tin hieu va he thong le vu ha

CHNG ITN HIU

L V H

I HC QUC GIA H NITrng i hc Cng ngh

2009

L V H (VNU - ColTech) Tn hiu v H thng 2009 1 / 27

Cc Loi Tn Hiu v Tnh Cht Tn hiu l g?

i lng vt l th hin mt qu trnh thng tinv mt hin tng.C th biu din di dng hm theo thi gianlin tc hay ri rc.Biu din ton hc: hm ca mt hay nhiu binc lp

m thanh: hm ca mt bin thi gian t.Hnh nh ng: hm ca ba bin x, y, t.


Cc Loi Tn Hiu v Tnh Cht Tn hiu lin tc v tn hiu ri rc

Tn hiu lin tc v ri rc theo thi gianTn hiu theo thi gian lin tc (tn hiu lin tc):

C th thay i ti bt c thi im no.Thng c bn cht t nhin.

Tn hiu theo thi gian ri rc (tn hiu ri rc):Ch thay i ti nhng thi im nht nh.C th c to ra bng cch ly mu mt tn hiulin tc ti nhng thi im nht nh.Thng lin quan ti cc h thng nhn to.


Cc Loi Tn Hiu v Tnh Cht Tn hiu lin tc v tn hiu ri rc

Tn hiu lin tc v ri rc theo gi trTn hiu c gi tr lin tc: gi tr ca tn hiuthay i mt cch lin tc.Tn hiu c gi tr ri rc: gi tr ca tn hiu thayi khng lin tc.

Tn hiu tng t v tn hiu sTn hiu tng t: tn hiu lin tc theo thi gianv c gi tr lin tc.Tn hiu s: tn hiu ri rc theo thi gian v cgi tr c lng t ha c gi tr ri rc.


Cc Loi Tn Hiu v Tnh Cht Tn hiu tun hon v tn hiu khng tun hon

Tn hiu tun hon: tn hiu c gi tr lp li theochu k, ngha l T > 0 : f (t + T ) = f (t).

Chu k c bn ca mt tn hiu tun hon: gi trnh nht ca T tha mn iu kin ni trn.

Tn hiu khng tun hon: gi tr ca tn hiukhng c lp li mt cch c chu k.


Cc Loi Tn Hiu v Tnh Cht Nhn qu, phn nhn qu v phi nhn qu

Tn hiu nhn qu: gi tr ca tn hiu lun bngkhng trn phn m ca trc thi gian, ngha lt < 0 : f (t) = 0.Tn hiu phn nhn qu: gi tr ca tn hiu lunbng khng trn phn dng ca trc thi gian,ngha l t > 0 : f (t) = 0.Tn hiu phi nhn qu: tn hiu c cc gi trkhc khng trn c phn m v phn dng catrc thi gian.


Cc Loi Tn Hiu v Tnh Cht Tn hiu chn v tn hiu l

Tn hiu chn: th biu din tn hiu c dngi xng qua trc tung, ngha l f (t) = f (t).Tn hiu l: th biu din tn hiu c dng ixng qua tm, ngha l f (t) = f (t).Bt c tn hiu no cng u c th biu dindi dng tng hp ca mt tn hiu chn vmt tn hiu l:

f (t) = feven(t) + fodd(t) :

feven(t) =12[f (t) + f (t)]

fodd(t) =12[f (t) f (t)]


Cc Loi Tn Hiu v Tnh Cht Tn hiu xc nh v tn hiu ngu nhin

Tn hiu xc nh: gi tr ca tn hiu ti bt cthi im no u c th tnh trc c bngbiu thc ton hc hay bng gi tr.Tn hiu ngu nhin: khng th d on chnhxc gi tr ca tn hiu ti mt thi im trongtng lai.

Cc tn hiu c ngun gc t nhin thng l tnhiu ngu nhin.


Cc Loi Tn Hiu v Tnh Cht Tn hiu a knh v tn hiu a chiu

Tn hiu a knh: thng c biu din didng vector m cc thnh phn l cc tn hiun knh:

F(t) = [f1(t) f2(t) ... fN(t)]

Tn hiu a chiu: thng c biu din didng hm ca nhiu bin c lp:

f (x1, x2, ..., xN)


Cc Loi Tn Hiu v Tnh Cht Tn hiu thun v tn hiu nghch

Tn hiu thun: gi tr ca tn hiu lun bngkhng k t mt thi im tr v trc, ngha lt < t0 t0 > : f (t) = 0.


Cc Loi Tn Hiu v Tnh Cht Tn hiu c di hu hn v tn hiu c di v hn

Tn hiu c di hu hn: tt c cc gi trkhc khng ca tn hiu u nm trong mtkhong hu hn trn trc thi gian, ngoikhong gi tr ca tn hiu lun bng khng,ngha l < t1 < t2

Nng Lng v Cng Sut Ca Tn Hiu Nng lng tn hiu

Nng lng ca mt tn hiu lin tc f (t) cnh ngha nh sau:

Ef = |f (t)|2dt

Nng lng ca mt tn hiu ri rc f (n) cnh ngha nh sau:

Ef =

n=|f (n)|2


Nng Lng v Cng Sut Ca Tn Hiu Norms ca tn hiu

Lp-norm ca mt tn hiu lin tc f (t) c nhngha nh sau:

||f (t)||p =[ |f (t)|pdt

]1/pLp-norm ca mt tn hiu ri rc f (n) c nhngha nh sau:

||f (n)||p =[ n=

|f (n)|p]1/p


Nng Lng v Cng Sut Ca Tn Hiu Norms ca tn hiu

Nng lng ca mt tn hiu chnh l bnhphng ca L2-norm ca tn hiu :

Ef = ||f ||22Khi p :

||f (t)|| = ess sup |f (t)|||f (n)|| = maxn {f (n)}


Nng Lng v Cng Sut Ca Tn Hiu Tn hiu nng lng

Tn hiu c nng lng hu hn c gi l tnhiu nng lng.Tn hiu tun hon khng phi l tn hiu nnglng: nng lng ca tn hiu tun hon lunlun v hn.Tn hiu xc nh c di hu hn l tn hiunng lng.


