9040 CHAPTER 9. NONLINEAR SYSTEMS
Fig 6
38:pdf
Phase-plane trajectory of limit cycle for Problem 9.22.
From the geometry of the limit cycle (see above Figure),
At point B,
We need to solve the above four equations for 0(for max =0,0= 0).
If we eliminate _
Busing Eq. (2), then,
9041
Solve for _
A,
Time history shown in the Figure below and shows a “nonlinear oscillator.”
Fig 7
39:pdf
Nonlinear oscillator for Problem 9.22.
23. Consider the point mass pendulum with zero friction as shown in Figure 9.67. Using the method
of isoclines as a guide, sketch the phase-plane portrait of the motion. Pay particular attention
to the vicinity of =. Indicate a trajectory corresponding to spinning of the bob around and
around rather than oscillating back and forth.
(a) Solution: The equations are:
I
=mgl sin ;
9042 CHAPTER 9. NONLINEAR SYSTEMS
M
Figure 9.67: Pendulum for Problem 9.23.
For ,sin  ;
=g
Using isoclines:
=_
d_
9043
x ‘ = y
y ‘ = sin(x)
-8 -6 -4 -2 0 2 4 6 8
-2
3
x
Phase portraits for Problem 9.23.
24. Draw the phase trajectory for a system
x= 106m/ sec2
between _x(0) = 0,x(0) = 0 and x(t) = 1mm. Find the transition time, tf, by graphical means
from the parabolic curve by comparing your solution with several different interval sizes and the
exact solution.
Solution: The phase portrait is shown in the figure below.
9044 CHAPTER 9. NONLINEAR SYSTEMS
00.2 0.4 0.6 0.8 1
0.035
0.045
x [mm]
Phase portrait
00.2 0.4 0.6 0.8 1
0.035
0.045
x [mm]
Phase portrait
Phase portrait for Problem 9.24.
x=a; _v
_x=a
v;
To obtain tgraphically, i.e., by graphical integration, we write,
9045
intervals and find (going from left to the right),
tf= t1+ t2+ t3+ t4+ t5
which compares well with the exact answer of t= 45 sec. Better approximation can be found
by finer division of x. Alternatively we can compute the time from,
25. Consider the system with equations of motion,
+_
+ sin = 0
a) What physical system does this correspond to?
b) Draw the phase portraits for this system.
c) Show a specific trajectory for 0= 0:5rad _
= 0.
Solution:
(a) Physical system is a pendulum with a hinge damping.
Let us define ,T !.
9046 CHAPTER 9. NONLINEAR SYSTEMS
Fig 5
(b) See phase portrait figure shown in the Figure below. Note the unstable equilibrium
(c) See trajectory corresponding to (x=0= 0:5rad; y =_
0= 0) in the phase portrait in
– 1 .5
-1
26. Consider the nonlinear upright pendulum with a motor at its base as an actuator. Design a
feedback controller to stabilize this system.
Solution:
= sin +u:
9047
27. Consider the system
_x=sin x
Prove that the origin is an asymptotically stable equilibrium point.
Solution:We wish to show that _
V(x) xTQx.
Select the Lyapunov function for
28. A first-order nonlinear system is described by the equation _x=f(x), where f(x)is a continuous
and differentiable nonlinear function that satisfies the following:
f(0) = 0;
f(x)>0 for x > 0;
f(x)<0 for x < 0:
Figure 9.68: Control system for Problem 29
Use the Lyapunov function V(x) = x2=2to show that the system is stable near the origin
(x= 0).
Solution:
_x=f(x);
29. Use the Lyapunov equation
ATP+PA =Q=I
to find the range of Kfor which the system in Fig. 9.68 will be stable. Compare your answer
with the stable values for Kobtained using Routh’s stability criterion.
Solution:
Our approach is to set up the continuous Lyapunov equation and check that Pis a positive
9049
Hence,
The two conditions for P>0are p > 0and pr q2>0, or,
and,
30. Consider the system
d
dt x1
x2=x1+x2u
x2(x2+u); y =x1:
Find all values of and for which the input u(t) = y(t)+will achieve the goal of maintaining
the output y(t)near 1.
