TRNG I HC QUY NHN

KHOA K THUT & CNG NGH

GIO TRNH

X L S TN HIU

Ngi bin son: Phm Hng Thnh

Quy Nhn 2009

1

MC LC

CHNG 1. BIU DIN TN HIU V H THNG RI RC

TRONG MIN THI GIAN RI RC n............................. 5

1.1. NHP MN............................................................................................. 5

1.1.1. nh ngha tn hiu .......................................................................... 5

1.1.2. Phn loi tn hiu ............................................................................ 5

1.1.3. H thng x l tn hiu ................................................................... 7

1.2. TN HIU RI RC ............................................................................... 8

1.2.1. Cc dng biu din ca dy s ....................................................8

1.2.2. Cc tn hiu ri rc c bn.................................................... 9

1.2.3. Cc php ton c bn ca dy ....................................................... 12

1.3. H THNG RI RC .......................................................................... 13

1.3.1. Khi nim....................................................................................... 13

1.3.2. Phn loi h thng ri rc.............................................................. 15

1.3.2.1. H thng khng nh (Memoryless systems) ........................... 15

1.3.2.2. H thng tuyn tnh (Linear systems) .................................... 15

1.3.2.3. H thng bt bin theo thi gian (Time-Invariant systems)... 16

1.3.2.4. H thng nhn qu (Causal systems) ..................................... 16

1.3.2.5. H thng n nh (Stable systems) ......................................... 17

1.3.3. H thng tuyn tnh bt bin theo thi gian .................................. 17

1.3.3.1. Khi nim............................................................................... 17

1.3.3.2. Tch chp................................................................................ 18

1.3.3.3. Cc tnh cht ca h thng tuyn tnh bt bin...................... 21

1.4. PHNG TRNH SAI PHN TUYN TNH H S HNG.............. 25

1.4.1. Khi nim....................................................................................... 25

1.4.2. Nghim ca PTSP-TT-HSH ........................................................... 25

1.5. H THNG RI RC QUY (RECURSIVE) V KHNG

QUY (NONRECURSIVE) ............................................................ 31

1.5.1. H thng khng quy FIR........................................................... 31

1.5.2. H thng quy IIR ...................................................................... 31

1.5.3. Thc hin h FIR v IIR............................................................... 34

1.6. HM TNG QUAN V HM T TNG QUAN .......................... 35

2

1.6.1. Hm tng quan ............................................................................ 35

1.6.2. Hm t tng quan........................................................................ 37

Chng 2. BIU DIN TN HIU V H THNG RI RC

TRONG MIN Z .................................................. 39

2.1. BIN I Z........................................................................................... 39

2.1.1 Bin i Z thun.............................................................................. 39

2.1.1.1. Bin i Z hai pha ................................................................. 39

2.1.1.2. Bin i Z mt pha................................................................ 40

2.1.2. Min hi t ca bin i Z ............................................................. 41

2.1.3. Cc tnh cht ca bin i z ........................................................... 45

2.1.4. Bin i z hu t ............................................................................ 47

2.2. BIN I Z NGC............................................................................ 49

2.2.1. nh l Cauchy ............................................................................... 49

2.2.2. Bin i z ngc............................................................................ 49

2.2.3. Cc phng php tm bin i z ngc ......................................... 50

2.2.3.1. Phng php thng d.......................................................... 50

2.2.3.2. Phng php khai trin X(z) thnh chui ly tha............... 51

2.2.3.3. Phng php phn tch X(z) thnh tng cc phn thc ti gin

53

2.3. PHN TCH H THNG RI RC TRN MIN Z........................... 60

2.3.1. Hm truyn t ca h thng TT-BB ............................................ 60

2.3.2. Hm truyn t ca h c m t bi PT SP TT HSH ........ 60

2.3.3. Gii phng trnh sai phn TT HSH s dng bin i z ............ 61

2.3.4. Phn tch h thng TT BB trn min z........................................ 64

CHNG 3. BIU DIN TN HIU V H THNG RI RC

TRONG MIN TN S LIN TC ...................................... 76

3.1. BIN I FOURIER....................................................................................... 77

3.1.1 Bin i Fourier thun.................................................................... 77

3.1.1.1. nh ngha.............................................................................. 78