Nng Lng v Cng Sut Ca Tn Hiu Cng sut ca tn hiu

Cng sut ca mt tn hiu l nng lng trungbnh ca tn hiu trong mt n v thi gian.Cng sut ca mt tn hiu lin tc f (t) ctnh nh sau:

Pf = limT

1T

T/2T/2|f (t)|2dt

Cng sut ca mt tn hiu ri rc f (n) ctnh nh sau:

Pf = limN

12N + 1

Ni=N

|f (n)|2


Nng Lng v Cng Sut Ca Tn Hiu Cng sut ca tn hiu

Cng sut ca mt tn hiu lin tc f (t) tunhon vi chu k T bng nng lng trung bnhca tn hiu c tnh trong mt chu k:

Pf =1T

T0|f (t)|2dt

Cng sut ca mt tn hiu ri rc f (n) tunhon vi chu k N cng bng nng lng trungbnh ca tn hiu c tnh trong mt chu k:

Pf =1N

Ni=0

|f (n)|2


Nng Lng v Cng Sut Ca Tn Hiu Tn hiu cng sut

Tn hiu c cng sut hu hn c gi l tnhiu cng sut.Mt tn hiu nu l tn hiu nng lng th khngth l tn hiu cng sut: cng sut ca tn hiunng lng lun bng khng.Mt tn hiu nu l tn hiu cng sut th khngth l tn hiu nng lng: nng lng ca tnhiu cng sut lun v hn. V d: tn hiu tunhon.


Bin i Bin Thi Gian Ca Tn Hiu Dch thi gian

Tr: dch tn hiu sang bn phi theo trc thigian, ngha l f (t) f (t T ) vi T > 0.Tin: dch tn hiu sang bn tri theo trc thigian, ngha l f (t) f (t + T ) vi T > 0.


Bin i Bin Thi Gian Ca Tn Hiu Nn/gin thi gian

Nhn bin thi gian vi mt h s t l s lmthay i b rng ca tn hiu.Nn tn hiu theo trc thi gian: f (t) f (at) via > 1.Gin tn hiu theo trc thi gian: f (t) f (at) vi0 < a < 1.


Bin i Bin Thi Gian Ca Tn Hiu o chiu thi gian

Trn th, php o chiu thi gian chnh lphp lt tn hiu qua trc tung ca th:

f (t) f (t)


Mt S Dng Tn Hiu Thng Dng Tn hiu xung n v

Tn hiu xung n v lin tc, k hiu (t), cnh ngha bi hm delta Dirac nh sau:

(t) =

0 (t 6= 0)6= 0 (t = 0) v

(t)dt = 1

Tn hiu xung n v ri rc, k hiu (n), cnh ngha nh sau:

(n) =

0 (n 6= 0)1 (n = 0)L V H (VNU - ColTech) Tn hiu v H thng 2009 22 / 27

Mt S Dng Tn Hiu Thng Dng Tn hiu nhy bc n v v tn hiu dc

Tn hiu nhy bc n v (lin tc), k hiu u(t),c nh ngha nh sau:

u(t) =

0 (t < 0)1 (t 0)Tn hiu dc (lin tc) c nh ngha nh sau:

r(t) =

0 (t < 0)t/t0 (0 t t0)1 (t t0)L V H (VNU - ColTech) Tn hiu v H thng 2009 23 / 27

Mt S Dng Tn Hiu Thng Dng Tn hiu sin

Mt tn hiu c dng hm sin gi tr thc thngc biu din nh sau:

s(t) = A cos(t + )

: A l bin , l tn s gc (rad/s) v lgc pha ca tn hiu. Chu k ca tn hiu nitrn c tnh bng cng thc T = 2pi/.Mt cch biu din khc ca tn hiu sin l biudin theo hm ca tn s f = 1/T (Hz) nh sau:

s(t) = A cos(2pift + )


Mt S Dng Tn Hiu Thng Dng Tn hiu dng hm m thc

Mt tn hiu c dng hm m gi tr thc thngc biu din nh sau:

f (t) = Aet

, A v l cc gi tr thc.Nu > 0, ta c mt hm tng; cn nu < 0,ta s c mt hm suy gim theo thi gian.


Mt S Dng Tn Hiu Thng Dng Tn hiu dng hm m phc

Mt tn hiu c dng hm m phc thng cbiu din nh sau:

f (t) = Ae(+j)t

p dng cng thc Euler cho ejt , tn hiu nitrn s biu din c di dng sau y:

f (t) = Aet [cos(t) + j sin(t)]


Mt S Dng Tn Hiu Thng Dng Tn hiu dng hm m phc

f (t) l mt hm c gi tr phc vi phn thc vphn o c tnh nh sau (nu A l gi trthc):

Re[f (t)] = Aet cos(t)Im[f (t)] = Aet sin(t)]

f (t) cn c gi l tn hiu dng sin phc vibin phc l Aet v tn s gc .Bin thc ca f (t) l |A|et v gc pha l , :

|A| =Re(A)2 + Im(A)2 v = arctan Im(A)

Re(A)


CHNG IIH THNG

L V H


2009


H Thng v Cc Thuc Tnh ca H Thng H thng l g?

