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Solution 4.1:

i) ( )n y is an odd memorylessnonlinear operator, so its describing function is independent

of . We consider two cases for the amplitude, a, of the sinusoidal input ras follow,

and compute the first harmonic of the steady-state part of n(r)

Let denote t . If 1a then3

3 33 3 2

33 1 34 2( ( )) sin sin ( )sin sin 3 ( )

2 4 2 4 2 4

a aa a a a

n r t a a aa

+= + = + = = +

If 1a> then

/21 3 3

0 0

/21 3 1

2 2

1

2 2 2

1 4 4sin : ( ) ( sin )sin ( sin sin )sin

2

1 4 1 1 1 1 3 1 1 1(2sin )sin sin 1 sin 1

2 2 2 2 8 2

1 1 2 1 1 11 (1 ) sin 1

8 2

aa n a d a d

a a a

ad a

a a a a a a a

aa a a a a a

= = = +

= +

+ +

2

2 1

2

1 11

2

1 1 1( ) (3 6)sin 3 1 4

2

a

a a aa a

= +

ii) ( )n x represents a hysteresis nonlinear operator which is not memoryless, thus its

describing function may depend on the frequency . However, through laborious but

routine calculations, it can be verified that ( )a is independent of and complex

indeed! In this case, if the input amplitude is less than , the output is simply equal to

zero

( ) 0a if a =

If the input amplitude a exceeds , then the steady-state output is described by

sin ,0 / 2

( ) , / 2( ) sin , 3 / 2

( ) ,3 / 2 2

sin ,2 2

ss

ma m

m az t ma m

m a

ma m

= +

where (0, / 2) be the unique number such that

2sin 1

a

=

88-87

:9/9/87

:


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and let denote t for convenience. After some character-building computations, one

finds that if a , then( )

re ima j = +

where

1 22 2 2( ) sin ( 1) ( 1) 1 ( 1)2

4( ) ( 1)

re

im

ma

a a a

ma

a a

=

=

You may also see Slotine (1991) or Vidyasagar (1993) for more detailed discussions.

Solution 4.2:

First we recall a few interesting points about the describing function of the hysteresis

nonlinearity in Exercise 4.1 part (ii) for 1m= = (See Slotine (1991), figures 5.17 and5.18)

1. ( ) 0 12. ( ) 1 1/ 0

a if aa as a

= =

and furthermore ( )a increases, when 1/ a decreases. Now, we plot the Nyquist diagram

of )(sG and the graph of1

( )a for 0a (See Fig. 1). One can see that the two plots

intersect at two points

1 2: ( 6.49, 30) : ( 1.02, 0.0766)L L

Fig. 1. Plots of )( jG and)(

1

a in Exercise 4.2

Two intersections and two limit cycles! However, note that one of them occurs at a

negative frequency which does not correspond to any other positive one!! This situation

arises due to the complexity of the given describing function. Now you can graphically find

the approximate frequency and amplitude corresponding to each limit cycle

1 2

1.025 16.34: :

1.33 / 7.38 /

a aL L

rad s rad s


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Hence, we predict two limit cycles: one at an angular frequency of 1.33 rad/sec and

amplitude of 1.025 (L1), and another at an angular frequency of -7.38 rad/sec and amplitude

of 16.34 (L2). See Fig. 2 which depicts these two L.C., separately.

About the stability of L1, since it passes through the stable region to the unstable one while

the amplitude, a, is increasing, thus L1 is an unstable L.C. Similarly, you can analyze the

stability of L2. Does it go form the stable region to the unstable one?!

Fig. 2. Two nontrivial limit cycles in Exercise 4.2

Solution 4.3:

First we can change the feedback system to the following form

Fig. 3. Position servo mechanism after composition of the two nonlinearities

Note that

It is straightforward to show that the describing function for the nonlinearity in Fig. 3 (relay

with a dead-zone a ) is given by

2

2

0

( ) 41

a

aa

a a


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2( )a

=

The Nyquist curve of )( jG crosses the negative real axis at 2=o , 0Im( ( )) 0G j = ,

for which 3/2)( =ojG (See Fig. 4). Thus, we should expect no oscillations if

4

3 >

Fig. 4. Nyquist curve of )( jG in Exercise 4.3

Solution 4.4:

The nonlinear system 2 ( ) 0x x px q x x+ + + = can be represented in the state space form

as follows

1 2

2 2 1 1 22 ( )

x x

x x px q x x

=

= + +

This can be rewritten as

1 2

2 1 2

1 2

3

2 ( )

( ) / 2

x x

x px x qu

y x x

u y y y

= = = + = =

The first part of this model, which has been asterisked, is a linear SISO system in which

[ ]

0 1 0

2

1 1

X X up q

y X

= +

=

and the corresponding transfer function is2

( )2

qs qG s

s s p

=

+ +.

Consequently, the given system can be put into the closed loop system shown in Fig. 5.

Now, it is sufficient to compute the describing function for the nonlinearity ( ) and also

the frequency response of ( )G s , to investigate the possible existence of periodic solutions.

3 3 33 23 3( sin ) sin sin ( )sin sin 3 ( ) 1

2 8 8 8

a a aa t a t t a t t a a = + = + = +


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Fig. 5. The closed loop system in Exercise 4.4

The frequency response of the linear part of the closed loop system is indicated by

( )2 2 2 2 2( ) 2 ( ) ( ) 4 ( )G j q p p j f f p = + = +

Clearly, is always Re( ) 0G and Im( ) 0G = has a nonzero solution equal to 2p= if

2p> . Therefore, the Nyquist diagram of ( )G j will exhibit one of the two different

general forms, which have been shown in Fig. 6, according to the value of p .

Fig. 6. Nyquist diagram of )( jG in Exercise 4.4 for (a) 2p< and (b) 2p>

Note:If 2p> then 2p= is the real axis crossover frequency and by replacing it into

the real part of ( )G j we have Re( )2

qG = . Moreover,

q

p is the value of ( 0)G j .

Here, ( )a is real function and clearly, its negative inverse is bounded via

1 1 ( ) 0a < < . Thus, according to Fig. 6 only in the second case where 2p> , it is

possible to have a nontrivial periodic solution with a nonzero frequency. This limit cycle

occurs at the intersection of ( )G j and 1 ( )a on the negative real axis in Fig. 6 (b).

In other words, if 1 22

qq < < and 2p> , we may predict a periodic solution at an

angular velocity of 2p= and an amplitude which is the positive solution to the

following equation

2

1 8 2( 1)

3 2 31

8

qa

qa

= = +

Note that this L.C. is unstable. What about the intersection at 0= ?!Good Luck

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