Unformatted text preview: EIGHT Root Locus Techniques
SOLUTIONS TO CASE STUDIES CHALLENGES
Antenna Control: Transient Design via Gain
a. From the Chapter 5 Case Study Challenge:
76.39K
G(s) = s(s+150)(s+1.32)
1
Since Ts = 8 seconds, we search along  2 , the real part of poles with this settling time, for 180o.
We find the point to be  0.5+j6.9 with 76.39K = 7194.23, or K = 94.18. Secondorder
approximation is OK since third pole is much more than 5 times further from the imaginary axis
than the dominant secondorder pair.
b.
Program:
numg= 1;
deng=poly([0 150 1.32]);
'G(s)'
G=tf(numg,deng)
rlocus(G)
axis([2,0,10,10]);
title(['Root Locus'])
grid on
[K1,p]=rlocfind(G)
K=K1/76.39 Computer response:
ans =
G(s)
Transfer function:
1
s^3 + 151.3 s^2 + 198 s
Select a point in the graphics window
selected_point =
0.5034 + 6.3325i 264 Chapter 8: Root Locus Techniques K1 =
6.0690e+003
p=
1.0e+002 *
1.5027
0.0052 + 0.0633i
0.0052  0.0633i
K=
79.4469
>>
ans =
G(s)
Transfer function:
1
s^3 + 151.3 s^2 + 198 s
Select a point in the graphics window
selected_point =
0.5000 + 6.2269i
K1 =
5.8707e+003
p=
1.0e+002 *
1.5026
0.0053 + 0.0623i
0.0053  0.0623i
K=
76.8521 Solutions to Case Studies Challenges 265 UFSS Vehicle: Transient Design via Gain
K(s+0.437)
a. Push K1 to the right past the summing junction yielding G(s) = s(s+2)(s+1.29)(s+0.193) , where
K = 0.25K1. Combine the parallel feedback paths and obtain H(s) = (s+1). Hence, G(s)H(s) =
K(s+0.437)(s+1)
s(s+2)(s+1.29)(s+0.193) . The root locus is shown below in (b). Searching the 10% overshoot line
(ζ = 0.591; θ = 126.24ο), we find the operating point to be 1.07 ± j1.46 where K = 3.389, or K1 =
13.556.
b.
Program:
numg= [1 0.437];
deng=poly([0 2 1.29 0.193]);
G=tf(numg,deng);
numh=[1 1];
denh=1;
H=tf(numh,denh);
GH=G*H;
rlocus(GH)
pos=(10);
z=log(pos/100)/sqrt(pi^2+[log(pos/100)]^2);
sgrid(z,0)
title(['Root Locus with ' , num2str(pos), ' Percent Overshoot Line'])
[K,p]=rlocfind(GH);
pause
K1=K/0.25
T=feedback(K*G,H)
T=minreal(T)
step(T) 266 Chapter 8: Root Locus Techniques title(['Step Response for Design of ' , num2str(pos), ' Percent']) Computer response:
Select a point in the graphics window
selected_point =
1.0704 + 1.4565i
K1 =
13.5093
Transfer function:
3.377 s + 1.476
s^4 + 3.483 s^3 + 6.592 s^2 + 5.351 s + 1.476
Transfer function:
3.377 s + 1.476
s^4 + 3.483 s^3 + 6.592 s^2 + 5.351 s + 1.476 Solutions to Problems 267 ANSWERS TO REVIEW QUESTIONS
1. The plot of a system's closedloop poles as a function of gain
2. (1) Finding the closedloop transfer function, substituting a range of gains into the denominator, and
factoring the denominator for each value of gain. (2) Search on the splane for points that yield 180 degrees
when using the openloop poles and zeros.
3. K = 1/5
4 . No
5. At the zeros of G(s) and the poles of H(s)
6. (1) Apply RouthHurwitz to the closedloop transfer function's denominator. (2) Search along the
imaginary axis for 180 degrees.
7. If any branch of the root locus is in the rhp, the system is unstable.
8.If the branch of the root locus is vertical, the settling time remains constant for that range of gain on the
vertical section.
9. If the root locus is circular with origin at the center
10. Determine if there are any breakin or breakaway points
11. (1) Poles must be at least five times further from the imaginary axis than the dominant second order
pair, (2) Zeros must be nearly canceled by higher order poles.
12. Number of branches, symmetry, starting and ending points
13. The zeros of the open loop system help determine the root locus. The root locus ends at the zeros.
Thus, the zeros are the closedloop poles for high gain. SOLUTIONS TO PROBLEMS
1.
a. No: Not symmetric; On real axis to left of an even number of poles and zeros 268 Chapter 8: Root Locus Techniques b. No: On real axis to left of an even number of poles and zeros
c. No: On real axis to left of an even number of poles and zeros
d. Yes
e. No: Not symmetric; Not on real axis to left of odd number of poles and/or zeros
f. Yes
g. No: Not symmetric; real axis segment is not to the left of an odd number of poles
h. Yes
2. jω jω
splane X
splane X O X σ σ X X (a) (b) jω X splane O jω X splane σ σ O
X X (c) (d) Solutions to Problems 269 jω X splane O jω X splane σ σ O
X X (c) (d) jω jω splane
splane
X X σ XX O (e) O X (f) 3. a. b. c. X σ 270 Chapter 8: Root Locus Techniques 4 1 d.
4. Breakaway: σ = 2.43 for K = 52.1 Solutions to Problems 271 5.
2 1.5 1 Imag Axis 0.5 0 0.5 1 1.5 2 7 6 5 4 3
Real Axis 2 1 0 1 Breakin: σ = 1.5608 for K = 61.986; Breakaway: σ = 5.437 for K = 1.613.