Mt h thng l mt thc th hot ng khi ctn hiu u vo (kch thch) v sinh ra tn hiuu ra (p ng).Ni cch khc, mt h thng c c trng bimi quan h gia tn hiu u vo v tn hiuu ra: y(t) = T[x(t)], x(t) l tn hiu vo,y(t) l tn hiu ra, v T l php bin i ctrng cho h thng.


H Thng v Cc Thuc Tnh ca H Thng M hnh ton hc ca h thng

Mi quan h gia tn hiu ra v tn hiu vo cah thng, ni cch khc l hnh vi ca h thng,c th c biu din bng mt m hnh tonhc.M hnh ton hc cho php xc nh h thng:xc nh tn hiu ra khi bit tn hiu vo.M hnh ton hc c s dng trong vic phntch v thit k h thng.


H Thng v Cc Thuc Tnh ca H Thng Cc thuc tnh ca h thng

Tnh tuyn tnhTnh bt binTnh nhn quTnh n nh


Cc V D v H Thng H thng truyn thng tng t


Cc V D v H Thng H thng truyn thng s


Cc V D v H Thng H thng iu khin


Cc Loi H Thng v Tnh Cht H thng lin tc v h thng ri rc

Cc h thng c tn hiu vo, tn hiu ra v cctn hiu s dng trong h thng u l cc tnhiu theo thi gian lin tc c gi l cc hthng lin tc.Cc h thng c tn hiu vo v tn hiu ra lcc tn hiu theo thi gian ri rc c gi lcc h thng ri rc.


Cc Loi H Thng v Tnh Cht H thng tnh v h thng ng

Cc h thng tnh, cn c gi l h thngkhng b nh, l nhng h thng trong gi trca tn hiu ra ch ph thuc gi tr ca tn hiuvo cng thi im.Cc h thng ng, cn c gi l h thng cb nh, l nhng h thng trong gi tr catn hiu ra ph thuc c vo gi tr trong qu khca tn hiu vo.


Cc Loi H Thng v Tnh Cht H thng n bin v h thng a bin

H thng SISO (Single-input single-output): mtbin vo v mt bin ra.H thng SIMO (Single-input multiple-output):mt bin vo v nhiu bin ra.H thng MISO (Multiple-input single-output):nhiu bin vo v mt bin ra.H thng MIMO (Multiple-input multiple-output):nhiu bin vo v nhiu bin ra.


Cc Loi H Thng v Tnh Cht H thng tuyn tnh v h thng phi tuyn

Mt h thng c trng bi mt php bin i Tc gi l h thng tuyn tnh khi iu kin sauy lun c tha mn:

T[x1(t) + x2(t)] = T[x1(t)] + T[x1(t)]

Cc h thng khng tha mn iu kin tuyntnh ni trn c gi l h thng phi tuyn.


Cc Loi H Thng v Tnh Cht H thng bt bin v h thng bin i theo thi gian

Mt h thng c gi l bt bin theo thi giankhi mi quan h gia tn hiu ra v tn hiu vokhng b ph thuc vo thi im bt u, nghal:

y(t) = T[x(t)] t0 : y(t t0) = T[x(t t0)]Cc h thng khng tha mn iu kin btbin ni trn c gi l h thng bin i theothi gian.


Cc Loi H Thng v Tnh Cht H thng nhn qu v h thng phi nhn qu

Mt h thng c gi l nhn qu nu tn hiura ca h thng ch c th ph thuc cc gi trca tn hiu vo hin ti v trong qu kh chkhng th ph thuc vo cc gi tr tng laica tn hiu vo.Mt h thng phi nhn qu l h thng m tnhiu ra c th ph thuc vo c cc gi tr tnglai ca tn hiu vo.


Cc Loi H Thng v Tnh Cht H thng n nh v h thng khng n nh

Mt h thng c gi l n nh nu tn hiu ralun c gii hn hu hn khi tn hiu vo c giihn hu hn, ngha l:

|x(t)|

CHNG IIIPHN TCH H THNG TRONG

MIN THI GIAN

L V H


2009


Phng Trnh Vi Phn ca H Thng Tuyn Tnh Bt Bin Biu din h thng bng phng trnh vi phn

M hnh phng trnh vi phn l loi m hnhton hc c s dng ph bin nht biudin cc h thng trong nhiu lnh vc khcnhau.i vi cc h thng vt l, phng trnh vi phnbiu din h thng c thit lp t cc phngtrnh ca cc nh lut vt l m hot ng cah thng tun theo.Cc h thng tuyn tnh bt bin c biu dinbi cc phng trnh vi phn tuyn tnh h shng.


Phng Trnh Vi Phn ca H Thng Tuyn Tnh Bt Bin V d: phng trnh vi phn ca mch RC

CdVradt +VraR =

VvoR


Phng Trnh Vi Phn ca H Thng Tuyn Tnh Bt Bin Phng trnh vi phn tuyn tnh h s hng

Dng tng qut ca cc phng trnh vi phntuyn tnh h s hng biu din cc h thngtuyn tnh bt bin:

Ni=0

aid iy(t)dt i =

Mj=0

bjd jx(t)dt j

vi x(t) l tn hiu vo v y(t) l tn hiu ra cah thng.Gii phng trnh vi phn tuyn tnh ni trn chophp xc nh tn hiu ra y(t) theo tn hiu vox(t).


