sổ tay cdt Chuong 27-pp dap ung f

8/6/2019 s tay cdt Chuong 27-pp dap ung f

1/17

27Phng php p ng tn s

Jyh-Jong SheenNational Taiwan Ocean University

27.1M u ........................................................... ......... ..1

27.2 th Bode ...................................................... ........ .3

27.3 th cc ................................................................ ..6

27.4Bin logarit i vi th pha ..............................7

27.5Xc nh hm truyn bng thc nghim ........ ........ ...8

27.6Tiu chun n nh Nyquist ............................ ........11

27.7S n nh tng i ...............................................14

27.1 M u

Vic phn tch v thit k nhng h thng iu khin cng nghip thng c thc hin bng phng php p ng tns. Bng phng php ny chng ta bit c p ng trng thi n nh ca mt h thng h s hng tuyn tnh c u vol tn hiu dng hnh sin. Chng ta bit rng p ng ca mt h thng vi u vo l tn hiu hnh sin th cng cho tn hiuu ra dng hnh sin vi cng tn s ca u vo. Tuy nhin bin v pha ca tn hiu u ra th khc u vo v s khc

bit ny l mt hm ca tn s. V vy chng ta s kho st mi quan h gia hm truyn v p ng tn s ca nhng hthng n nh tuyn tnh.

Xt mt h thng h s hng n nh tuyn tnh nh trn hnh 27.1. S dng cng thc Euler, ejt= cost + jsint, vi githit tn hiu u vo dng hnh sin c cho bi cng thc:

0 0 0( ) cos sinj tu t U e U t jU t = = + (27.1)

Bin i Laplace ca u(t) ta c

0 0 0

0 2 2 2 2 2 2( )

U U s U s jU s U j

s j s s s

+= = = +

+ + +(27.2)

Biu thc th nht trong phng trnh 27.2 l bin i Laplace ca U0cost, trong khi biu thc th hai, b i s o j, lbin i Laplace ca U0sint.

Gi s rng hm truyn G(s) c vit di dng sau:

1 2

( ) ( )( )

( ) ( )( ) ( )n

n s n sG s

d s s p s p s p= =

+ + +L(27.3)

Hnh 27.1 H thng h s hng tuyn tnh n nh

Trong pi, i=1,2,..,n l cc im cc phn bit. Bin i Laplace ca u ra Y(s) l

0( ) ( ) ( ) ( ) UY s G s U s G ss j

= =

(27.4)

Phng trnh 27.4 tng ng vi1


2/17

S tay C in t

1

1

( ) n

n

kkY s

s p s p s j

= + + +

+ + L (27.5)

H s c xc nh bi cng thc

0 0[( ) ( )] [ ( )] ( )

s js js j Y s U G s U G j

==

= = =

T tnh c bin i laplace ngc ca Y(s) :

1

1 0( ) ( ) , 0np tp t j tny t k e k e U G j e t

= + + + L (27.6)

h thng n nh, tt c cc im cc phi c phn thc m, do vy tt c cc s hng ip tik e , i=1,2n, tin ti khng

khi t tin ti v cc. Trng thi n nh ca u ra y(t) l

( )

0 0( ) lim ( ) ( ) ( ) j t j t

sst

Y t y t U G j e U G j e +

= = = (27.7)

Hm truyn dng hnh sin, ( )G j , c vit li di dng hm m

( ) ( ) jG j G j e =

Trong

2 2( ) {Re[ ( ]} {Im[ ( ]}G j G j G j = + (27.8a)

V

1 Im[ ( )]( ) tanRe[ ( )]

G jG j

G j

= = (27.8b)

Phng trnh (27.7) cho ta bit rng i vi mt h thng n nh, u vo dng hnh sin th p ng trng thi n nh u ra cng dng hnh sin vi cng tn s u vo. Bin u ra gp G(j) ln u vo v gc pha khc nhau mt lng = ( )G j .

