TEST BANK FOR Nonlinear Systems 3rd Edition By Hassan K. Khalil

,Nonlinear Systems
Third Edition




非线性系统
（第三版）
Solutions Manual
Hassan K. Khalil




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, 《非线性系统（第三版）》习题解答




Chapter 1


• 1.1 Let Zt = 11, %2 = 11<1), •.• , Zn= 1l(n-l)_




Xn-1 = Zn
Zn = g(t,x1, ... ,xn, u)
11 = %1


•
.,. -_ fl (1) , • • •,
1 •2 Let Zl -_ Y, ..,2 _
Xn-1 - 'I/ (n-3) , •n
,.. _ 1 ,(n-1) _
- 11 fh (t , 1/tJ/{1) , • • • 1 'I/ (n-2))
· U,




Zn-2 =
Zn-1 =
Zn =
=

'I/ = Z1


.,. -- 1/ I ...
• 1 •8 Let *l ,.. - 11 (1)
2 - 0
.,. - y(n-1) ...
• • • 7 '"'" -
_ ..
0 *n+l - '"'1 • •' 0
.. _ z(m-1)
'"'f'l+na - •

:i1 = Z2



Xn-1 = Zn
Zn = g(z1,.,. , Xn,Zri+t,, ., , Zn+m, U)
tn+1 = Zn+2



Zn+m-1 = Zn+m
Zn+m = u
'JJ = :i:1




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, 《非线性系统（第三版）》习题解答

• 1.4 Let z1 = q, z2 = q,· z =[ z.z12 ] E Ram ·

Z1 = Z2
z2 = q = M-1(:i:i)(u - C(.z1,.z2):t:2 - D:t:2 - g(.zi))



%1 = x2
MgL. k
z2 = - -smz1
I
- -(z1 - :t:3)
I
%3 = x,
k 1
x, = -(.z1 - zs) + -u
J J


:i:1 = z2
:i:2 = -M-1(.z1)[h(z1, z2) + K(z1 - za)]
%3 = z,
%4 = J- 1 K(z1 - za) + 1-1 v
• 1.7 Let
:i: = Az + Bu, fl = Cz
be a state model of the linear system.

u = r-,J,(t,11) = r- ¢(t, Cz)

Bence
x = h - B1/,(t,Cz) + Br, i =Cz
• 1.8 (a) Let 2:1 = 6, z2 =6, zs =E9 , and u = E,o.
z1 = z2
p D 'h .
= -M - -2:2
:i:2
M - -z1smz1
M
'1'l 'II 1
., +-cosz1 + -u
%3 = - -2:3 ., .,
(b) The equilibrium points are the roots of the equatiom

0 = .Z:z
0 = 0.815- Dz2 - 2.0zs sinz1
0 :: -2.7.zs + 1.7cosz1 + 1.22
0.4076
Z2 = 0 => Zs = -:---
IID %1
Substituting 2:3 in the third equation yields

(1.22+ l.7cosz1)sinz1 -1.l~ =0

The foregoing equation has two roots z1 = 0.4067 and z1 = 1.6398 in the interval -1t S z1 ::s;: ,r. Due to
periodicity, 0.4067 + 2mr and 1.6398 + 2mr are also roots for n = :H,:1:2, .... Ea.ch root z 1 = z gives an





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