E I G H T

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 - , the real part of poles with this settling time, for 180 o.
2

We find the point to be - 0.5+j6.9 with 76.39K = 7194.23, or K = 94.18. Second-order

approximation is OK since third pole is much more than 5 times further from the imaginary axis

than the dominant second-order 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
8-2 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 8-3

UFSS Vehicle: Transient Design via Gain

K(s+0.437)
a. Push -K1 to the right past the summing junction yielding G(s) = , where
s(s+2)(s+1.29)(s+0.193)
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)
. The root locus is shown below in (b). Searching the 10% overshoot line (
s(s+2)(s+1.29)(s+0.193)

= 0.591;  = ), 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)
8-4 Chapter 8: Root Locus Techniques

T=minreal(T)
step(T)
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 Design Problems 8-5

ANSWERS TO REVIEW QUESTIONS

1. The plot of a system's closed-loop poles as a function of gain
2. (1) Finding the closed-loop transfer function, substituting a range of gains into the denominator, and
factoring the denominator for each value of gain. (2) Search on the s-plane for points that yield 180 degrees
when using the open-loop 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 Routh-Hurwitz to the closed-loop 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 break-in 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 closed-loop 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
8-6 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.
j j

s-plane X
s-plane

X X O  X 

(a) (b)

j j

X s-plane X s-plane

O O  

X X

(c) (d)
 Solutions to Design Problems 8-7

j j

X s-plane X s-plane

O O  

X X

(c) (d)
j j

s-plane
s-plane

X X X X  O O X X 

(e) (f)

3.
 3
a. There are two asymptotes with a = , and real axis intersection
2 2
2
0 + 0 − 6 − (− )
a = 3 = −2.67 . The break-in and breakaway points are obtained by finding
3 −1
 ( + 6)
2
 3 + 6 2
K =− =− . Obtaining
2 2
+ +
3 3
 2 2
(
  +  3 + 12 −  + 6
3
) ( 2

2 ( + 2)
2 )
dK
=−  3
=− and solving for the roots of the
d  2
2
 2
2

 +   + 
 3  3
numerator we get:  = 0, − 2, − 2 . So the root locus looks as follows:
8-8 Chapter 8: Root Locus Techniques

 2
K s + 
 3
b. We can obtain K from 1 + 2 s = −2 = 0 , resulting in K = 12 . Note that the open loop
s ( s + 6)
 2
12 s + 
 3
zero will appear as a closed loop zero, so the closed loop transfer function is T ( s ) =
( s + 2) 3

4.

Breakaway:  = -2.43 for K = 52.1

5.
The root locus is:
 Solutions to Design Problems 8-9

There are no break-in or breakaway points which can be verified by computing

𝑁(𝑠)𝐷 ′ (𝑠) − 𝑁 ′ (𝑠)𝐷(𝑠) = 0, or (𝑠 2 + 11𝑠 + 10)(2𝑠 − 2) − (2𝑠 + 11)(𝑠 2 − 2𝑠 −
24) = 0, or 13𝑠 2 + 68𝑠 + 244 = 0 which has only complex conjugate roots not on the
locus.

To find the range of 𝐾 for closed loop stability we use the characteristic equation
𝐾(𝑠 + 1)(𝑠 + 10)
1+ =0
(𝑠 + 4)(𝑠 − 6)
Which can be rewritten as (1 + 𝐾)𝑠 2 + (11𝐾 − 2)𝑠 + (10𝐾 − 24) = 0. The Routh array
is

𝑠2 1+𝐾 10𝐾 − 24

𝑠 11𝐾 − 2

1 10𝐾 − 24

2 12
The first row demands 𝐾 > −1; the second 𝐾 > 11 and the third 𝐾 > 5
. The
12
intersection of the three requirements is 𝐾 > 5
= 2.4.
8-10 Chapter 8: Root Locus Techniques

6.
20 K ( s + 5)
Convert the denominator to the following form: D( s ) = 1 + and thus identify
s 3 + 2s 2 + 7 s
20 K ( s + 5) 20 K ( s + 5)
G( s) = = .
s + 2s + 7 s s( s 2 + 2s + 7)
3 2

Plotting the root locus yields

Im Im
s-plane s-plane

Re Re
 Solutions to Design Problems 8-11

7.

Closed-loop poles will be in the left-half-plane when rhp pole reaches the origin,

(3)(5) 15
or K  = .
( 2 )( 2 ) (2) 8

8.
8-12 Chapter 8: Root Locus Techniques

(3)(3) 9
Closed-loop poles will be in the right-half-plane for K = (gain at the origin).
(4)(4) 16
9 9
Therefore, stable (although marginally stable) for 𝐾 < ; unstable for 𝐾 > .
16 16

9.
System 1:

a. Breakaway:  = 1.41 for K = 0.03; Break-in:  = -1.41 for K = 33.97.

b. Imaginary axis crossing at j1.41 for K = 1. Thus stable for K > 1.

c. At break-in point, poles are multiple. Thus, K = 33.97.

 Solutions to Design Problems 8-13

d. Searching along 1350 line for 1800, K = 5 at 1.414  1350.

System 2:

a. Break-in:  = -1.41 for K = 28.14.

b. Imaginary axis crossing at j1.41 for K = 0.67. Thus stable for K > 0.67.

c. At break-in point, poles are multiple. Thus, K = 28.14.

d. Searching along 1350 line for 1800, K = 4 at 1.414  1350.

10.
a.

Root locus crosses the imaginary axis at the origin for K = 6. Thus the system is stable for K > 6.

b.
8-14 Chapter 8: Root Locus Techniques

Root locus crosses the imaginary axis at j0.65 for K = 0.79. Thus, the system is stable for K < 0.79.

11.

The system is closed loop unstable for all 𝐾 > 0. The value of 𝐾 at which there will be only
two roots in the RHP can be found by calculating the point at which the real RHP pole
crosses the origin. Namely
𝐾(𝑠 + 6)
1+ 2 | =0
(𝑠 + 1)(𝑠 − 2)(𝑠 + 4) 𝑠=0
4
It can readily be seen that there will be only two RHP roots for 𝐾 < 3.
 Solutions to Design Problems 8-15

12.
System: G
Gain: 517
Root Locus for Prob. 8-16 Pole: 0.00268 + 6.3i
8 Damping: -0.000425
Overshoot (%): 100
Frequency (rad/s): 6.3

6
System: G
Gain: 74.2
Pole: -1.66 + 2.21i
4 Damping: 0.602
Overshoot (%): 9.38
Imaginary Axis (seconds-1)

Frequency (rad/s): 2.76

-2

System: G
-4 Gain: 121
Pole: -1.37 - 3.14i
Damping: 0.401
Overshoot (%): 25.3
-6 Frequency (rad/s): 3.43

-8
-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1
-1
Real Axis (seconds )

Root locus crosses the imaginary axis at j6.3 with a gain of 517. Real axis breakaway is at –
− 13 1
2.0 at a gain of 36.0. Real axis intercept for the asymptotes is = −4 . The angles of
3 3
 5
the asymptotes are: 3 , , 3 . Some other points on the root locus are (see figure):
 = 0.401: -1.37 - j3.14, K = 121
 = 0.602: -1.66 + j2.21, K = 74.2

13.
a. Root locus crosses the imaginary axis at  j3.162 at K = 52.
8-16 Chapter 8: Root Locus Techniques

b. Since the gain is the product of pole lengths to -5, K = (1) ( 42 + 12 )( )

42 + 12 = 17 .

14.
a.

b. 𝜎𝑎 = 0−1−4−6−(−3)
4−1
8
=− ;𝜃=
3
(2𝑘+1)𝜋
3
𝜋
= , 𝜋,
3
5𝜋
3

c. The characteristic equation is 𝑠 4 + 11𝑠 3 + 34𝑠 2 + (24 + 𝐾)𝑠 + 3𝐾 = 0

The corresponding Routh array is:

𝑠4 1 34 3𝐾

𝑠3 11 24 + 𝐾
350 − 𝐾
𝑠2 11 3𝐾
(−𝐾 + 75)(𝐾 + 112)
𝑠 350 − 𝐾

1 3𝐾

The system will have poles on the 𝑗𝜔 axis when 𝐾 = 75.

d. We calculate the breakaway points from 𝑁(𝑠)𝐷′ (𝑠) − 𝑁 ′ (𝑠)𝐷(𝑠) = 0 where 𝑁(𝑠) = (𝑠 + 3);

𝐷(𝑠) = 𝑠 4 + 11𝑠 3 + 34𝑠 2 + 24𝑠; 𝑁 ′ (𝑠) = 1; 𝐷′ (𝑠) = 4𝑠 3 + 33𝑠 2 + 68𝑠 + 24; giving 3𝑠 4 +

34𝑠 3 + 133𝑠 2 + 204𝑠 + 72 = 0 which has roots at −0.4915, −2.93 ± 𝑗1.099, −4.973. The only

viable breakpoint is -0.4915. We use the characteristic equation to solve for 𝐾 = 1.9256.
 Solutions to Design Problems 8-17

15.
(0 - 3 - 6) - (-)
Assume that root locus is epsilon away from the asymptotes. Thus, a = ≈ -1;
2
(2k+1)  3
Angle = = , . Hence  = 7. Checking assumption at –1 ± j100 yields -180o with K =
2 2 2

9997.02.