Phng Trnh Vi Phn ca H Thng Tuyn Tnh Bt Bin Gii phng trnh vi phn tuyn tnh

Nghim ca phng trnh vi phn tuyn tnh hs hng c dng nh sau:

y(t) = y0(t) + ys(t)

y0(t): p ng khi u, cn gi l p ng khikhng c kch thch, l nghim ca phng trnhthun nht

Ni=0

aid iy(t)dt i = 0 (1)

ys(t): p ng trng thi khng, l nghim cbit ca phng trnh i vi tn hiu vo x(t).


Phng Trnh Vi Phn ca H Thng Tuyn Tnh Bt Bin Xc nh p ng khi u

y0(t) l p ng ca h thng i vi iu kinca h thng ti thi im khi u (t = 0),khng xt ti tn hiu vo x(t).Phng trnh thun nht (1) c nghim dng estvi s l mt bin phc, thay vo phng trnh tac:

Ni=0

aisiest = 0

s l nghim ca phng trnh i s tuyntnh bc N sau y:

Ni=0

aisi = 0 (2)



Phng trnh (2) c gi l phng trnh ctrng ca h thng.Gi cc nghim ca phng trnh (2) l{sk |k = 1..N}, nghim tng qut ca phngtrnh thun nht (1) s c dng nh sau nu cc{sk} u l nghim n:

y0(t) =N

k=1

ckesk t

Gi tr ca cc h s {ck} c xc nh t cciu kin khi u.



Trong trng hp phng trnh (2) c nghimbi, nghim tng qut ca phng trnh thunnht (1) s c dng nh sau:

y0(t) =k

(ckesk t

pk1i=0

t i)

trong pk s ln bi ca nghim sk .


Phng Trnh Vi Phn ca H Thng Tuyn Tnh Bt Bin Xc nh p ng trng thi khng

ys(t) l p ng ca h thng i vi tn hiuvo x(t) khi cc iu kin khi u u bngkhng.ys(t) cn c gi l nghim c bit caphng trnh vi phn tuyn tnh biu din hthng. xc nh ys(t), thng thng ta gi thit ys(t)c dng tng t tn hiu vo x(t) vi mt vi hs cha bit, sau thay vo phng trnh xc nh cc h s.


Phng Trnh Vi Phn ca H Thng Tuyn Tnh Bt Bin Xc nh p ng trng thi khng

Ch khi gi thit dng ca ys(t): ys(t) phi clp vi tt c cc thnh phn ca y0(t).V d, nu x(t) = et , ta c th gp mt strng hp nh sau:

Nu et khng phi l mt thnh phn ca y0(t), tac th gi thit ys(t) c dng cet .Nu l mt nghim n ca phng trnh ctrng (2) et l mt thnh phn ca y0(t) ys(t)phi c dng ctet .Nu l mt nghim bi bc p ca phng trnh ctrng (2) et , tet ,...,tp1et l cc thnh phn cay0(t) ys(t) phi c dng ctpet .


Biu Din H Thng Bng p ng Xung nh ngha tch chp ca hai tn hiu

Tch chp ca hai tn hiu f (t) v g(t), k hiuf (t) g(t), c nh ngha nh sau:

f (t) g(t) = +

f ()g(t )d


Biu Din H Thng Bng p ng Xung Cc tnh cht ca tch chp

Tnh giao hon:

f (t) g(t) = g(t) f (t)Tnh kt hp:

[f (t) g(t)] h(t) = f (t) [g(t) h(t)]Tnh phn phi:

[f (t) + g(t)] h(t) = f (t) h(t) + g(t) h(t)


Biu Din H Thng Bng p ng Xung Cc tnh cht ca tch chp

Dch thi gian: nu x(t) = f (t) g(t), ta cx(t t0) = f (t t0) g(t) = f (t) g(t t0)

Nhn chp vi tn hiu xung n v:

f (t) (t) = f (t)Tnh nhn qu: nu f (t) v g(t) l cc tn hiunhn qu th f (t) g(t) cng l tn hiu nhnqu.


Biu Din H Thng Bng p ng Xung p ng xung ca h thng tuyn tnh bt bin

Cho mt h thng tuyn tnh bt bin c biudin bng mi quan h y(t) = T[x(t)], ta c thbin i biu din nh sau:

y(t) = T[x(t) (t)] = T[

x()(t )d]

=

x()T[(t )]d = x(t) h(t)

, h(t) = T[(t)] c gi l p ng xungca h thng tuyn tnh bt bin biu din bi T.Mt h thng tuyn tnh bt bin l xc nh khip ng xung ca h thng xc nh.


Biu Din H Thng Bng p ng Xung Phn tch p ng xung ca h thng tuyn tnh bt bin

H thng tnh (h thng khng b nh): p ngxung ch c gi tr khc khng ti t = 0.H thng nhn qu: p ng xung l tn hiunhn qu.H thng n nh: khi v ch khi iu kin sauy i vi p ng xung c tha mn

|h(t)|dt

Biu Din H Thng Bng p ng Xung p ng xung ca cc h thng ghp ni

Ghp ni tip hai h thng:

p ng xung tng hp h(t) = h1(t) h2(t)Ghp song song hai h thng:

p ng xung tng hp h(t) = h1(t) + h2(t)L V H (VNU - ColTech) Tn hiu v H thng 2009 16 / 21

M Hnh Bin Trng Thi Bin trng thi ca h thng

Trng thi ca mt h thng c m t bngmt tp hp cc bin trng thi.M hnh bin trng thi ca mt h thng tuyntnh bt bin l tp hp cc phng trnh viphn ca cc bin trng thi, cho php xc nhtrng thi trong tng lai ca h thng khi bittrng thi hin thi v tn hiu vo h thnghon ton xc nh khi trng thi khi u cah thng l xc nh.M hnh bin trng thi rt thun tin biudin h thng a bin.


