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Similarly, t h e program f o r t h e fourth-order transfer function approximation with
T = 0.1 sec is
[num,denl = pade(0.1, 4);
printsys(num, den, ' s t )
numlden = sA4 - 2O0sA3 + 1 80O0sA2 - 840000~+ 16800000 sA4 + 200sA3 + 1 8000sA2 + 840000s + 16800000
Notice that the pade approximation depends o n the dead time T a n d the desired order
for t h e approximating transfer function.
EXAMPLE PROBLEMS AND SOLUTIONS
A-6-1. Sketch the root loci for the system shown in Figure 6-39(a). (The gain K is assumed to be posi-
tive.) Observe that for small or large values of K the system is overdamped and for medium val-
ues of K it is underdamped.
Solution. The procedure for plotting the root loci is as follows:
1. Locate the open-loop poles and zeros on the complex plane. Root loci exist on the negative
real axis between 0 and -1 and between -2 and -3.
2. The number of open-loop poles and that of finite zeros are the same.This means that there
are no asymptotes in the complex region of the s plane.
(2)
Figure 6-
(a) Control system; (b) root-locus plot.
Chapter 6 / Root-Locus Analysis
3. Determine the breakaway and break-in points.The characteristic equation for the system IS
The breakaway and break-in points are determined from
d K - (^) - - - ( 2 s^ +^ l ) ( s^ +^ 2 ) ( s^ +^ 3 )^ -^ s(s^ +^ 1 ) ( 2 > +^ 5 ) rls (^) [ ( s + 2 ) ( s + 3)
as follows:
Notice that both points are on root loci. Therefore, they are actual breakaway or break-in points. At point s = -0.634, the value of K is
Similarly, at s = -2.36h,
(Because points = -0.634 lies between two poles,it is a breakaway point, and because point
s = -2.366 lies between two zeros, it is a break-in point.)
4. Determine a sufficient number of points t h d satisfy the angle condition. (It can he found
that the root loci involve a circle with center at -1.5 that passes through the breakaway and break-in points.) The root-locus plot for this system is shown in Figure 6-3Y(h).
Note that this system is stable for m y positive value of K since all the root loci lie in the left-
half s plane. Small kalues of I*: (0 c K < 0.0718) correspond to an overdampcd system. Medium value
01' I< (0.0718 .-: K .;14) correspond to an underdamped system. Finally. large values ol K ( 14 = K ) correspond to an overdamped systern. With a large value of K , the steady state can be I-eachcdin much shorter time than with a \mall value o f I<.
The value of K should be adjusted so thal system performance is optimum according to ;I
given performance index.
Example Problems and Solutions 385
Notice that at points s = -2 * ~2.0817the ang!e condition is not satisfied. Hence, they are nei-
ther breakaway nor break-in points. In fact, if we calculate the value of K, we obtain
(To be an actual breakaway or break-in point, the corresponding value of K must be real and
positive.) The angle of departure from the complex pole in the upper half s plane is
The points where root-locus branches cross the imaginary axis may be found by substituting
s = j w into the characteristic equation and solving the equation for w and K as follows: Noting
that the characteristic equation is
we have
which yields
Root-locus branches cross the imaginary axis at w = 5 and w = -S.The value of gain K at the crossing points is 150. Also, the root-locus branch on the real axis touches the imaginary axis at
w = 0. Figure 6-40(b) shows a root-locus plot for the svstern.
It is noted that if the order of the numerator of G ( s ) H ( s ) is lower than that of the denomi-
nator by two or more, and if some of the closed-loop poles move on the root locus toward the right
as gain K is increased, then other closed-loop poles must move toward the left as gain K is in-
creased.This fact can be seen clearly in this problem. If the gain K is increased from K = 34 to K = 68, the complex-conjugate closed-loop poles are moved from s = -2 + 13.65 to s = -1 + j4:
the third pole is moved from s = -2 (which corresponds to K = 34) to s = -4 (which corre-
sponds to K = 68).Thus, the movements of two complex-conjugate closed-loop poles to the right
by one unit cause the remaining closed-loop pole (real pole in this case) to move to the left by two units.