3.1.1.2. S tn ti ca bin i Fourier............................................... 78

3.1.1.3. Cc dng biu din ca hm X(ej) ........................................ 79

3.1.1.4 Quan h gia bin i Fourier v bin i Z........................... 81

3

3.1.2. Bin i Fourier ngc .................................................................. 82

3.1.3. Cc tnh cht ca bin i Fourier................................................. 83

3.2. PH CA TN HIU S ...................................................................... 88

3.2.1. Cc c trng ph ca tn hiu s.................................................. 88

3.2.2. Ph ca tn hiu lin tc x(t) v tn hiu ly mu x(n.T) ................ 90

3.3. C TNH TN S V HM TRUYN T PHC CA H X

L S TUYN TNH BT BIN NHN QU .......................... 93

3.3.1 c tnh tn s v hm truyn t phc H(ej)............................... 93

3.3.2. Phn tch h x l s theo hm truyn t phc H(ej) ................. 96

3.4. CC B LC S L TNG ............................................................ 98

3.4.1. B lc thng thp l tng............................................................. 98

3.4.2. B lc thng cao l tng............................................................. 100

3.4.3. B lc di thng l tng ............................................................. 102

3.4.4. B lc di chn l tng............................................................... 104

3.4.5. B lc s thc t ........................................................................... 107

CHNG 4. BIU DIN TN HIU V H THNG RI RC

TRONG MIN TN S RI RC (MIN K) ................... 108

4.1. BIN I FOURIER RI RC CA DY TUN HON ................ 108

4.2. BIN I FOURIER RI RC CA DY KHNG TUN

HON C DI HU HN (DFT) .................................. 110

4.2.1. Bin i Fourier ri rc (DFT) .................................................... 110

4.2.2. Quan h gia DFT vi FT v ZT ................................................. 114

4.3. PHP DCH VNG, TCH CHP VNG V CC TNH CHT

CA DFT...................................................... 116

4.3.1. Php dch vng v tch chp vng ca DFT ................................. 116

4.3.1.1. Php dch vng ..................................................................... 116

4.3.1.1. Php dch vng ..................................................................... 119

4.3.2. Cc tnh cht ca DFT................................................................. 122

4.4. TNH TRC TIP DFT V IDFT...................................................... 126

4.4.1. S lng php ton khi tnh trc tip DFT v IDFT ................... 126

4.4.2. Tnh DFT v IDFT ca dy x(n)N thc, i xng, N l ................ 127

4.4.3. Tnh DFT v IDFT ca dy x(n)N thc, i xng, N chn ........... 132

4

4.4.4. Tnh DFT v IDFT ca dy x(n)N thc, phn i xng, N l ....... 134

4.4.5. Tnh DFT v IDFT ca dy x(n)N thc, phn i xng, N chn .. 137

Chng 5. TNG HP B LC S C P NG XUNG CHIU

DI HU HN............................................ 141

5.1. PHN TCH B LC S FIR PHA TUYN TNH........................... 141

5.1.1. c tnh xung h(n) ca cc b lc s FIR pha tuyn tnh ............ 141

5.1.2. c tnh tn s ca b lc s FIR pha tuyn tnh ........................ 145

5.1.2.1. c tnh tn s ca b lc FIR pha tuyn tnh loi 1 ........... 146

5.1.2.2. c tnh tn s ca b lc FIR pha tuyn tnh loi 2 ........... 149

5.1.2.3. c tnh tn s ca b lc FIR pha tuyn tnh loi 3 ........... 149

5.1.2.4. c tnh tn s ca b lc FIR pha tuyn tnh loi 4 ........... 151

5.2. CC PHNG PHP TNG HP B LC S FIR PHA TUYN

TNH............................................................. 152

5.2.1. Phng php ca s..................................................................... 152

5.2.1.1. Cc bc chnh thit k b lc s bng phng php ca s

150

5.2.1.2. Mt s hm ca s thng dng .......................................... 153

5.2.2. Phng php ly mu tn s........................................................ 160

5.2.2.1. C s ca phng php ly my tn s ............................... 160

5.2.2.2. Cc bc tng hp b lc s theo phng php ly mu

tn s ................................................... 163

CHNG 6. THIT K B LC S C P NG XUNG

C CHIU DI V HN IIR................................................. 165

6.1. C S TNG HP B LC S IIR........................................................... 165

6.2. PHNG PHP BT BIN XUNG............................................................ 166

6.3. PHNG PHP BIN I SONG TUYN............................................... 170

6.4. PHNG PHP TNG NG VI PHN............................................ 175

6.5. B LC TNG T BUTTERWORTH .......................................................175

6.6. B LC TNG T CHEBYSHEP........................................................... 176

6.7. B LC TNG T ELIP (CAUER).............................................................178

5

Chng 1

BIU DIN TN HIU V H THNG RI RC TRONG MIN THI

GIAN RI RC n

1.1. Nhp mn

1.1.1. nh ngha tn hiu

Tn hiu l mt i lng vt l cha thng tin (information). V mt ton hc, tn hiu c biu din bng mt hm ca mt hay nhiu bin c lp.

V d 1.1. - Tn hiu m thanh l dao ng c hc lan truyn trong khng kh, mang thng tin truyn n tai. Khi bin thnh tn hiu in (in p hay dng in)

th gi tr ca n l mt hm theo thi gian.

- Tn hiu hnh nh tnh hai chiu c c trng bi mt hm cng

sng ca hai bin khng gian. Khi bin thnh tn hiu in, n l hm mt bin thi gian.

thun tin, ta qui c (khng v th m lm mt tnh tng qut) tn hiu l mt hm ca mt bin c lp v bin ny l thi gian (mc d c khi khng phi

nh vy, chng hn nh s bin i ca p sut theo cao).

Gi tr ca hm tng ng vi mt gi tr ca bin c gi l bin

(amplitude) ca tn hiu. Ta thy rng, thut ng bin y khng phi l gi tr cc i m tn hiu c th t c.

1.1.2. Phn loi tn hiu

Tn hiu c phn loi da vo nhiu c s khc nhau v tng ng c cc

cch phn loi khc nhau. y, ta da vo s lin tc hay ri rc ca thi gian v bin phn loi. C 4 loi tn hiu nh sau:

- Tn hiu tng t (Analog signal): thi gian lin tc v bin cng lin tc.

- Tn hiu lng t ha (Quantified signal): thi gian lin tc v bin ri rc. y l tn hiu tng t c bin c ri rc ha.

- Tn hiu ri rc (Discrete signal): L tn hiu c biu din bi hm ca cc

bin ri rc.

+ Tn hiu ly mu: Hm ca tn hiu ri rc l lin tc (khng c lng t ho)

+ Tn hiu s: Hm ca tn hiu ri rc l ri rc. Tn hiu s l tn hiu c ri

rc c bin v bin s

Cc loi tn hiu trn c minh ha trong Hnh 1.1.

6

Trn Hnh 1.2 m t qu trnh s ha cc tn hiu tng t v tn hiu xung

thnh tn hiu s 4 bt. Khi s ha tn hiu tng t s gy ra sai s lng t (xem Hnh 1.2a), nhng khi s ha tn hiu xung th ngoi sai s lng t cn c sai s

v pha (xem Hnh 1.2b).

a. S ha tn hiu tng t. b. S ha tn hiu xung.

Hnh 1.2: Qu trnh s ha tn hiu lin tc.

t

n

nT

nT

nT

nT

nT

Bt 3

Bt 2

Bt 1

Bt 0

2

4

0

2

4

0

2

4

0 t

nT

nT

nT

nT

nT

nT

x(t)

x(nT)

x(nT)

Bt 3

Bt 2

Bt 1

Bt 0

2

4

0

2

4

0

2

4

0

x(t)

x(nT)

x(nT)

0

1

0

0 0

1

1

1

7

Nhn xt: Do tn hiu s l mt trng hp c bit ca tn hiu ri rc nn cc phng php x l tn hiu ri rc u hon ton c p dng cho x l tn

hiu s. Trong chng trnh chng ta s tm hiu cc phng php x l tn hiu ri rc.

1.1.3. H thng x l tn hiu

a) H thng tng t

b) H thng s

c) H thng x l tn hiu tng qut

Tn hiu x(t) u vo c chuyn thnh tn hiu s nh ADC, qua DSP a vo DAC ta c y(t).

Hold Quantizer DSP DAC

ADC

Sample Signal

x(t) y(t)

Digital Signal

xa(t) ya(t) HT

xd(nTs) yd(nTs) HT

8

1.2. Tn hiu ri rc

1.2.1. Cc dng biu din ca dy s

Mt tn hiu ri rc c th c biu din bng mt dy cc gi tr (thc hoc phc). Phn t th n ca dy (n l mt s nguyn) c k hiu l x(n) v mt dy

c k hiu nh sau:

x = {x(n)} vi - < n < . (1.1.a)

x(n) c gi l mu th n ca tn hiu x.

Dy s c th c biu din di cc dng hm s, bng s liu, th, hoc dy s liu. Di dng hm s, dy s x(n) ch xc nh vi i s l cc s nguyn n, dy s khng xc nh ngoi cc gi tr nguyn n ca i s. V d 1.2. Dy s x(n) c biu din bng hm s :

[ ][ ]

=.,

,)(

30,0

30,1

n

nnx

- Biu din dy s x(n) di dng bng s liu Bng 1.1.

Bng 1.1

th dy x(n)

n - ... -3 -2 -1 0 1 2 3 4 5 ... x(n) 0 0 0 0 0 1 1 1 1 0 0 0 0

- Biu din th ca dy x(n) trn Hnh 1.6, - Biu din dy x(n) di dng dy s liu : { }...,0,0,1,1,1,1,0,...)(

=nx , trong k

hiu ch s liu ng vi im gc n = 0.

Ta cng c th biu din theo kiu lit k. V d:

x = { ..., 0, 2, -1, 3, 25, -18, 1, 5, -7, 0,...}. (1.1.b)

Trong , phn t c ch bi mi tn l phn t tng ng vi n = 0, cc phn t tng ng vi n > 0 c xp ln lt v pha phi v ngc li.

Nu x = x(t) l mt tn hiu lin tc theo thi gian t v tn hiu ny c ly mu cch u nhau mt khong thi gian l Ts, bin ca mu th n l x(nTs).

Ta thy, x(n) l cch vit n gin ha ca x(nTs), ngm hiu rng ta chun ho trc thi gian theo Ts.

Ts gi l chu k ly mu (Sampling period).

Fs = 1/Ts c gi l tn s ly mu (Sampling frequency).

Ghi ch:

31 2

1

40-1

x(n)

n

9

- T y v sau, trc thi gian s c chun ha theo Ts, khi cn tr v thi gian

thc, ta thay bin n bng nTs. - Tn hiu ri rc ch c gi tr xc nh cc thi im nguyn n. Ngoi cc thi

im ra tn hiu khng c gi tr xc nh, khng c hiu chng c gi tr bng 0.

- n gin, sau ny, thay v k hiu y , ta ch cn vit x(n) v hiu y l dy x = {x(n)}.

1.2.2. Cc tn hiu ri rc c bn

a/. Tn hiu xung n v (Unit inpulse sequence)

y l mt dy c bn nht, k hiu l (n), c nh ngha nh sau:

=

=,0,0

0,1)(

n

nn

hay

{ }...,0,...,0,1,0...,0,...)(

=n .

