Design and Flight Control of Miniature Air Vehicle

7/31/2019 Design and Flight Control of Miniature Air Vehicle

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A T HESIS

SUBMIT TED FOR TH E DEGREE OF

Master of Engineering

IN TH E FACULTY OF ENGINEERING


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, ,


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( ) = + + + < 0

= [ ]

=

( ) < 0 ( ) < 0 0

(+

(1

) )=

( )+

(1

) ( )< 0

( ) = + ( + + ) < 0

= [ ] = [ ]


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=


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=

=

=

=

= =

=

,

=


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=


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1.8 10


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/

,


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=1

2

= .


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2 4

=

=


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/

/

2


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+ (

cos)

=

+ + (sin ) =

+ + =

+ + =

=

=

= [ ]

= [ ]

= [ ]

= +

=


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0 0 1 0

0 0

0 0 1 00 0 0 1

= + = +

= + = +

= + + = + +

= cos + = sin

= + = +

= + = +

= = ( )

= =

= + =

= [ ]

= [ ]

= [ ]

= +


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=

0 0 1 0

0

0

0 0

0 1 0 00 0 1 00 0 0 1

= = +

= = = +

= = +

= 1 = +

= = +

=

= 1 + + = 1 + +

= 1 + + = 1 + +

= 1 + +


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= [( . ) + ( . ) ]

= [( . ) + ( . ) ]

= [( . ) + ( . ) ]

= [( . ) ( . )]

= [( . ) ( . )]


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= [( . ) ( . )]


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, , ,

, , ,

, ,

=

0.19 1.65

0.05

0.54 6.41

3.77

0.000 9.81 0.10710.36 0.00 6.453

3.04 0.00 32.630.00 0.00 1.00 0.00 0.000.00 0.00 0.00 0.00 8.00

=

00

008

=

0.276 0.02 12.39 9.81 0.256 1.9030.00 0.00 1.00 0.00 20.35 0.325

1.0560.0440.00

0.00

10.44 0.900.00

0.00

3.27440.0906

0.00

0.00

0.000.000.00

0.00

0.2150.00

8.00

0.00

7.0450.000.00

8.00

=

0 00 0008

0

000

8


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= 0.002088 0.1315 + 3.128 33.92 + 141.8

= (7.82 10 ) + (7.624 10 ) (9.925 10 ) 0.03464 0.1434= (46.25 10 ) + (36.23 10 ) + (287.2 10 ) + 0.0716 + 0.3961

= 0

= 9.81

= (1.584 10 ) + (5.242 10 ) 0.01196 1.503

= + (39.59 10 ) (90.63 10 ) + 0.7357 7.2

= (57.78 10 ) + (137.3 10 ) 1.149 + 11.6

= 0

= (9.664 10 ) + (1.959 10 ) (1.922 10 ) 0.02219 0.02376

= + (2.646 10 ) 0.06582 + 0.4759 4.267

= + (1.993 10 ) 0.04688 + 0.3632 3.431

= 0


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= (2.938 10 ) + (2.606 10 ) 0.09362 + 0.005573

= + (1.566 10 ) 0.3446 + 2.204 8.682

= + (79.66 10 ) 1.761 + 11.38 44.31

= (8.142 10 ) (3.394 10 ) + 0.006069 0.03468

= (5.488 10 ) + (1.501 10 ) (1.523 10 ) + 0.01201 0.03316

= (4.855 10 ) + (13.28 10 ) 0.428 + 1.063 13.08

= 9.81

= (1.806 10 ) + (4.64 10 ) (4.261 10 ) + 0.006566 1.059

= (4.238 10 ) + (1.144 10 ) 0.1211 + 0.9659 11.48= (1.639 10 ) (6.174 10 ) + 0.2012 + 3.031

= 0

= (1.256 10 ) + (2.82 10 ) (7.224 10 ) + 0.001988 + 0.04248

= (2.246 10 ) (3.798 10 ) + 0.01615 0.915

= (7.479 10 ) (1.005 10 ) (1.626 10 ) 0.01952 + 0.1177

= 0

= (1.953 10 ) + (5.647 10 ) (6.251 10 ) + (3.964 10 )

