ELG4152DS11____dp11.4

ELG 4152 DGD Notes (Week 6)

1

AP11.3 A system has a matrix differential equation ubb

+

=

2

1x2001

x& , what values for b1 and b2 are

required so that the system is controllable?

Solution: As stated in chapter 11.2, a system is controllable if and only if the determinant

of the controllability matrix is non-zero. So we first get the controllability matrix from the

given information, then select the right b1 and b2 to make the determinant of the

controllability matrix to be non-zero.

In this problem, n=2, controllability matrix Pc=[B AB],

AB=

=

2

1

2

1

22001

bb

bb

, Pc=

2

1

2

1

2bb

bb

; 2121212)det( bbbbbbPc =−= ;

So for the system to be controllable, we require that 01 ≠b and 02 ≠b .

DP11.4 A high performance helicopter has a model as

δαθσθ ndtdx

dtd

dtd

+−−= 112

2

δσθαθ gdtdx

dtdg

dtxd

+−−= 222

2

.

The goal is to control the pitch angle θ by adjust the rotor angle δ, where x is the

translation in the horizontal direction.

Given: σ1 =0.415 α1 =1.43 n=6.27

σ2 =0.0198 α2 =0.0111 g=9.8

Find a) the state variable representation of the system;

b) the transfer function for )(/)( ss δθ ;

c) using the state variable feedback to achieve adequate performance for the

controlled system. The performance requirements are Ts<1.5s, overshoot <20%, and

steady-state error less than 20% of the input magnitude.


2

Solution:

As stated in the problem, the input of the system is δ, the output of the system is θ. To be

clear in notation, we use z instead x for the translation in the horizontal direction. Then

the given system is δαθσθ ndtdz

dtd

dtd

+−−= 112

2

δσθαθ gdtdz

dtdg

dtzd

+−−= 222

2

a) Set the state variable as

=

z&

&θθ

x ,

=z&&

&&

&

& θθ

x , [ ]x001== θy .

[ ]x010=θ&

[ ] δασδαθσθ nx0 1111 +−−=+−−= nz&&&& ;

[ ] δσαδσθαθ gx2222 +−−=+−−= ggzgz &&&&

so the state space representation is

δσαασ

+

−−−−=

gn

g

0x0

010x

22

11& = δ

+

−−−−

8.927.60

x0198.043.18.90111.0415.00010

[ ]x001== θy

b) Find the transfer function:

From Chapter 3 Eq 3.71, we knew that transfer function of a system is

BAsICsXsYsG 1][

)()()( −−== .


3

1

22

111 0

01][

−

−

+−+−

=−σα

ασsg

ss

AsI

∆

+−−−++

+−−++

=

T

ssgssssgsgss

)(0)()(

)())((

11

222

112121

σαασσσααασσ

[ ]∆

−+=

−=−= −− gns

gnAsIBAsICsG 1211 )(0

][001][)( ασ , where

gasaasssgsssAsI 121212

213

21121 )()())((]det[ +−+++=−+++=−=∆ σσσσααασσ

i.e. 109.00077.0435.0

0154.027.6)( 23 +−++

=sss

ssG

c) state variable feedback design:

Let [ ]x321 k-k-k −=δ + rk4 , then the closed-loop system is

[ ] rgnk-k-k

gn

ggn

g4321

22

11

22

11 k0

x0

x00100

x0010

x

+−

+

−−−−=

+

−−−−=

σαασδ

σαασ&

rgknk

gkgkgkgnknknk

+

−−−−−−−−−−=

4

4

32221

31211

0x

010

σαασ

−−−−−−−−−−=

32221

31211

010

gkgkgkgnknknkH

σαασ

)01

det()det(

32221

31211

++++−+++

−=−

gksgkgkgnknksnk

sHsI

σαασ


4

= ))(())(())(( 223211313221 agkgksnkggknkagksnkss ++++−+−++++ σσσ

= +−−++++++++ saagkankkgknsnkgks )()( 212113122212

21233 σσσσσσ

0)1)(2( 22123111 =+++=++− sssnkgnkgkaga nn γωζωσ

to achieve the design specification, we need Ts<1.5 P.O<20, i.e.

