ANALIZA SIGNALA I SISTEMA II Auditorne vježbe Vježba broj 1

DISKRETNI SISTEMI U PROSTORU STANJA

( ) ( ) ( ) ninfbnxanx i

m

jjiji ,....2,1 ,1

1=+=+ ∑

=

promjenjive stanja

( ) ( ) ( )∑=

⋅+=m

iii nfdnxcng

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

mmmmm

m

b

bb

B

aaa

aaa

A.

,

.................

.......,........

2

1

21

1121

[ ] dDcccC m == ,..........,, 21

( ) ( ) ( )( ) ( ) ( ) IZLAZANA -J

STANJANA -J 1nDfnCXng

nBfnAXnx+=

+=+

Zadatak 1: Sistem je opisan jednačinama u prostoru stanja. Naći diferentnu j-nu koja opisuje ovaj sistem. Nacrtati blok dijagram sistema.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )⎪

⎩

⎪⎨

⎧

++=++−=+

++=+

nfnxnxngnfnxnxnx

nfnxnxnx

21

212

211

2211

Rješenje:

( ) ( ) ( )( ) ( ) ( )⎩

⎨⎧

+=+=+

nDfnCXngnBfnAXnX

1

[ ] 1 ,1 1 ,21

,2 1- 1 1

==⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡= DCBA

( ) ( ) ( )[ ] ( ) ( )( ) [ ] ( )zBFzAzIzX

zBFzzXAzIzBFzzAXzzXI

111

11

11

−−−

−−

−−

−=

=−

+=⋅

( ) ( ) ( ) [ ] ( ) ( )zDFzBFzAzICzDFzCXzG +−=+= −−− 111

( ) [ ][ ] ( )zFDBzAzICzG +−= −−− 111

Funkcija sistema: ( ) [ ][ ]DBzAzICzH +−= −−− 111

2-1

- 12

1 0 0 1

11

11

11

111 K

ZZZZ

ZZZZ

AZI =⎥⎥⎦

⎤

⎢⎢⎣

⎡ −=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−⎥

⎦

⎤⎢⎣

⎡=−

−−

−−

−−

−−−

1−K : K detK 1 adjK =−

( )( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−=⎥

⎦

⎤⎢⎣

⎡ −⋅⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−

−=⋅

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−

−+−

=

+−=+−−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−

−−=

−

−

−−−

−−−−

−−

−−

−−−

−−−−−

−−

−−

−−

−−

12

1

111

1111

11

11

121

12211

11

11

11

11

23det1

21

1 21

det1

1 21

1331

133211det

1 21

1 21

ZZZ

KZZ

ZZZZ

KBZK

ZZZZ

ZZK

ZZZZZK

ZZZZ

ZZZZ

adjKT

[ ] ( )

1331)(

33det

123

1 1det

1

1211

2112

111

+−=+=

−⋅=⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−⋅=

−−−−

−−

−−

−−−

ZZDBZCKzH

ZZKZZ

ZK

BZCK

( ) ( )zFZZ

zG ⋅+−

= −− 1331

12

( ) ( ) ( ) ( )zFzGZzGZzG =+− −− 12 33 ( ) ( ) ( ) ( ),1323 nfngngng =+−−−

( ) ( ) ( ) ( )nfngngng =−+−− 2313

Zadatak 2: Sistem je zadan jednačinama u prostoru stanja. Naći diferentnu jednačinu koja opisuje ovaj sistem i nacrtati blok – dijagram.

( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )⎪⎩

⎪⎨

⎧

++=+−=+

=+

nfnxnxngnfnxnxnx

nxnx

21

212

21

51

1

Rješenje:

[ ] 1 ,11 ,10

,51

10==⎥

⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡−

= DCBA

[ ] [ ]