Dy )(n c biu din bng th nh hnh 1.3(a)

b/. Dy ch nht Dy ch nht c k hiu l rectN(n) v c nh ngha nh sau:

=.,0

10,1)(

Nn

NnnrectN

c/. Tn hiu nhy bc n v (Unit step sequence)

Dy ny thng c k hiu l u(n) v c nh ngha nh sau:

0 th dy c cc gi tr dng v gim khi n tng, Hnh 1.3(d). Nu 1< < 0 th cc gi tr ca dy s ln lc i du v c ln gim khi n tng. Nu | |>1 th

ln ca dy s tng khi n tng.

e/. Tn hiu tun hon (Periodic sequence)

11

Mt tn hiu x(n) c gi l tun hon vi chu k N khi: x(n+N) = x(n), vi

mi n. Mt tn hiu tun hon c chu k N=8 c biu din bng th Hnh 1.3(e). D nhin, mt tn hiu hnh sin cng l mt tn hiu tun hon.

V d: l mt tn hiu tun hon c chu k l N=5, xem Hnh1.3(f)

f/. Dy c chiu di hu hn

Dy c xc nh vi s mu N hu hn (N im trn trc honh) gi l dy

c chiu di hu hn. N c gi l chiu di ca dy, k hiu l:

L[x(n) ] = N.

V d 1.3. L[rectN(n) ]=N.

g/. Nng lng v cng xut ca dy

Nng lng ca mt dy c nh ngha nh sau:

,)(2

=

=n

x nxE

trong )(nx l modul ca x(n).

V d 1.4. .1)(1

0

22

)( NnxEN

nnnrectN

===

=

=

Cng xut trung bnh ca dy:

.)(12

1lim

2=

+=

N

NnN

x nxNP

Nng lng ca dy x(n) trong khong NnN :

.)(2

=

=N

NnxN nxE

Vy, ,lim+

=N

xNx EE

.12

1xNx EN

P+

=

Dy nng lng: nu nng lng ca dy x(n) l hu hn th x(n) c gi l dy nng lng.

Dy cng xut: nu cng xut trung bnh ca x(n) l hu hn th x(n) c gi l dy cng xut.

12

1.2.3. Cc php ton c bn ca dy

Cho 2 dy x1 = {x1(n)} v x2 = {x2(n)} cc php ton c bn trn hai dy c nh ngha nh sau:

1/. Php nhn 2 dy: y = x1 . x2 = {x1(n).x2(n)} (1.8) 2/. Php nhn 1 dy vi 1 h s: y = a.x1 = {a.x1(n)} (1.9)

3/. Php cng 2 dy: y = x1 + x2 = {x1(n) + x2(n)} (1.10) 4/. Php dch mt dy (Shifting sequence):

- Dch phi: Gi y l dy kt qu trong php dch phi n0 mu mt dy x ta c:

y(n) = x(n-n0), vi n0 > 0 . (1.11)

- Dch tri: Gi z l dy kt qu trong php dch tri n0 mu dy x ta c:

z(n) = x(n+n0), vi n0 > 0. (1.12)

Php dch phi cn gi l php lm tr (delay). Php lm tr mt mu thng c k hiu bng ch D hoc Z-1 . Cc php dch tri v dch phi c minh ha

trong cc Hnh 1.4.

Hnh 1.4: (a) Dy x(n)

(b) Php dch phi 4 mu trn tn hiu x(n)

(c) Php dch tri 5 mu trn tn hiu x(n)

Nhn xt: Ta thy, mt tn hiu x(n) bt k c th biu din bi tn hiu xung n

v nh sau:

+

=

=n

knkxnx ).()()(

Cch biu din ny s dn n mt kt qu quan trng trong phn sau.

Ghi ch:

Cc php tnh thc hin trn cc tn hiu ri rc ch c ngha khi tn s ly

mu ca cc tn hiu ny bng nhau.

13

1.3. H thng ri rc

1.3.1. Khi nim

a. H thng thi gian ri rc (gi tt l h thng ri rc):

H thng thi gian ri rc l mt thit b (device) hay l mt thut ton

(algorithm) m n tc ng ln mt tn hiu vo (dy vo) cung cp mt tn hiu ra (dy ra) theo mt qui lut hay mt th tc (procedure) tnh ton no . nh

ngha theo ton hc, l mt php bin i hay mt ton t (operator) m n bin mt dy vo x(n) thnh dy ra y(n).

K hiu: y(n) = T{x(n)}. (1.14)

Tn hiu vo c gi l tc ng hay kch thch (excitation), tn hiu ra c

gi l p ng (response). Biu thc biu din mi quan h gia kch thch v p ng c gi l quan h vo ra ca h thng.

Quan h vo ra ca mt h thng ri rc cn c biu din nh Hnh 1.5.

V d 1.5. H thng lm tr l tng c nh ngha bi phng trnh:

y(n) = x(n nd) , vi - < n < (1.15)

nd l mt s nguyn dng khng i gi l tr ca h thng.

V d 1.6. H thng trung bnh ng (Moving average system) c nh ngha bi phng trnh:

vi M1 v M2 l cc s nguyn dng.

H thng ny tnh mu th n ca dy ra l trung bnh ca (M1 + M2 + 1) mu ca dy vo xung quanh mu th n, t mu th n-M2 n mu th n+M1 .

b. p ng xung (impulse response) ca mt h thng ri rc

p ng xung h(n) ca mt h thng ri rc l p ng ca h thng khi kch

thch l tn hiu xung n v (n), ta c:

14

Trong cc phn sau, ta s thy, trong cc iu kin xc nh p ng xung ca

mt h thng c th m t mt cch y h thng .

V d 1.7. p ng xung ca h thng trung bnh cng l

c. Biu din h thng bng s khi

c th biu din mt h thng bng s khi, ta cn nh ngha cc phn

t c bn. Mt h thng phc tp s l s lin kt ca cc phn t c bn ny.

c1/. Phn t nhn dy vi dy (signal multiplier), tng ng vi php nhn hai dy,

c s khi nh sau:

a. y(n) = x1(n) . x2(n) b. =

=M

ii nxny

1

)()(

c2/. Phn t nhn mt dy vi mt hng s (Constant multiplier), tng ng vi php nhn mt h s vi mt dy

c3/. Phn t cng (Adder), tng ng vi php cng hai dy, c s khi nh sau:

a. y(n) = x1(n) + x2(n) b. =

=M

ii nxny

1

)()(

c4/. Phn t lm tr mt mu (Unit Delay Element), tng ng vi php lm tr mt mu, c s khi nh sau:

X X y(n y(nx1(n)

x2(n)

x1(n)

x2(n)

xi(n)

xM(n)

+ + y(n y(nx1(n)

x2(n)

x1(n)

x2(n)

xi(n)

xM(n)

x(n) y(n) = a.x(n) a

x(n) y(n) = x(n - 1) D

15

Trong cc phn sau, ta s thnh lp mt h thng phc tp bng s lin kt

cc phn t c bn ny.

1.3.2. Phn loi h thng ri rc

Cc h thng ri rc c phn loi da vo cc thuc tnh ca n, c th l cc thuc tnh ca ton t biu din h thng (T).

1.3.2.1. H thng khng nh (Memoryless systems)

H thng khng nh cn c gi l h thng tnh (Static systems) l mt h

thng m p ng y(n) mi thi im n ch ph thuc vo gi tr ca tc ng x(n) cng thi im n .

Mt h thng khng tha mn nh ngha trn c gi l h thng c nh hay h thng ng (Dynamic systems).

V d 1.8. - H thng c m t bi quan h vo ra nh sau: y(n) = [x(n)]2, vi

mi gi tr ca n, l mt h thng khng nh. - H thng lm tr trong V d 1.5, ni chung l mt h thng c nh khi nd>0.

- H thng trung bnh ng trong V d 1.6 l h thng c nh, tr khi M1=M2=0.

1.3.2.2. H thng tuyn tnh (Linear systems)

Mt h thng c gi l tuyn tnh nu n tha mn nguyn l chng cht (Principle of superposition). Gi y1(n) v y2(n) ln lt l p ng ca h thng

tng ng vi cc tc ng x1(n) v x2(n), h thng l tuyn tnh nu v ch nu:

vi a, b l 2 hng s bt k v vi mi n.