6.
Convert the denominator to the following form: D ( s ) = 1 + G ( s) = 20 K ( s + 5)
20 K ( s + 5)
.
=
3
2
s + 2 s + 7 s s ( s 2 + 2 s + 7) Plotting the root locus yields 20 K ( s + 5)
and thus identify
s3 + 2s 2 + 7 s 272 Chapter 8: Root Locus Techniques 7. Im Im splane Re Re 8.
2.5 2 1.5 1 Imag Axis 0.5 0 0.5 1 1.5 2 2.5
8 6 4 2
Real Axis 0 2 4 Closedloop poles will be in the lefthalfplane when rhp pole reaches the origin,
or K > ( (3)(3)
9
=.
2 )( 2 )(1) 2 splane Solutions to Problems 273 9. Closedloop poles will be in the righthalfplane for K > (2)(2) 4
= (gain at the origin).
(3)(3) 9 Therefore, stable for K < 4/9; unstable for K > 4/9 .
10. 2.5
2
1.5
1 Imag Axis 0.5
0
0.5
1
1.5
2
2.5
5 4 3 2
Real Axis 1 0 1 Breakaway: σ = 3.436 for K = 1.781. System is never unstable. System is marginally stable for K = ∞. 274 Chapter 8: Root Locus Techniques 11.
System 1: (a) a. Breakaway: σ = 1.41 for K = 0.03; Breakin: σ = 1.41 for K = 33.97.
b. Imaginary axis crossing at j1.41 for K = 1. Thus stable for K > 1.
c. At breakin point, poles are multiple. Thus, K = 33.97.
d. Searching along 1350 line for 1800, K = 5 at 1.414 ∠ 1350.
System 2: (b) a. Breakin: σ = 1.41 for K = 28.14. Solutions to Problems 275 b. Imaginary axis crossing at j1.41 for K = 0.67. Thus stable for K > 0.67.
c. At breakin point, poles are multiple. Thus, K = 28.14.
d. Searching along 1350 line for 1800, K = 4 at 1.414 ∠ 1350.
12.
a. Root locus crosses the imaginary axis at the origin for K = 6. Thus the system is stable for K > 6. b. Root locus crosses the imaginary axis at j0.65 for K = 0.79. Thus, the system is stable for K < 0.79.
13.
6 4 Imag Axis 2 0 2 4 6
7 6 5 4 3 2
1
Real Axis 0 1 2 3 4 276 Chapter 8: Root Locus Techniques
There will be only two righthalfplane poles when pole at +2 moves into the lefthalfplane at the
origin. Thus K = (5)( 2 )( 2 )(2 )
= 6.67 .
3 14.
15 10 Imag Axis 5 0 5 10 15
12 10 8 6 4
Real Axis 2 0 2 Root locus crosses the imaginary axis at j7.348 with a gain of 810. Real axis breakaway is at –2.333
at a gain of 57.04. Real axis intercept for the asymptotes is − 15
= − 5 . The angle of the asymptotes
3 5π
π
is = 3 , π, 3 . Some other points on the root locus are:
ζ = 0.4: 1.606 + j3.68, K = 190.1 ζ = 0.6: 1.956 + j2.6075, K = 117.8
ζ = 0.8: 2.189 + j1.642, K = 79.55 Solutions to Problems 277 15.
a.
θ θ1 3 θ2 90 Imaginary axis crossing: j1.41 at K = 1.5. Stability: K < 1.5. Breakaway: 1.41 at K = 0.04. Points on
root locus: 1.5 ± j0, K = 0.0345; 0.75 ± j1.199, K = 0.429; 0 ± j1.4142, K = 1.5;
0.75 ± j1.1989, K = 9. Finding angle of arrival: 90  θ1  θ2 + θ3 = 90o  tan1(1/3)  tan1(1/2) +
θ3 = 180o. Thus, θ3 = 135o.
b. Imaginary axis crossing: j1.41 at K = 1. Stability: K < 1. Breakaway: 1.41 at K = 0.03. Breakin:
1.41 at K = 33.97. Points on root locus: 1.5 ± j0, K = 0.02857; 0.75 ± j1.199, K = 0.33;
0 ± j1.4142, K = 1; 0.75 ± j1.1989, K = 3.
16. a. Root locus crosses the imaginary axis at ± j 3.162 at K = 52.
b. Since the gain is the product of pole lengths to 5, K = (1) ( 42 + 12 )( ) 42 + 12 = 17 . 278 Chapter 8: Root Locus Techniques 17.
a. b. σ a = 8
(2k+1)π
5π
(0  2  3  4)  (1)
π
=  3 ; Angle =
= 3 , π, 3
3
3 c. Root locus crosses imaginary axis at j4.28 with K = 140.8.
d. K = 13.125
18.
Assume that root locus is epsilon away from the asymptotes. Thus, σa =
Angle = (0  3  6)  (α)
≈ 1 ;
2 (2k+1)π
π 3π
= 2 , 2 . Hence α = 7. Checking assumption at –1 ± j100 yields 180o with K =
2 9997.02.
19. a. Breakaway: 0.37 for K = 0.07. Breakin: 1.37 for K = 13.93
b. Imaginary axis crossing: ±j0.71 for K = 0.33
c. System stable for K < 0.33 Solutions to Problems 279
d. Searching 120o find point on root locus at 0.5∠120ο =  0.25 ± j0.433 for K = 0.1429
20.
a.
20 15 10 Imag Axis 5 0 5 10 15 20
40 30 20 10
Real Axis 0 10 20 b. 0 < K < 23.93
c. K = 81.83 @ 13.04 ± j13.04
d. At the breakin point, s = 14.965, K = 434.98.
