bai 3 - chuong 3 - dklt mien t - thiet ke bdk

8/10/2019 Bai 3 - Chuong 3 - DKLT Mien t - Thiet Ke Bdk

1/13

III. !I"U KHI#N LIN T$CTRONG MI"N TH%I GIAN

BM!i"u Khi#n T$!%ngTh.S.!&ng V'n M(

3.4 THI&T K&B'!I"U KHI#N


2/13

III. !I"U KHI#N LIN T$C TRONG MI"N TH%I GIAN


Ph!n h"i tr#ng thi Ph!n h"i $%u ra&x = Ax + Bu

= Ax + B w ! Rx( )

= A!BR( )x + Bw

&x(t) =Ax +Bu

=Ax +B w !Ry( ) = Ax + B w ! RCx( )

= A!BRC( )x(t)+ Bw

det sI! A+ BR( ) = (s ! s1)(s ! s2 )K det sI! A+ BRC( ) = (s ! s1)(s ! s2 )K

C%n xc $&nh ma tr'n R th(a mn:
mailto:[email protected]


3/13


4/13

3.4 THI&T K&B'!I"U KHI#NPH()NG PHP ACKERMANN

B*+c 2:)*a MHTT v+d#ng chu,n $i+u khi-n

sT= (0,K,0,1)(B,AB,K,A

n!1B)

!1

"sT(B,AB,K,A

n!1B) = (0,K,0,1)

S =

sT

sTA

M

sTA

n!1

"

#

$$$$

%

&

''''

(SB =

sTB

sTAB

M

sTA

n!1B

!

"

####

$

%

&&&&

=

0

0

M

1

!

"

##

##

$

%

&&

&&

SA =

sTA

sTA

2

M

sTA

n

!

"

####

$

%

&&&&

=

sTA

sTA

2

M

!a0sT! a

1sTA !K! a

n!1sTA

n!1

"

#

$$$$

%

&

''''

0 1 0 K 0

0 0 1 K 0

M M M O 0

0 0 0 K 1

!a0 !a

1 !a

2 K !a

n!1

"

#

$$$$$

%

&

'''''

S =

0 1 0 K 0

0 0 1 K 0

M M M O 0

0 0 0 K 1

!a0 !a

1 !a

2 K !a

n!1

"

#

$$$$$

%

&

'''''

sTA

sTA

2

M

sTA

n

!

"

####

$

%

&&&&

=

sTA

sTA

2

M

!a0sT! a

1sTA!K! a

n!1sTA

n!1

"

#

$$$

$

%

&

'''

'

=SA

z = Sx! &x =S!1

&z

&x = Ax +Bu! &z = SAS!1

x+ SBu =

0 1 0 K 0

0 0 1 K 0

M M M O 0

0 0 0 K 1

!a0 !a

1 !a

2 K !a

n!1

"

#

$$$$$

%

&

'''''

x+

0

0

M

1

!

"

####

$

%

&&&&

u

)4t

)4t ma tr'n

Ta c

)5i bi6n


5/13

III. !I"U KHI#N LIN T$C TRONG MI"N TH%I GIAN

Hy tm ma tr!n ph"n h#i tr$ng thi K sao cho h%kn c &'(c cc &i)mc*c mong mu+n l


6/13


Dnh cho -1i t*2ng MIMO

PH()NG PHP ROPPENECKER

B'!I"U KHI#N PH4N H5I TR6NG THI GN !I#M C7C

(Tham kh8o)


7/13


PH()NG PHP MODAL

&x = Ax +Bu

y = Cx + Du!"#

B*+c 1:Xc $&nh ma tr'n ModalM $-$*a ma tr'nAv+d#ng $*8ng cho

B'!I"U KHI#N PH4N H5I TR6NG THI GN !I#M C7C

)1i t*3ng MIMO c m hnh:

M!1AM=

!1 0 L 0

0 !2 L 0

M M O M

0 0 L !n

"

#

$$$$

%

&

''''

=diag(!i )

V9i l cc gi tr&ring c7aA!i

M = a1 K a

n( )!

i

I! A

( )a

i

= 0

V d9:Bi6n $5i ma tr'n A thnh d#ng $*8ng cho:

A=1 2

3 2

!

"#$

%&

l cc vec t.bn ph!i c7aAa

i


8/13


PH()NG PHP MODAL

B*+c 2:)5i bi6n $-thu $*3c MHTT t*.ng $*.ng

G = M!1

AM=

!1

0 L 0

0 !2

L 0

M M O M

0 0 L !n

!

"

####

$

%

&&&&

=diag(!i )

x = Mz ! z = M"1x

! &z = M!1AMz + M

!1Bu = Gz + M

!1Bu

S=

s1 0 L 0

0 s1 L 0

M M O M

0 0 L sn

!

"

####

$

%

&&&&

=diag(si )

B*+c 3:Xc $&nh ma tr'n $*8ng cho S ch:a t;t c!$i-m c0c mong mu1n c7a h/

& & " #


9/13

3.4 THI&T K&B'!I"U KHI#NPH()NG PHP MODAL

T = M!1

B( )!1

= B!1

M

" R = !T S!G( )M!1 B2$i+u khi-n ph!n h"i m

B*+c 4:Xc $&nh b2$i+u khi-n

& & " #


10/13

3.4 THI&T K&B'!I"U KHI#NPH()NG PHP MODAL

Tr*:ng h2p ma tr;n B suy bi


11/13


Tnh ton b2)K ph!n h"i $%u ra dnh cho $1i t*3ng MIMO

PH()NG PHP MODALB'!I"U KHI#N PH4N H5I !=U RA GN !I#M C7C

& & ' " #


12/13


T=m hnh )K ph!n h"i tr#ng thi v9iRxl b2)K, c%n tm m2tma tr'n Q$-c th-chuy-n $i-m tr#ng thixv+y, ngh?a l:

PH()NG PHP MODALB'!I"U KHI#N PH4N H5I !=U RA

QC = I

!Q = C"1

Tuy nhin khng ph!i lc no c@ngtm $*3c Qdo ma tr'n Csuy bi6n.Do $, ta tnh nghi/m Q theo ph*.ngphp sau:

QC =I

!CQC = C

!Q = CT(CC

T)"1

& & ' " #


13/13


Cc b*9c thi6t k6b2)K R ph!n h"i $%u ra theo ph*.ng php Modal:

PH()NG PHP MODALB'!I"U KHI#N PH4N H5I !=U RA

Q = CT(CC

T)!1

B*+c 1:Thi6t k6B)K ph!n h"i tr#ng thi ModalRx

B*+c 2:Xc $&nh ma tr'n Q

B*+c 3:Tnh B)KR R =RxQ

bai 3 - chuong 3 - dklt mien t - thiet ke bdk

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