Ta thy, i vi mt h thng tuyn tnh, th p ng ca mt tng cc tc

ng bng tng p ng ca h ng vi tng tc ng ring l.

Mt h thng khng tha mn nh ngha trn c gi l h thng phi tuyn

(Nonliear systems).

V d 1.9. Ta c th chng minh c h thng tch ly (accumulator) c nh

ngha bi quan h:

+

=

=n

kxny )()( (1.20)

l mt h thng tuyn tnh. H thng ny c gi l h thng tch ly v mu th n

ca p ng bng tng tch ly tt c cc gi tr ca tn hiu vo trc n thi

im th n.

16

= a.y1(n) + b.y2(n) vi a v b l cc hng s bt k.

Vy h thng ny l mt h thng tuyn tnh.

1.3.2.3. H thng bt bin theo thi gian (Time-Invariant systems)

Mt h thng l bt bin theo thi gian nu v ch nu tn hiu vo b dch nd mu th p ng cng dch nd mu, ta c:

Nu y(n) =T{x(n)} v x1(n) = x(n-nd) th y1(n) = T{x1(n)} = {x(n-nd)} = y(n - nd). (1.21)

Ta c th kim chng rng cc h thng trong cc v d trc u l h thng bt bin theo thi gian.

V d 1.10. H thng nn (compressor) c nh ngha bi quan h:

y(n) = x(M.n) (1.22)

vi - < n < v M l mt s nguyn dng.

H thng ny c gi l h thng nn bi v n loi b (M-1) mu trong M mu (n sinh ra mt dy mi bng cch ly mt mu trong M mu). Ta s chng

minh rng h thng ny khng phi l mt h thng bt bin.

Chng minh: Gi y1(n) l p ng ca tc ng x1(n), vi x1(n) = x(n nd), th

y1(n) = x1(Mn) = x(Mn nd),

nhng

y(n-nd) = x[M(n-nd)] y1(n).

Ta thy x1(n) bng x(n) c dch nd mu, nhng y1(n) khng bng vi y(n)

trong cng php dch . Vy h thng ny khng l h thng bt bin, tr khi M = 1.

1.3.2.4. H thng nhn qu (Causal systems)

Mt h thng l nhn qu nu vi mi gi tr n0 ca n, p ng ti thi im

n=n0 ch ph thuc vo cc gi tr ca kch thch cc thi im n n0. Ta thy, p ng ca h ch ph thuc vo tc ng qu kh v hin ti m khng ph

thuc vo tc ng tng lai. Ta c

17

y(n) = T{x(n)} = F{x(n), x(n-1), x(n-2),...}

vi F l mt hm no .

H thng trong v d 1 l nhn qu khi nd 0 v khng nhn qu khi nd < 0.

V d 1.11. H thng sai phn ti (Forward difference systems) c nh ngha bi quan h

y(n) = x(n+1) - x(n) . (1.23)

R rng y(n) ph thuc vo x(n+1), v vy h thng ny khng c tnh nhn

qu.

Ngc li, h thng sai phn li (Backward difference systems) c nh

ngha bi quan h: y(n) = x(n) x(n-1). (1.24)

l mt h thng nhn qu.

1.3.2.5. H thng n nh (Stable systems)

Mt h thng n nh cn c gi l h thng BIBO (Bounded-Input

Bounded-Output) nu v ch nu vi mi tn hiu vo b gii hn s cung cp dy ra gii hn.

Mt dy vo x(n) b gii hn nu tn ti mt s dng hu hn Bx sao cho:

|x(n)| Bx < +, vi mi n. (1.25)

Mt h thng n nh i hi rng, ng vi mi dy vo hu hn, tn ti mt s dng By hu hn sao cho:

|y(n)| By < + , vi mi n. (1.26)

Ghi ch: Cc thuc tnh phn loi h thng trn l cc thuc tnh ca h

thng ch khng phi l cc thuc tnh ca tn hiu vo. Cc thuc tnh ny phi tha mn vi mi tn hiu vo.

1.3.3. H thng tuyn tnh bt bin theo thi gian

(LTI: Linear Time-Invariant System)

1.3.3.1. Khi nim

H thng tuyn tnh bt bin theo thi gian l h thng tha mn ng thi hai tnh cht tuyn tnh v bt bin.

Gi T l mt h thng LTI, s dng cch biu din (1.13) v (1.14), ta c th vit:

vi k l s nguyn.

18

p dng tnh cht tuyn tnh, pt(1.27) c th c vit li:

p ng xung ca h thng l: h(n) = T{(n)}, v h thng c tnh bt bin,

nn:

h(n - k) = T{(n - k)} (1.29)

Thay pt(1.29) vo pt(1.28) ta c

T pt(1.30), ta thy mt h thng LTI hon ton c th c c t bi p

ng xung ca n v ta c th dng pt(1.30) tnh p ng ca h thng ng vi mt kch thch bt k. H thng LTI rt thun li trong cch biu din cng nh

tnh ton, y l mt h thng c nhiu ng dng quan trng trong x l tn hiu.

1.3.3.2. Tch chp

* nh ngha: Tch chp ca hai dy x1(n) v x2(n) bt k, k hiu: *, c nh

ngha bi biu thc sau:

(1.30) c vit li: y(n) = x(n)*h(n). (1.32)

Vy, p ng ca mt h thng bng tch chp tn hiu vo vi p ng xung

ca n.

Nh vy, vi mi mt gi tr ca n ta phi tnh 1 tng theo k ca tch x(k).h(n-

k) nh sau:

V d 1.12. ..

=

==k

khkxyn )1()()1(1

=

==k

khkxyn )()()0(0

=

==k

khkxyn )1()()1(1

=

==k

khkxyn )2()()2(2

19

=

==k

khkxyn )3()()3(3

..

Tp hp cc gi tr ca y(n) ta s c y.

* Phng php tnh tch chp bng th

Tch chp ca hai dy bt k c th c tnh mt cch nhanh chng vi s

tr gip ca cc chng trnh trn my vi tnh. y, phng php tnh tch chp bng th c trnh by vi mc ch minh ha. Trc tin, d dng tm dy

x2(n-k), ta c th vit li:

x2 (n-k) = x2 [-(k - n)]. (1.33)

T pt(1.33), ta thy, nu n>0, c x2(n-k) ta dch x2(-k) sang phi n mu, ngc li, nu n 0 v |a|

20

@ Vi 0 n < N-1: Hnh 1.5(b). trnh by hai dy x(k) v h(n-k), trong trng

ny, ta thy x(k).h(n-k) = ak nn

.)(0

=

=n

kany (1.36)

Ta thy, y(n) chnh l tng (n+1) s hng ca mt chui hnh hc c cng bi l a, p dng cng thc tnh tng hu hn ca chui hnh hc, l:

Hnh 1.5: Cc dy xut hin trong qu trnh tng chp. (a);(b);(c)Cc dy x(k) v h(n-k) nh l mt hm ca k vi cc gi tr khc nhau ca n (ch cc mu khc 0

mi c trnh by ); (d) Tng chp y(n) = x(n) * h(n).

@ Vi (N-1) < n: Hnh 1.5(b). trnh by hai dy x(k) v h(n-k), tng t nh trn ta c: x(k).h(n-k) = ak.

V d ny tnh tch chp trong trng hp n gin. Cc trng hp phc tp

hn, tch chp cng c th tnh bng phng php th, nhng vi iu kin l 2 dy phi c mt s hu hn cc mu khc 0.

21

Ch : Vic thc hin php chp 2 chui c chiu di hu hn: L[x1(n) ]=L1,

L[x2(n) ]=L2 th:

+ L = L [y(n) ] = L1+L2 1

+ Nu cc mu ca x nm trong khong [Mx, Nx], nu cc mu ca h nm trong khong [Mh, Nh] th cc mu ca y nm trong khong [Mx+Mh, Nx+Nh].