A-6-3. Consider the system shown in Figure 6-41(a). Sketch the root loci for the system. Observe that
for small or large values of K the system is underdamped and for medium values of K it is
overdamped. Solution. A root locus exists on the real axis between the origin and -m. The angles of asymp- totes of the root-locus branches are obtained as
+180°(2k + 1) Angles of asymptotes = 3
The intersection of the asymptotes and the real axis is located on the real axis at
Example Problems and Solutions
Figure 6-
(a) Control system;
(bj root-locus plot. (b)
The breakaway and break-in points are found from d K / d s = 0. Since the characteristic equation is
s3 + 49' + 5s + K = 0
we have
K = -( s3 + 4s2 + 5s).
Now we set
which yields
s = -1, s = -1.
Since these points are on root loci, they are actual breakaway or break-in points. (At points = -1,
the value of K is 2, and at point s = -1.6667, the value of K is 1.852.)
The angle of departure from a complex pole in the upper half s plane is obtained from
e = 1800 - 153.430 - go
or
The root-locus branch from the complex pole in the upper half s plane breaks into the real axis
at s = -1.6667.
Next we determine the points where root-locus branches cross the imaginary axis. By substi-
tuting s = jw into the characteristic equation, we have
( j ~ + ) 4(jw)'~ + 5 ( j w ) + K = 0
or
( K - 4w2) + jo(5 - w2) = 0
from which we obtain
w = r t f l , K = 2 0 or w = O , K = O
Chapter 6 / Root-Locus Analysis
The intersection of the asymptotes and the real axis is found from
The breakaway and break-in points are found from d K / d s = 0. Noting that
K = -s(s + l ) ( s 2 + 4s + 13) = -(s4 + 5s3 + 17s2 + 13s)
we have
dK = -(4s3 + 15s2 + 34s + 13) = 0 ds
from which we get
Point s = -0.467 is o n a root locus.Tl~erefore,it is an actual breakaway point.The gain values K
corresponding to points s = -1.642 f 12.067 are complex quantities. Since the gain values are
not real positive, these points are neither breakaway nor break-in points. The angle of departure from the complex pole in the upper half s plane is
Next we shall find the points where root loci may cross the jw axis. Since the characteristic equation is
by substituting s = jw into it we obtain
from which we obtain
w = f 1.6125, K = 37.44 o r w = 0, K = 0
The root-locus branches that extend to the right-half s plane cross the imaginary axis at
w = 11.6125. Also, the root-locus branch o n the real axis touches the imaginary axis at w = 0. Fig-
ure 6-42(b) shows a sketch of the root loci for the system. Notice that each root-locus branch that
extends to the right half s plane crosses its own asymptote.
Chapter 6 / Root-Locus Analysis
Ad-5. Sketch the root loci for the system shown in Figure 6-43(a).
Solution. A root locus exists on the real axis between points s = -1 and s = -3.6. The asymp-
totes can be determined as follows:
+180°(2k + 1)
Angles of asymptotes = = 90°, -90"
The intersection of the asymptotes and the real axis is found from
Since the characteristic equation is
we have
The breakaway and break-in points are found from
dK (^) - (3s'^ +^ 7.2s)(s^ +^ 1)^ -^ ( s 3^ +^ 3.6s')
ds ( S +
Figure 6-
(a) Control system; (b) root-locus plot.
Example Problems and Solutions
The intersection of the asymptotes and the real axis is obtained from
Next we shall find the breakaway points. Since the characteristic equation is
we have
The breakaway and break-in points are found from
from which we get
Thus, the breakaway or break-in points are at s = 0 and s = -1.2. Note that s = -1.2 is a double root. When a double root occurs in dK/ds = 0 at point s = -1.2, d2K/(ds2) = 0 at this point.The
value of gain K at point s = -1.2 is
This means that with K = 4.32 the characteristic equation has a triple root at points = -1.2.This
can be easily verified as follows:
Hence, three root-locus branches meet at point s = -1.2. The angles of departures at point
s = -1.2 of the root locus branches that approach the asymptotes are f180°/3, that is, 60" and
-60". (See Problem A-6-7.)