Solution:
(a) It is desired to maintain the output y(t)of the system,
9050 CHAPTER 9. NONLINEAR SYSTEMS
the stability of the system by linearizing the nonlinear state equations near these equilibria.
The nonlinear, closed-loop system equations are,
To find the equilibrium points for the desired output of y= 1, we set x1= 1,_x1= _x2= 0, to
get,
which can be solved for the equilibrium values of x2and the necessary relationship between
and . Simultaneous solution yields,
Substituting these into the nonlinear closed-loop equations, we get,
There are no values of which produce stable roots. So we conclude x1= 1 and x2= 1 is an
unstable equilibrium point.
31. Consider the nonlinear autonomous system
d
dt 2
4
x1
x2
x33
5=2
4
x2(x3x1)
x2
11
x1x33
5:
a) Find the equilibrium point(s).
b) Find the linearized system about each equilibrium point.
c) For each case in part (b), what does Lyapunov theory tell us about the stability of the
nonlinear system near the equilibrium point?
Solution:
(a) Setting _x1= _x2= _x3= 0 and solving the nonlinear equations, we obtain [1;0;0]Tand
Then the nonlinear equations become,
Thus, the linearized system is _y =Ay where,
Then the nonlinear equations become,
(i) The characteristic equation is (s2+2)(s+1) = 0. The linear system is neutrally (marginally)
32. Consider the circuit shown in Figure 9.69. For what diode characteristics will this system be
stable?
Solution:
The system equations are:
33. Van der Pol’s equation: Consider the system described by the nonlinear equation
x+(1 x2) _x+x= 0
with the constant ” > 0.
9053
Figure 9.70: Diode characteristics.
(a) Show that the equations can be put in the form [Liénard or (x; y) plane]:
_x=y+x3
3x
_y=x.
(b) Use the Lyapunov function V=1
2(x2+ _x2)and sketch the region of stability as predicted
by this Vin the Liénard plane.
(c) Plot the trajectories of part (b) and show the initial conditions that tend to the origin. Sim-
ulate the system in Simulink R
using various initial conditions on x(0) and _x(0). Consider
two cases with = 0:5, and = 1:0.
Solution:
(a) If we differentiate the first Liénard equation, we obtain
or,
9054 CHAPTER 9. NONLINEAR SYSTEMS
(b) If we linearize the system, we obtain
Then,
Fig 18
48:pdf
Stability region in the (x,_x) plane.
9055
-2 1.5 -1 0.5 00.5 11.5 2
1.5
1
1.5
x
Stability reg ion
Figure 9.71: Stability region in the Liénard (x,y) plane.
The stability region may be mapped into the Liénard plane. The circular boundary in the
(x,_x) plane can be mapped into the (x,y) plane:
9056 CHAPTER 9. NONLINEAR SYSTEMS
Fig 19
(c) Using pplane7.m software we see that the limit cycle is nearly circular with radius 2:
-2 – 1 .5 -1 – 0 .5 00 .5 11 .5 2
-3
x
Phase portraits for van der Pol equation in Liénard (x,y) form for = 0:5:
9057
x ‘ = y + (0.33333 x x x x)
y ‘ = – x
-2 -1.5 -1 -0.5 00.5 11.5 2
-3
2
3
x
Phase portraits for van der Pol equation in the Liénard (x,y) plane for = 1:0.
x ‘ = y
y ‘ = – x – 0.5 (1 – x x) y
-2 1.5 -1 0.5 00.5 11.5 2
-3
3
x
Phase portraits for van der Pol equation in the (x,_x) plane for = 0:5.
9058 CHAPTER 9. NONLINEAR SYSTEMS
x ‘ = y
y ‘ = – x – (1 – x x) y
-2 -1.5 -1 -0.5 00.5 11.5 2
-3
3
x
Phase portraits for van der Pol equation in the (x,_x) plane for = 1:0.
The Simulink simulations are shown in the attached figures.
Simulink simulation diagram for van der Pol’s equation for (x,_x) plane.
9059
Simulink simulation of van der Pol’s equation in the Liénard (x,y) plane.
Sample trajectories are shown in the attached figures.
0.6 –0.4 0.2 00.2 0.4 0.6 0.8 11.2
0.8
0.6
x