M Hnh Bin Trng Thi Phng trnh trng thi

Gi {u1(t),u2(t)...} l cc tn hiu vo,{y1(t), y2(t)...} l cc bin ra, v {q1(t),q2(t)...}l cc bin trng thi ca mt h thng tuyntnh bt bin.Phng trnh trng thi ca h thng l ccphng trnh vi phn tuyn tnh bc nht:dqi(t)dt =

j

aijqj(t) +k

bikuk(t) (i = 1,2, ...)

Cc tn hiu ra c xc nh t bin trng thiv cc tn hiu vo nh sau:

yi(t) =j

cijqj(t) +k

dikuk(t) (i = 1,2, ...)


M Hnh Bin Trng Thi Phng trnh trng thi

M hnh trng thi ca mt h thng tuyn tnhbt bin thng c biu din di dng matrn nh sau:

dq(t)dt = Aq(t) + Bu(t)

y(t) = Cq(t) + Du(t)

, u(t), y(t) v q(t) l cc vector ct vi ccphn t ln lt l cc tn hiu vo, tn hiu rav cc bin trng thi ca h thng; A, B, C vD l cc ma trn h s.


M Hnh Bin Trng Thi Thit lp phng trnh trng thi

Thit lp cc phng trnh trng thi t phngtrnh vi phn biu din h thng tuyn tnh btbin sau y:

Ni=0

aid iy(t)dt i =

Mj=0

bjd jx(t)dt j

t uj(t) = d jx(t)/dt j (j = 0..M) l cc tn hiuvo ca h thng v vit li phng trnh trndi dng:

Ni=0

aid iy(t)dt i =

Mj=0

bjuj(t)


M Hnh Bin Trng Thi Thit lp phng trnh trng thi

Chn cc bin trng thi nh sau:

q1(t) = y(t),q2(t) =dy(t)dt , ...,qN(t) =

dN1y(t)dtN1

Cc phng trnh trng thi:

dq1(t)dt = q2(t),

dq2(t)dt = q3(t), ...

dqN1(t)dt = qN(t)

dqN(t)dt =

1aN

N10

aiqi+1(t) +Mj=0

bjuj(t)


CHNG IVBIU DIN TN HIU BNG

CHUI FOURIER

L V H


2009


Tn Hiu Dng Sin v H Thng Tuyn Tnh Bt Bin p ng ca h thng tuyn tnh bt bin vi tn hiu dng sin

Xem xt mt h thng tuyn tnh bt bin c png xung h(t) v tn hiu vo x(t) = ejt . png ca h thng c tnh nh sau:

y(t) = h(t) x(t) =

h()ej(t)d

= ejt

h()ejd = H()ejt

, H() l p ng tn s:

H() =

h()ejd

c trng cho p ng ca h thng vi tn s ca tn hiu vo dng sin.



Tn hiu ra c cng tn s vi tn s ca tnhiu vo dng sin.S thay i v bin v pha ca tn hiu ra sovi tn hiu vo c c trng bi p ng tns H() vi hai thnh phn sau y:

|H()| =Re[H()]2 + Im[H()]2

c gi l p ng bin , v

() = arctanIm[H()]Re[H()]

c gi l p ng pha ca h thng.L V H (VNU - ColTech) Tn hiu v H thng 2009 3 / 13


Khi , ta c th biu din tn hiu ra di dngsau y:

y(t) = |H()|ej()ejt = |H()|ej[t+()]

ngha l, so vi tn hiu vo th tn hiu ra cbin ln gp |H()| ln v lch pha i mtgc l ().


Biu Din Chui Fourier ca Tn Hiu Lin Tc Tun Hon Biu din chui Fourier ca tn hiu tun hon

Mt tn hiu x(t) tun hon vi chu k T c thbiu din c mt cch chnh xc bi chuiFourier di y:

x(t) =

k=ckejk0t

, 0 = 2pi/T l tn s c bn ca tn hiux(t).Ni cch khc, mi tn hiu tun hon u cth biu din nh mt t hp tuyn tnh ca cctn hiu dng sin phc c tn s l mt snguyn ln tn s c bn.


Biu Din Chui Fourier ca Tn Hiu Lin Tc Tun Hon iu kin hi t

iu kin sai s bnh phng trung bnh giax(t) v biu din chui Fourier ca x(t) bngkhng l x(t) phi l tn hiu cng sut, ngha l:

1T

T0|x(t)|2dt

Biu Din Chui Fourier ca Tn Hiu Lin Tc Tun Hon Biu din p ng ca h thng tuyn tnh bt bin

p ng ca mt h thng tuyn tnh bt binc p ng tn s l H() vi mi thnh phnejk0t l H(k0)ejk0t p ng ca h thng vi tn hiu vo x(t) s biu din c nh sau:

y(t) =

k=ckH(k0)ejk0t


Biu Din Chui Fourier ca Tn Hiu Lin Tc Tun Hon Tnh trc giao ca cc thnh phn {ejk0 t}