16.

a. The breakaway points are found from 𝑁(𝑠)𝐷 ′ (𝑠) − 𝑁 ′ (𝑠)𝐷(𝑠) = 0 or

(𝑠 2 − 5𝑠 + 6)(3𝑠 2 + 10𝑠 + 6) − (2𝑠 − 5)(𝑠 3 + 5𝑠 2 + 6𝑠) = 0
or
𝑠 4 − 10𝑠 3 + 13𝑠 2 + 60𝑠 + 36 = 0

whose roots are 10.662, -2.5, 2.4, -0.56; the latter two being on the root locus.

b. The characteristic equation is

𝐾(𝑠 − 2)(𝑠 − 3)
1+ =0
𝑠(𝑠 + 2)(𝑠 + 3)
8-18 Chapter 8: Root Locus Techniques

or

𝑠 3 + (5 + 𝐾)𝑠 2 + (6 − 5𝐾)𝑠 + 6𝐾 = 0

The Routh array is

𝑠3 1 6 − 5𝐾

𝑠2 3+𝐾 6𝐾
−(𝐾 − 1)(𝐾 + 6)
𝑠 5+𝐾

1 6𝐾

The intersection with the j-axis will occur when the third row is 0, namely K=1. At this point we

can find the intersections by solving 6𝑠 2 + 6 = 0; which gives 𝑗𝜔 = ±𝑗.

c.
From the Routh Hurwitz table., the last row demands 𝐾 > 0, which in turn satisfies the
second row. Thus the third row will be positive if 𝐾 < 1. The range for closed loop
stability is 0 < 𝐾 < 1.

d. Searching 120, we find the point -0.366j0.637 for K=0.41

17.
 Solutions to Design Problems 8-19

(-1 -2 -3 -4) - (0) 5 (2k+1)  3 5 7

a. Asymptotes: int = =- ; Angle = = , , ,
4 2 4 4 4 4 4
b. Breakaway: -1.38 for K = 1 and -3.62 for K = 1
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 180 0. Point is 1.4171 =
- 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 = , where - zc is the location of the
zc

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.

8-20 Chapter 8: Root Locus Techniques

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.

18.

19.
a.

b. Root locus crosses 20% overshoot line at 1.8994  117.126o = - 0.866 ± j1.69 for K = 9.398.
4 
c. Ts = = 4.62 seconds; Tp = = 1.859 seconds
0.866 1.69

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:  = -4.27

 Solutions to Design Problems 8-21

20.

b.
(−5−4−3−1)−(2)
𝜎𝒂 = = −7.5
4−2

(2𝑘 + 1)𝜋 𝜋 3𝜋
𝜃𝑎 = = ,
4−2 2 2
c. The characteristic equation is 𝑠 4 + 13𝑠 3 + (𝐾 + 59)𝑠 2 + (107 − 2𝐾)𝑠 + (60 + 2𝐾) = 0

The corresponding Routh array is:

𝑠4 1 K+59 60 + 2𝐾

𝑠3 13 107 − 2𝐾
660 + 15𝐾
𝑠2 13 60 + 2𝐾
−(𝐾 + 45.792)(𝐾 − 44.025)
𝑠 22 + 0.5𝐾

1 60 + 2𝐾

So the system is closed loop stable when 0 < 𝐾 < 44.0252.

d. The breakaway points are found from 𝑁(𝑠)𝐷 ′ (𝑠) − 𝑁 ′ (𝑠)𝐷(𝑠) = 0 or
2𝑠 5 + 7𝑠 4 − 44𝑠 3 − 147𝑠 2 + 166𝑠 + 334 = 0
8-22 Chapter 8: Root Locus Techniques

Which has the roots: 4.324, 1.6573, -3.4207, -4.5698 and -1.4908. Only the last two roots
are in the root locus.

e. For 20% overshoot, the Equation yields  = 0.456. Searching along this damping ratio line, we find
the 1800 point at –0.566 + j1.09 where K = 12.5.
f. –0.566 + j1.09
g. Second-order approximation not valid because of the existence of closed-loop zeros in the rhp.

h.
Program:
s=tf('s');
G=12.5*(s^2-2*s+2)/(s^4+13*s^3+59*s^2+107*s+60);
T=G/(1+G);
step(T)
Computer response:

Simulation shows over 35% overshoot and nonminimum-phase behavior. Second-order

approximation not valid.

21.
a. Draw root locus and minimum damping ratio line.
 Solutions to Design Problems 8-23

Minimum damping ratio is  = cos (180 - 145.55) = cos 34.45o = 0.825. Coordinates at tangent point

of  = 0.825 line with the root locus is approximately –1 + j0.686. The gain at this point is 0.32.

b. Percent overshoot for  = 0.825 is 1.019%.

4 
c. Ts = = 4 seconds; Tp = = 4.57 seconds
1 0.6875

d. Second-order approximation is not valid because of the two zeros and no pole-zero cancellation.

22.

Since the problem stated the settling time at large values of K , assume that the root locus is
−25+𝛼 4
approximately close to the vertical asymptotes. Hence, 𝜎𝑎 = = − . Since specified 𝑇𝑠 = 1sec,
2 𝑇𝑠

∝= 17. The root locus is shown below

8-24 Chapter 8: Root Locus Techniques

23.
The design point is – 0.6 ± j1.2. Excluding the pole at – , the sum of angles to the design point is –
156.08o. Thus, the contribution of the pole at –  is 156.08 – 180 = – 23.92o.

The following geometry applies:

j
x j1.2

x  σ
–α – 0.6

1.2
Hence, tan = = tan 23.92 = 0.4436. Thus  = 3.305. Adding this pole at – 3.305
 − 0.6
yields 180o at – 0.6 ± j1.2 with K = 51.8.
 Solutions to Design Problems 8-25

24.
a.

15

10

5
Imag Axis

-5

-10

-15
-10 -8 -6 -4 -2 0 2
Real Axis

b. Searching along the 10% overshoot line (angle = 126.239o), the point - 0.7989 + j1.0898 yields
180o for K = 81.74.
c. Higher-order 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 second-order
approximation is valid.
d. Searching along the imaginary axis yields 180 o at j2.53, with K = 394.2.
Hence, for stability, 0 < K < 394.2.

25.
a. For a peak time of 1s, search along the horizontal line, Im =  Tp=  to find the point of

intersection with the root locus. The intersection occurs at –2 ± j at a gain of 11.
8-26 Chapter 8: Root Locus Techniques

10

Imag Axis
0

-2

-4

-6

-8

-10
-5 -4 -3 -2 -1 0 1 2
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)
 Solutions to Design Problems 8-27

26.
a.

b. Searching the j 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 break-in point, -2.56.

d.
j

j2


x j1

    
x x x O O 
-6 -5 -4 -3 -2 -1

90o
x -j1
8-28 Chapter 8: Root Locus Techniques

To find the angle of departure from the poles at -1±j1: - - 2 - 3 + 4 + 5 - 6 - 900

= - tan-1(1/5) - tan-1(1/4) - tan-1(1/3) + tan-1(1/2) + tan-1(1/1) - 6 - 90o = 1800 . Thus, 6 = - 242.22o

e. Searching along the  = 0.3 line ( = 180 - cos-1() = 107.458o) for 180o we locate the point
3.96  107.458o = -1.188±j3.777. The gain is 127.133.
27.

a.

b. To find the j crossing we will use Routh Hurwitz. The characteristic equation can be written
as

s 4 + 13s3 + 32s 2 + (20 + K )s + 4K = 0

The Routh array is

s4 1 32 4K
𝑠3 13 20 + K
396 − K
𝑠2 4K
13
(K − 24.4)( K + 324.4)
𝑠
K − 396

1 4K

For closed loop stability, the fifth row demands K  0 , the 3d row K  396 ; satisfying these
requirements the 4th row will be positive as long as K  24.4 . Thus the system will have roots in the
 Solutions to Design Problems 8-29

j axis when K = 24.4 . Substituting this value, we get a row of zeros so the poles are located at the
solution of 28.6s + 97.6 = 0 ; namely s =  j1.487 .
2

c. The breakaway points are obtained from the solution of 𝑁(𝑠)𝐷 ′ (𝑠) − 𝑁 ′ (𝑠)𝐷(𝑠) = 0 or
3s 4 + 42s3 + 188s 2 + 256s + 80 = 0 , which has roots −5.974  j1.322 , −1.61 and
−0.443 the breakaway point occurring at the latter. At this point K = 1.0323 .
e. Searching for the crossing of the =0.3 line (  = 180 − cos−1  = 107.5 )for 180 we locate
the point −0.292  j 0.926 , when K = 5.61.

28.
a.

