since the third
pole is more than 20 times away from the j!axis than the
dominant poles, the design is satisfactory.
The design can also be veri ed using step response of the
closedloop system:
K z
Tcs = ss + 1s +sp++K s + z ;
with MATLAB by the fact 0:5 Mp 16:3, and
!n 7 ts 0:86 sec.
13 Lag Compensator:
In time domain response, steadystate response is also important. The purpose of lag compensator is to improve steadystate
errors. Indeed with lag compensator Ds, we have that
lim Ds =
s!0 z 1:
p The above implies that
1
1
Ep = 1 +1K ; Ev = K ; Ea = K
p
v
a
Kp = slim0 GsKz=p; Kp = slim sGsKz=p;
!
!0
2GsKz=p;
Ka = slim0 s
!
and thus tracking error is reduced by lag compensator. 14 1 Lead and lag compensators are used to improve the time response
of the system.
Transient response: the dominant poles are in the desired
location, with rest of poles at least 3 5 times away from
the j!axis as the dominant ones.
Lead Compensator is an approximation of PD controller
p 0 s + z 1 = 1 + s=z:
@
A
plim z s + p
!1
A consequence is that it is likely increasing the overshot. We
note that PD controller is not realizable. Even it can be
realized, we would not use because it ampli es the noise or
disturbance. A compromise is to use lead compensator with
large p.
Steadystate error: the gain at ! = 0 is enlarged so that the
steadystate error is reduced. Recall that the steadystate
errors are:
1
1
Ep = 1 +1K ; Ev = K ; Ea = K
p
v
a
Kp = slim Gs; Kp = slim sGs; Ka = slim s2Gs:
!0
!0
!0 15 1
Example: Gs =
ss + 1 with performance requirement
0:5, and !n 2rad=sec, and steadystate error for unit ramp
input ess 3:33 Kv 30.
The root locus is skipped that has three branches. = 0:5 implies gives the dashed ray with = 30o, and
!n = 2 gives a dashed circle of radius 2. The intersection is
so = ,1 1:7321j that is the desired pole location.
Compute the argument
m m as = 1 + 2 + 180o = 120o + 90o + 180o 30o: where 1 = 120o and 2 = 90o.
ζ=0.5 φ ψ φ2 φ1
ωn =2 16 The desired poles are so = ,1 1:7321j by = 30o. Since
= 0 is the lower bound, 30o. To ensure the dominance
of so, we set = o
m+5 = 35o; = m
p = 5o: The distance from so to the real axis is 3 = 1:7321. Thus p
z , 1 = 3ctan35o = 2:4736;
p
p , 1 = 3ctan5o = 19:7974: that gives z = 3:4736, and p = 20:7974.
Compute K as s
K = ss +s1z + p
+ s=,1+1:7321j = 22:7974: For steadystate performance, we have that
Kv = slim0 s ssK s +sz+ p = Kz = 3:8077:
!
+ 1
p
But the desired performance is Kv 1=ess = 1=:033 = 30.
Hence lag compensator is required:
30
Dlags = s + z ; z = 3:8077 = 7:8788:
s+p p
Set z=p = 7:8788. A general rule is to choose z ten time
smaller or equal to the distance of dominant pole to the j!axis This gives z = 1=10 = 0:1; p = z=8 = 0:0127:
17 Thus the dynamic compensator is s
+3
KDleadsDlags = 22:7974 ss+ 20::4736 s ++:0:1
7974 0 0127 Veri cation:
The closedloop poles are the roots of ss +1s +20:7974s +0:0127+22:7974s +3:4736s +0:1
that are ,19:7915; ,1:0578 1:7337j , 0:097 which satisfy
the transient requirement by p !n = 1:73372 + 1:05782 = 2:0309 2; = 1:0578=2:0309 = 0:5209 0:5:
The steadystate error is
20:7974 0:0127
ess =
= 0:0328
22:7974 3:4736 0:1 18 0:0333: 8:7s + 0:5
Example: Gs = 2
s + 2s + 26 with performance requirement
P:O: 16:3; ts 1:333s;
and steadystate error for unit step input ess 0:05.