21. a. Asymptotes: σint = (2k+1)π
π 3π 5π 7π
(1 2 3 4)  (0)
5
=  2 ; Angle =
=4 , 4 , 4 , 4
4
4 b. Breakaway: 1.38 for K = 1 and 3.62 for K = 1 280 Chapter 8: Root Locus Techniques c. Root locus crosses the imaginary axis at ±j2.24 for K = 126. Thus, stability for K < 126.
d. Search 0.7 damping ratio line (134.427 degrees) for 1800. Point is 1.4171∠134.427ο =
 0.992 ± j1.012 for K = 10.32.
e. Without the zero, the angles to the point ±j5.5 add up to 265.074o. Therefore the contribution of
5.5
the zero must be 265.074  180 = 85.074o. Hence, tan 85.074o = z , where  zc is the location of the
c
zero. Thus, zc = 0.474. f. After adding the zero, the root locus crosses the imaginary axis at ±j5.5 for K = 252.5. Thus, the
system is stable for K < 252.5.
g. The new root locus crosses the 0.7 damping ratio line at 2.7318∠134.427o for K = 11.075
compared to 1.4171∠134.427o for K = 10.32 for the old root locus. Thus, the new system's settling
time is shorter, but with the same percent overshoot. Solutions to Problems 281 22. 23. 1
1
T(s) = 2
=
s + αs + 1 s2 + 1
αs 1+ 2
s +1 1 . Thus an equivalent system has G(s) = 2
s +1 αs Plotting a root locus for G(s)H(s) = 2
, we obtain,
s +1 and H(s) = αs. 282 Chapter 8: Root Locus Techniques 24.
a. b. Root locus crosses 20% overshoot line at 1.8994 ∠ 117.126o =  0.866 ± j1.69 for K = 9.398.
π
4
c. Ts = 0.866 = 4.62 seconds; Tp = 1.69 = 1.859 seconds
d. Root locus crosses imaginary axis at ±j3.32 for K = 60. Therefore stability for K < 60.
e. Other poles with same gain as dominant poles: σ = 4.27
25.
a.
20 15 10 Imag Axis 5 0 5 10 15 20
8 b. 7 6 5 4 3
2
Real Axis 1 0 1 2 3 Solutions to Problems 283 ( −6 − 5 − 4 − 2) − (2 )
= − 9.5
4−2
(2k + 1)π π 3π
θa =
=,
4−2
22 σa = c. At the jω axis crossing, K = 115.6. Thus for stability, 0 < K < 115.6.
d. Breakaway points at σ = 2.524 @ K = 0.496 and σ = 5.576 @ K = 0.031.
e. For 25% overshoot, Eq. (4.39) yields ζ = 0.404. Searching along this damping ratio line, we find
the 1800 point at –0.6608 + j1.496 where K = 35.98.
f. –7.839 ± j7.425
g. Secondorder approximation not valid because of the existence of closedloop zeros in the rhp.
h.
Program:
numg=35.98*[1 2 2];
deng=poly([2 4 5 6]);
G=tf(numg,deng);
T=feedback(G,1)
step(T) Computer response:
Transfer function:
35.98 s^2  71.96 s + 71.96
s^4 + 17 s^3 + 140 s^2 + 196 s + 312 Simulation shows over 30% overshoot and nonminimumphase behavior. Secondorder
approximation not valid.
26.
a. Draw root locus and minimum damping ratio line. 284 Chapter 8: Root Locus Techniques Minimum damping ratio 145.55 o Minimum damping ratio is ζ = cos (180  145.55) = cos 34.45o = 0.825. Coordinates at tangent point
of ζ = 0.825 line with the root locus is approximately –1 + j0.686. The gain at this point is 0.32.
b. Percent overshoot for ζ = 0.825 is 1.019%.
π
4
c. Ts = 1 = 4 seconds; Tp = 0.6875 = 4.57 seconds
d. Secondorder approximation is not valid because of the two zeros and no polezero cancellation.
27.
The root locus intersects the 0.55 damping ratio line at –7.217 + j10.959 with K = 134.8. A
justification of a secondorder approximation is not required. The problem stated the requirements in
terms of damping ratio and not percent overshoot, settling time, or peak time. A secondorder
approximation is required to draw the equivalency between percent overshoot, settling time, and peak
time and damping ratio and natural frequency.
28.
Since the problem stated the settling time at large values of K, assume that the root locus is
4
11 + α
=  T . Since Ts is given
approximately close to the vertical asymptotes. Hence, σint =
2
s
as 4 seconds, σint = 1 and α = 9. The root locus is shown below. Solutions to Problems 285 29. The design point is  0.506 ± j1.0047. Excluding the pole at α , the sum of angles to the design point
is 141.37o. Thus, the contribution of the pole at α is 141.37  180 =  38.63o. The following
geometry applies: Hence, tan θ = 1.0047
= tan 38.63 = 0.799. Thus α = 1.763. Adding this pole at 1.763 yields
α  0.506 180o at  0.506 ± j1.0047 with K = 7.987.
30.
a. 286 Chapter 8: Root Locus Techniques 15 10 Imag Axis 5 0 5 10 15 10 8 6 4
Real Axis 2 0 2 b. Searching along the 10% overshoot line (angle = 126.239o), the point  0.7989 + j1.0898 yields
180o for K = 81.74.
c. Higherorder poles are located at approximately –6.318 and –7.084. Since these poles are more than
5 times further from the imaginary axis than the dominant pole found in (b), the secondorder
approximation is valid.
d. Searching along the imaginary axis yields 180o at j2.53, with K = 394.2.
Hence, for stability, 0 < K < 394.2.
31.