Ts= 5.14

n

<ζω

, i.e. 6667.25.1/4 =>nζω , pick 3=nζω ;

P.O=21100 ζ

ζπ

−−

e <20, i.e. 4558.0>ζ , pick 46.0=ζ ,

then 52.6>nω , pick 7.6=nω .

According to the explanation in section 5.4, the real part of the third root should greater

than 10 times of the real part of the dominant roots (the roots of the second order system).

So 30||10|1| =≥ nζωγ

Then 135022536)30)(456()/1)(2( 23222 +++=+++=+++ sssssssss nn γωζω

Comparing coefficients yields

−++−−

=

+−+−−

gaaa

kkk

gnngagnagann

gn

1

21

21

3

2

1

21

1212

135022536

0

0 σσ

σσσ

so K=[ 53.1116 -28.6441 21.9555].

(the results may be a little bit different because different ζ and ωn were picked up. It is

OK as long as the performance requirements are achieved )

MP11.1 consider the system u

−+

−−

−=

101

x1086

304011

x& ,

[ ]x121y = .


5

Using the ctrb and obsv functions show that the system is controllable and observable.

Solution: the Matlab script is the following:

>> det_con = 118 the system is controllable. det_obs = -19 the systemis observable MP11.2 Consider the system

u

+

−−

=10

x5.42

10x&

[ ]x01y = .

Determine if the system is controllable and observable. Compute the transfer function from u to

y.


6

Solution: the Matlab script is the following.

>> det_con = -1 the system is controllable. det_obs = 1 the systemis observable Transfer function: 1 --------------- s^2 + 4.5 s + 2

MP11.8 A system is described by a single-input state equation with

=

−

=10

,x0100

A B .

Using the method if section 11.4(eq 11.19) and a negative unity feedback, determine the

optimal system when [ ]01(0)xT = .


7

Solution: set the state variable feedback as [ ]x212211 -k-kx-kxku =−= , where k1,k2>0

Then [ ] x-

00x

0110

x10

x0110

10

x0110

x21

21

−

+

−

=−−

+

−

=

+

−

=kk

kku&

x1

10

21

−−−

=kk

;

−−−

=211

10kk

H

to achieve a good feedback contol, we need to minimize the performance index J, i.e. to

design the symmetric matrix P to keep -IPHP =+TH .

−

−=

−−−

+

−−−

1001

110

1212110

2

1

22

1211

22

1211

2

1

kkpppp

pppp

kk

−

−=

−−−−−−

+

−−−−−−

1001)1()1()1()1(

2221212211

221121

2221212211

221121

pkppkppkpk

pkppkppkpk

−

−=

−++−

−+−−−10

01)(2)1(

)1()1(2

2221212222111

12222111121

pkppkpkppkpkppk

so )1(2

1)1(2

1

1112 kk

p+

=−−

−=

21

112

222 )1(2

2)21(1

kkkp

kp

++

=+=

21

12

122

2212211 )1(223)11(

kkkkkpkpkp

++++

=++=

the performance index [ ]21

12

122

112212

121100 )1(2

2301

01xxkkkkkp

pppp

PJ T

++++

==

== .

To minimize J as a function of k2, we take the derivative with repect to k2 and set it equal to zero.

221

112

122212

2 ])1(2[)1(2)23())1(2(2

kkkkkkkkk

kJ

+++++−+

=∂∂ = 2

21

12

122

1 )1(223

)1(1

kkkkk

k ++++

−+

02

2)1(2

12

2

1

1

=+

−+

=k

kk

, i.e. 22

1

1

2)1(

1k

kk

+=

+

)2)(1( 112

2 ++= kkk


8

)2)(1( 112 ++= kkk

1

1

1

11

1

2

21

22

21

12

122

12

)1()2)(1(

)1()1(22

)1(223

kk

kkk

kk

kkk

kkkkkJ

++

=+

++=

+=

+=

++++

=

use Matlab to plot the relationship between k1 and k2, J and k1 to help us pick a proper set

of [k1 k2].

The figure shows that as k1 increase, the performance index J decreases. However, we see

that the rate of decrease slows considerably after k1>20. Also k2 increase as k1 increase. We want

to keep both gains as small as possible, while still having a small J. A reasonable selection is

k1=20 and k2=21.5.


9

ELG4152DS11____dp11.4

Documents

Transcript of ELG4152DS11____dp11.4