15

10

5
Imag Axis

-5

-10

-15
-5 -4 -3 -2 -1 0 1 2 3 4
Real Axis

b. Searching the j 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.
8-30 Chapter 8: Root Locus Techniques

j

s-plane

 
X  j2

 X 
X 
-3 -2 -1 2

 
X -j2

To find the angle of arrival to the zero at 2 + j2:

4 2 2
 1 +  2 − 3 − 4 −  5 −  6 = 1 + 90 − 0 − tan−1   − tan−1   − tan−1   = 180
3 4 5
Solving for 1, the angle of arrival is 1 = –191.50.

e. The closed-loop zeros are the poles of H(s), or –1 ± j2.

f. Searching the  =  ( = ) for 180o, we find the point

= -0.6537+j1.705. The gain, K = 0.8764.

g. Higher-order poles are at –2.846 ± j1.731. These are not 5 times further than the dominant poles.
Further, there are closed-loop zeros at –1 ± j2 that are not cancelled any higher-order poles. Thus, the
second-order approximation is not valid.
 Solutions to Design Problems 8-31

29.
Root Locus for Problem 8-39
40

30

20

Imaginary Axis (seconds-1)

10

-10

-20

-30

-40
-40 -35 -30 -25 -20 -15 -10 -5 0
Real Axis (seconds -1)
8-32 Chapter 8: Root Locus Techniques

Zoomed Root Locus for 8-39

5
System: untitled1
Gain: 5.94e+03
4 Pole: 0.00403 + 2.76i
Damping: -0.00146
Overshoot (%): 100
Frequency (rad/s): 2.76
3 System: untitled1
Gain: 1.09e+03
Pole: -0.978 + 0.979i
2 0.707 Damping: 0.707
Overshoot (%): 4.33
Imaginary Axis (seconds-1)

Frequency (rad/s): 1.38

1

0
System: untitled1
Gain: 690
-1 Pole: -1.13 - 0.0584i
Damping: 0.999
Overshoot (%): 0
-2 0.707 Frequency (rad/s): 1.13

-3

-4

-5
-2 -1.5 -1 -0.5 0 0.5
Real Axis (seconds -1)

a. As could be seen from the zoomed graph, the root locus crosses the imaginary axis at j2.76 with K
= 5940. Therefore, the system is stable for 0 < K < 5940.

b. Search the 0.707 damping ratio line for 180 o and find –0.978 + j0.978 with K = 1090.
c. Assume critical damping where root locus breaks away from the real axis. Locus breaks away at –
1.13 with K = 690.

30.
Program:
numg=1;
deng=poly([0 -3 -7 -9]);
numh=[1 30];
denh=[1 20 225];
G=tf(numg,deng)
Gzpk=zpk(G)
H=tf(numh,denh)
rlocus(G*H)
pause
K=0:10:1e4;
rlocus(G*H,K)
sgrid(0.707,0)
axis([-2,0.5,-5,5]);
pause
for i=1:1:3;
[K,P]=rlocfind(G*H)
end
 Solutions to Design Problems 8-33

T=feedback(K*G,H)
step(T)

Computer response:
G=
1
------------------------------
s^4 + 19 s^3 + 111 s^2 + 189 s
Continuous-time transfer function.
Gzpk =
1
-------------------
s (s+9) (s+7) (s+3)
Continuous-time zero/pole/gain model.
H=
s + 30
----------------
s^2 + 20 s + 225
Continuous-time transfer function.
Root Locus
100

80

60

40
Imaginary Axis (seconds-1)

20

-20

-40

-60

-80

-100
-100 -80 -60 -40 -20 0 20 40 60 80
Real Axis (seconds -1)

Select a point in the graphics window

8-34 Chapter 8: Root Locus Techniques

Root Locus
5

2 0.707
Imaginary Axis (seconds-1)

-1

-2 0.707

-3

-4

-5
-2 -1.5 -1 -0.5 0 0.5
Real Axis (seconds -1)

selected_point =
-0.9729 + 0.9783i
K=
1.0891e+03

P=
-9.9610 +11.1994i
-9.9610 -11.1994i
-8.5608 + 1.6058i
-8.5608 - 1.6058i
-0.9782 + 0.9799i
-0.9782 - 0.9799i

Select a point in the graphics window

selected_point =
0.0030 + 2.7484i
K=
5.9000e+03
P=
-9.7971 +11.2966i
-9.7971 -11.2966i
-9.7016 + 3.2277i
-9.7016 - 3.2277i
-0.0013 + 2.7518i
-0.0013 - 2.7518i
Select a point in the graphics window
selected_point =
-1.1295 + 0.0155i
K=
 Solutions to Design Problems 8-35

688.7886
P=
-9.9753 +11.1923i
-9.9753 -11.1923i
-8.3945 + 1.2181i
-8.3945 - 1.2181i
-1.1302 + 0.0155i
-1.1302 - 0.0155i
T=
688.8 s^2 + 1.378e04 s + 1.55e05
---------------------------------------------------------------------
s^6 + 39 s^5 + 716 s^4 + 6684 s^3 + 28755 s^2 + 4.321e04 s + 2.066e04
Continuous-time transfer function.

The step response for the critically damped case is shown below:

Step Response
8

5
Amplitude

0
0 2 4 6 8 10 12
Time (seconds)

31.

a. Search j=j5 for 180and find −3.41 + j5 with K = 58.5 .

Kz
b. K a = = 11.7
10

c. A settling time of 0.8 sec yield a real part of -5. Thus if the zero is at the origin
K
G( s) = , which yields complex poles with -5 as the real part. At the design point
s( s + 10)

−5 + j5 , K = 50 .
8-36 Chapter 8: Root Locus Techniques

32.
a. Searching along n = -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.

33.

The damping ratio corresponding to the required percent overshoot is:

%OS
- ln ( )
100 − ln 0.205
= = = 0.45;
%OS
2 + ln2 (
100
)  2 + ln 2 0.205

The following MATLAB M-file was written to plot the root locus; zoom it to find the gain K at that

value of  , then plot the closed-loop unit-step response, c(t).

numg = [1 0.02];

deng = poly([0 0 -4 -10 -25]);

G = tf(numg, deng);

rlocus(G);

title('Full Root Locus')

pause

axis ([-5, 0, -1, 6]);

z=0.45;

sgrid(z,0)

title('Root Locus Zoomed-in for Dominant Poles at a 0.45 Damping

Line')

[K1,p]=rlocfind(G);

pause

K = K1;

T=feedback(K*G,1); %T is the closed-loop TF of the system

 Solutions to Design Problems 8-37

T=minreal(T);

step(T);

grid

ylabel('Unit-Step Response, c(t)')

title('Unit-Step Response at K = K1')

The full root-locus obtained is shown below, followed by the zoomed-in root locus and the unit-step

response with all of the requested important characteristics marked on it.

For  = 0.45, K = 2,183.8 and the dominant poles were found to be at –1.2263 ± 2.4239i.
8-38 Chapter 8: Root Locus Techniques

Root Locus Zoomed-in for Dominant Poles at a 0.45 Damping Line

6
0.45

4
Imaginary Axis (seconds-1)

-1
-5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0
Real Axis (seconds -1)

Full Root Locus

80

60

40
Imaginary Axis (seconds-1)

20

-20

-40

-60

-80
-100 -80 -60 -40 -20 0 20 40 60 80
Real Axis (seconds -1)
 Solutions to Design Problems 8-39

Unit-Step Response at K = K1
1.4 System: T
Peak amplitude: 1.21
Overshoot (%): 20.5
At time (seconds): 1.41
1.2
System: T
Final value: 1
1
Unit-Step Response, c(t)

System: T
Settling time (seconds): 3.04
System: T
0.8 Rise time (seconds): 0.592

0.6

0.4

0.2

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
Time (seconds)

34.
K ( s − 1)
a. The characteristic equation is 1 + = 0 or s 2 + (5 + K )s + (6 − K ) = 0
( s + 2)(s + 3)

The Routh array is

s2 1 6−K

s 5+ K

1 6−K

It follows that −5  K  6

b. The locus for K  0 is

8-40 Chapter 8: Root Locus Techniques

( + 1)( + 2)  2 + 5 + 6
c. To find the break-in, breakaway points let K = = . Then
 −1  −1
calculate =
( =
)
dK ( − 1)(2 + 5) −  2 + 5 + 6  2 − 2 − 11
d ( − 1)2 ( − 1)2
The roots of the numerator are  = −2.4641, 4.4641

The locus for K  0 is

 Solutions to Design Problems 8-41

d. The smallest settling time for the system will occur when both roots are as far away as possible
to the left of the j axis. This will happen when the system has identical roots at -2.4641. To find
the value of K we use the characteristic equation
K ( s − 1)
1− s = −2.4641 = 0 . Solving for K we get K = −0.0718
( s + 2)(s + 3)

K ( s − 1) K
e. The proportional error constant K p = Lim =−
s →0 ( s + 1)(s + 2) 6
1 6
Then ess = = = 0.9882
1+ K p 6 − K
4
f. The system in this case is critically damped, the settling time Ts  = 1.6 sec .The step
2.4641
response will approximately be:

35.
1300K
a. The open loop transmission is L( s ) = .
s − 739600 2

− 860 + 860  3
There are two asymptotes with  a = = 0 and angles  a = ,
2−0 2 2
 − 860
2 2
dK 2
To find the breakaway points let K = − . Then =− = 0 , so the
1300 d 1300
breakaway points occur when  = 0 . The gain at this point is given by the solution of
1300K
1+ 2 s =0 = 0 or K = 568.9
s − 860 2
It is obvious from the figure that the system is unstable for all values of K  0.
8-42 Chapter 8: Root Locus Techniques