Program:
pos=10;
z=log(pos/100)/sqrt(pi^2+[log(pos/100)]^2)
numg=1;
deng=poly([0 3 4 8]);
G=tf(numg,deng)
Gzpk=zpk(G)
rlocus(G,0:1:100)
pause
axis([2,0,2,2])
sgrid(z,0)
pause
[K,P]=rlocfind(G)
T=feedback(K*G,1)
pause
step(T) Computer response:
z=
0.5912
Transfer function:
1 Solutions to Problems 287 s^4 + 15 s^3 + 68 s^2 + 96 s
Zero/pole/gain:
1
s (s+8) (s+4) (s+3)
Select a point in the graphics window
selected_point =
0.7994 + 1.0802i
K=
81.0240
P=
7.1058
6.2895
0.8023 + 1.0813i
0.8023  1.0813i
Transfer function:
81.02
s^4 + 15 s^3 + 68 s^2 + 96 s + 81.02 288 Chapter 8: Root Locus Techniques 32. a. For a peak time of 1s, search along the horizontal line, Im = π/ Tp= π, to find the point of
intersection with the root locus. The intersection occurs at –2 ± jπ at a gain of 11. Solutions to Problems 289 10
8
6
4 Imag Axis 2
0
2
4
6
8
10
5 4 3 2 1
Real Axis b.
Program:
numg=11*[1 4 5];
deng=conv([1 2 5],poly([3 4]));
G=tf(numg,deng);
T=feedback(G,1);
step(T) 0 1 2 290 Chapter 8: Root Locus Techniques 33.
a. b. Searching the jω axis for 180o, we locate the point j6.29 at a gain of 447.83.
c. Searching for maximum gain between 4 and 5 yields the breakaway point, 4.36. Searching for
minimum gain between 2 and 3 yields the breakin point, 2.56.
d. jω j2
θ6 x x 6 θ1
x
5 θ2
x
4 θ3
O
3 θ4
O
2 j1 θ5 σ 1 x 90o j1 Solutions to Problems 291
To find the angle of departure from the poles at 1±j1: θ1  θ2  θ3 + θ4 + θ5  θ6  900
=  tan1(1/5)  tan1(1/4)  tan1(1/3) + tan1(1/2) + tan1(1/1)  θ6  90o = 1800 . Thus, θ6 = 242.22o
e. Searching along the ζ = 0.3 line (θ = 180  cos1(ζ) = 107.458o) for 180o we locate the point
3.96 ∠ 107.458o = 1.188±j3.777. The gain is 127.133.
34.
a. b. Searching the jω axis for 180o, we locate the point j2.56 at a gain of 30.686.
c. Searching for maximum gain between 0 and 2 yields the breakaway point, 0.823. Searching for
maximum gain between 4 and 6 yields the breakaway point, 5.37. Searching for minimum gain
beyond 8 yields the breakin point, 9.39.
e. Searching along the ζ = 0.3 line (θ = 180  cos1(ζ) = 107.458o) for 180o we locate the point
1.6 ∠ 107.458o = 0.48 ± j1.53. The gain is 9.866. 292 Chapter 8: Root Locus Techniques 35. a. Searching the 15% overshoot line (ζ = 0.517; θ = 121.131ο) for 180o, we find the point 2.404
∠ 121.131ο = 1.243 + j2.058.
b. K = 11.09.
c. Another pole is located left of 3. Searching for a gain of 11.09 in that region, we find the third
pole at 4.514.
d. The third pole is not 5 times farther than the dominant pair from the jω axis. the secondorder
approximation is estimated to be invalid. Solutions to Problems 293 36.
a.
15 10 Imag Axis 5 0 5 10 15
5 4 3 2 1
0
Real Axis 1 2 3 4 b. Searching the jω axis for 180o, we locate the point j1.69 at a gain of 4.249.
c. Searching between 2 and 3 for maximum gain, the breakaway is found at 2.512.
d. jω
splane X X
3 θ6
X
2 θ3 θ1 j2 θ5 X σ 2 1 θ4 θ2 j2 To find the angle of arrival to the zero at 2 + j2: ⎛4
⎛2
⎛2
θ 1 + θ 2 − θ3 − θ4 − θ 5 − θ 6 = θ1 + 90 − 0 − tan−1 ⎝ ⎞ − tan−1 ⎝ ⎞ − tan −1 ⎝ ⎞ = 180
⎠
⎠
⎠
Solving for θ1, the angle of arrival is θ1 = –191.5 .
0 3 4 5 294 Chapter 8: Root Locus Techniques e. The closedloop zeros are the poles of H(s), or –1 ± j2.
f. Searching the ζ = 0.358; (θ = 110.97ο) for 180o, we find the point
= 0.6537+j1.705. The gain, K = 0.8764.
g. Higherorder poles are at –2.846 ± j1.731. These are not 5 times further than the dominant poles.
Further, there are closedloop zeros at –1 ± j2 that are not cancelled any higherorder poles. Thus,
the secondorder approximation is not valid.
37. 50
40
30
20 Imag Axis 10
0
10
20
30
40
50 40 35 30 25 20
15
Real Axis 10 5 0 5 a. The root locus crosses the imaginary axis at j2.621 with K = 4365. Therefore, the system is stable
for 0 < K < 4365.
b. Search the 0.707 damping ratio line for 180o and find –0.949 + j0.949 with K = 827.2.
c. Assume critical damping where root locus breaks away from the real axis. Locus breaks away at –
1.104 with K = 527.6.
38.