V d 1

B lc thng thp bc nht trn hnh Fig. 27.2. c hm truyn t:

0 ( ) 1( )( ) 1i

V sG s

V s RCs= =

+

Hm truyn dng sin c cho bi2

Hnh 27.2 B lc thngthp bc nht

Hnh 27.3 p ng tn sca G(s)=1/(0.5s+1) vi

u(t) = sin2t


3/17

Phng php p ng tn s

1

1 1( )

( ) 1 ( / ) 1G j

j RC j

= =

+ +

Trong 1=1/RC, bin v gc pha ca p ng tn s l:

1

211

1( ) ( ) tan

1 ( / )G j

= = +

Hnh 27.3 minh ha p ng tn s ca h thng khi RC=5 vi u vo u=sin2t. N cng cho thy rng p ng trng thin nh khng lin quan n iu khin u vo v bin trng thi n nh ca u ra l 1/ 2 v gc pha l -45o.

27.2 th Bode

C 3 cch thng thng biu din p ng tn s ca mt h thng, l:

1. Biu Bode hoc th logarit

2. th cc

3. th bin logarit i vi pha hoc biu Nichols

Trong phn ny, chng ta s trnh by biu Bode ca mt hm truyn dng sin, sau l th cc v th bin logarit i vi pha.

u im chnh khi s dng th logarit l kh nng th hin c tnh tn s c bc cao v thp ca hm truyn trn cngmt biu v khng b rng buc khi thm vo nhng s hng khc nhau ca mt hm truyn bc cao. Cc kiu tha strong hm truyn l:

1. H s khuch i K

2. im cc (hoc im khng) ti gc ( )n

j

3. im cc (hoc im khng) trn trc thc ( )1

1j

+

4. im cc lin hp phc (hoc im khng) ( ) ( )1

2/ 2 / 1n nj j

+ +

ng cong bin logarit v gc pha cho bn tha s trn c th d dng v c v thm vo ng thi trn th t c nhng ng cong cho mt hm truyn. Qu trnh v th logarit c th c n gin ha bng vic s dng

phng php xp x tim cn cho nhng ng cong ny v c c ng cong thc t ti nhng tn s quan trng c th.

H s khuch i K

H s khuch i logarit i vi h s khuch i K l:

0 , 020 log ,

180 , 0

if KK consta nin decibel K

if K >= =

. V vy c hai ng cong tim cn chomt tha s im cc:

3


4/17

S tay C in t

10 ,

120log

1 1120log 20 log log ,

dB

j

+ =

=

?

dc ca ng cong tim cn khi 1/ >> l -20 dB/decade cho mt tha s im cc. Hai tim cn giao nhau ti1/ = , gi l tn s gc hoc tn s gy. H s logarit thc t ti 1/ = l -3 dB. Gc pha l 1( ) tan = .

Biu Bode ca mt tha s khng ( )1j+ cngc c bng cch nh trn. Tuy nhin dc ng cong tim cn

bin khi 1/ >> l +20 dB/decade v gc pha l 1( ) tan = + . th Bode ca tha s bc nht c th hintrn hnh 27.4. S xp x tuyn tnh ng cong gc pha cng c th hin.

Hnh 27.4 Biu Bode cho biu thc bc nht ( )1

1

+j

Hnh 27.5 Biu Bode cho biu thc bc hai ( ) ( )1

2/ 2 / 1

+ + n nj j

im cc lin hp phc (hoc im khng) ( ) ( )1

2/ 2 / 1

+ + n n

j j

Bin v gc pha ca im cc lin hp phc ( ) ( )1

2/ 2 / 1n nj j

+ + l

1/ 2122 2

2

2

12

1

2 2

2 1 1 2

2 /2 1 tan1 /

n n nn

n

n n n

j j

j j

+ + = +

+ + =

4


5/17

Phng php p ng tn s

Bin ca tha s im cc lin hp phc l 1 khi n > . V vy, hai ng congtim cn cho tha s im cc lin hp phc l

12

0 ,20log 2 1

40(log log ),n

n nn n

dBj j

+ +

=

?