Finally, we shall examine if root-locus branches cross the imaginary axis. By substituting s = jw
into the characteristic equation, we have
This equation can be satisfied only if w = 0, K = 0. A t point w = 0, the root locus is tangent to
the j o axis because of the presence of a double pole at the origin. There are no points that root- locus branches cross the imaginary axis. A sketch of the root loci for this system is shown in Figure 6-44(b).
Example Problems and Solutions
A-6-7. Referring to Problem A-6-6, obtain the equations for the root-locus branches for the system
shown in Figure 6-44(a). Show that the root-locus branches cross the real axis at the breakaway
point at angles f 60".
Solution. The equations for the root-locus branches can be obtained from the angle condition
which can be rewritten as
/ s + 0.4 - 2b - /s + 3.6 = *180°(2k + 1)
By substituting s = u + jw, we obtain
By rearranging, we have
tan-' (-)^ W - tan-' )(: = tan-' )(: + tan-' (L) *l8O0(2k + 1 ) u + 0.4 (^) u + 3.
Taking tangents of both sides of this last equation, and noting that
we obtain
which can be simplified to
which can be further simplified to
For u f -1.6, we may write this last equation as
Chapter 6 / Root-Locus Analysis
Figure 6-
Unit-step response curve for the system shown in Figure 6-45 (a).
The unit-step response of this system is
The inverse Laplace transform of C ( s ) gives
c(t) = 1 + 1. 6 6 6 ~ - ~ ' - 2.666e-", fort 2 0
The unit-step response curve is shown in Figure 6-46. Although the system is not oscillatory, the
unit-step response curve exhibits overshoot. (This is due to the presence of a zero at s = -1.)
A-6-9. Sketch the root loci of the control system shown in Figure &47(a). Determine the range of gain
K for stability.
Solution. Open-loop poles are located at s = 1, s = -2 + j d , and s = -2 - j d. A root locus
exists on the real axis between points s = 1 and s = -03. The asymptotes of the root-locus
branches are found as follows:
*180°(2k + 1 ) Angles of asymptotes =
The intersection of the asymptotes and the real axis is obtained as
The breakaway and break-in points can be located from d K / d s = 0. Since
K = -( r - l ) ( s 2 + 4s + 7) = -(s3 + 3s2 + 3s - 7)
we have
which yields
(s + I ) =~ 0
Chapter 6 / Root-Locus Analysis
(a)
Figure 6- (a) Control system; (b) root-locus plot.
Thus the equation d K / d s = 0 has a double root at 3 = -1. (This means that the characteristic equation has a triple root at s = -1.) The breakaway point is located at s = -1. Three root-locus branches meet at this breakaway point.The angles of departure of the branches at the breakaway point are ilX0°/3, that is. 60" and -60". We shall next determine the points where root-locus branches may cross the imaginary axis. Noting that the characteristic equation is
(.r - l)(.s2 + 4s + 7 ) + K = 0
or
.r+ 3 , + ~ 3. ~~ - 7 + K = o
we substitute s = j w into it and obtain
(jw)' + 3 ( j ~ ) ~+ 3 ( j w ) - 7 + K - O
By rewriting this last equation, we have
( K - 7 - 3w2) + ,043 - w 2 ) = 0
This equation is satisfied when
= K = 7 + 3 w " l 6 or w = 0 ,
Example Problems and Solutions
Thus the root-locus branches consist of three lines. Note that the root loci for K > 0 consist of
portions of the straight lines as shown in Figure 6-47(b). (Note that each straight line starts from
an open-loop pole and extends to infinity in the direction of 180°, 60°, or -60" measured from the
real axis.) The remaining portion of each straight line corresponds to K < 0.
A-6-10. Consider the system shown in Figure 6-48(a). Sketch the root loci
Solution. The open-loop zeros of the system are located at s = fj. The open-loop poles are lo-
cated at s = 0 and s = -2. This system involves two poles and two zeros. Hence, there is a possi-
bility that a circular root-locus branch exists. In fact, such a circular root locus exists in this case,
as shown in the following. The angle condition is
By substituting s = u + jw into this last equation, we obtain
Taking tangents of both sides of this equation and noting that
Figure 6-
(a) Control system; (b) root-locus plot.