Program:
numg=1;
deng=poly([0 3 7 8]);
numh=[1 30];
denh=[1 20 200];
G=tf(numg,deng)
Gzpk=zpk(G)
H=tf(numh,denh) Solutions to Problems 295 rlocus(G*H)
pause
K=0:10:1e4;
rlocus(G*H,K)
sgrid(0.707,0)
axis([2,2,5,5]);
pause
for i=1:1:3;
[K,P]=rlocfind(G*H)
end
T=feedback(K*G,H)
step(T) Computer response:
Transfer function:
1
s^4 + 18 s^3 + 101 s^2 + 168 s
Zero/pole/gain:
1
s (s+8) (s+7) (s+3)
Transfer function:
s + 30
s^2 + 20 s + 200
Select a point in the graphics window
selected_point =
0.9450 + 0.9499i
K=
828.1474
P=
9.9500
9.9500
8.1007
8.1007
0.9492
0.9492 +10.0085i
10.0085i
+ 1.8579i
 1.8579i
+ 0.9512i
 0.9512i Select a point in the graphics window
selected_point =
0.0103 + 2.6385i
K=
4.4369e+003
P=
9.7320 +10.0691i 296 Chapter 8: Root Locus Techniques 9.7320
9.2805
9.2805
0.0126
0.0126 10.0691i
+ 3.3915i
 3.3915i
+ 2.6367i
 2.6367i Select a point in the graphics window
selected_point =
1.0962  0.0000i
K=
527.5969
P=
9.9682
9.9682
7.9286
7.9286
1.1101
1.0962 +10.0052i
10.0052i
+ 1.5303i
 1.5303i Transfer function:
527.6 s^2 + 1.055e004 s + 1.055e005
s^6 + 38 s^5 + 661 s^4 + 5788 s^3 + 23560 s^2
+ 3.413e004 s + 1.583e004 Solutions to Problems 297 39.
a. Search jω = j10 line for 180o and find 4.533 + j10 with K = 219.676.
b. K a = 219.676 x 6
20 c. A settling time of 0.4 seconds yields a real part of 10. Thus if the zero is at the origin, G(s)
K
s(s+20) , which yields complex poles with 10 as the real part. At the design point, 10 + j10, K =
200. 298 Chapter 8: Root Locus Techniques 40.
a. Searching along ζωn = 1 for 180o, find –1 + j2.04 with K = 170.13.
b. Assume critical damping when root locus breaks away form the real axis. Searching for maximum
gain, the breakaway point is at 1.78 with K = 16.946.
41.
K
3
2
T(s) = 3
2 + 5s + K . Differentiating the characteristic equation, s + 6s + 5s + K = 0, yields,
s + 6s
δs
δs
δs
+ 12s
+5
+ 1 = 0.
3 s2
δK
δK
δK
δs
,
Solving for
δK
1
δs
=
δK 3s2 + 12s + 5
The sensitivity of s to K is
K δs
K
1
=s
Ss:K = s
δK
3s2 + 12s + 5
a. Search along the = 0.591 line and find the root locus intersects at s = 0.7353∠126.228ο =
 0.435 + j0.593 with K = 2.7741. Substituting s and K into Ss:K yields
Ss:K = 0.487  j0.463 = 0.672∠43.553o
b. Search along the ζ = 0.456 line and find the root locus intersects at s = 0.8894∠117.129ο =
 0.406 + j0.792 with K = 4.105. Substituting s and K into Ss:K yields
Ss:K = 0.482j0.358 = 0.6∠36.603o
c. Least sensitive: ζ = 0.456.
42.
0.00076s3
The sum of the feedback paths is He(s) = 1 + 0.02s + s+0.06 . Thus, and Solutions to Problems 299
Plotting the root locus, Searching vertical lines to calibrate the root locus, we find that 0.00076K is approximately 49.03 at
10 ± j41.085. Searching the real axis for 0.00076K = 49.03, we find the third pole at 36.09.
41.085
a. ζ = cos (tan1 ( 10 )) = 0.236
b. %OS = e −ζπ / 1−ζ 2 x100 = 46.63% 102 + 41.0852 = 42.28 rad/s
4
4
d. Ts =
= 10 = 0.4 seconds
ζωn c. ωn = e. Tp = π
ωn 1ζ2 π = 41.085 = 0.076 seconds 43.
K 1s
K2
Push K2 to the right past the summing junction and find, T(s) = (1 + K ) ( 2
)
2
s + K3s + K2
K2
K1(s+K )
1
K2
K 1 (s + K )
s2 + K 2
K 3s
1
. Changing form, T(s) =
. Thus, G(s)H(s) = 2
. Sketching the
=2
K3s
s + K2
s + K 3s + K 2
1+ 2
s + K2
root locus, 300 Chapter 8: Root Locus Techniques a.  K2
K1(s+K )
1 s2 + K b. T(s) = 2 K3s 1+ 2
s + K2 K2
K1(s+K ) K2
1
=2
. Therefore closedloop zero at  K . Notice that the zero
1
s + K3s + K2 at the origin of the root locus is not a closedloop zero.
K1s
K2
c. Push K2 to the right past the summing junction and find, T(s) = (1 + K ) ( 2
)
2
s + K3s + K2
K2
K1(s+K )
1
K2
K 1 (s + K )
2+K s
s
K2
1
3
. Changing form, T(s) =
. Thus, G(s)H(s) = 2
. Sketching the
=2
K2
s + K3s
s + K3s + K2
1+ 2
s + K3s
root locus, K2
The closedloop zero is at  K .
1 Solutions to Problems 301 44. a b c d 45.
a. Using Figure P8.15(a),
[Ms2+(D+Dc)s+(K+Kc)]X(s)  [Dcs+Kc]Xa(s) = 0
Rearranging,
[Ms2+Ds+K]X(s) = [Dcs+Kc](X(s)Xa(s))
where [Dcs+Kc](X(s)Xa(s)) can be thought of as the input to the plant.