dc ca ng cong tim cn khi n

>> l -40 dB/decade cho tha s im cc lin hp phc. Nhng ng timcn bin giao nhau ti n = , tn s t nhin. H s khuch i thc t ti n = l ( ) 1/ 2nG j = . Biu Bode catha s im cc lin hp phc c th hin trn hnh 27.5. Nhn trn hnh 27.5 ta thy rng s khc nhau gia ng cong

bin thc t v ng xp x tim cn l mt hm ca h s suy gim (damping ratio). Tn s cng hng r c nh

ngha l tn s m c gi tr nh ca p ng tn s Mr. Khi h s suy gim tin ti khng, r tin ti n . Tn s cnghng c th c xc nh bng cch ly o hm ca bin i vi tn s v cho n bng khng. Tn s cng hng vgi tr nh ca bin c miu t bi

21 2 , 0.707r n = < (27.9a)

V

2

1

, 0.7072 1 2rM =


6/17

S tay C in t

Hnh 27.6 th Bode ca hm truyn trong v d 2

Bin chnh xc t c bng cch tnh ton bin thc t ti nhng tn s quan trng nh tn s t nhin hoc tns gc ca mi tha s. ng cong pha c th t c bng cch thm vo pha ca mi tha s. Mc d xp x tuyn tnhca c tnh pha cho im cc hoc im khng n gin ph hp cho vic phn tch lc u nhng li gia ng cong phachnh xc v xp x tuyn tnh ca im cc lin hp phc c th l ln, nh trn hnh 27.6. V vy, nu yu cu c mtng cong gc pha chnh xc th mt chng trnh my tnh nh Matlab hoc C c s dng tm ra ng cong phathc t.

27.3 th cc

th cc ca mt hm truyn dng sin ( )G j l mt th ca c bin v pha ca p ng tn s trong h ta cc khi tn s thay i t 0 ti v cng. V vy, hm truyn dng sin c th c miu t di dng sau:

[ ] [ ]( ) Re ( ) Im ( ) ( ) jG j G j j G j G j e = + =

th cc ca ( )G j l th Re[ ( )G j ] trn trc honh (trc nm ngang) i vi Im[ ( )G j ] trn trc tung (trcthng ng) trong mt phng phc khi thay i t 0 ti v cc. V vy, vi mi gi tr ca , mt th cc ca ( )G j

c nh ngha bi mt vc t c ln l ( )G j v gc pha l ( )G j = , nh phng trnh 27.8.

Chng ta c th kho st hnh dng tng qut ca th cc ty thuc vo kiu ca h thng v bc tng i ca hmtruyn. Bc tng i ca hm truyn c nh ngha l s khc nhau gia bc ca a thc mu s v t s. Xt mt hmtruyn c dng sau:

1 2

1

0 1

1

0 1

(1 )(1 )( )

( ) (1 )(1 )

( ) ( )

( ) ( )

+ +=

+ +

+=

+

L

L

L

L

a b

N

m m

n n

K j jG j

j j j

b j b j

a j b j

Trong K>0 v bc tng i 0n m . Bin v gc pha ca ( )G j khi tin ti 0 hoc v cng c th hintrong bng 27.2. Hnh dng tng qut ca th cc vi cc h thng khc nhau trong on tn s thp c th hin trnhnh 27.7. on tn s cao ca th cc vi cc bc tng i khc nhau c th hin trn hnh 27.8. T ta thy rngqu o nghim ca ( )G j song song vi trc honh hoc trc tung, ng thi bin tin ti v cng khi tin ti 0+ cho

nhng h thng ln hn khng. Nu bc tng i ln hn khng, qu o nghim ( )G j hi t v gc theo chiu kim ngh v tip xc vi trc ta . Ch rng ng cong ta cc c th rt phc tp do c tnh ng hc ca t thc v muthc vt qu di tn s. Do vy, th cc ca ( )G j trong di tn s quan tm phi c xc nh chnh xc.