2-1-

1-1-

2-1-

2-1-1-

2-1-

Z5Z1ZZ

Z5Z1ZZZ

Z5Z1detK

−++

=+

−++

=⋅⎥⎦

⎤⎢⎣

⎡⋅=⋅⎥

⎦

⎤⎢⎣

⎡⋅⎥⎦

⎤⎢⎣

⎡ +⋅=

−+=⎥⎦

⎤⎢⎣

⎡ +=

=⎥⎦

⎤⎢⎣

⎡

+−−

=⎥⎦

⎤⎢⎣

⎡

−+

−⎥⎦

⎤⎢⎣

⎡=−

−

−

−

−−

−−−

−

−−

−−

−

−−

−−

61

det111

det10

151

11

,1

51

511

50

1001

1

1

2

11

111

1

11

11

1

11

11

DBCK

KZZ

KZZZZ

BCK

ZZZ

adjK

KZZ

ZZZZ

AZI

( )( ) ( ) ( ) ( ) ( ) ( )16215

5161

21

1

−+=−−−+⇒−+

+= −−

−

nfnfngngngZZ

ZzFzG

Blok dijagram:

( ) ( ) ( ) ( ) ( )16215 −++−+−−= nfnfngngng

Z-1 Z-1

+5

g(n)

f(n)

_

6

Zadatak 3: Odrediti funkciju sistema H(z) i impulsni odziv h(n) sistema koji generira Fibonacciev niz: {1,2,3,5,8,13,…….}. Ovaj sistem može biti predstavljen slijedećim j-nama u prostoru stanja:

( ) ( ) ( )

( ) [ ] ( ) ( )nfnxng

nfnxnx

+=

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡=+

1 110

1 1 1 0

1

Rješenje:

[ ]

2111

21

12

1

211

1

2

11

11

2111

1

11

211

211

11

11

1

11

11

11

1 11

det1

det

10

11

11

11

11

-1det,1

1

10

1 00 1

−−−−

−−

−−

−

−−−

−

−

−−

−−

−−−−

−

−−

−−−

−−−

−−

−−

−

−−

−−

−−=+

−−+

=⎥⎦

⎤⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⋅⎥

⎦

⎤⎢⎣

⎡ −−−

=

⎥⎦

⎤⎢⎣

⎡ −−−

=

−=⎥⎦

⎤⎢⎣

⎡ −=

=⎥⎦

⎤⎢⎣

⎡

−−−

=⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡=−

ZZDBZCK

ZZZZ

ZZ

KBZCK

ZZ

KZZZZ

ZZBZK

ZZZ

ZZK

ZZKZ

ZZadjK

KZZZ

ZZZ

AZI

( ) 2111

−− −−=

ZZzH

( )

( ) ( )

.1 ,2

512

515

1

251

251

251

251

251

251

251

2512

51

251

2512

51Re

251

2511

11

11

12

251

12

251

1

2

2

2

lim

lim

−≥⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −−⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=

=−

−+

⎟⎟⎠

⎞⎜⎜⎝

⎛ +

++

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛ −

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛ +−+

+

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

⋅⋅⎟⎟⎠

⎞⎜⎜⎝

⎛ −−==

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

=−−

=

++

++

−

+→

−

−→

−∑

n

ZZ

ZZZ

ZZ

ZZZZzHznh

ZZ

ZZZ

ZzH

nn

nn

n

z

n

z

n

Zadatak 4: Sistem je zadan diferentnom jednačinom:

( ) ( ) ( ) ( ) ( ) ( )3111392716 −+−−=−+−+−− nfnfngngngng a) Odrediti H(z) b) Nacrtati blok dijagram sistema c) Odrediti j-ne u prostoru stanja za ovaj sistem d) Odrediti j-ne tako da on bude zadan u pratećoj formi. Rješenje:

a) ( ) ( ) ( ) ( ) ( ) ( )zFZzFZzGZzGZzGZzG 31321 11976 −−−−− +−=++−

( ) ( )( ) 321

31

976111

−−−

−−

++−+−

==ZZZ

ZZzFzGzH

b) ( ) ( ) ( ) ( ) ( ) ( )3111392716 −+−−−−−−−= nfnfngngngng

c) Sistem može imati različite opise u prostoru stanja u zavisnosti od načina odabira promjenljivih stanja. U ovom slučaju odabraćemo promjenljive na sljedeći način:

( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )⎪

⎩

⎪⎨

⎧

+−=++−=+

−+=+⇒=

nfnxnxnxnxnx

nfnxnxnxngnx

119171

61

13

312

211

1

( ) ( )nxng 1=

[ ] 0 ,001 ,1101

,009-107-016

==⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡= DCBA

d) Za izvođenje prateće forme prostora stanja polazi se od uopštenog oblika prikaza diferentne j-ne.