Example Problems and Solutions
we obtain
which is equivalent to
These two equations are equations for the root 1oci.The first equation corresponds to the root locus
on the real axis. (The segment between s = 0 and s = -2 corresponds to the root locus for
0 5 K < m.The remaining parts of the real axis correspond to the root locus for K < 0.) The second equation is an equation for a circle. Thus, there exists a circular root locus with center at u = i , w = 0 and the radius equal to a / 2. The root loci are sketched in Figure 6-48(b). [That part of the circular locus to the left of the imaginary zeros corresponds to K > 0. The portion of
the circular locus not shown in Figure 6-48(b) corresponds to K < 0.
A-6-11. Consider the control system shown in Figure 6-49. Plot the root loci with MATLAB.
Solution. MATLAB Program 6-11 generates a root-locus plot as shown in Figure 6-50.The root loci must be symmetric about the real axis. However, Figure 6-50 shows otherwise. MATLAB supplies its own set of gain values that are used to calculate a root-locus plot. It does so by an internal adaptive step-size routine. However, in certain systems, very small changes in the gain cause drastic changes in root locations within a certain range of gains.Thus,MATLAB takes too big a jump in its gain values when calculating the roots, and root locations change by a relatively large amount. When plotting, MATLAB connects these points and causes a strange-looking graph at the location of sensitive gains. Such erroneous root-locus plots typically occur when the loci approach a double pole (or triple or higher pole), since the locus is very sensitive to small gain changes.
MATLAB Program 6-1 1
Figure 6 4 9
Control system.
num = [O 0 1 0.41;
den = [ I 3.6 0 01;
rlocus(num,den);
v = [-5 1 -3 31; axis(v)
grid title('Root-Locus Plot of G(s) = K(s + 0.4)/[sA2(s+ 3.6))')
Chapter 6 / Root-Locus Analysis
Figure 6 5 1
Root-locus plot.
Figure 6 5 2
Root-locus plot.
Root-Locus Plot o f G(s) = K(s+O 4)/[s2(s+3.6)] (^5 )
-5 1 6
0 -5 -4 -3 -2 - 1 0 Real A X I S
Root-Locus Plot of G(s) = K(s+0.4)/[s2(s+3.6)]
Real Axis
A-6-12. Consider the system whose open-loop transfer function G ( s ) H ( s ) is given by
Using MATLAB, plot root loci and their asymptotes.
Solution. We shall plot the root loci and asymptotes on one diagram. Since the open-loop trans-
fer function is given by
G ( s ) H ( s ) = -
K
s ( s + l ) ( s + 2 )
- K s3 + 3s2 + 2s the equation for the asymptotes may be obtained as follows: Noting that
lim
K
= lim
K =- K
3+m s3 + 3~~ + 2~ S-'m S~ + 3 ~ 2 + 3~ + 1 ( S + q
Chapter 6 / Root-Locus Analysis
the equation for the asymptotes may be given by
num = [O O O 11
den = [ I 3 2 01 and for the asymptotes, numa = [O O O 11
dena = [ I 3 3 11
In using the following root-locus and plot commands
the number of rows of r and that of a must be the same. To ensure this, we include the gain con-
stant K in the commands. For example,
MATLAB Program 6-1 3
num = [O O O I ] ;
den = [ I 3 2 01;
numa = [O 0 0 1 I ; dena = [ I 3 3 1 I ;
K1 = 0:0.1:0.3;
K2 = 0.3:0.005:0.5;
K3 = 0.5:0.5:10;
K4 = 1O:S:I 00;
K = [Kl K2 K3 K4];
r = rlocus(num,den,K);
a = rlocus(numa,dena,K);
y = [r a];
plot(y,'-'
v = [-4 4 -4 41; axis(v)
grid
title('Root-Locus Plot of G(s) = K/[s(s + 1 )(s + 2)) and Asymptotes')
xlabel('Rea1Axis')
ylabeU1lmagAxis')
***** Manually draw open-loop poles in the hard copy *****
Example Problems and Solutions