Bng 27.2 ( )G j i vi cc kiu h thng v bc tng i khi 0 + v

Kiu h thng

N

0 + Bc tng i

n - m

0 00K 0 00 0/ 0b a

1 090 1 00 90 6


7/17

Phng php p ng tn s

2 0180 2 00 180

3 0270 3 00 270

Hnh 27.7 th cc vi cc kiu h thng khc nhau khi 0

Hnh 27.8 th cc vi cc bc tng i khc nhau khi

Chng ta thy rng i vi mt h kn, th cc ca mt hm truyn c ngha trong vic xc nh s n nh ca hthng. th cc ca mt s h thng n gin c th hin trn hnh 27.9.

27.4 Bin logarit i vi th pha

Mt hng khc miu t p ng tn s ca h thng bng mt th n gin l v th bin logarit ca n ivi gc pha trn di tn s quan tm. ng cong c c l mt hm ca tn s . Bin logarit i vi th pha cgi l biu Nichols.

u im ca biu Nichols l s n nh tng i ca h kn c th c xc nh rt nhanh v qu trnh b vng lpkn c tm ra d dng. Biu Nchols ca mt h thng trn hnh 27.9 c miu t trong hnh 27.10 tin so snh.Hnh 27.11 hin th 3 ng cong p ng tn s khc nhau ca mt h thng bc hai.

2

2 2( )

2n

n n

G ss s

=

+ +

7


8/17

S tay C in t

Hnh 27.9 th cc ca mt s hm truyn n gin

27.5 Xc nh hm truyn bng thc nghim

Chng ta c th thu c m hnh hm truyn t vic o p ng tn s ca mt h thng n nh. Trc tin, biu Bode ca p ng tn s c v t cc gi tr o. Sau hm truyn h c th suy ra t th bin v pha da trn squan h gia cc tha s im cc v im khng c bn.

Mt my phn tch sng l thit b o bin v pha ca p ng trng thi n nh khi tn s ca dng sng hnh sin u vo thay i. Mt my phn tch hm truyn c th c s dng o hm truyn h kn hoc h.

Chng ta s s dng mt chng trnh my tnh c kt hp cng vi mt card bin i s - tng t v tng t - s pht ra tn hiu u vo dng sin v o p ng tn s ca h thng. Xt mt b lc thng thp SallenKey bc hai nh trnhnh 27.12. Hm truyn ca b lc c cho bi

0

2 2

( )( )

( ) / 2 ( / ) 1i n n

V s KG s

V s s s = =

+ +(27.10)

Hnh 27.10 Biu Nichols ca mt s hm truyn n gin

Trong

1 2

2

1n

A B A B

R R

K R R R C C

+

= =

V

1(1 )

2 B A B A A

A B BA B

C R R R C K

C R CR R

+= +

Tn hiu thi gian thc trn Windows trong Matlab c s dng cng vi card DA v AD Advantech PCL-818L. Thi

gian ly mu l 0.001s. Bin o c v gc pha c th hin trn hnh 27.13. T th Bode, chng ta tm ra h skhuch i dc bng 1.995 v tn s t nhin 17.90n = rad/s. T phng trnh 27.9b v M r= 1.993, chng ta c 0.26 = .

Mt cch nh gi hm truyn l s dng tn hiu kch thch mnh trong di tn s quan tm v o u ra tngng. K thut nhn dng h thng c p dng tm ra bc v tham s ca hm truyn. Tn hiu kch thch ph hp l tnhiu xung, tn hiu qut hnh sin, chui ngu nhin v Hnh 27.14 th hin tn hiu qut hnh sin v u ra tng ng. Cngc nhn dng h thng trn Matlab c s dng nh gi hm truyn.

8


9/17

Phng php p ng tn s

Hnh 27.11 Ba dng p ng tn s ca 2 2 2( ) /( 2 ) = + +n n nG s s s : (a) Biu Bode, (b) th cc, (c) Biu Nichols

Hnh 27.12 B lc thng thp Sallen-Key

9


10/17


11/17

Phng php p ng tn s

27.6 Tiu chun n nh Nyquist

Tiu chun n nh Nyquist cung cp mt phng php th xc nh s n nh ca h kn t ng cong p ngtn s h h. Tiu chun ny c da trn kt qu t l thuyt v bin s phc nh l mt nguyn tc i s Cauchy.