1.3.3.3. Cc tnh cht ca h thng tuyn tnh bt bin

V tt c cc h thng LTI u c th biu din bng tch chp, nn cc tnh cht

ca tng chp cng chnh l cc tnh cht ca h thng LTI.

Cc tnh cht ca tch chp

a) Tnh giao hon (Commutative): cho 2 dy x(n) v h(n) bt k, ta c

y(n) = x(n)*h(n) = h(n)*x(n). (1.41)

Chng minh: Thay bin m=n-k vo pt (1.33), ta c:

Hnh 1.6: Minh ha tnh cht giao hon

b) Tnh phi hp (Associative): Cho 3 dy x(n), h1 (n) v h2(n), ta c

y(n) = [x(n)*h1(n)]*h2 (n) = x(n)*[h1(n)*h2(n)]. (1.44)

vi

vi

vi

x(n) y(nh(n)

h(n) x(n) y(n

22

Hnh 1.7: Minh ha tnh cht phi hp

Tnh cht ny c th chng minh mt cch d dng bng cch da vo biu thc nh ngha ca tng chp.

H qu 1: Xt hai h thng LTI c p ng xung ln lc l h1(n) v h2(n) mc lin tip (cascade), ngha l p ng ca h thng th 1 tr thnh kch thch ca h

thng th 2 (Hnh 1.6(a)). p dng tnh cht phi hp ta c:

y(n) = x(n)*h(n) = [x(n)*h1(n)]*h2(n) = x(n)*[h1(n)*h2(n)]

hay h(n) = h1(n)*h2(n) = h2(n)*h1(n) (tnh giao hon). (1.45)

c) Tnh cht phn b vi php cng (Distributes over addition): tnh cht ny

c biu din bi biu thc sau:

y(n) = x(n)*[h1(n) + h2(n)] = x(n)*h1(n) + x(n)*h2(n). (1.46)

Hnh 1.8: Minh ha tnh cht phn b

H qu 2: Xt hai h thng LTI c p ng xung ln lt l h1(n) v h2(n) mc

song song (parallel), p dng tnh cht phn b ta c p ng xung ca h

thng tng ng l

h(n) = h1(n) + h2(n). (1.47)

Cc tnh cht khc

a./ H thng LTI n nh:

nh l: Mt h thng LTI c tnh n nh nu v ch nu

(1.48)

vi h(n) l p ng xung ca h thng.

h(n) = h1(n) + h2(n) x(n y(n

h1(n) x(n y(n

h2(n)

+

h(n) = h1(n) * h2(n) y(nx(n

h1(n) h2(n) x(n y(n

23

Chng minh:

iu kin : Xt mt tn hiu vo hu hn, ngha l

Vy |y(n)| hu hn khi iu kin pt(1.48) tha mn, hay pt(1.48) l iu kin h thng n nh.

iu kin cn: chng minh iu kin cn ta dng phng php phn chng. Trc tin ta gi s rng h thng c tnh n nh, nu ta tm c mt tn hiu vo

no tha mn iu kin hu hn v nu tng S phn k (S ) th h thng s khng n nh, mu thun vi gi thit.

Tht vy, ta xt mt dy vo c ngha nh sau:

y, h*(n) l lin hp phc ca h(n), r rng |x(n)| b gii hn bi 1, tuy nhin,

nu s , ta xt p ng ti n = 0:

Ta thy, kt qu ny mu thun vi gi thuyt ban u (h thng n nh).

Vy, s phi hu hn.

b./ H thng LTI nhn qu

nh l: Mt h thng LTI c tnh nhn qu nu v ch nu p ng xung h(n) ca n tha mn iu kin:

h(n) = 0, vi mi n < 0. (1.49)

Chng minh:

iu kin : T

=

=k

knhkxny )()()( , kt hp vi (1.49) ta c

=

=n

k

knhkxny )()()( . (1.50)

vi

vi

24

T pt(1.50), ta thy gii hn trn ca tng l n, ngha l y(n) ch ph thuc

vo x(k) vi k n, suy ra h thng khng c tnh nhn qu. V vy, iu kin cn v h thng c tnh nhn qu phi l:

h(n)=0 khi n < 0.

V d 1.14. H thng tch lu c nh ngha bi

T (1.51) ta thy h(n) ca h thng ny khng tha iu kin (1.48) nn

khng n nh v h(n) tha iu kin (1.49) nn n l mt h thng nhn qu.

Dy nhn qu: Dy x c gi l nhn qu nu

x(n) = 0 vi n

25

1.4. Phng trnh sai phn tuyn tnh h s hng

(LCCDE: Linear Constant-Coefficient Difference Equations)

1.4.1. Khi nim

Mt h thng LTI m quan h gia tc ng x(n) v p ng y(n) ca n tha

mn phng trnh sai phn truyn tnh h s hng bc N di dng:

c gi l h thng c phng trnh sai phn truyn tnh h s hng (LCCDE).

Trong , cc h s ak v br l cc thng s c trng cho h thng.

H thng LTI c LCCDE l mt lp con quan trng ca h thng LTI trong

x l tn hiu s. Ta c th so snh n vi mch R_L_C trong l thuyt mch tng t (c c trng bng phn trnh vi tch phn tuyn tnh h s hng).

V d 1.16. Xt h thng tch ly y(n) - y(n-1) = x(n). (1.56)

Phng trnh (1.56) chnh l LCCDE ca mt h thng tch ly, vi N=1, a0 =1, a1=-1, M=0 v b0 =1. Ta vit li y(n) = y(n-1) + x(n) . (1.57)

T (1.57), ta thy, vi mi gi tr ca n, phi cng thm vo x(n) mt tng c tch ly trc y(n-1). H thng tch ly c biu din bng s khi Hnh 1.9

v (1.57) l mt cch biu din quy ca h t.hng.

1.4.2. Nghim ca PTSP-TT-HSH

Phng trnh sai phn tuyn tnh h s hng l mt dng quan h vo ra m t h thng LTI. Trong phn ny, ta s tm biu thc tng minh ca p ng y(n)

bng phng php trc tip. Cn mt phng php khc tm nghim ca phng trnh ny l da trn bin i z s c trnh by trong chng sau, ta gi

l phng php gin tip.

Tng t nh phng trnh vi tch phn tuyn tnh h s hng ca h thng

lin tc theo thi gian. Trc tin, ta tm nghim ca phng trnh sai phn thun nht (homogeneous diference equation), l phng trnh (1.55) vi v phi bng

0. y chnh l p ng ca h thng vi tn hiu vo x(n) = 0. Sau , ta tm mt nghim ring (particular solution) ca (1.55) vi x(n). Cui cng, nghim tng qut

26

(total solution) ca LCCDE (1.55) l tng nghim ca phng trnh sai phn thun

nht vi nghim ring ca n. Th tc tm nghim nh sau:

a./ Bc 1 Tm nghim ca phng trnh sai phn thun nht (p ng ca h

thng khi tn hiu vo bng 0)

Phng trnh sai phn thun nht c dng:

(Bng cch chia 2 v cho a0 c dng (1.58) vi a0 = 1).

Ta bit rng, nghim ca phng trnh vi phn thng c dng hm m, v

vy, ta gi s nghim ca phng trnh sai phn thun nht c dng:

y0(n) = n. (1.59)

Ch s y0(n) c dng ch rng l nghim ca phng trnh thun nht.

Thay vo (1.58) ta thu c mt phng trnh a thc:

hay: n N (N + a1N-1 + a2

N-2 + + aN-1 + aN) = 0. (1.60)

a thc trong du ngoc n c gi l a thc c trng (characteristic polynomial) ca h thng.