1300K ( s + 200)
b. The open loop transmission is L( s ) = .
( s + 1000)(s 2 − 739600)
− 1000 − 860 + 860 − (−200)
There are two asymptotes with  a = = −400 and angles
3 −1
 3
a = ,
2 2

To find the breakaway points let

( + 1000)( 2 − 860 2 )  3 + 1000 2 − 739.6  10 3  − 739.6  10 6
K =− =− .
1300( + 200) 1300( + 200)

dK 2 3 + 1600 2 + 400  10 3  + 591.68  10 6

Then =− = 0 The numerator has two
d 1300( + 200)

complex conjugate solutions, and a real  = −928 . The gain at this point is given by the solution
1300K ( s + 200)
of 1 + s = −928 = 0 or K = 9.25
( s + 1000)(s 2 − 860 2 )

We use Routh-Hurwitz to find the range of K for which the system is closed loop stable. Let
1300K ( s + 200)
1+ = 0 or
( s + 1000)(s 2 − 860 2 )

s 3 + 1000s 2 + (1300K − 739600)s + (260000K − 739.6 106 ) = 0

The Routh array is:

 Solutions to Design Problems 8-43

s3 1 1300K − 739600

s2 1000 260000K − 739.6 106

s 1040K

1 260000K − 739.6 106

The dominant requirement given by the fourth row: K  2844.6

36.
V
s+

a. After substituting numerical values G ( s) = ( s) = 0.5V 2 0.6
 s − 12.25

The system’s characteristic equation is:

V
s+
1 + 0.5V 2 0.6 = 0 or s 2 + 0.5Vs + (0.833V 2 − 12.25) = 0
s − 12.25
8-44 Chapter 8: Root Locus Techniques

The Routh array is

s2 1 0.833V 2 − 12.25

s 0.5V

1 0.833V 2 − 12.25
m
From which we get V  3.83
s
b. The characteristic equation cannot be written in the form 1 + VG eq ( s ) = 0

− 0.5  − 3.08V 2 + 49
c. Solving for the two roots we get s1, 2 =
2

A simple script that will plot the root locus is

>> v=linspace(0,4,100000);

>> s1 = (-0.5+sqrt(-3.08*v.^2+49))/2;

>> s2 = (-0.5-sqrt(-3.08*v.^2+49))/2;

>> plot(real(s1),imag(s1),real(s2),imag(s2))
 Solutions to Design Problems 8-45

0.3

0.2

0.1

-0.1

-0.2

-0.3

-0.4
-4 -3 -2 -1 0 1 2 3 4

37.
a. The system’s characteristic equation is found by calculating det(sI − A) = 0 . This results in
s 4 − 12.3415s 3 + (54.5414K − 256.9538)s 2 + 1250.2Ks − 1995.2K = 0

Which can be manipulated into

54.5414s 2 + 1250.2s − 1995.2 54.5414( s − 1.498)( s + 24.42)
1+ K = 0 or 1 + K =0
s − 12.3415s − 256.9538s
4 3 2
s 2 ( s + 11)(s − 23.3472)
 3
b. There are two asymptotes with  a = , and real axis intersection
2 2
0 + 0 − 11 + 23.3472 − 1.498 + 24.42
a = = 17.63 .
4−2

To find the break in-breakaway points let

 4 − 12.3415 3 − 256.9538 2
K =− , then
54.5414 2 + 1250.2 − 1995.2
dK
=−
(54.5414 2 + 1250.2 − 1995.2)(4 3 − 37.0245 2 − 513.91 )
d (54.5414 2 + 1250.2 − 1995.2)2
−
( 4
)
− 12.3415 3 − 256.9538 2 (109.0828 + 1250.2)
(54.5414 2
+ 1250.2 − 1995.2 )
2
8-46 Chapter 8: Root Locus Techniques

109.0828 5 + 3077.5 4 − 38840 3 − 247380 2 + 1025400

=−
(54.5414 2
+ 1250.2 − 1995.2 )2

The numerator of this expression has roots at  = 0,− 36.13,11.95, − 7.1, 3.0664

The root locus is:

c. For any value of K  0 there are always closed loop poles in the RHP.

38.
a. Using Mason’s rule it can readily be found that (with d=0) the open loop transmission from u r to
u o is
(K s + 1)
1 1
KT Km
u0 s Ls Cs = K T K m ( K s + 1)
=
ur
1+ Km
1
+
1
+

C L s L(C + C L ) s 2 + K mCs + 1 
Ls LCs 2 C

So the system’s characteristic equation is:

K T K m ( K s + 1)
1+  =0

s L(C + C L ) s 2 + K mCs + 1 
or
 Solutions to Design Problems 8-47

L(C + C L ) s 3 + K mCs 2 +  +  K T K m K  s + K T K m = 0

b. Substituting numerical values the characteristic equation becomes

15.2 (11 + CL )s 3 + 14.52s 2 + 73.7375m s + 2.39125 = 0

The Routh array is

s3 15.2 (11 + CL ) 73.7375m

s2 14.52 2.39125
36.347 (11 + C L ) − 1.07067
s
− 14.52

1 2.39125

To have all the first column positive it is required that

C L  −11F and C L  29.45mF . It follows that

0  CL  29.45mF due to physical restrictions in the capacitance.

The characteristic equation can also be expressed as

90909s 3
1 + CL 3 = 0 or
s + 86842s 2 + 4.4101 108 s + 1.5212  1010
90909s 3
1 + CL = 0.
(s + 35)(s + 5379)(s + 81428)
There are no asymptotes. The break-in and breakaway points are calculated by first obtaining
 3 + 86842 2 + 4.4101  108  + 1.5212  1010
CL = − . Then differentiating,
90909 3
dC L  3 (3 2 + 173684 + 4.4101 108 ) − ( 3 + 86842 2 + 4.4101 108  + 1.5212  1010 )3 2
=−
d 90909 6
8-48 Chapter 8: Root Locus Techniques

86842 4 + 7.893  108  3 + 4.564  1010  2

= The roots of the numerator are
90909 6
 = 0, 0, − 9030.7, − 58.2
The value of CL at -9030.7 is obtained from
90909s 3
1 + CL s = −9030.7 = 0 giving
s 3 + 86842s 2 + 4.4101 108 s + 1.5212  1010
C L = 3.5523 10 −5

The root locus is:

 Solutions to Design Problems 8-49

39.
a. Substituting values we have
𝐾 𝑠−0.5 0.1538
𝐺𝑑 (𝑠) = 0.003(1 + ) , and 𝐺𝑡 (𝑠) =
𝑠 𝑠+0.5 𝑠(𝑠+0.833)

The systems characteristic equation is

𝐾 𝑠 − 0.5
1 + 𝐺𝑑 𝐺𝑡 = 1 + 4.614 × 10−4 (1 + ) =0
𝑠 𝑠(𝑠 + 0.833)(𝑠 + 0.5)

Which can be rewritten as:

4.614 × 10−4 (𝑠 − 0.5)
1+𝐾 =0
𝑠4 + 1.333𝑠 3 + 0.417𝑠 2 − 2.308 × 10−4 𝑠

or
4.614 × 10−4 (𝑠 − 0.5)
1+𝐾 =0
𝑠(𝑠 − 0.0006)(𝑠 + 0.5029)(𝑠 + 0.8306)

There are 3 asymptotes with a real axis intersection at

0 + 0.0006 − 0.5029 − 0.8306 − 0.5
𝜎𝑎 = = −0.61
4−1

The breakaway points are found by defining

𝑁(𝑠) = 𝑠 − 0.5 and 𝐷(𝑠) = 𝑠 4 + 1.333𝑠 3 + 0.417𝑠 2 − 2.308 × 10−4 𝑠. Then 𝑁′(𝑠) = 1; 𝐷′(𝑠) =

4𝑠 3 + 4𝑠 2 + 0.834𝑠 − 2.308 × 10−4

Then 𝑁(𝑠)𝐷′ (𝑠) − 𝑁 ′ (𝑠)𝐷(𝑠) = 0 becomes 3𝑠 4 + 0.667𝑠 3 − 1.583𝑠 2 − 0.417𝑠 + 1.154 × 10−4 =

That has roots at 𝑠 = 0.7417, −0.694, −0.2703,0.0003. The only solution in the locus is -0.2703.

The roots locus is:

8-50 Chapter 8: Root Locus Techniques

b. The system is unstable for all values of K, although with relatively large time constants
for small K. It is very unlikely that the driver-train system is unstable.

40.