Hai tn hiu f (t) v g(t) tun hon vi cng chuk T c gi l trc giao nu iu kin sau yc tha mn: T

0f (t)g(t)dt = 0

Hai tn hiu ejk0t v ejl0t vi tn s c bn0 = 2pi/T trc giao nu k 6= l :

k 6= l : T0ejk0tejl0tdt = 0


Biu Din Chui Fourier ca Tn Hiu Lin Tc Tun Hon Tnh cc h s ca chui Fourier

Cc h s ca chui Fourier ca tn hiu tunhon x(t) c tnh bng cch s dng tnhcht trc giao ca cc tn hiu thnh phn{ejk0t} nh sau: T

0x(t)ejk0tdt =

T0

l=

clejl0tejk0tdt

=

l=cl T0ejl0tejk0tdt

= ckT

ck = 1T T0x(t)ejk0tdt


Biu Din Chui Fourier ca Tn Hiu Lin Tc Tun Hon Cc tnh cht ca biu din chui Fourier

Tnh tuyn tnh:

x(t) =

k=ckejk0t v z(t) =

k=

dkejk0t

x(t) + z(t) =

k=(ck + dk)ejk0t

Dch thi gian:

x(t) =

k=ckejk0t

x(t t0) =

k=

(ckejk0t0

)ejk0t



o hm:

x(t) =

k=ckejk0t dx(t)dt =

k=

(jk0ck)ejk0t

Tch phn:

x(t) =

k=ckejk0t

t

x()d =

k=

ckjk0

ejk0t



Cng thc Parseval:

1T

T0|x(t)|2dt =

k=

|ck |2

Gi tr |ck |2 c th coi nh i din cho cngsut ca tn hiu thnh phn ejk0t trong tn hiux(t) hm biu din gi tr |ck |2 theo tn sk = k0 (k Z ) cho ta bit phn b cng sutca tn hiu x(t) v c gi l ph mt cngsut ca x(t).Ch : ph mt cng sut ca tn hiu tunhon l mt hm theo tn s ri rc.



Tnh i xng: vi tn hiu tun hon x(t) cbiu din chui Fourier

x(t) =

k=ckejk0t

ph mt cng sut ca x(t) l mt hm chn,ngha l: k : |ck |2 = |ck |2. Ngoi ra:

Nu x(t) l tn hiu thc: k : ck = ck .Nu x(t) l tn hiu thc v chn: k : ck = ck .Nu x(t) l tn hiu thc v l: k : ck = ck .


CHNG VBIN I FOURIER CA TN

HIU

L V H


2009


Bin i Fourier ca Tn Hiu Khng Tun Hon M rng biu din chui Fourier

Xem xt mt tn hiu lin tc khng tun honx(t), ta c th coi x(t) nh mt tn hiu tunhon c chu k T (hay 0 0), khi x(t)c th biu din c bng chui Fourier nhsau:

x(t) = lim00

+k=

ckejk0t

:

ck = lim00

1T

+T/2T/2

x(t)ejk0tdt

= lim00

02pi

+pi/0pi/0

x(t)ejk0tdt


Bin i Fourier ca Tn Hiu Khng Tun Hon M rng biu din chui Fourier

V 0 0 nn = k0 l mt bin lin tc, ta cth vit li cc biu thc trang trc nh sau:

x(t) = lim00

10

+

c()ejtd

= lim00

+

c()0

ejtd

, c() l mt hm theo tn s lin tc vc xc nh nh sau:

c() = lim00

02pi

+pi/0pi/0

x(t)ejtdt


Bin i Fourier ca Tn Hiu Khng Tun Hon Bin i Fourier

t X () = 2pic()/0, chng ta c c cngthc ca bin i Fourier ca tn hiu x(t):

X () = F [x(t)] = +

x(t)ejtdt

v cng thc ca bin i Fourier nghch:

x(t) = F1[X ()] = 12pi

+

X ()ejtd



Cch biu din khc ca bin i Fourier ca tnhiu x(t), vi bin tn s f thay cho tn s gc:

X (f ) = +

x(t)ej2piftdt

v cng thc ca bin i Fourier nghch tngng:

x(t) = +

X (f )ej2piftdf



Hm X () c gi l ph (Fourier) ca tn hiux(t) theo tn s.Hm biu din|X ()| =

Re[X ()]2 + Im[X ()]2 c gi l

ph bin ca tn hiu x(t) theo tn s.Hm () = arctan[Im[X ()]/Re[X ()]] cgi l ph pha ca tn hiu x(t) theo tn s.


Bin i Fourier ca Tn Hiu Khng Tun Hon iu kin hi t

iu kin cc bin i Fourier thun vnghch ca tn hiu x(t) tn ti l x(t) phi l tnhiu nng lng, ngha l: +

|x(t)|2dt

Bin i Fourier ca Tn Hiu Khng Tun Hon Cc tnh cht ca bin i Fourier

Tnh tuyn tnh:

F [x1(t) + x2(t)] = X1() + X2()Dch thi gian:

F [x(t t0)] = X ()ejt0

Dch tn s:

F [x(t)ejt ] = X ( )



Co gin trc thi gian:

F [x(at)] = 1|a|X(a

)o hm:

F[dx(t)dt

]= jX ()

Tch phn:

F[ t

x()d]=X ()j



Bin i Fourier ca tch chp:

F [f (t) g(t)] = F ()G()Bin i Fourier ca tch thng (iu ch):

F [f (t)g(t)] = 12pi

F () G()



Cng thc Parseval: +|x(t)|2dt = 1

2pi

+|X ()|2d

Gi tr |X ()|2 c th coi nh i din cho nnglng ca tn hiu thnh phn ejt trong tn hiux(t) hm biu din |X ()|2 theo tn s chota bit phn b nng lng ca tn hiu x(t) vc gi l ph mt nng lng ca x(t).Ch : ph mt nng lng ca tn hiukhng tun hon l mt hm theo tn s lintc.