0.002547

= (3.647 10 ) + (1.067 10 ) (1.196 10 ) + 0.07574 0.4344

= (5.849 10 ) + (3.18 10 ) + (8.305 10 ) + (1.812 10 )

( 1.042 10 )

= (

3.542 10)

(

1.024 10)

+(

1.133 10)

(

7.179 10)

+

0.04069

= (7.366 10 ) + (2.127 10 ) (2.352 10 ) + (1.486 10 )

( 7.155 10 )

= (1.313 10 ) (3.793 10 ) (4.196 10 ) 0.0266 + 0.1507


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=

( ) ( , , , )

( , , , )

= + =

= 1 + =


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+ < 0 = > 0

=

+ < 2 = > 0

= +

( + ) + ( + ) < 2

= > 0


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+ + + +( + )

< 2 00 0

> 0

= +


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( ) =( )

( ) =

( ) = + ( )

( ) = + ( ) + ( )

, , ,

, ,

( , , , ) , , , , , ,

= + = [ ] +

= + = +

= + ( ) = + ( )

= =

= 00

+ 00

+0

0


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=

0 00 0

0 00 0 1

+

00

00

+

00

00

= + [ ]

= + [ ]

= [ ]

= [ ]

= + = =

= +

00

+

0

= +

, ,

= +

, , ,


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= = [ ] = =

= [ ]

=

=

0 0 1

1 0 0 0

0 0 00 0 00 0 0

00

det( ) =

, , ,


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= + +

= + +

= + +

= +

= [0 0]

[ 0 0]0

0

00

00

0

000 +

[0 0]

[ 0]

[0 ]


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= [ 0 0][ 0 0]

0

0

00

00

0

00

0

+ [0 0][ 0][0 ]

+ + + +

( + ) 0

( ) <


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1. Find out the SOF representation of the system as in Eqs. 4.18, 4.19.

2. Augment the SOF output to obtain the generalized plant . Define , (gain and

phase margin constraints) and any damping constraints on the control design.

3. Compute the optimal guaranteed performance index, , using function of

MATLAB. This serves as the starting point.

4. Initialize and also initialize the incremental step in as d.

5. For > 0, obtain the initial X from the riccati equation below.

+ + = 0 (4.36)

where = > 0

The Q is taken as Identity matrix.

Initialize j=1

6. Using X from previous step, solve for P, , to minimize amd satisfy the LM Is

below,

+ + + +

( + ) 0 (4.39)


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( ) < (4.40)

7. If < 0 then put =

j=j+1

Goto step 5

8. Else if j=1 or if n>0 and n 0 (4.43)

( ) < (4.44)

i. I f P has a feasible solution then

Put X=P , n=n+1 , j=j+1

Goto step 6

ii. Else

Put = +

Goto step 6

9. Else if j 1 and > 0 Put = +

j=j+1, Goto step 6

10. Else if j 1 and < 0

Form the closed loop system with feedback gain .

= +

Find using normhinf function in Matlab.

Find the Gain M argin (GM) and Phase Margin (PM) and the damping (Z) of the closed

loop system.

i. I f < 1 and > and > and > then Goto step 11.

ii. Else = +


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j=j+1, Goto step 5

11. Final Gain =

12. Find = =

END


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ln

1

1. Sample number of plantsm from plant parameter space

1

2ln

2

where1 is the confidence and is the accuracy parameter

2. For each plant

Apply modified ILMI algorithm for a plant from the plant spaceP.

i. I f a controller ( ) is feasible for then store both .

i=i+1, count=count+1

ii. Else i=i+1

3. For each controller

Compute the closed loop system for each and find the norm.4. Find the average norm for each controller and find the controller which gives the

minimum value.

5. END


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=

00

00

10

000

0

+

0000

20%

20%

40%


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, , ,

W =0.1

( + 0.21)

W =0.2

( + 0.21)


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W =( + 0.15)

( + 10)

= 5 = 50

1.0611 + 0.0210.

( . )

1.5481 + 1.1673( . )

0.1797.

( . )

( ) =( + )

( ) =+

, ,

= 0.02 , = 50 , = 2.83, = 2.913


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= ( ( ) ( ))


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sin


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Documents

Design and Flight Control of Miniature Air Vehicle