1
Rr 
Cr s Rr 5
a. Z r = = = &
eq
Rr +
1 Rr  C r s + 1 0.04 s + 1
Cr s
Z cable = Rcable + Lcable s = 0.06 + 0.00005s

Hence:
1 1 0.04 s + 1
= =
Z cable + Z req 0.06 + 0.00005 s + 5 (0.06 + 0.00005s )(0.04 s + 1) + 5
0.04 s + 1
0.04 s + 1 20000 s + 500000
= =
2 E − 6  s + 0.00245 s + 5.06
2
s + 1225 s + 2530000
2

Vs ( s) s 2 + 1225 s + 2530000
Thus: G( s) =
(
= 2
)
I s ( s) s + 1225 s + 2530000 0.008 s + (20000 s + 500000)
=
 Solutions to Design Problems 8-51

s 2 + 1225s + 2530000
=
0.008 s3 + 9.8 s 2 + 40240 s + 500000
b. The transfer function of the forward loop, KG(s):

KG( s) =
Vs ( s)
=
(
K s 2 + 1225s + 2530000 )
Ev ( s) 0.008 s3 + 9.8 s 2 + 40240 s + 500000

The transfer function of the feedback loop is given by: H ( s ) = 200 .

s + 200

The system’s characteristic equation is:


1 + 
(
K s 2 + 1225s + 2530000 )
 200 
 =0
 0.008 s + 9.8 s + 40240 s + 500000  s + 200 
3 2

Or equivalently

( ) (
s 4 + 1425s3 + (5275000 + 25000 K )s 2 + 1.0685109 + 3.0625107 K s + 1.25 1010 + 6.3251010 K )
=0

The following MATLAB M-file was written to plot the root locus for the system and to find the

required-above operational parameters and functions:

numg = [1 1225 2.53E6];

deng = [0.008 9.8 40240 500000];

G = tf(numg, deng);

numh = 200;

denh = [1 200];

H = tf(numh, denh);

rlocus(G*H);

title('Full Root Locus for DC Bus Voltage Control System')

pause

axis ([-150, 0, -150, 150]);

z=0.707;

sgrid(z,0)
8-52 Chapter 8: Root Locus Techniques

title('Root Locus Zoomed-in around Dominant Poles with a 0.707

Damping Line')

[K1,p]=rlocfind(G*H);

pause

K = K1;

T=feedback(K*G,H); %T is the closed-loop TF of the system

T=minreal(T);

step(750*T);

grid

title('Step Response of DC Bus Voltage')

The first figure shown below is the full root locus for that system.

4
x 10 Full Root Locus for DC Bus Voltage Control System
1
T(s)

0.8

0.6

0.4

0.2
Imaginary Axis

-0.2

-0.4

-0.6

-0.8

-1
-1200 -1000 -800 -600 -400 -200 0 200 400 600
Real Axis
 Solutions to Design Problems 8-53

Zooming into the locus by setting the x-axis (real-axis) limits to -150 to 0 and the y-axis (imaginary-

axis) limits to -150 to 150, we get the following Root locus with the line corresponding to a damping

ratio, ζ = 0.707. That plot was used to find the gain, K, at which the system has complex-conjugate

closed-loop dominant poles with a damping ratio ζ = 0.707.

Root Locus Zoomed-in around Dominant Poles with a 0.707 Damping Line
150
0.707 T
Dominant Poles

100

50
Imaginary Axis

-50

-100

0.707
-150
-150 -100 -50 0
Real Axis

From the locus we found that:

i. The gain, K, at which the system would have complex-conjugate closed-loop dominant poles with a
damping ratio ζ = 0.707 is: K = 1.6832;
ii. The coordinates of the corresponding point selected on the root-locus are:

-1.085 E+002 +1.090 E+002i

iii. The corresponding values of all closed-loop poles are:

p= 1.0 E+003 *

-0.6040 + 2.1602i
8-54 Chapter 8: Root Locus Techniques

-0.6040 - 2.1602i

-0.1085 + 0.1090i

-0.1085 - 0.1090i
iv. The output voltage vs(t) for a step input voltage vdc-ref (t) = 750 u(t) was plotted in (c) below.

c. The output response is shown below with the required characteristics noted on the graph and listed
below.

Step Response of DC Bus Voltage

800
T
System: T
Settling Time (sec): 0.0338
700 System: T
Peak amplitude: 721
Overshoot (%): 7.41 System: T
At time (sec): 0.0206 Final Value: 671
600
System: T
Rise Time (sec): 0.00996
DC Bus Voltage, Volts

500

400

300

200

100

0
0 0.01 0.02 0.03 0.04 0.05 0.06
Time (sec)

i.The actual percent overshoot = 7.41% and the corresponding peak time, Tp = 0.0206 sec;
ii.The rise time, Tr = 0.00996 sec, and the settling time, Ts = 0.0338 sec;
iii.The final steady-state value is 671 volts.

41.
a. The open loop transfer function has an excess of 1 pole over zeros, so there is 1
asymptote at 𝜃 = 180°.
The breakaway points are calculated from 𝑁(𝑠)𝐷 ′ (𝑠) − 𝑁 ′ (𝑠)𝐷(𝑠) = 0 or
 Solutions to Design Problems 8-55

(𝑠 2 + 0.342𝑠 + 0.023)(3𝑠 2 + 0.428𝑠 + 0.001449)

− (2𝑠 + 0.342)(𝑠 3 + 0.214𝑠 2 + 0.001449) = 0
which can be simplified to
𝑠 4 + 0.684𝑠 3 + 0.140739𝑠 2 + 0.009844𝑠 + 3.3327 × 10−5 = 0
The roots for this polynomial are -0.1487j0.048, -0.0036, and -0.3831; the latter two
being the breakaway points since they are on the root locus. The values for gain at these
points are found solving:

0.0415𝑘𝑐 (𝑠 + 0.092)(𝑠 + 0.25)

1+ | =0
𝑠(𝑠 + 0.007)(𝑠 + 0.207) 𝑠=−0.0036
giving 𝑘𝑐 = 2.754 × 10−3 . And
0.0415𝑘𝑐 (𝑠 + 0.092)(𝑠 + 0.25)
1+ | =0
𝑠(𝑠 + 0.007)(𝑠 + 0.207) 𝑠=−3831
giving 𝑘𝑐 = 15.78.
The corresponding root locus is:

b. The characteristic equation can be rewritten as

𝑠 3 + (0.214 + 0.0415𝑘𝑐 )𝑠 2 + (0.001449 + 0.01419𝑘𝑐 )𝑠 + 0.0009545𝑘𝑐 = 0
Substituting 𝑘𝑐 = 5.35 we get
𝑠 3 + 0.436𝑠 2 + 0.0774𝑠 + 0.0051 = 0
The roots of this polynomial are -0.1457j0.1185 and -0.1447. They are marked in the
root locus above. Note that all roots have the same real value so the system will be the
fastest possible; as per the overshoot the complex conjugate pair results in a damping
0.1457
factor  = cos 𝜃 = = 0.76, with negligible overshoot.
√0.11852 +0.14572
8-56 Chapter 8: Root Locus Techniques

42.

 L ( s) 250 ( s 2 + 1.2 s + 12500)

GP ( s) = = K M D( s ) = 3
M (s) s + 8.1s 2 + 62003 s + 31250

The MATLAB M-file is:

num1=250;

num2=[1 1.2 12500];

num = conv(num1, num2);

den=[1 8.1 62003 31250];

G=tf(num, den);

rlocus(G)

pause

K=0:0.1:5;

rlocus(G,K)

sgrid(0.4,0.2)

axis([-50,0.5,100,250]);

pause

for i=1:1:3;

[K,P]=rlocfind(G)

end

T=260*feedback(K*G,1)

step(T)

grid;

The MATLAB full root-locus obtained is shown below. After a few attempts, a step

response with an overshoot of 14.6% was achieved. The figure, shown below the root locus,
 Solutions to Design Problems 8-57

illustrates the step-response obtained with all of the requested important characteristics

marked on it.
Root Locus
300

200

100
Imaginary Axis (seconds-1)

-100

-200

-300
-250 -200 -150 -100 -50 0 50
Real Axis (seconds -1)
8-58 Chapter 8: Root Locus Techniques

System: T
Peak amplitude: 298
Overshoot (%): 14.8
At time (seconds): 0.0423
Step Response
300
System: T
Settling time (seconds): 0.102

250 System: T
Final value: 259

System: T
Rise time (seconds): 0.0272
Output Angular Speed, rad/sec
200

150

100

50

0
0 0.05 0.1 0.15
Time (seconds)

This output response was obtained at

K=

3.1319

P=

1.0e+02 *

-7.2280 + 0.0000i

-0.3414 + 1.1144i

-0.3414 - 1.1144i
 Solutions to Design Problems 8-59

T=

2.036e05 s^2 + 2.443e05 s + 2.545e09

--------------------------------------------------

s^3 + 791.1 s^2 + 6.294e04 s + 9.819e06

Continuous-time transfer function.