Ni chung, a thc ny c N nghim, k hiu l 1, 2,, N, c gi tr thc hoc phc. Nu cc h s a1, a2,, aN c gi tr thc, thng gp trong thc t, cc

nghim phc nu c s l cc cp lin hp phc. Trong N nghim cng c th c mt s nghim kp (mutiple-order roots).

a.1/ Trng hp, tt c cc nghim l phn bit, khng c nghim kp, th nghim tng qut ca phng trnh sai phn thun nht l:

y0(n) = A1 n1 + A2

n2 + + AN

nN =

=

N

k

nkkA

1

(1.61)

y, A1 , A2 ,, A N l cc hng s tu nh. Cc hng s ny c xc

nh da vo cc iu kin u ca h thng.

a.2/ Trng hp c nghim bi, gi s a thc c trng c nghim bi bc m

ti 2 th ta c:

y0(n) = A1 n1 + (A20 + A21n + A22n

2 + +A2(m-1)nm-1) n2 + + AN

nN

V d 1.17. Xc nh p ng vi tn hiu vo x(n) = 0 ca mt h thng c m

t bi pt bc 2 nh sau:

y(n) - 3y(n-1) - 4y(n-2) = 0. (1.62)

Gii:

Ta bit nghim ca (1.62) c dng: y0n) = n, thay vo (1.62), ta thu c:

(1.58)

27

n - 3n-1 - 4n-2 = 0 hay n -2 (2 - 3 - 4) = 0

v phng trnh c tnh l: (2 - 3 - 4) = 0.

Ta c 2 nghim 1 = -1 v 2 = 4, nghim ca phng trnh thun nht c dng tng qut l:

y0(n) = A1n1 + A2

n2 = A1(-1)

n + A2(4)n. (1.63)

p ca h thng vi tn hiu vo bng 0 c th thu c bng cch tnh gi

tr cc hng s C1 v C2 da vo cc iu kin u. Cc iu kin u c cho thng l gi tr ca p ng cc thi im n=-1; n = -2;...; n = -N. y, ta c

N=2, v cc iu kin u c cho l y(- 1) v y(-2). T (1.62) ta thu c:

y(0) = 3y(-1) + 4y(-2)

y(1) = 3y(0) - 4y(-1) = 13y(-1) + 12y(-2).

Mt khc, t pt(1.63) ta c:

y(0) = A1 + A 2

y(1) = - A 1 + 4 A 2.

Suy ra A 1 + A 2 = 3y(-1) + 4y(-2)

- A 1 + 4 A 2 = 13y(-1) + 12y(-2).

Gii h 2 phng trnh trn ta c:

A 1 = (-1/5)y(-1) + (4/5)y(-2)

A 2 = (16/5)y(-1) + (16/5)y(-2).

Vy p ng ca h thng khi tn hiu vo bng 0 l

y0(n) = [(-1/5)y(-1) + (4/5)y(-2)](-1)n + [(16/5)y(-1) + (16/5)y(-2)](4)n

(1.64)

Gi s, y(-2)=0 v y(-1)=5, th A1=-1 v A2 =16. Ta c

y0(n) = (-1)n+1 + (4)n+2, vi n 0.

b./ Bc 2: Nghim ring ca phng trnh sai phn Tng t nh cch tm nghim ca phng trnh thun nht, tm nghim

ring ca phng trnh sai phn khi tn hiu vo x(n)0, ta on rng nghim ca phng trnh c mt dng no , v th vo PT-SP-TT-HSH cho tm mt

nghim ring, k hiu yp(n). Ta thy cch lm ny c v m mm!. Nu tn hiu

vo x(n) c cho bt u t thi im n 0 (ngha l x(n)=0 khi n

28

y(n) - 3y(n-1) - 4y(n-2) = x(n) + 2x(n-1). (1.67)

tn hiu vo l: x(n) = 4nu(n). Hy xc nh nghim ring ca (1.67).

Gii:

Trong V d 1.17, ta xc nh nghim ca phng trnh sai phn thun nht cho h thng ny, l (1.63), ta vit li:

y0(n) = A1(-1)n + A 2(4)

n. (1.68)

Nghim ring ca (1.63) c gi thit c dng hm m: yp(n) = K(4)nu(n) .

Tuy nhin chng ta thy dng nghim ny c cha trong nghim thun nht (1.68). V vy, nghim ring ny l tha (th vo (1.67) ta khng xc nh c K).

Ta chn mt dng nghim ring khc c lp tuyn tnh vi cc s hng cha trong nghim thun nht. Trong trng hp ny, ta x l ging nh trng hp c

nghim kp trong phng trnh c tnh. Ngha l ta phi gi thit nghim ring c dng: yp(n) = Kn(4)

nu(n). Th vo (1.67):

Kn(4)nu(n) - 3K(n-1)(4)n-1u(n-1) - 4 K(n-2)(4)n-2u(n-2) = (4)nu(n) + 2(4)n-1u(n-1). xc nh K, ta c lng phng trnh ny vi mi n 2, ngha l vi

nhng gi tr ca n sao cho hm nhy bc n v trong phng trnh trn khng b trit tiu. n gin v mt ton hc, ta chn n = 2 v tnh c K = 6/5. Vy:

yp(n) = (6/5)n(4)nu(n) . (1.69)

c./ Bc 3: Nghim tng qut ca phng trnh sai phn:

Tnh cht tuyn tnh ca LCCDE cho php ta cng nghim thun nht v nghim ring thu c nghim tng qut. Ta c nghim tng qut l:

y(n) = y0 (n) + yp (n). (1.70)

V nghim thun nht y0(n) cha mt tp cc hng s bt nh {Ai}, nn

nghim tng qut cng cha cc hng s bt nh ny, xc nh cc hng s ny,

ta phi c mt tp cc iu kin u tng ng ca h thng. Ch rng y0(n) v

yp(n) phi l c lp tuyn tnh vi nhau.

V d 1.19. Gii phng trnh sai phn )()()( 12 += nynxny , vi tc ng

)()( nunx = v iu kin ban u 01)( =y .

Gii : - Bc 1: Tm nghim y0(n) ca phng trnh thun nht : 012 )()( = nyny

Th nAny .)(0 = vo phng trnh thun nht:

2020.2 )(.. 11 === nnn AAA

Theo (1.7-13) nhn c nghim t do: )(..)( 20 nuAnyn=

- Bc 2: Tm nghim cng bc di dng )(.)(.)( nunxny BBp == . Th yp(n)

vo phng trnh sai phn cho nhn c:

29

)()()(. 1.2 nununu BB =

Phng trnh trn ng vi mi 1n , xc nh B chn n = 1 v c:

)()()(. 10.21 uuu BB = 11)2( == BBB .

Vy nghim cng bc l: )()( nuny p = .

- Bc 3: Nghim tng qut ca phng trnh sai phn cho l

)()(..)()()( 20 nunuAnynynyn

p =+=

- Bc 4: Xc nh hng s sai phn t iu kin ban u. Theo phng trnh

sai phn v iu kin ban u u bi xc nh c:

10.211200 )()()( ==+= yuy .

Do nghim tng qut c gi tr y(0) l: 10020 )()(..)( 0 == uuAy .

Vy 211 == AA . Nghim tng qut ca phng trnh sai phn:

)()(..)( 22 nununy n = , hay )(].1[)( )1(2 nuny n = +

V d 1.20. Tm phn ng y(n) ca h x l s c phng trnh sai phn

)()()()()( 122312 +=+ nxnxnynyny , vi tc ng )()( nunx = v iu kin ban

u y(-1) = y(-2) = 0. Cho bit tnh n nh ca h cho.

Gii : - Bc 1: Tm nghim y0(n) ca phng trnh thun nht :

02312 )()()( =+ nynyny .

Th nAny .)(0 = vo phng trnh thun nht :

0320.3.2 )(... 2221 =+=+ nnnn AAAA .

Gii phng trnh c trng 032 )( 2 =+ nhn c cc nghim:

11 = v 32 =

Theo (1.7-13) nghim t do l: )(].)([)( 3210 nuAAnyn+= .