Design Procedure:
We consider dynamic compensator of the form Ds = DleadsDlags 8:7sK 0:5 :
+ It follows that the loop transfer function is
0
10
1
0
s + zd 1 B s + zg C B
K
A
DsGs = @ s + p A @ s + p A @ s2 + 2s + 26 C :
d
g
Hence it is type 0 that has nite essvalue for step input.
Our objective is to design lead lag compensator
s+z
s+z
Dleads = s + pd ; Dlags = s + pg ;
d
g
and the gain K such that specs. are met. 19 We rst design the lead compensator. Percentage overshot
requirement gives
j
= 0:5: r 2 logP:O:j 2
+ logP:O: P:O:=0:163
Settling time requirement gives
!n 4:6 = 6:1333: ts Thus the dominant poles are at
r ,
!n j!n 1 , 2 = ,3:0667 j 5:3116:
We choose so = ,4 5j to have some design margin =
p2 2
!n = 4 + 5 = 6:4031 6:1333; = 4=!n = 4=6:4031 = 0:6247 0:5: We now compute angles generated by p1 = ,1 + 5j , and
p1 = ,1 , 5j to the dominant pole at so = ,4 + 5j :
1 = 180 o; 2 = 90 o + arctan3=10 = 106:7o: Hence the required phase compensation is see the gure next
page
o
o
m = 1 + 2 + 180 = 106:7 :
Thus two lead compensator is required. 20 We take two identical lead compensators:
0
s + zd 12
Dleads = @ s + p A
d
Each needs compensate for 53:4o. We take = 58o to ensure
the dominance of so. from the gure below we obtain zd = 4 + 5= tan58o = 7:1243;
pd = 4 + 5= tan4:6o = 66:1442:
The required gain is
0
s + pd 12 s2 + 2s + 26
K = @s + z A
= 3502:
d
s=,4+5j
φ1
5j ψ φ
p z φ2 21 Design of lag compensator: the position constant is now
1
0
s + zd 12 0 K
Kp = slim @ s + p A @ s2 + 2s + 26 A = 1:5626 19:
!0
d
Therefore we need lag compensator
s+z z
19
Dlags = s + pg ; pg = 1:5626 = 12:16:
g
g
Set zg = 0:4 yields pg = 0:4=12:16 = 0:0329.
The dynamic compensator is
0
402:5287 1 0 s + 7:1242 12 0 s + 0:4 1
Ds = @ s + 0:5 A @ s + 66:1442 A @ s + 0:0329 A :
Veri cation: Closedloop poles are ,63:07 55:84j , ,3:97
5:10j , and ,0:25 which satisfy the requirement. MATLAB
is used to simulate the closedloop step response:
0.8 1
0.9 0.7 0.8
0.6
0.7
0.5 0.6 0.4 0.5
0.4 0.3 0.3
0.2
0.2
0.1 0
0 0.1 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 0
0 2 22 2 4 6 8 10 12 14 16 18 20 Analysis: With lead compensator only, the step response satis es transient requirement left gure. With both lead and
lag compensator, esserror is satis ed, but the ts spec. is not
met right gure. In some design case, lag compensator may
give such problem. A common way to remedy it is design of
pre lter F s such that ts is improved.
r(t) F(s) + D(s)  G(s) y(t) Note that the closedloop pole at ,0:25, and the zero at ,0:4
are the main source of causing large ts. Thus we choose
F s = 11+ s=00::25 :
+ s= 4
The step response is plotted in next page 23 1.4 1.2 1 0.8 0.6 0.4 0.2 0
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Both steadystate error and settling time now satisfy the design
requirement. But the percentage overshot exceeds the speci cation. Why? Because the zeros of lead compensator are also zeros
of the closedloop system that cause the increase of the overshot.
Hence we need readjust the dominant pole position using the
knowledge learned from Section x3.5, and do a second design.
However we skip in the class lecture, and leave you to work out
in the project design assignment. 24 ...
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