SOLUTIONS TO DESIGN PROBLEMS

43.
61.73K
G(s) =
(s+10)3 (s2 + 11.11s + 61.73)

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 =
4 4
23323.61. Thus, K = 377.83 and Ts = = = 1.462 seconds.
n 2.736
8-60 Chapter 8: Root Locus Techniques

44.
 3
a. There are two asymptotes with a = , and real axis intersection
2 2
0 + 0 − 10 − (−1)
a = = −4.5 . To find the breakaway and break-in points, write
3 −1
 2 ( + 10)
K= . The derivative of this expression is
3.333  10 4 ( + 1)
dK ( + 1)(3 2 + 20 ) − ( 3 + 10 2 )  (2 2 + 13 + 20)
=− = − . The denominators
d 3.3333  10 4 ( + 1) 2 3.3333  10 4 ( + 1) 2

roots for this expression are  = 0, − 2.5,− 4

The values ofK at -2.5 and -4 are calculated from

3.333  104 ( s + 1) −4
1+ K s =−2.5 = 0 giving K = 9.375  10 and
s 2 ( s + 10)
3.333  10 4 ( s + 1)
1+ K s = −4 = 0 giving K = 9.6  10
−4

s ( s + 10)
2

The root locus is:

b. The line corresponding to  = 0.7 must be at an angle  = cos −1 0.7 = 45.57  so the line
must lie along the points s = a + jb = a − ja tan = a(1 − j1.0202) . The angle condition for
 Solutions to Design Problems 8-61

the root locus must be satisfied. Namely

 3.333  10 4 ( s + 1)   3.333  10 4 (1 + a − j1.0202) 
arg 
s = a (1− j1.0202 )  = arg
 
 s ( s + 10)
2
  a (1 − j1.0202) (10 + a − j1.0202) 
2 2

1.0202 1.0202
= − tan −1 + tan −1 − 2(−45.57  ) = 180 
1+ a 10 + a
A numerical search gives a = −1.136 . So the root locus and the line intersect when
s = −1.136(1 − j1.0202) = −1.136 + j1.1589 . Then the value of K can be found from
3.333  10 4 ( s + 1)
1+ K s = −1.136+ j1.1589 = 0 giving K = 6.05  10 −4
s 2 ( s + 10)

.45.
a. Using MATLAB and the Symbolic Math Toolbox, the open-loop expression that yields a root
locus as a function of N2 is

0.2284x107N2 (s2 + 3.772e-05s + 66.27) (s2 + 49.99s + 8789)

Gdt(s) = ________________

s(s+45.12) (s2 + 4.893s + 8.777e04)

8-62 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=20e-3
numGdt=3.92*N^2*KLSS*KHSS*KG*s;
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)
 Solutions to Design Problems 8-63

'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

KG =

668

JR =

190120

JG =

3.8000

tel =

0.0200
8-64 Chapter 8: Root Locus Techniques

ans =

Gdt with Values Substituted

16 2 /
.9931 10 N s / (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 / 10 2 5 12 2 4 14 2 3
.9931 10 N s / (.4349 10 N s + .2174 10 N s + .3851 10 N s
/

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 / 10 5 12 4 14 3
.9931 10 N s / ((.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 =

Gdt Different Form 3

7 2 /
.2284 10 N s / (
/

5 4 3 2 2
(1.000 s + 49.99 s + 8855. s + 3313. s + 582400. s) N

9 2 5 4 7 3
+ .1658 10 s + 41.87 s + 2094. s + .3684 10 s )

ans =
 Solutions to Design Problems 8-65

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

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 +90.3548i
-24.9950 -90.3548i
-0.0000 + 8.1404i
-0.0000 - 8.1404i

ans =

Poles of GH

rootsdenGH =

1.0e+002 *

0
-0.0245 + 2.9624i
-0.0245 - 2.9624i
-0.4512

Select a point in the graphics window

selected_point =

-3.8230 + 6.5435i
8-66 Chapter 8: Root Locus Techniques

K =

51.5672

P =

-21.1798 +97.6282i
-21.1798 -97.6282i
-3.8154 + 6.5338i
-3.8154 - 6.5338i

b. From the computer response, K = 0.2284x107N2 = 49.6. Therefore, N is approximately 5/1000.

46.
a. The characteristic equation is given by:
𝑠4
1+𝐾 =0
(𝑠 2 + 𝑠 + 1)2
or
(1 + 𝐾)𝑠 4 + 2𝑠 3 + 3𝑠 2 + 2𝑠 + 1 = 0

The Routh table is

𝑠4 1+𝐾 3 1
𝑠3 2 2
𝑠2 2−𝐾 1
𝑠 2(1 − 𝐾)
2−𝐾
1 1

Clearly for closed loop stability 𝐾 < 1.

b. There is no locus on the real axis, and no asymptotes. The root locus starts at the loci of the

complex poles and ends at the zeros in the origin:

 Solutions to Design Problems 8-67

Root Locus
1.5

0.5
Imaginary Axis

-0.5

-1

-1.5
-0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3
Real Axis

47.
% Parameters
Jl=10;Bl=1;k=100;Jm=2;Bm=0.5;a=0.25;%a is the location of the zero
%numerator and denominator of the open loop transfer function
numo=k*[1 a];
deno=[Jl*Jm (Jl*Bm+Jm*Bl) (k*(Jl+Jm)+Bl*Bm) k*(Bl+Bm) 0];
syso=tf(numo,deno);

%Pole-Zero map for the open loop transfer function

pzmap(syso);
%Root Locus
rlocus(syso);axis([-1 0 -0.3 0.3]);
zgrid(0.707,[]); %grid for zeta=0.707 for approx. 5% overshoot
[KD,poles]=rlocfind(syso);
%Choose the appropriate location of the poles in the window and multiply the factor by the open
loop
%transfer function
syso=KD*syso;
%Close the loop
sysc=feedback(syso,1);
figure;
%Obtain the response
step(sysc)
8-68 Chapter 8: Root Locus Techniques

Root Locus

0.707
0.25

0.2

0.15

0.1
Imaginary Axis

0.05

-0.05

-0.1

-0.15

-0.2

-0.25
0.707
-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0
Real Axis

Select a point in the graphics window

selected_point =

-0.1031 + 0.0978i
 Solutions to Design Problems 8-69

Step Response
1.4
System: sysc
Peak amplitude: 1.05
1.2 Overshoot (%): 5
At time (sec): 25.8

0.8
Amplitude

0.6

0.4

0.2

0
0 10 20 30 40 50 60
Time (sec)

Note the 5 % overshoot specified.

8-70 Chapter 8: Root Locus Techniques

48.

Front Panel

Block Diagram

Details of the Case Structure

 Solutions to Design Problems 8-71

49.

The following MATLAB M-file was written to plot the root locus for the system and to find the

required-above operational parameters and functions:

numg = poly ([-0.071-6.25i -0.071+6.25i]);

deng = poly ([-0.047 -2 -0.262+5.1i -0.262-5.1i]);

G = tf(numg, deng);

rlocus(G);

pos=(16);

z=-log(pos/100)/sqrt(pi^2+(log(pos/100))^2);

sgrid(z,0)

title(['Root Locus with ', num2str(pos), ' Percent Overshoot Line

for Synchronous Machine with Te = 0.5 sec'])

[K1,p]=rlocfind(G);

pause

K=0.936*K1;

T=feedback(K*G,1); %T is the closed-loop TF of the system

T=minreal(T);

step(T);

grid
8-72 Chapter 8: Root Locus Techniques

Root Locus w ith 16 Percent Overshoot Line for Synchronous Machine w ith Te = 0.5 sec
20

15

10

5 0.504
Imaginary Axis

-5 0.504

-10

-15

-20
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2
Real Axis

Root Locus with 16% Overshoot Line for Synchronous Machine

a. The gain at which the system becomes marginally stable is:

K = 7.1045

b.The closed-loop poles, p, and transfer function, T(s), corresponding to a 16% overshoot
are:

p= -0.2300 + 4.9446i

-0.2300 - 4.9446i

-1.0555 + 1.8634i

-1.0555 - 1.8634i

2.633 s^2 + 0.3739 s + 102.9

T (s) = --------------------------------------------------------

s^4 + 2.571 s^3 + 29.88 s^2 + 53.81 s + 105.3

 Solutions to Design Problems 8-73

c. The coordinates of the point selected on the root-locus were: -1.0557 + 1.8634i and the
corresponding unit-step response, with  (t) in p.u, was found.
d.MATLAB was used to plot that unit-step response and to note on that curve the
required characteristics:

Step Response
1.4 System: T
Peak amplitude: 1.13
Overshoot (%): 15.3
1.2 At time (sec): 1.46
System: T
Settling Time (sec): 5.17
1

System: T
Final Value: 0.977
0.8 System: T
Amplitude

Rise Time (sec): 0.703

0.6

0.4

0.2

0
0 2 4 6 8 10 12
Time (sec)

50.
a. The open loop transfer function can be described as:
0.5232𝑏𝑝 (𝑠 3 + 1.3179𝑠 2 + 1.2354 × 10−4 )
𝐺(𝑠) =
𝑠 3 + 3.373𝑠 2 + 2.321𝑠 + 8.715 × 10−4
The system has open loop poles at -0.0004, -0.9625 and -2.4101; open loop zeros at 0, -
0.001, -1.3178
8-74 Chapter 8: Root Locus Techniques

