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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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Tnh hm truyn ca h ri rc t s khi. Th d 1
Hm truyn kn ca h thng:
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Tnh hm truyn ca h ri rc t s khi. Th d 2
Tnh hm truyn kn ca h thng:
Bit rng:
Gii:
Hm truyn kn ca h thng:
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Tnh hm truyn ca h ri rc t s khi. Th d 2
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Tnh hm truyn ca h ri rc t s khi. Th d 2
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Tnh hm truyn ca h ri rc t s khi. Th d 2
Hm truyn kn ca h thng:
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Tnh hm truyn ca h ri rc t s khi. Th d 3
Tnh hm truyn kn ca h thng:
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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Tnh hm truyn ca h ri rc t s khi. Th d 3
Gii:
Hm truyn kn ca h thng:
Ta c:
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Tnh hm truyn ca h ri rc t s khi. Th d 3
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Tnh hm truyn ca h ri rc t s khi. Th d 3
Hm truyn kn ca h thng:
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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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Th d thnh lp PTTT h ri rc t PTTT h lin tc
Gii:
Bc 1:
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Th d thnh lp PTTT h ri rc t PTTT h lin tc
Bc 2: Tnh ma trn qu
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Th d thnh lp PTTT h ri rc t PTTT h lin tc
Bc 3: Ri rc ha
PTTT ca h lin tc
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Th d thnh lp PTTT h ri rc t PTTT h lin tc
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