Tnh i xng:Ph mt nng lng ca x(t) l mt hm chn,ngha l: : |X ()|2 = |X ()|2.Nu x(t) l tn hiu thc: : X () = X ().Nu x(t) l tn hiu thc v chn: X () cng l hmchn, ngha l : X () = X ().Nu x(t) l tn hiu thc v l: X () cng l hm l,ngha l : X () = X ().


CHNG VIBIN I LAPLACE V P DNGTRONG PHN TCH H THNG

L V H


2009


Bin i Laplace ca Tn Hiu Bin i Laplace

Bin i Laplace ca mt tn hiu x(t) c nhngha nh sau:

X (s) = +

x(t)estdt

vi s l mt bin phc: s = + j.Bin i Laplace nghch:

x(t) = 1j2pi

+jj

X (s)estds


Bin i Laplace ca Tn Hiu Min hi t ca bin i Laplace

Min hi t (ROC) ca bin i Laplace l mtvng trong mt phng s sao cho vi cc gi trca s trong min ny th bin i Laplace hi t.V d:

Min hi t ca bin i Laplace ca tn hiu u(t) lna bn phi trc j ca mt phng s.Min hi t ca bin i Laplace ca tn hiux(t) = u(t) l na bn tri trc j ca mt phngs.

Hai tn hiu khc nhau c th c bin iLaplace ging nhau, nhng khi min hi tca chng phi khc nhau.



Min hi t ca bin i Laplace ch ph thucvo phn thc ca bin s.Min hi t ca bin i Laplace phi khngcha cc tr cc.Nu mt tn hiu c di hu hn v tn ti tnht mt gi tr ca s bin i Laplace catn hiu hi t th min hi t ca bin iLaplace khi l ton b mt phng s.



Nu mt tn hiu thun c min hi t ca bini Laplace cha ng = 0 th min hi t phi cha ton b phn bn phi 0 trongmt phng s.Nu mt tn hiu nghch c min hi t ca bini Laplace cha ng = 0 th min hi t phi cha ton b phn bn tri 0 trong mtphng s.


Bin i Laplace ca Tn Hiu Cc tnh cht ca bin i Laplace

Tnh tuyn tnh:

L[x1(t) + x2(t)] = L[x1(t)] + L[x2(t)]vi min hi t cha ROC[X1(s)]

ROC[X2(s)].

Dch thi gian:

L[x(t t0)] = est0X (s)vi min hi t l ROC[X (s)].Dch trong min s:

L[es0tx(t)] = X (s s0)vi min hi t l ROC[X (s)] dch i mt khongbng s0.



Co gin trc thi gian:

L[x(t)] = 1|a|X(sa

)vi min hi t l ROC[X (s)] b co gin vi hs .o hm:

L[dx(t)dt

]= sX (s)

vi min hi t cha ROC[X (s)].



Tch phn:

L[ t

x()d]=1sX (s)

vi min hi t cha ROC[X (s)]{ > 0}.

Bin i Laplace ca tch chp:

L[x1(t) x2(t)] = X1(s)X(s)

vi min hi t cha ROC[X1(s)]ROC[X2(s)].



nh l v gi tr khi u: nu x(t) l mt tnhiu nhn qu v lin tc ti t = 0, ta c

x(0) = lims sX (s)

nh l v gi tr cui: nu x(t) l mt tn hiunhn qu v lin tc ti t = 0, ta c

limt

x(t) = lims0

sX (s)


Bin i Laplace ca Tn Hiu Tnh bin i Laplace nghch

Phng php khai trin phn thc ti gin

Khng gim tng qut, gi s X (s) c biudin di dng phn thc N(s)/D(s), N(s)v D(s) l cc a thc vi bc ca N(s) bcca D(s).Gi s {spk} l cc tr cc ca X (s): {spk} l ccnghim ca phng trnh D(s) = 0.



Phng php khai trin phn thc ti gin (tip)

Nu tt c {spk} u l cc tr cc n, X (s)khai trin c thnh tng ca cc phn thc dng ti gin:

X (s) =k

Aks spk

, cc h s {Ak} c tnh nh sau:Ak = (s spk)X (s)|s=spk



Phng php khai trin phn thc ti gin (tip)

Trng hp tng qut (cc bi): t mk l bcbi ca tr cc spk , X (s) s c khai trin nhsau

X (s) =k

mkm=1

Akm(s spk)s

, cc h s {Akm} c tnh nh sau:

Akm =1

(mk m)!dmkm(s spk)mkX (s)

dsmkm

s=spk



Bin i Fourier nghch ca cc phn thc ti gin

L1[

1s

]=

etu(t) ( > )

etu(t) ( < )

L1[

1(s )n

]=

tn1

(n1)!etu(t) ( > )

tn1(n1)!etu(t) ( < )


Hm Chuyn (Truyn) ca H thng Tuyn Tnh Bt Bin nh ngha hm chuyn

Xem xt mt h thng tuyn tnh bt bin c png xung h(t), ngha l:

y(t) = h(t) x(t)Ly bin i Laplace ca c hai v ca phngtrnh trn v p dng tnh cht bin i Laplaceca tch chp:

Y (s) = H(s)X (s) H(s) = Y (s)X (s)

H(s) c gi l hm chuyn ca h thng.