Gi s F(s) l mt hm hu t ca s vi h s thc dng phn tch mi im trn mt phng phc (s-plane) ngoi tr

nhng im cc ca n. tF

2.5 l ng cong khp kn theo chiu kim ng h trong mt phng s v khng i qua im

cc v im khng ca F(s). ng binF

c c nh ngha l s thay th gi tr ca s trn bin s cho s trong F(s).Kt qu l c mt ng bin lin tc khp kn trong mt phng F(s).

Nguyn tc ca phng php ny c th bt u nh sau:

Biu ng binF

ca hm phc F(s) c nh ngha trn s trong mt phng s s ch c bao quanh im gc

ca mt phng F(s) nu cha im cc v im khng ca hm. S vng mF

bao quanh im gc theo chiu kim ng hl:

N Z P = (27.11)

Trong Z v P tng ng l s im khng v im cc ca F(s) c bao quanh bi ng bin s khp kn theochiu kim ng h trong mt phng s.

V d 3 minh ha phng php, xt hm truyn hu t sau:

( 3)( 4)( )

( 1)( 2)

s sF s

s s

+ +=

+ +

Trong im khng l s = -3, -4 v im cc l s = -1, -2. Cc biu ng bin ca F(s) c th hin trn hnhFig.27.15, r l biu ng bin ca ng bin trn theo chiu kim ng h vi bn knh l r trong mt phng skhng bao quanh im cc v im khng. Chng ta rt ra cc nhn xt sau y t th Fig.27.15:

1. ng bin 0.5 khng bao quanh gc ca mt phng F(s) khi ng bin trong mt phng s khng bao quanhim cc v im khng.

2. 1.99 bao quanh gc mt ln theo chiu ngc kim ng h khi ng bin bao im cc ti s = -1 theo chiu kimng h trong mt phng s, v t phng trnh 27.11, N = Z P = 0 1 = -1. Ch rng 1.99 l ng bin khpkn vi 2 vng lp v ch mt bao quanh gc nh trn hnh 27.15.

Hnh 27.15 Bn ng bin ca F(s) trong v d 3

11


12/17

S tay C in t

Hnh 27.16 H thng kn

3. 2.5 bao quanh gc 2 ln theo chiu ngc kim ng h khi ng bin cha 2 im cc ti s = -1, -2 v N = Z P = 0 2 = -2.

4. Khi bn knh ca ng bin c tng ln cha im cc ti s =-1, -2 v im khng ti s = -3 th N = Z P =1-2 = -1 v biu ng bin nh 3.5 bao quanh gc mt ln theo hng ngc chiu kim ng h.

5. Khi bn knh ca ng bin c tng xa hn bao quanh c hai im cc v im khng th N = 2 2 = 0 vbiu ng bin nh 4.5 khng bao quanh im gc.

By gi chng ta s p dng nguyn tc i s Cauchy pht trin tiu chun n nh Nyquist. Gi s rng phng trnhc tnh ca h thng vng lp h trn hnh 27.16 l

( ) 1 ( ) ( ) 0F s G s H s= + =

t L(s) = G(s)H(s), hm truyn ca vng lp. S dng nguyn tc i s, gi s rng khng c im cc hoc imkhng ca F(s) nm trn trc o trong mt phng s. Chng ta nh ngha mt qu o Nyquist, s , c to ra bi trc o vna ng trn bn knh v cc. ng bin ny hon thnh s bao gm ton b na mt phng phc bn phi nh miu ttrn hnh 20.17 (a). Biu ng bin tng ng F nh trn hnh 27.17(b). Nguyn tc i s cho php rng N tng ngvi s vng bao quanh chiu kim ng h ca im gc trong mt phng 1 + L(s) bi F . P l s im cc ca ca F(s)trong na mt phng s bn phi v v vy l s im cc ca hm truyn L(s) trong na mt phng s bn phi. Z l s im