Since the number of open loop poles equals the number of open loop zeros there are no
asymptotes.
To calculate the breakaway points we form 𝑁(𝑠)𝐷 ′ (𝑠) − 𝑁 ′ (𝑠)𝐷(𝑠) = 0 which in this
case is
(𝑠 3 + 1.3179𝑠 2 + 1.2354 × 10−4 )(3𝑠 2 + 6.746𝑠 + 2.321)
− (3𝑠 2 + 2.6358𝑠 + 1.2375 × 10−4 )(𝑠 3 + 3.373𝑠 2 + 2.321𝑠
+ 8.715 × 10−4 ) = 0
Simplifying this results in
2.0551𝑠 4 + 4.6418𝑠 3 + 3.0610𝑠 2 + 0.0023𝑠 + 1.0786 × 10−7 = 0
The solutions for this equation are -1.129j4.618, -0.0007, -0.00005033 the latter two
points being on the root locus.
From
0.5232𝑏𝑝 (𝑠 3 + 1.3179𝑠 2 + 1.2354 × 10−4 )
1+ | =0
𝑠 3 + 3.373𝑠 2 + 2.321𝑠 + 8.715 × 10−4 𝑠=−0.00005073
𝑏𝑝 = 5.0092983 × 105 and similarly for s=-0.0007, 𝑏𝑝 = 2570.
The sketch for the roots locus is:

b. The system is overdamped when 0 < 𝑏𝑝 < 2570 and 5.0092983 × 105 < 𝑏𝑝 < ∞.
c. The system is critically damped when 𝑏𝑝 = 2570 and 𝑏𝑝 = 5.0092983 × 105 .
d. The system is underdamped when 2570 < 𝑏𝑝 < 5.0092983 × 105.
e. For larger values of 𝑏𝑝 the system has dominant closed loop poles very close to the j
axes.
 Solutions to Design Problems 8-75

51.
a.

The breakaway points are found from 𝑁(𝑠)𝐷 ′ (𝑠) − 𝑁 ′ (𝑠)𝐷(𝑠) = 0 or

𝑠(2𝑠 − 𝑏) − (𝑠 2 − 2𝑏𝑠 + 𝑏 2 ) = 0
Which can be simplified to 𝑠 2 − 𝑏 2 = 0, from which 𝑠 = ±𝑏, only the negative solution
being on the root locus. To find the value of gain at this point we use the characteristic
equation
𝑘𝑠
1+ | =0
(𝑠 − 𝑏)2 𝑠=−𝑏
From which we get that at the breakaway point 𝑘 = 𝑏.

b. The characteristic equation can be written as 𝑠 2 + (𝑘 − 2𝑏)𝑠 + 𝑏 2 = 0 from the

quadratic equation solution is can be readily observed that the roots will be in the left
half plane as long as 𝑘 > 2𝑏.

c. From the quadratic equation it can also be observed that the two roots will be complex
when 𝑘 = 2𝑏. At this point the characteristic equation becomes 𝑠 2 + 𝑏 2 = 0 the roots
are 𝑠 = ±𝑗𝑏 so the system will oscillate with a frequency of b rad/sec.

52.

Substituting the values given for the capacitance, inductances and resistances and reducing the

internal loops to single blocks, yields the following feedback loop:

8-76 Chapter 8: Root Locus Techniques

0.01616( s + 114.3)
( s + 2200)

-
Mα(s) 17636700 Vcα (s) 0.0577( s + 2200) Vα(s)
+ ( s + 114.3s + 17636700)
2
( s + 220)

The transfer function of this feedback loop is:

V ( s) 1017637.5 ( s + 2200) 2
GP ( s) = = =
M  ( s) 16445.0 ( s + 114.3) ( s + 2200) + ( s + 2200)((s + 220)(s 2 + 114.3s + 17636700)

1017637.5 ( s + 2200) 1017637.5 ( s + 2200)

= =
s + 334.3 s + 17678291s + 3.88195  10
3 2 9
( s + 219.9) ( s 2 + 114.4s + 17653251.48)

The MATLAB M-file is:

numg = [1017637.5 2238802500];

deng1 = [1 219.9];

deng2 = [1 114.4 17653251.48];

deng = conv(deng1, deng2);

G = tf(numg, deng);

rlocus(G);

title('Full Root Locus for Grid Connected VSC')

pause

axis ([-300, 0, -50, 5000]);

z=0.012;

sgrid(z,0)

title('Root Locus Zoomed-in for Dominant Poles at a 0.012 Damping

Line')
 Solutions to Design Problems 8-77

[K1,p]=rlocfind(G);

pause

K = K1;

T=feedback(K*G,1); %T is the closed-loop TF of the system

T=minreal(T);

step(208*T);

grid

ylabel('Output Voltage of the VSC, Va(t), in volts')

title('Output Response for a Grid Connected VSC')

The full root-locus obtained is shown below. The close-up of that locus (from -300 to 0 on the real

axis and from -50 to 5000 on the imaginary axis) follows. From that close-up, K and the coordinates

of the dominant poles corresponding to ζ = 0.012 were found to be:

Selected_point = -5.0590e+01 + 4.2172e+03i; K = 0.1278;

p= 1.0e+03 *

-0.0500 + 4.2169i

-0.0500 - 4.2169i

-0.2344 + 0.0000i

The corresponding closed-loop transfer function is:

1.3e05 s + 2.86e08

T= ---------------------------------------

s^3 + 334.3 s^2 + 1.781e07 s + 4.168e09

8-78 Chapter 8: Root Locus Techniques

The figure, shown below the close-up, illustrates the step-response obtained at that value of the gain

(with all of the requested important characteristics marked on it) when a step input, r(t)= 208 u(t),

volts, was applied at t = 0.

4 Full Root Locus for Grid Connected VSC
x 10
2.5

1.5

1
Imaginary Axis (seconds-1)

0.5

-0.5

-1

-1.5

-2

-2.5
-3000 -2500 -2000 -1500 -1000 -500 0 500 1000 1500
Real Axis (seconds -1)
 Solutions to Design Problems 8-79

Root Locus Zoomed-in for Dominant Poles at a 0.012 Damping Line

5000
0.012

4500

4000

3500
Imaginary Axis (seconds-1)

3000

2500

2000

1500

1000

500

0
-300 -250 -200 -150 -100 -50 0
Real Axis (seconds -1)

53.
a. The open loop transfer function is

K (520s + 10.3844)
KG ( s ) P( s ) = −
s + 2.6817s 2 + 0.11s + 0.0126
3

− 520K ( s + 0.02)
=
( s + 0.02 + j 0.0661)( s + 0.02 − j 0.0661)( s + 2.6419)

There are two asymptotes with a real axis intersection given by

− 2.6419 − 2(−0.02) − (−0.02) 2k
a = − 1.33 and angles  a = . For k = 0 ,  a = 0 and
3 −1 2
for k = 1, a =  .

To obtain the breakaway points let

1  3 + 2.6817 2 + 0.11 + 0.0126
K =− =
G ( ) H ( ) 520 + 10.3844

And calculate and solve

8-80 Chapter 8: Root Locus Techniques

dK (520 + 10.3844)(3 2 + 5.3634 + 0.11) − 520( 3 + 2.6817 2 + 0.11 + 0.0126)

=
d (520 + 10.3844) 2
1040 3 + 1426 2 + 55.7 − 5.41
= =0
(520 + 10.3844) 2

Giving  = −1.33, − 0.0879, 0.0446 with only the latter in the root locus. The value of K at

 = 0.0446 is given by:

 3 + 2.6817 2 + 0.11 + 0.0126
K= = 6.82  10 −4
520 + 10.3844  = 0.0446

−4
It was already found in Problem 6.? That the system is closed loop stable for K  2.04  10 . The

root locus is:

b. Now the open loop transfer function is:

K (520s + 10.3844)
KG ( s ) P( s ) =
s + 2.6817s 2 + 0.11s + 0.0126
3

520K ( s + 0.02)
=
( s + 0.02 + j 0.0661)( s + 0.02 − j 0.0661)( s + 2.6419)
(2k + 1)
There are two asymptotes with a real axis intersection as in part a. but with angles a =
2
 3 
. For k = 0 , a = and for k = 1 , a = =− .
2 2 2

The breakaway point calculation is similar to the one in part a. giving

 = −1.33, − 0.0879, 0.0446 with the first two points in the root locus.
 Solutions to Design Problems 8-81

 = −1.33 is given by:

The value of K at
 + 2.6817 + 0.11 + 0.0126
3 2
K =− = 0.0033
520 + 10.3844  = −1.33

 = −0.0879 is given by:

The value of K at
 + 2.6817 2 + 0.11 + 0.0126
3
K =− = 6.5  10 −4
520 + 10.3844  = −0.0879

We use Routh-Hurwitz to show that the system is closed loop stable for all K  0 . The

characteristic equation is:

520s + 10.3844
1+ K =0
s + 2.6817s 2 + 0.11s + 0.0126
3

or

s 3 + 2.6817s 2 + 0.11s + 0.0126 + K (520s + 10.3844) = 0

or

s 3 + 2.6817s 2 + (0.11 + 520K )s + (0.0126 + 10.3844K ) = 0

The Routh Array is:

s3 1 0.11 + 520K

s2 2.6817 0.0126 + 10.3844K

0.2824 + 1384.1K
s
2.6817

1 0.0126 + 10.3844K

It can easily be verified that all the entries in the first column are positive for all K  0.