Mt h thng tuyn tnh bt bin biu din cbng mt phng trnh vi phn tuyn tnh h shng vi dng tng qut nh sau:

Ni=0

aid iy(t)dt i =

Mj=0

bjd jx(t)dt j

Ly bin i Laplace ca c hai v ca phngtrnh trn, ta thu c:

Ni=0

aisiY (s) =Mj=0

bjsjX (s)



Hm chuyn ca h thng khi c xc nhnh sau:

H(s) = Y (s)X (s) =M

j=0 bjsjNi=0 aisi

Hm chuyn cho php xc nh h thng, datrn vic gii phng trnh vi phn tuyn tnhbng bin i Laplace v bin i Laplacenghch:

y(t) = L1[H(s)X (s)]


Hm Chuyn (Truyn) ca H thng Tuyn Tnh Bt Bin Hm chuyn ca cc h thng ghp ni

Ghp ni tip hai h thng:

Hm chuyn tng hp H(s) = H1(s) H2(s)Ghp song song hai h thng:

Hm chuyn tng hp H(s) = H1(s) + H2(s)L V H (VNU - ColTech) Tn hiu v H thng 2009 17 / 21


H thng vi phn hi m:

Hm chuyn tng hpH(s) = H1(s)/[1+ H1(s)H2(s)]



H thng vi phn hi dng:

Hm chuyn tng hpH(s) = H1(s)/[1 H1(s)H2(s)]


Bin i Laplace Mt Pha nh ngha bin i Laplace mt pha

Bin i Laplace mt pha cho tn hiu x(t)c nh ngha nh sau:

X 1(s) = L1[x(t)] = 0

x(t)estdt

Nu x(t) l tn hiu nhn qu: bin i Laplacemt pha v hai pha ca x(t) l nh nhau.


Bin i Laplace Mt Pha Cc tnh cht ca bin i Laplace mt pha

Phn ln cc tnh cht ca bin i Laplace mtpha ging vi bin i hai pha. Khc bit cbn l tnh cht ca o hm:

L[dx(t)dt

]= sX (s) X (0)

L[d2x(t)dt2

]= s2X (s) sX (0) dX (s)ds

s=0

p dng: gii phng trnh vi phn tuyn tnh ciu kin khi u p dng cho h thngtuyn tnh bt bin nhn qu.


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Bin i Bin Thi Gian Cua Tn HiuDich thi gianNn/gin thi gianao chiu thi gian

Mt S Dang Tn Hiu Thng DngTn hiu xung n viTn hiu nhay bc n vi v tn hiu dcTn hiu sinTn hiu dang hm mu thcTn hiu dang hm mu phc

slides2_4929.pdfH Thng v Cc Thuc Tnh cua H ThngH thng l g?M hnh ton hoc cua h thngCc thuc tnh cua h thng

Cc V Du v H ThngH thng truyn thng tng tH thng truyn thng sH thng iu khin

Cc Loai H Thng v Tnh ChtH thng lin tuc v h thng ri racH thng tinh v h thng ngH thng n bin v h thng a binH thng tuyn tnh v h thng phi tuynH thng bt bin v h thng bin i theo thi gianH thng nhn qua v h thng phi nhn quaH thng n inh v h thng khng n inh

chuong3_8679.pdfPhng Trnh Vi Phn cua H Thng Tuyn Tnh Bt BinBiu din h thng bng phng trnh vi phnV du: phng trnh vi phn cua mach RCPhng trnh vi phn tuyn tnh h s hngGiai phng trnh vi phn tuyn tnhXc inh p ng khi uXc inh p ng trang thi khng

Biu Din H Thng Bng p ng Xunginh nghia tch chp cua hai tn hiuCc tnh cht cua tch chpp ng xung cua h thng tuyn tnh bt binPhn tch p ng xung cua h thng tuyn tnh bt binp ng xung cua cc h thng ghp ni

M Hnh Bin Trang ThiBin trang thi cua h thngPhng trnh trang thiThit lp phng trnh trang thi

chuong4_0307.pdfTn Hiu Dang Sin v H Thng Tuyn Tnh Bt Binp ng cua h thng tuyn tnh bt bin vi tn hiu dang sin

Biu Din Chui Fourier cua Tn Hiu Lin Tuc Tun HonBiu din chui Fourier cua tn hiu tun honiu kin hi tuBiu din p ng cua h thng tuyn tnh bt binTnh trc giao cua cc thnh phn {ejk0 t}Tnh cc h s cua chui FourierCc tnh cht cua biu din chui Fourier

slide5_9048.pdfBin i Fourier cua Tn Hiu Khng Tun HonM rng biu din chui FourierBin i Fourieriu kin hi tuCc tnh cht cua bin i Fourier

chuong_6_signal_and_system_lt_8441.pdfBin i Laplace cua Tn HiuBin i LaplaceMin hi tu cua bin i LaplaceCc tnh cht cua bin i LaplaceTnh bin i Laplace nghich

Hm Chuyn (Truyn) cua H thng Tuyn Tnh Bt Bininh nghia hm chuynHm chuyn cua cc h thng ghp ni

Bin i Laplace Mt Phainh nghia bin i Laplace mt phaCc tnh cht cua bin i Laplace mt pha

tin hieu va he thong le vu ha

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