khng ca phng trnh c tnh F(s) ca h thng kn trong na mt phng s bn phi. Do vy Z phi l 0 cho h kn nnh. Trong thc t, mt s sa i c to ra n gin ha vic vic ng dng tiu chun Nyquist. Thay v v th F trong mt phng 1+L(s), chng ta ch v th L(s), dc theo ng bin s . Mt phng ng bin sinh ra L trong mt

phng L(s) v c cng hnh dng vi F nhng b dch chuyn 1 n v v pha bn tri nh trn hnh 27.17(c). V vy N ls vng bao quanh im -1 trong mt phng L(s).

Hnh 27.17 Biu Nyquist

Hnh 27.18 Biu Nyquist v qu o nghim trong v d 4

n y, tiu chun n nh Nyquist c bt u nh sau:

iu kin cn v cho s n nh ca h kn c nh ngha bi hm truyn L(s) l

Z N P = + (27.12)

phi bng 0, trong N l s vng bao quanh im -1 trong mt phng L(s) v P l s im cc khng n nh ca hmtruyn L(s).12


13/17

Phng php p ng tn s

V d 4

Xt mt h thng vi hm truyn l

2 2 2( ) ( ) ( )

( 1)( 2)

s sKL s KG s H s K

s s s

+= =

+ +(27.13)

Chng ta hy xc nh di gi tr ca h s K h kn n nh. Do c mt im cc ti s=0, nn cn sa i qu o

Nyquist i vng qua gc ta . ng bin c v trn hnh 27.18(a), ng bao c la chn l na ng trnc bn knh tin gn ti 0 trong mt s gii hn. Chng ta thc hin cc bc sau y phc tho th Nyquist trn hnh27.18(b):

1. Xc nh ( )L j khi 0 + : L(s) thuc h thng kiu 1 v v vy

1( ) 90L j

j

= o

Tng ng vi bng 27.2

2. Xc nh ( )L j khi : L(s) c bc tng i l 1 v

1( ) 0 90L j

j

= o

Tng ng vi bng 27.2

3. T th Bode, v th cc ca ( )L j khi thay i t 0+ ti . Mc d ng cong bin ca (2 2 2s s + ) ging vi ( 2 2 2s s+ + ), pha ca 2 2 2s s + thay i t 0o ti 180o . V vy mt s phc tho biu

Bode cho thy ng cong bin thay i t v cc v 0 v gc pha thay i t 90o ti 450o . V vy c 2

im ti pha bng 180o v 360o , s c hai s giao nhau ca qu o ( )L j vi trc thc trong mtphng L(s).

4. V th cc ca ( )L j khi thay i t 0 ti tng ng vi ng cong ( )L j trong bc 3 v tngng vi trc thc trong mt phng L(s).

5. Xc nh biu ng bin ca ng vng nh quanh gc ca mt phng s hon thin th. ng bao

0lim , 90 90j

s e

= o o

Biu ng bin c xc nh bi

2

0 0 00

( ) 2 2 1 1lim ( ) lim lim lim

( 1)( 2)

j jj

j j j j

e eL e

e e e e

+= = =

+ +

Biu to c l mt na ng trn ln c bn knh tin gn ti v cc. Na ng trn ny bt u ti im ( 0 )L j

v di chuyn 180o theo hng ngc chiu kim ng h kt ni vi im ( 0 )L j+ trong mt phng L(s).