For the active absorber,
(Mcs2+Dcs+Kc)Xa(s)  (Dcs+Kc)X(s) = 0
or
Mcs2Xa(s)+Dcs(Xa(s)X(s))+Kc(Xa(s)X(s)) = 0
Adding Mcs2X(s) to both sides,
Mcs2(Xa(s)X(s))+Dcs(Xa(s)X(s))+Kc(Xa(s)X(s)) = Mcs2X(s)
Let Xa(s)X(s) = Xc(s) and s2X(s) = C(s) = plant output acceleration. Therefore, (1) 302 Chapter 8: Root Locus Techniques
Mcs2Xc(s)+DcsXc(s)+KcXc(s) = McC(s)
or
(Mcs2+Dcs+Kc)Xc(s) = McC(s) (2) Using Eqs. (1) and (2), and Xa(s)X(s) = Xc(s),
Mc
Dcs+Kc
Xc(s)
X(s)
C(s) = Mcs2+Dcs+Kc ; Xc(s) = Ms2+Ds+K
which suggests the following block diagram: Structure Input force
F(s) + 1
Ms + Ds + K X(s) s2 2 Output structure
acceleration
C(s) force feedback from absorber
Fc(s) Dc s+ Kc Xc(s) Mc
M cs + Dcs + K c
2 Active vibration absorber b. Substituting M = D = K = Dc = Kc =1 and redrawing the block diagram above to show X(s) as the
output yields a block diagram with G s = Mc s 2 s + 1
1
. To study the steadyand H s =
s2 + s + 1
Mc s2 + s + 1 state error, we create a unityfeedback system by subtracting unity from H(s). Thus He(s) = H(s)1 =
G
1
. The equivalent G(s) for this unityfeedback system is Ge s =
Mc s3 − s − 1
2 +s+1
1 + G He
Mc s
= Mc s 2 + s + 1
Mc s4 + 2 Mc s3 + s 3 + Mc s 2 + 2 s 2 + s . Hence the equivalent unityfeedback system is Type 1 and will respond with zero steadystate error for a step force input. Solutions to Design Problems 303 c. Using Ge(s) in part b, we find T s = Ge
1 + Ge = Mc s 2 + s + 1
s 2 + 2 s + 2 Mc s2 + s 3 + 2 s 2 + 2 s + 1
M c s2 + s + 1 numerator and denominator by s3 + 2 s2 + 2 s + 1, T s = s3 + 2 s2 + 2 s + 1
s2 + 2 s + 2 M c s 2 . Dividing . Thus, the system +1
s3 + 2 s2 + 2 s + 1
s2 + 2 s + 2 M c s 2
s2 + 2 s + 2 M c s 2
has the same root locus as a system with G(s)H(s) =
=
.
s3 + 2 s2 + 2 s + 1
s + 1 s2 + s + 1 Sketching the root locus, jω
o
x x oo σ x
o SOLUTIONS TO DESIGN PROBLEMS
46. 4
a. For a settling time of 0.1 seconds, the real part of the dominant pole is  0.1 =  40. Searching
along the σ =  40 line for 180o, we find the point –40 + j57.25 with 20,000K = 2.046 x 109, or K =
102,300.
57.25
b. Since, for the dominant pole, tan1 ( 40 ) = 55.058o, ζ = cos (55.058o) = 0.573. Thus,
%OS = e −ζπ / 1−ζ 2 x100 = 11.14%. c. Searching the imaginary axis for 180o, we find ω = 169.03 rad/s for 20,000K = 1.43 x 1010. Hence, K = 715,000. Therefore, for stability, K < 715,000.
47. G(s) = 61.73K
(s+10)3 (s2 + 11.11s + 61.73) 304 Chapter 8: Root Locus Techniques a. Root locus crosses the imaginary axis at ±j6.755 with 61.73K equal to 134892.8. Thus for oscillations, K = 2185.21.
b. From (a) the frequency of the oscillations is 6.755 rad/s.
c. The root locus crosses the 20% overshoot line at 6∠117.126o =  2.736 + j5.34 with 61.73K = 23323.61. Thus, K = 377.83 and Ts = 4
4
= 2.736 = 1.462 seconds.
ζωn 48.
a. Finding the transfer function with Ca as a parameter,
2 .. Y m (s ) Y (s )
G Plotting the root locus, 2 s (s + 1 ) 2 = s (2 s + 2 )
2 (C a + 1 ) s + 4 s + 2 = s2 + 4s+ 2 1+ C as
2 2 s + 4s+ 2 Solutions to Design Problems 305 4
2
2
b. Since 2ζωn = C +1 ; ωn2 = C +1 , ζ2 = C +1 = 0.692. Hence, Ca = 3.2.
a
a
a
49.
a. b. The pole at 1.8 moves left and crosses the origin at a gain of 77.18. Hence, the system is stable for K > 77.18, where K = 508K2. Hence, K2 < 0.152.
c. Search the ζ = 0.5 (θ = 120o ) damping ratio line for 180o and find the point, 8.044 + j13.932 = 16.087 ∠ 120ο with a K = 508K2 = 240.497. Thus, K2 = 0.473.
d. Search the real axis between 1.8 and 1.6 for K = 240.497 and find the point 1.01. 240.497K1(s+1.6)
Thus Ge(s) = s(s+1.01)(s+8.044+j13.932)(s+8.044j13.932) = 240.497K1(s+1.6)
s(s+1.01)(s2+16.088s+258.8066) . Plotting the root locus and searching the jω axis for 180o we find j15.792 with 240.497K1 = 4002.6, or K1
= 16.643.
e. Root Locus with 0.45 Damping Ratio Line 30 20 Imag Axis 10 0 10 20 30
30 20 10 0
Real Axis 10 20 30 306 Chapter 8: Root Locus Techniques Search the ζ = 0.45 (θ = 116.744o ) damping ratio line for 180o and find the point, 6.685 + j13.267
= 14.856 ∠ 116.744o with a K = 240.497K1 = 621.546. Thus, K1 = 2.584.