- Bc 2: Tm nghim cng bc di dng )(..)(..)( nunnxnny BBp == . Th yp(n)

vo phng trnh sai phn cho nhn c :

)()()()()()()(. 1.222.311.2. +=+ nunununnunnu BBnB .

Phng trnh trn ng vi mi 2n , xc nh B chn n = 2 v c:

)()()()(. 1221.222 uuuu BB +=+ 4

3)21)22 (( =+=+ BBB .

Vy nghim cng bc l )(.)( .4

3nunny p = .

- Bc 3: Nghim tng qut ca phng trnh sai phn cho l

)(..)(..()(.)()()(4

3)3210 nunnuAnuAnynynyn

p ++=+= .

- Bc 4: Xc nh hai hng s sai phn t iu kin ban u. Theo phng trnh sai phn v iu kin ban u u bi, xc nh c

)()()()()( 12023120 +=+ uuyyy 100.210.30.20 )()( =+=+ yy

v )()()()()( 02113021 uuyyy +=+

30

111.210.31.21 )()( =+=+ yy

Theo nghim tng qut xc nh c bc 3 c h phng trnh

=++=

=++=

1114

31)311

1004

30)300

)(..)(..()(.)(

)(..)(..()(.)(

121

021

uuAuAy

uuAuAy

=+

=+

14

33

1

21

21

AA

AA.

Gii h phng trnh trn tm c: 16

131 =A v

16

32 =A .

Vy nghim tng qut ca phng trnh sai phn cho l

)(..)(..()(.)(4

3)3

16

3

16

13nunnununy n ++= .

Hay )(...()(4

3)3

16

3

16

13nunny n

++= .

Trong thnh phn dao ng t do l nghim ca phng trnh sai phn

thun nht : )(..()( )316

3

16

130 nuny

n

+= .

H s l s cho c dao ng t do y0(n) - khi n , nn theo nh l n nh 1, h khng tha mn iu kin n nh.

Cc v d trn cho thy rng, gii phng trnh sai phn tuyn tnh h s hng bng phng php tm nghim tng qut l kh phc tp, khi phng trnh sai

phn c bc N > 2 s cng phc tp hn v phi gii phng trnh bc cao. Nh vy, c hai phng php gii phng trnh sai phn tuyn tnh h s hng

c trnh by trn u phc tp, v th ngi ta s tm phng php khc gii phng trnh sai phn d dng hn, vn s c nghin cu chng

hai.

31

1.5. H thng ri rc quy (recursive) v khng quy (nonrecursive)

1.5.1. H thng khng qui (H c p ng xung c chiu di hu hn FIR)

Mt h thng m p ng y(n) ch ph thuc vo kch thch thi im hin

hnh v cc thi qu kh l mt h thng khng qui.

Ta thy mt h thng khng qui c biu din bi mt PT-SP-TT-HSH c bc

N = 0, l

(H s a0 c a vo cc h s br, bng cch chia 2 v cho a0 ).

p ng xung ca h thng l

Ta thy y l mt h thng LTI c p ng xung di hu hn (Finite duration Impulse Response system -FIR) v nhn qu.

H thng FIR (H thng vi p ng xung c chiu di hu hn) l mt h thng m p ng xung ca n tn ti mt s hu hn cc mu khc 0.

Ta thy, h thng FIR lun lun n nh nu tt c cc mu trong p ng

xung ca n c ln hu hn. V d 1.21. Tm p ng xung ca h c m t bi pt sau:

y(n) = x(n) + 4x(n-1) + 5x(n-2) x(n-3) t phng trnh ta thy: b0= 1, b1=4, b2=5, b3=-1.

Suy ra h(n)=(n) + 4(n-1) + 5(n-2) (n-3) v h thng ny lun n nh.

1.5.2. H thng qui (H c p ng xung c chiu di v hn IIR)

H thng c biu din bi phng trnh SP-TT-HSH bc N>0 c gi l h qui. p ng ca h thng ph thuc vo kch thch thi im hin ti v

qu kh v c p ng thi m qu kh.

0)()()( 01 00 0

= ==

avoiknya

arnx

a

bny

N

k

kM

r

r

hay 1)()()( 010

== ==

avoiknyarnxbnyN

kk

M

rr .

Nhn xt:

- Do ak, br l cc h s do h thng qui ph thuc vo c ak, ln br.

(1.71)

(1.72)

32

- Vi x(n)= (n) th y(n) = h(n) L p ng xung ca h qui. Ta thy rng

h(n) ca h qui c chiu di v hn. Vy h thng qui l h thng c p ng xung c chiu di v hn (Infinite duration Impulse Response system IIR)

V d 1.22. Tm p ng xung v xt s n nh ca h thng sau:

y(n) - ay(n-1) = x(n); y(n)=0 vi n0 th

x(n)=0 do vy yp(n) = 0 vi n>0 vy:

Khi x(n)= (n) th y(n)=y0(n) = h(n):

V vy ta c: h(n)=y0(n) ==

N

k

nkkA

1

trong k l cc nghim n ca phng trnh

.0

0=

=N

k

kNkka

33

Cn cc h s Ak c xc nh t cc iu kin u.

S n nh ca h IIR nhn qa

=

=

==n n

nhnhS0

)()(

= =

= =

=0 10 1 n

N

k

nkk

n

N

k

nkk AAS

Suy ra

=

= ==

=00 11 n

n

kn

N

kk

nk

N

kk AAS do

=

N

kkA

1

l hng s nn nu 1

34

Xt s n nh ca h:

. |3)u(n) -(2|)( 2n === +

=

=nn

nhS

Vy h ny khng n nh

1.5.3. Thc hin h FIR v IIR

H FIR:

i vi h thng khng qui FIR, vi phng trnh sai phn biu din h thng l:

Ta c s nh sau:

Trong thc t, i vi cc mch qui, t khi ngi ta thc hin c mt s c bc N > 2, v khi mch d mt tnh n nh do sai s. Mt khc, thit k cc

khu bc 2 c phn thun li hn. V vy, ngi ta chia h thng ra thnh nhiu mch con c bc ln nht l 2 mc lin tip hoc song song vi nhau.

H IIR

Phng trnh ca h IIR c vit li di dng cng thc truy hi

(1.73)

Hnh 1.10: H thng khng quy FIR

(1.74)

35

S khi

1.6. Hm tng quan v hm t tng quan

1.6.1. Hm tng quan

Khi x l tn hiu s, trong nhiu trng hp, cn so snh hai tn hiu s hoc hai dy s liu. so snh hai tn hiu s hoc hai dy s, ngi ta s dng hm

tng quan )(mrxy , vi bin m l khong cch gia cc mu ca hai tn hiu s hoc

hai dy s c so snh.

nh ngha: Hm tng quan )(mrxy ca dy y(n) i vi dy x(n) l dy

)(mrxy c xc nh bng biu thc

=

=n

xy mnynxmr )().()( (1.75)

hoc

=

+=n

xy nymnxmr )().()( (1.76)

y ch s di xy xc nh hng tng quan, vi x(n) l dy gc cn y(n) l dy c so snh. Bin m l khong cch gia hai dy tnh bng s mu. Cc

biu thc (1.75) v (1.76) l nh nhau v s dch chm m mu ca dy y(n) so vi

dy x(n) hon ton tng ng vi s dch nhanh m mu ca dy x(n) so vi dy

y(n).

Hnh 1.11: H thng quy IR

36

so snh dy x(n) vi dy y(n) ta dng hm tng quan )(mryx

=

=n

yx mnxnymr )().()( (1.77)

hoc:

=

+=n

yx nxmnymr )().()( . (1.78)

Nu thay m = - m vo (1.77) s nhn c (1.78), v tng t, nu thay m = -

m vo (1.76) s nhn c (1.77), do c

)()( mrmr yxxy = . (1.79)

Nh vy, )(mryx l i xng ca )(mrxy qua trc tung v chng u mang thng

tin nh nhau v s tng quan gia hai dy x(n) v y(n).