( ) ( ) ( ) ( ) ( ) ( )0 i 1

.....b21......1

00

m211

==−+−+−=−++−+

bamnfnfbnfbmngangang m

PRATEĆA FORMA:

[ ] 0 ,.... ,

1..000

,

-..---1..000............0..1000..010

111

121

==

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

= −

−−

DbbbCB

aaaa

A xmmm

mxmmmm

Jordanova matrica

[ ] 0 ,1- 0 11 ,100

,67-9-100010

==⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡= DCBA

Zadatak 5: Dat je sistem u prostoru stanja. Odrediti diferentnu j-na koja opisuje ovaj sistem, funkciju sistema H(z) i impulsni odziv h(n), te nacrtati blok – dijagram sistema.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )( ) ( )⎪

⎪⎩

⎪⎪⎨

⎧

=+=+

++−=+++−=+

nxngnfnxnx

nfnxnxnxnfnxnxnx

1

13

312

211

7101231

61

Rješenje:

[ ] 0 ,001 ,721 ,

0010103-016-

==⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡= DCBA

( ) ( )( ) ( ) DBZAZICzFzGzH +−== −−− 111

K 1010

13061

0010

0306

100010001

1

11

11

1

11

11

1 =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−+=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=−

−

−−

−−

−

−−

−−

−

ZZZ

ZZ

ZZZ

ZZHZI

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++++−=

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++++−

=

−−−−

−−−−−

−−

−−−−−

−−−

−−−

2121

21121

21

21212

211

121

3611010661103

1

36161061101031

ZZZZZZZZZ

ZZ

ZZZZZZZZZZZ

adjK

T

321231 1036131061det −−−−−− −++=+−+= ZZZZZZK

[ ]

[ ] 10361

72det

1

721

72

3611010661103

1

det001

321

321

1

1

1

21

1

1

1

2121

21121

21

11

−−−

−−−

−

−

−

−−

−

−

−

−−−−

−−−−−

−−

−−

−++++

=⋅⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⋅=

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⋅⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++++−=

ZZZZZZ

KZZZ

ZZ

ZZZ

ZZZZZZZZZ

ZZ

KBZCK

( )1036

7210361

7223

2

321

321

−++++

=−++

++= −−−

−−−

ZZZZZ

ZZZZZZzH

Diferentna j-na:

( ) ( ) ( ) ( ) ( ) ( ) ( )372213102316 −+−+−=−−−+−+ nfnfnfngngngng Blok dijagram:

( ) ( ) ( ) ( ) ( ) ( ) ( )372213102316 −+−+−+−+−−−−= nfnfnfngngngng

h(n):

( )1036

7223

2

−++++

=ZZZ

ZZzH ,

pojas konvergencije:

5Z 5 ,2 ,1 Z 321 f−=−== ZZ

Pošto je sistem kanzalan, polovi s nalaze unutar prstena konvergencije, pa se h(n) računa preko izraza:

( ) ( ) ( )( )( ) ( )

( )( )( ) ( ) ( )( )( ) ( )

( ) ( ) ( ) ( ) ( ) 1 59

11297

955

18222

971

1810

5521

72lim2521

72lim

1521

72limRe

11111

12

5

12

2

12

,, 1

1

321

≥−−−−=⇒−+−−

+=

=+++−

++++

++−++

+

+−++−

++==

−−−−−

−

−→

−

−→

−

→

−∑

nnh

ZZZZZ

ZZZZZZZ

ZZ

ZZZZZ

ZZZzzHnh

nnnnn

n

Z

n

Z

n

ZZZ Z

n

Zadatak 6: Sistem je zadan diferentnom j-nom:

( ) ( ) ( ) ( ) ( ) ( )2132212 −+−=−+−+−+ nfnfngngngng a) Nacrtati blok – dijagram sistema b) Naći j-ne u prostoru stanja c) Naći j-ne u prostoru stanja u pratećoj formi. Rješenje:

a) ( ) ( ) ( ) ( ) ( ) ( )2132212 −+−+−−−−−−= nfnfngngngng

b) ( ) ( )ngnx =1

( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )nxnx

nfnxnxnxnfnxnx

13

312

11

211

21

−=+++−=+

+−=+

[ ] 0 ,001 ,011

B ,002101012

==⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

= DCA

c) Prateća forma:

[ ] 0 ,110 ,100

B ,212

100010

==⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−= DCA