The root locus is:

8-82 Chapter 8: Root Locus Techniques

54.

a. With the speed controller configured as a proportional controller [KISC= 0 and GSC(s) =
KPSC], the open-loop transfer function is:
0.11 K Psc ( s + 0.6)
GSC ( s )Gv ( s ) = .
s ( s + 0.5173) + 5 (s + 0.6)  (s + 0.01908)

Expanding the denominator of this transfer function, gives: DG ( s ) = 6 s + 3.613 s + 0.05724 .

2

Solving for the roots shows that there are two open-loop poles: – 0.5858 and – 0.0163. Thus, the

open-loop transfer function may be re-written as:

0.11 K Psc ( s + 0.6) K1 ( s + 0.6)

GSC ( s )Gv ( s ) = = = −1 (1)
6 s + 3.613 s + 0.05724 ( s + 0.5858)( s + 0.016 3)
2

K P sc  0.11
In this equation: K1 = (2)
6

The following MATLAB M-file was written to plot the root locus for the system and to find the

value of the proportional gain, K1, at the breakaway or break-in points.

 Solutions to Design Problems 8-83

numg = [1 0.6];

deng = poly ([-0.0163 -0.5858]);

G = tf(numg, deng);

rlocus(G);

pos=(0);

z=-log(pos/100)/sqrt(pi^2+(log(pos/100))^2);

sgrid(z,0)

title(['Root Locus with ', num2str(pos) , ' Percent Overshoot

Line'])

[K1,p]=rlocfind(G);

pause

T=feedback(K1*G,1); %T is the closed-loop TF of the system

T=minreal(T);

step(T);

axis ([0, 8, 0, 1]);

grid

The root locus shown below was obtained. Using MATLAB tools, the gain at the break-in point was

found to be larger and, hence, would yield a faster closed-loop unit-step response. The following

repeated real poles were found, which indicated that the step response is critically damped: p = -

0.6910, - 0.6910. These poles corresponded to: K1 = 0.78 (which corresponds to KPSC = 42.54). The

closed-loop transfer function, T(s), was found to be:

0.78 s + 0.468

T(s) = -----------------------

s2 + 1.382 s + 0.4775
8-84 Chapter 8: Root Locus Techniques

Therefore, it was used to find the closed-loop transfer function of the system, to plot its unit-step

response, c(t), shown below, and to find the rise-time, Tr, and settling time, Ts.

Root Locus with 0 Percent Overshoot Line

0.1
0.992 0.982 0.966 0.93 0.86 0.6

0.08

0.997
0.06

0.04
System: G
0.999 Gain: 0.78
Pole: -0.691
0.02
Imaginary Axis

Damping: 1
Overshoot (%): 0
Frequency (rad/sec): 0.691
0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
0

-0.02
0.999
-0.04

-0.06
0.997

-0.08

0.992 0.982 0.966 0.93 0.86 0.6

-0.1
-0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1
Real Axis
 Solutions to Design Problems 8-85

System: T
Step Response Final Value: 0.98
1

System: T
0.9 Settling Time (sec): 4.69
System: T
0.8 Rise Time (sec): 2.69

0.7

0.6
Amplitude

0.5

0.4

0.3

0.2

0.1

0
0 1 2 3 4 5 6 7 8
Time (sec)

As could be seen from the graph, these times are:

Tr = 2.69 sec and Ts = 4.69 sec

b. When integral action was added (with KISC/KPSC = 0.4), the transfer function of the speed
K I sc K P sc (s + 0.4)
controller became: GSC ( s) = K P sc + = and the open-loop transfer
s s
function obtained was:

0.11K Psc ( s + 0.6)( s + 0.4) K1 ( s + 0.6) (s + 0.4)

GSC ( s )Gv ( s ) = = = −1
s(6s + 3.613 s + 0.05724)
2
s( s + 0.5858)( s + 0.016 3)

0.11K P sc 6 K1
Where K1 = or K pSC = = 54.5455  K1
6 0.11
8-86 Chapter 8: Root Locus Techniques

The following MATLAB M-file was written to plot the root locus for the system and to find the

gain, K1, which could result in a closed-loop unit-step response with 10% overshoot.

numg = poly ([-0.4 -0.6]);

deng = poly ([0 -0.0163 -0.5858]);

G = tf(numg, deng);

rlocus(G);

pos=(10);

z=-log(pos/100)/sqrt(pi^2+(log(pos/100))^2);

axis ([-1, 0, -0.5, 0.5]);

sgrid(z,0)

title(['Root Locus with ', num2str(pos) , ' Percent Overshoot

Line'])

[K1,p]=rlocfind(G);

pause

T=feedback(K1*G,1); %T is the closed-loop TF of the system

T=minreal(T);

step(T);

axis ([0, 20, 0, 1.5]);

grid

The root locus shown below was obtained. Using MATLAB tools, the gain at the point selected on

the locus (- 0.275 + j 0.376) was found to be K1 = 0.526 (which corresponds to KPSC = 28.7). The

corresponding closed-loop transfer function, T(s), is:

0.526 s 2 + 0.526 s + 0.1262

T (s) =
s 3 + 1.128 s 2 + 0.5355 s + 0.1262
 Solutions to Design Problems 8-87

T(s) has the closed-loop poles: p = – 0.580, – 0.275 ± j 0.376 and zeros at – 0.4 & – 0.6. Thus, the

complex conjugate poles are not dominant, and hence, the output response, c(t), obtained using

MATLAB, does not match that of a second-order underdamped system. Note also that the settling

time, Ts = 15 sec, , the rise time, Tr = 2 sec, the peak time, Tp = 5.03 sec, and the overshoot is 24.5%

(higher than the 10% corresponding to the dominant poles).

Root Locus with 10 Percent Overshoot Line

0.5
0.591

0.4
System: G
0.3 Gain: 0.527
Pole: -0.275 + 0.376i
Damping: 0.591
0.2 Overshoot (%): 9.98
Frequency (rad/sec): 0.466
0.1
Imaginary Axis

-0.1

-0.2

-0.3

-0.4

0.591
-0.5
-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0
Real Axis
8-88 Chapter 8: Root Locus Techniques

Step Response
1.5
System: T
Peak amplitude: 1.24
Overshoot (%): 24.5
At time (sec): 5.03

System: T
Final Value: 1
1

System: T
Output, c(t), volts

Settling Time (sec): 15

System: T
Rise Time (sec): 2

0.5

0
0 2 4 6 8 10 12 14 16 18 20
Time (sec)

It should be mentioned that since we applied 1 volt-unit-step inputs (as compared to 4 volts in the

Hybrid vehicle progressive problem in Chapter 5) in both parts (a) and (b) above, we should not be

surprised that the final (steady-state) value of output voltage of the speed transducer was 1 volt,

which corresponds to a change in car speed of only 5 km/hr.

55.
a. After substitution of the delay approximation the resulting open loop transfer
function is:

137.2 10−6 K (0.0513 − s)

G( s) =
( s 2 + 0.0224s + 196 10−6 )( s + 0.0513)

There are 2 asymptotes, using equation 8.66 we obtain that the angles are 0 and
180 . The intersection with the real axis (although irrelevant given the asymptote
angles) is
−0.0112 + j 0.0084 − 0.0112 − j 0.0084 − 0.0513 − 0.0513
a = = 0.0625
3 −1
 Solutions to Design Problems 8-89

b. The break-in and breakaway points can be obtained from

N (s) D(s) − N (s) D(s) = 0 where N (s) = (0.0513 − s) ,
D(s) = s3 + 0.0737s 2 + 0.0013s + 1.0055 10−5 , N (s) = −1 and
D(s) = 3s 2 + 0.1474s + 0.0013 . After substituting and simplifying one gets
−2s3 + 0.0802s 2 + 0.0076s + 7.6745 10−5 = 0 . This polynomial has solutions
at s = 0.0881, −0.0359, −0.0121 . Only first root is in the root locus so it’s a
break-in point.
c. The characteristic equation is:

137.2 10−6 K (0.0513 − s )

1+ =0
s 3 + 0.0737 s 2 + 0.0013s + 1.0055  10−5

or

s3 + 0.0737s 2 + (0.0013 −137.2 10−6 K ) s + (1.0055 10−5 + 7.03836 10 −6 K ) = 0

The corresponding Routh array is:

s3 1 0.0013 −137.2 10−6 K

s2 0.0737 1.0055 10−5 + 7.03836 10−6 K

s 1.2086 10−3 − 2.3727 10−4 K

1 1.0055 10−5 + 7.03836 10−6 K

It is readily seen that the fourth row entry is positive as long as K  0 . The third row
entry is positive as long as K  5.1922 . Therefore the range for closed loop stability is

0  K  5.1922 .
d. The system will oscillate when K = 5.1922 . At this point the auxiliary equation
becomes:

Qa ( s ) = 0.0737 s 2 + 4.65996  10−5

Equating to zero the equation has a solution at s =  j 0.025 . So the oscillation

frequency is 0.025 rad/sec.
The resulting root locus is shown below:
8-90 Chapter 8: Root Locus Techniques