6. Tnh ton im giao nhau ca qu o ( )L j cng vi trc thc, nhng im ny quan h vi s n nh tng

i ca h thng. Gi s rng qu o ( )L j giao vi trc thc nhng tn s ti hn cr . Th

180 360 , 0( )0 360 , 0

crk KL j

k K + + >=

+ +


14/17

S tay C in t

Trong

( 3)(2 2 ) 2

3

K K K c

K

+ =

+

t c=0 v gii tm K, chng ta c h s ti hn

3 210.79, 3.79

2crK

= =

Thay th gi tr ca crK vo trong phng trnh ph thuc crK

2( 3) 2 0cr cr K s K + + =

Chng ta t c tn s ti hn

0.65, 0.792

3.10, 3.793crcr

cr

crcr

KK

KK

== = = +

Ti tn s ti hn, chng ta c phng trnh c tnh

1 ( ) 0cr cr K L j+ =

Do nhng im ca qu o ( )L j ct trc thc l

1 1 1( ) ,

0.79 3.79cr crL j

K = =

th Nyquist hon thin c th hin khng theo t l trn hnh 27.18(b). Di gi tr ca h s K cho h thng n nhc th c xc nh nh tiu chun Nyquist. i vi nhng gi tr khc nhau ca K, biu Nyquist cn v li m svng bao quanh im -1. Chng ta c th trnh vic ny bng cch m s ln bao quanh im -1/K. T tiu chun Nyquist,Z=N+P, P=0. T hnh 27.18(b), chng ta thy rng c bn trng hp bao quanh im -1/K.

1. K>0 v -1/K0.79, v N=2. Chng ta c Z=2 v h thng c hai im cc khng n nh

3. K


15/17

Phng php p ng tn s

Hnh 27.19 Bin pha v bin khuch i (gain and phase margin)

hoc1 ( ) 1 ( ) ( ) 0cr cr cr L j G j H j + = + =

iu ny ch ra rng h thng kn c mt cp im cc phc ti crs = . Do vy, h thng bin gii n nh v daong vi vi tn s cr , c qui nh l tt c cc im cc ca h thng vng lp kn th bn na tri mt phng s. Mt

cch tng qut qu o ( )L j cng gn n im -1+j0 trong th Nyquist th p ng ca h thng cng dao ng. V l

do ny m s tip xc ca qu o nghim ( )L j ti im -1 c th c s dng o bin gii n nh. Hai cch otruyn thng bin gii n nh l bin khuch i v bin pha.

Bin khuch i v bin pha c nh ngha thng cho h thng kn n nh c c trng bi pha nh nht, hmtruyn vng lp l L(s). Bin khuch i l tha s m khuch i h h ca h thng kn n nh c th c tng ln trckhi i n khng n nh. Bin pha l lng pha thm vo ti tn s giao vi khuch i c i hi to cho h thngvng lp kn n nh ti n nh bin. V vy chng ta c cc nh ngha sau y:

Bin khuch i (GM): Bin khuch i l s nghch o ca ln ( )L j ti tn s giao vi pha , pha ca

( )L j ti -180o. Bin khuch i c cho bi

1( ) 20log ( )

( )GM GM dB L j

L j

= =

Bin pha (PM): Bin pha c nh ngha l gc gia pha ca hm truyn ti tn s ct vi khuch i ( ) 1L j = vgc -180o hoc 0.7 < .

Bin pha v bin khuch i c th hin trn hnh 27.19. Bin pha v bin khuch i l bin n nh cho h thng mtu vo mt u ra(SISO). Chng khng th p dng cho h thng nhiu u vo nhiu u ra (MIMO). Thm vo na l sth hin bin n nh km hn trong mt phng c s kt hp khc nhau gia pha v khuch i, nh trn hnh 27.20.

iu ny dn n mt thc t l bin pha v bin khuch i l thc o bin n nh trong gii hn ch c s khc nhau giapha v khuch i, nhng khng c s kt hp c hai. Kt qu l, mt h thng c bin pha v bin khuch i ttnhng n li gn st vi trng thi khng n nh, nh ch ra trn hnh 27.20. to ra s thiu bin khuch i v pha,

bin n inh th ba, return difference c s dng trong l thuyt iu khin hin i. Chng ch c p dng cho h thngSISO.

