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Mn hc

L THUYT IU KHIN T NG

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Chng 6

M T TON HC

H THNG IU KHIN RI RC

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Ni dung chng 6

Khi nim

Php bin i Z Hm truyn Phng trnh trng thi

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Khi nim

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My tnh s = thit b tnh ton da trn c s k thut vi x

l (vi x l, vi iu khin, my tnh PC, DSP,).

u im ca h thng iu khin s:

Linh hot

D dng p dng cc thut ton iu khin phc tp

My tnh s c th iu khin nhiu i tng cng mt lc

H thng iu khin dng my tnh s

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H thng iu khin ri rc l h thng iu khin trong c

tn hiu ti mt hoc nhiu im l (cc) chui xung.

H thng iu khin ri rc

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Ly mu d liu

Ly mu l bin i tn hiu lin tc theo thi gian thnh tn hiu

ri rc theo thi gian.

Biu thc ton hc m t qu

trnh ly mu:

nh l Shannon

Nu c th b qua c sai s lng t ha th cc khu chuyn

i A/D chnh l cc khu ly mu.

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Khu gi d liu

Khu gi d liu l khu chuyn tn hiu ri rc theo thi gian

thnh tn hiu lin tc theo thi gian

Khu gi bc 0 (ZOH): gi tn

hiu bng hng s trong thi

gian gia hai ln ly mu.

Hm truyn khu gi bc 0.

Nu c th b qua c sai s lng t ha th cc khu chuyn

i D/A chnh l cc khu gi bc 0 (ZOH).

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Php bin i Z

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Trong :

Min hi t (Region Of Convergence ROC)

ROC l tp hp tt c cc gi tr z sao cho X(z) hu hn.

(s l bin Laplace)

Nu

nh ngha php bin i Z

Cho x(k) l chui tn hiu ri rc, bin i Z ca x(k) l:

- X(z) : bien oi Z cua chuoi x(k). Ky hieu:

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ngha ca php bin i Z

Gi s x(t) l tn hiu lin tc trong min thi gian, ly mu x(t)

vi chu k ly mu T ta c chui ri rc x(k) = x(kT).

Biu thc ly mu tn hiu x(t)

Biu thc bin i Z chui x(k) = x(kT).

Do

l nh nhau, do bn cht ca vic bin i Z mt tn hiu

chnh l ri rc ha tn hiu .

z = eTs nn v phi ca hai biu thc ly mu v bin i Z

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Tnh cht ca php bin i Z

Cho x(k) v y(k) l hai chui tn hiu ri rc c bin i Z l:

Z {x(k )} = X ( z ) Z {y(k )} = Y ( z )

Tnh tuyn tnh:

Tnh di trong min thi gian:

T l trong min Z:

o hm trong min Z:

nh l gi tr u:

nh l gi tr cui:

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Bin i Z ca cc hm c bn

Hm nc n v:

Hm dirac:

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Bin i Z ca cc hm c bn

Hm m:

Hm dc n v:

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Hm truyn ca h ri rc

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Tnh hm truyn t phng trnh sai phn

Bin i Z hai v phng trnh trn ta c:

trong n>m, n gi l bc ca h thng ri rc

Quan h vo ra ca h ri rc c th m t bng phng trnh

sai phn

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Tnh hm truyn t phng trnh sai phn

Lp t s C(z)/R(z) , ta c hm truyn ca h ri rc:

Hm truyn trn c th bin i tng ng v dng:

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Tnh hm truyn t phng trnh sai phn - Th d

Tnh hm truyn ca h ri rc m t bi phng trnh sai phn:

c(k + 3) + 2c(k + 2) 5c(k + 1) + 3c(k ) = 2r (k + 2) + r (k )

Gii: Bin i Z hai v phng trnh sai phn ta c:

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Tnh hm truyn ca h ri rc t s khi

Cu hnh thng gp ca cc h thng iu khin ri rc:

Hm truyn kn ca h thng:

trong :

GC ( z) : hm truyn ca b iu khin, tnh t phng trnh sai phn

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Tnh hm truyn ca h ri rc t s khi. Th d 1

Tnh hm truyn kn ca h thng:

Giai:

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Bit rng:

Gii:


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Bit rng:

B iu khin Gc(z) c quan h vo ra m t bi phng trnh:

u (k ) = 10e(k ) 2e(k 1)

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Gii:


Ta c:

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Phng trnh trng thi

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Khi nim

Phng trnh trng thi (PTTT) ca h ri rc l h phng trnh

sai phn bc 1 c dng:

trong :

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Thnh lp PTTT t phng trnh sai phn (PTSP)

Trng hp 1: V phi ca PTSP khng cha sai phn ca tn

hiu vo

t bin trng thi theo qui tc: Bin u tin t bng tn hiu ra: Bin th i (i=2..n) t bng cch lm

sm bin th i1 mt chu k ly mu

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Thnh lp PTTT t PTSP

Trng hp 1 (tt)

Phng trnh trng thi:

trong :

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t cc bin trng thi:

Thnh lp PTTT t PTSP

Th d trng hp 1

Vit PTTT m t h thng c quan h vo ra cho bi PTSP sau:2c(k + 3) + c(k + 2) + 5c(k + 1) + 4c(k ) = 3r (k )

Phng trnh trang thai:

trong o:

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Thnh lp PTTT t PTSP

t bin trng thi theo qui tc: Bin u tin t bng tn

hiu ra Bin th i (i=2..n) t bng

cch lm sm bin th i1mt chu k ly mu v tr 1lng t l vi tnh hiu vo

Trng hp 2: V phi ca PTSP c cha sai phn ca tn hiu

vo

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Thnh lp PTTT t PTSP

Trng hp 2 (tt)

Phng trnh trng thi:

trong :

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Thnh lp PTTT t PTSP

Trng hp 2 (tt)

Cc h s trong vector Bd xc nh nh sau:

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t cc bin trng thi:

Thnh lp PTTT t PTSP

Th d trng hp 2

Vit PTTT m t h thng c quan h vo ra cho bi PTSP sau:

2c(k + 3) + c(k + 2) + 5c(k + 1) + 4c(k ) = r (k + 2) + 3r (k )

Phng trnh trng thi:

trong :

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Thnh lp PTTT t PTSP

Th d trng hp 2 (tt)

Cc h s ca vector Bd xc nh nh sau:

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Thnh lp PTTT t PTSP dng phng php ta pha

Xt h ri rc m t bi phng trnh sai phn

t bin trng thi theo qui tc: Bin trng thi u tin l nghim ca phng trnh:

Bin th i (i=2..n) t bng cch lm sm bin th i1 mt

chu k ly mu:

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Thnh lp PTTT t PTSP dng phng php ta pha

Phng trnh trng thi:

trong :

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Th d thnh lp PTTT t PTSP dng PP ta pha

Vit PTTT m t h thng c quan h vo ra cho bi PTSP sau:

2c(k + 3) + c(k + 2) + 5c(k + 1) + 4c(k ) = r (k + 2) + 3r (k )

t bin trng thi theo phng php ta pha, ta c phngtrnh trng thi:

trong :

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Thnh lp PTTT h ri rc t PTTT h lin tc

Thnh lp PTTT m t h ri rc c s khi:

Bc 1: Thnh lp PTTT m t h lin tc (h):

Bc 2: Tnh ma trn qu

vi

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Thnh lp PTTT h ri rc t PTTT h lin tc

Bc 3: Ri rc ha PTTT m t h lin tc (h):

vi

Bc 4: Vit PTTT m t h ri rc kn (vi tn hiu vo l r(kT))

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Th d thnh lp PTTT h ri rc t PTTT h lin tc

Thnh lp PTTT m t h ri rc c s khi:

Vi a = 2, T = 0.5, K = 10

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Gii:

Bc 1:

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Bc 2: Tnh ma trn qu

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Bc 3: Ri rc ha

PTTT ca h lin tc

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Bc 4: PTTT ri rc m t h kn

Vy phng trnh trng thi ca h ri rc cn tm l:

vi

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Tnh hm truyn t PTTT

Cho h ri rc m t bi PTTT

Hm truyn ca h ri rc l:

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Th d tnh hm truyn t PTTT

Tnh hm truyn ca h ri rc m t bi PTTT

Gii: Hm truyn cn tm l

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