50.
a. Update the block diagram to show the signals that form Hsys(s). Perform block diagram reduction of the parallel paths from TW(s). Reduce the momentum wheel assembly to a single block. Solutions to Design Problems 307 Substitute values and find Ge(s) = 4.514x10 −6 K(s + 0.01)
. Plotting the root locus yields
2
s (s + 0.043)
Root Locus with 0.404 Damping ratio Line 0.06 0.04 Imag Axis 0.02 0 0.02 0.04 0.06
0.05 0.04 0.03 0.02 0.01 0
Real Axis 0.01 0.02 0.03 0.04 0.05 b. Searching the 25% overshoot line (ζ = 0.404; θ = 113.8o) for 180o yields 0.0153 + j0.0355 with a gain = 4.514E6K = 0.0019. Thus, K = 420.9.
c. Searching the real axis between –0.025 and –0.043 for a gain of 0.0019. we find the third pole at  0.0125. Simulate the system. There is no polezero cancellation. A simulation shows approximately
95% overshoot. Thus, even though the compensator yields zero steadystate error, a system redesign
for transient response is necessary using methods discussed in Chapter 9. 308 Chapter 8: Root Locus Techniques 51.
a.
1 x 104 Configuration A 0.8 0.6 0.4 Imag Axis 0.2 0 0.2 0.4 0.6 0.8 1 1 2000 1500 x 104 1000 500
Real Axis 0 500 1000 Configuration B 0.8
0.6
0.4 Imag Axis 0.2
0
0.2
0.4
0.6
0.8
1 2000 1500 1000
Real Axis 500 0 500 Solutions to Design Problems 309 1 x 104 Configuration C 0.8 0.6 0.4 Imag Axis 0.2 0 0.2 0.4 0.6 0.8 1 2000 1500 1000 500 0
500
Real Axis 1000 1500 2000 2500 b.
Configuration A: System is always unstable.
Configuration B: root locus crosses jω axis at j2897 with a gain of 3.22 x 106. Thus, for stability, K < 3.22 x 106.
Configuration C: root locus crosses jω axis at j1531 with a gain of 9.56 x 105. System is unstable at high gains. Thus, for stability, 9.56 x 105 > K.
52.
a. Using MATLAB and the Symbolic Math Toolbox, the openloop expression that yields a root locus as a function of N2 is
0.2284x107N2 (s2 + 3.772e05s + 66.27) (s2 + 49.99s + 8789)
Gdt(s) =
________________
s(s+45.12) (s2 + 4.893s + 8.777e04) 310 Chapter 8: Root Locus Techniques Program:
syms s N KLSS KHSS KG JR JG tel s
numGdt=3.92*N^2*KLSS*KHSS*KG*s;
denGdt=(N^2*KHSS*(JR*s^2+KLSS)*(JG*s^2*[tel*s+1]+KG*s)+JR*s^2*KLSS*[(JG*s^2
+KHSS)*(tel*s+1)+KG*s]);
Gdt=numGdt/denGdt;
'Gdt in General Terms'
pretty(Gdt)
'Values to Substitute'
KLSS=12.6e6
KHSS=301e3
KG=668
JR=190120
JG=3.8
tel=20e3
numGdt=3.92*N^2*KLSS*KHSS*KG*s; Solutions to Design Problems 311 numGdt=vpa(numGdt,4);
denGdt=(N^2*KHSS*(JR*s^2+KLSS)*(JG*s^2*[tel*s+1]+KG*s)+JR*s^2*KLSS*[(JG*s^2
+KHSS)*(tel*s+1)+KG*s]);
denGdt=vpa(denGdt,4);
'Gdt with Values Substituted'
Gdt=numGdt/denGdt;
pretty(Gdt)
Gdt=expand(Gdt);
Gdt=vpa(Gdt,4);
'Gdt Different Form 1'
pretty(Gdt);
denGdt=collect(denGdt,N^2);
'Gdt Different Form 2'
Gdt=collect(Gdt,N^2);
pretty(Gdt)
[numGdt,denGdt]=numden(Gdt);
numGdt=numGdt/0.4349e10;
denGdt=denGdt/0.4349e10;
denGdt=expand(denGdt);
denGdt=collect(denGdt,N^2);
Gdt=vpa(numGdt/denGdt,4);
'Gdt Different Form 3'
pretty(Gdt)
'Putting into Form for RL as a Function of N^2 using previous results'
numGH=[1 49.99 8855 3313 582400];
denGH=[41.87 2094 0.3684e7 0.1658e9 0];
denGH=denGH/denGH(1)
GH=tf(numGH,denGH)
GHzpk=zpk(GH)
'Zeros of GH'
rootsnumGH=roots(numGH)
'Poles of GH'
rootsdenGH=roots(denGH)
K=0:1:10000;
rlocus(GH,K)
sgrid(0.5,0)
pause
axis([10,0,20,20])
[K,P]=rlocfind(GH) Computer response:
ans =
Gdt in General Terms
98 2
/
2
2
2
 N KLSS KHSS KG s / (N KHSS (JR s + KLSS) (JG s (tel s + 1) + KG s)
25
/
2
2
+ JR s KLSS ((JG s + KHSS) (tel s + 1) + KG s))
ans =
Values to Substitute
KLSS =
12600000
KHSS =
301000 312 Chapter 8: Root Locus Techniques KG =
668
JR =
190120
JG =
3.8000
tel =
0.0200
ans =
Gdt with Values Substituted
16 2
.9931 10
Ns /
/ (301000.