Minimum return difference: The minimum return difference l gi tr nh nht ca ( ) 1L j + , khi 0 < < . Nhntrn hnh 27.20 ta thy rng the minimum return difference l khong cch nh nht t th Nyquist ti im -1. Do vy

bin pha v bin khuch i l trng hp c bit ca the minimum return difference. Bin khuch i c quan htrc tip ti trng hp khi the minimum return difference xy ra ti tn s ct vi pha, v bin pha th tng ng vitrng hp khi the minimum return difference xy ra ti tn s ct vi khuch i.

Mc d the minimum return difference l thc o tt cho bin n nh hn so vi bin pha v bin khuchi, nhng n him khi c s dng trong l thuyt iu khin c in. y l l do m khi phn tch v thit k h thng

iu khin c in thng s dng biu Bode v biu Nichols thay v th Nyquist. Bin pha v khuch i cxc nh d dng hn t biu Bode hoc biu Nichols so vi th Nyquist. Mc d vi thc t l the minimum returndifference c th c nh gi mt cch d dng t th Nyquist nhng li kh xc nh the minimum returndifference t th Bode v biu Nichols.

15


16/17

S tay C in t

By gi chng ta hy tm quan h gia bin pha v h s suy gim ca mt h thng bc hai. Xt mt h thng bc haichun

2

2 2( )

2n

n n

T ss s

=

+ +(27.14)

Chng ta gi s rng hm truyn T(s) l ca mt h thng c phn hi v c vit li nh sau:

( )( )1 ( )

G sT sG s

=+

hm truyn ca h h G(s) c cho bi:

2

( )( 2 )

n

n

G ss s

=

+(27.15)

Bin pha xy ra ti tn s ct khuch i c khi ( ) 1cG j = , hoc:

2

2 2 2 1/ 21

( 4 )n

c c n

=

+

Phng trnh ny c th c vit li nh sau:2 2 2 2 2 4( ) 4 ( ) 0c n c n + =

Gii phng trnh, vi c dng ta c:

2

4 1/ 2 2

2(4 1) 2c

n

= +

Thay cs j= vo phng trnh 27.15, bin pha ca h thng l:

1

1 4 1/ 2 2 1/ 2

1

4 1/ 2 2

180 ( )

180 90 tan 2

190 tan [(4 1) 2 ]

2

1tan 2

(4 1) 2

c

c

n

PM G j

= +

=

= +

= +

o

o o

o (27.16)

Phng trnh (27.16) ch ra quan h gia h s suy gim ca mt h thng bc hai chun vi bin pha ca n tng ngvi h thng h (27.15) trong mt h thng c phn hi. Phng trnh ny cho bit mi quan h gia p ng tn s v png thi gian. th ca i vi PM c th hin trn hnh 27.21. ng cong i vi PM c th c xp x bngng thng nt t trn 27.21. S xp x tuyn tnh c th c din t bng phng trnh

0.01PM =

16


17/17

Phng php p ng tn s

Hnh 27.21 H s suy gim i vi bin pha ca mt h thng bc hai

S xp x tuyn tnh ny c chnh xc hp l khi 0.7 < v c ngha trong quan h gia p ng tn s v s biudin tc thi ca mt h thng bc hai. Phng trnh 27.17 cng c th c s dng cho h thng bc cao nu h thngc gi s c mt cp im cc phc khng suy gim.

Ti liu tham kho

[1] Dorf, R.C., and Bishop, R.H., Modern Control Systems, 9th ed., Prentice-Hall, 2001.

[2] Ogata, K., Modern Control Engineering, 2nd ed., Prentice-Hall, 1990.

[3] Kuo, B.C., Control Systems, 7th ed., Prentice-Hall, 1995.

[4] Franklin, G.F., Powell, J.D., and Emami-Naeini, A.,Feedback Control of Dynamic Systems, 3rd ed., Addison-Wesley,1994.

[5] Phillips, C.L., and Harbor, R.D.,Feedback Control Systems, 4th ed., Prentice-Hall, 2000.

[6] Wolovich, W.A., Automatic Control Systems: Basic Analysis and Design, Harcourt Brace College Publishing, 1994.

17

Documents

sổ tay cdt Chuong 27-pp dap ung f