/ 2
2
8
2
N (190100. s + .1260 10 ) (3.800 s (.02000 s + 1.) + 668. s)
13 2
2
+ .2396 10
s ((3.800 s + 301000.) (.02000 s + 1.) + 668. s))
ans =
Gdt Different Form 1
16 2
.9931 10
Ns /
10 2 5
12 2 4
14 2 3
/ (.4349 10
N s + .2174 10
N s + .3851 10
Ns
/ 14 2 2
16 2
12 5
13 4
+ .1441 10
N s + .2533 10
N s + .1821 10
s + .9105 10
s
17 3
18 2
+ .1602 10
s + .7212 10
s)
ans =
Gdt Different Form 2
16 2
.9931 10
Ns /
10 5
12 4
14 3
/ ((.4349 10
s + .2174 10
s + .3851 10
s
/ 14 2
16
2
18 2
12 5
+ .1441 10
s + .2533 10
s) N + .7212 10
s + .1821 10
s
13 4
17 3
+ .9105 10
s + .1602 10
s)
ans = Solutions to Design Problems 313 Gdt Different Form 3
72
.2284 10 N s /
/(
/ 5
4
3
2
2
(1.000 s + 49.99 s + 8855. s + 3313. s + 582400. s) N
92
5
4
73
+ .1658 10 s + 41.87 s + 2094. s + .3684 10 s )
ans =
Putting into Form for RL as a Function of N^2 using previous results
denGH =
1.0e+006 *
Columns 1 through 4
0.0000 0.0001 0.0880 3.9599 Column 5
0
Transfer function:
s^4 + 49.99 s^3 + 8855 s^2 + 3313 s + 582400
s^4 + 50.01 s^3 + 8.799e004 s^2 + 3.96e006 s
Zero/pole/gain:
(s^2 + 66.27) (s^2 + 49.99s + 8789)
s (s+45.12) (s^2 + 4.893s + 8.777e004)
ans =
Zeros of GH
rootsnumGH =
24.9950
24.9950
0.0000
0.0000 +90.3548i
90.3548i
+ 8.1404i
 8.1404i ans =
Poles of GH
rootsdenGH =
1.0e+002 *
0 314 Chapter 8: Root Locus Techniques 0.0245 + 2.9624i
0.0245  2.9624i
0.4512
Select a point in the graphics window
selected_point =
3.8230 + 6.5435i
K=
51.5672
P=
21.1798
21.1798
3.8154
3.8154 +97.6282i
97.6282i
+ 6.5338i
 6.5338i b. From the computer response, K = 0.2284x107N2 = 49.6. Therefore, N is approximately 5/1000.
53.
a.
YhYcat
Spring
displacement
Desired
force 1
100 Input
voltage+ 1
1000 Controller  Input
transducer F up K 0 .7883 ( s + 53 .85 )
( s2 + 15 .47 s + 9283 )( s 2 + 8 .119 s + 376 .3 ) Actuator 82300 Pantograph
dynamics F out Spring 1
100
Sensor Y hYcat
Spring
displacement
Desired
force + Fup 100 1
1000 Controller  G(s) = 0 .7883 ( s + 53 .85 )
(s 2 + 15 .47 s + 9283 )( s 2 + 8 .119 s + 376 .3 ) Actuator K Pantograph
dynamics Yh ( s) − Ycat (s)
0.7883( s + 53.85)
=2
2
Fup ( s)
(s + 15.47s + 9283)(s + 8.119 s + 376.3)
Ge(s)=(K/100)*(1/1000)*G(s)*82.3e3
0.6488K (s+53.85)
Ge(s) = 2 __________________________ (s + 8.119s + 376.3) (s^2 + 15.47s + 9283) 82300
Spring Fout Solutions to Design Problems 315 100
80
60
40 Imag Axis 20
0
20
40
60
80
100 100 80 60 40
Real Axis 20 0 b. 38% overshoot yields ζ = 0.294. The ζ = 0.294 line intersects the root locus at –9 + j27.16. Here, Ke = 7.179 x 104. Thus K = Ke/0.6488, or K = 1.107 x 105.
c. Ts = 4/Re = 4/9 = 0.44 s; Tp = π/Im = π/27.16 = 0.116 s
d. Nondominant closedloop poles are located at –3.4 ± j93.94. Thus poles are closer to the imaginary axis than the dominant poles. Second order approximation not valid.
e.
Program:
syms s
numg=(s+53.85);
deng=(s^2+15.47*s+9283)*(s^2+8.119*s+376.3);
numg=sym2poly(numg);
deng=sym2poly(deng);
G=tf(numg,deng)
K=7.179e4
Ke=0.6488*K
T=feedback(Ke*G,1)
step(T) Computer response:
Transfer function:
s + 53.85
s^4 + 23.59 s^3 + 9785 s^2 + 8.119e004 s
+ 3.493e006 K=
71790
Ke = 316 Chapter 8: Root Locus Techniques 4.6577e+004
Transfer function:
4.658e004 s + 2.508e006
s^4 + 23.59 s^3 + 9785 s^2 + 1.278e005 s
+ 6.001e006 Tp = 0.12 s, Ts = 0.6 s, %OS = 0.66 − 0.42
= 57.1% .
0.42 ...
View
Full
Document
This note was uploaded on 04/04/2008 for the course MECH COntrol Sy taught by Professor Khurshid during the Spring '08 term at Michigan State University.
 Spring '08
 khurshid

Click to edit the document details