Biu thc hm tng quan )(mrxy c dng gn ging vi biu thc tch chp v

r rng c lin quan vi biu thc tch chp. Tht vy, bin i biu thc (1.76) s thy c s lin quan

=

=

==+=nn

xy mxmynmxnynymnxmr )(*)()]([).()().()(

Vy )(*)()(*)()( mymxmxmymrxy == (1.80)

Tng t )(*)()(*)()( mxmymymxmryx == . (1.81)

V th, mi thut ton v chng trnh dng tnh tch chp )(*)( nynx u c

th s dng tnh hm tng quan )(mrxy , ch cn thay cc dy vo x(n) v y(n)

bng cc dy vo x(-m) v y(m).

tm hm tng quan )(mrxy ca cc dy c di hu hn vi N nh, c th

tnh tng mu ca )(mrxy tng t nh tnh tch chp.

V d 1.24. Hy xc nh hm tng quan )(mrxy ca hai dy hu hn:

=

2,1,2,1)(nx v { }.)( 2,1,3,2,1

=ny

Gii : Dng cng thc (1.75) ln lt tnh cc gi tr ca )(mrxy :

01231221100 ..).().()().()(1

2

=+++== =n

xy nynxr .

tnh )(mrxy vi m < 0, ln lt dch tri dy y(n) so vi dy x(n)

132211322111 ...)).(()().()(1

2

=+++=+= =

n

xy nynxr

10221123122 ...).()().()(1

2

=+++=+= =

n

xy nynxr

30201221133 ...).()().()(1

2

=+++=+= =

n

xy nynxr

37

20201022144 ...).()().()(1

2

=+++=+= =n

xy nynxr

.00201020155 ...).()().()(1

2

=+++=+= =

n

xy nynxr

Tnh tip s c 0)( =mrxy vi mi m -5.

tnh )(mrxy vi m > 0 , ln lt dch phi dy y(n) so vi dy x(n)

63221120111 .).(.).()().()(1

2

=+++== =n

xy nynxr

32211020122 )(...).()().()(1

2

=+++== =n

xy nynxr

21201020133 ...).()().()(1

2

=+++== =n

xy nynxr

.00201020144 ...).()().()(1

2

=+++== =n

xy nynxr

Tnh tip s c 0)( =mrxy vi mi m 4

T cc kt qu tnh ton trn, nhn c dy tng quan )(mrxy l

=

2,3,6,0,13,1,3,2)(mrxy

V d 1.25. Hy xc nh hm tng quan )(mrxy ca hai dy

4)()( nrectnx = v )()( 2 nunyn= .

Gii : C =

=

==3

0

)()(4 )()(.)()( 22

n

mn

n

mnxy mnumnunrectmr .

C th thy ngay rng khi n[ 0 , 3 ] th )( mnu = 1 vi mi m 0 nn :

mm

n

mnxy mr

=

=

== 21521

2122 .)(

43

0

)( vi mi m 0

712121202121 ....)()( 21013

0

)1( =+++== =

n

nxy nur

312120202222 ....)()( 10123

0

)2( =+++== =

n

nxy nur

112020202323 ....)()( 01233

0

)3( =+++== =

n

nxy nur .

Tnh tip s c 0)( =mrxy vi mi m 4

1.6.2. Hm t tng quan Hm t tng quan )(mrx dng xc nh quan h ti cc thi im khc

nhau ca dy x(n).

nh ngha: Hm t tng quan )(mrx ca dy x(n) l dy c xc nh

bng biu thc sau:

38

=

==n

x nxnxmnxnxmr )(*)()().()( . (1.82)

i chiu cc biu thc (1.82) v (1.75), th hm t tng quan )(mrx l

trng hp ring ca hm tng quan )(mrxy khi y(n) = x(n), tc l khi so snh dy

x(n) vi chnh n ti hai thi im cch nhau m mu.

Hm t tng quan )(mrx t gi tr cc i ti m = 0 v )(0xr l gi tr tng

quan ca x(n) ti cng mt thi im v c :

.)().()(0

=

==n

xx Enxnxr (1.83)

Vy )(0xr chnh l nng lng ca tn hiu x(n).

V d 1.26. Hy xc nh hm t tng quan )(mrx ca dy:

).()( 42 nrectnxn=

Gii: Theo cng thc (1.82) c

=

=

==3

04

244 )()().()( 2222

n

nm

n

mnnx mnrectmnrectnrectmr

64

8522222220 6420

3

0

23

04

20 )()( =

=

=

+++=== n

n

n

nx nrectr

8

210212121221221 ).....()()( 6420

3

04

21 =

=

+++=+= n

nx nrectr

5.....()()( )0202121222222 642023

04

22 =

=

+++=+= n

nx nrectr

8.....()()( )0202021223223 642033

04

23 =

=

+++=+= n

nx nrectr .

Tnh tip s c 0)( =mrxy vi mi m -4

128

691212120221221 )( .....)()( 64201

3

04

21 =

=

+++== n

nx nrectr

256

51212020222222 ).....()()( 64202

3

04

22 =

=

+++== n

nx nrectr

512

11202020223223 ).....()()( 64203

3

04

23 =

=

+++== n

nx nrectr .

Tnh tip s c 0)( =mrxy vi mi m 4.

39

Chng 2

BIU DIN TN HIU V H THNG RI RC TRONG MIN Z

Chng 1 trnh by cch tnh p ng ca mt h thng trc tip t p ng xung ca n, bng cch tnh tng chp ca kch thch vi p ng xung. Cch

tnh tng chp trc tip da vo cng thc nh ngha nh lm tn rt nhiu thi gian v cng sc. Hn na, trong thc t s mu khc khng ca kch thch v p

ng xung l rt nhiu nn ta khng th tnh bng tay. Tuy nhin, phng php tnh tng chp bng th nh trnh by cho ta mt thut ton ca chng trnh

tnh tng chp bng my tnh. Vic gii phng trnh sai phn tuyn tnh h s hng bng phng php qui cng ch c ngha khi s dng my tnh.

K thut bin i l mt cng c hu hiu phn tch h thng LTI. Bin i Z i vi tn hiu ri rc c vai tr tng t nh bin i Laplace i vi tn

hiu lin tc, v chng c quan h ging nhau vi bin i Fourier. Tng chp ca hai dy trong min thi gian s bin thnh tch ca hai bin i Z tng ng trong

min bin phc z. Tnh cht ny s lm n gin ha vic tnh p ng ca h thng vi cc tn hiu vo khc nhau. Phng trnh sai phn tuyn tnh h s hng

cng c gii mt cch d dng hn khi dng cng c bin i Z.

Nh ta s thy trong cc chng sau, bin i Fourier gia vai tr cha kha

trong trong vic biu din v phn tch cc h thng ri rc. Tuy nhin, trong mt s trng hp cn phi s dng dng tng qut ha ca bin i Fourier, l bin

i Z.

2.1. Bin i z 2.1.1. Bin i Z thun 2.1.1.1. Bin i Z hai pha

nh ngha: Bin i Z hai pha ca dy x(n) l chui ly tha ca bin s

phc z :

=

=n

nznxzX ).()( .

(2.1)

Min xc nh ca hm X(z) l cc gi tr ca z chui (2.1) hi t.

Dy x(n) c gi l hm gc, cn X(z) c gi l hm nh Z. Bin i Z hai pha thng c gi vn tt l bin i Z. Chui (2.1) l biu thc bin i Z

thun v c k hiu nh sau:

)()]([ znxZT X= .

Hay: )()( znx XZT

( ZT l ch vit tt ca thut ng ting Anh : Z - Transform).

40

V d 2.1. Hy xc nh bin i Z hai pha ca cc dy sau:

a. )(n b. )( kn c. )( kn + d. }{)( 1,5,2,3

=nx

e. )(nu f. )( 3nu g. )( 3+nu h. )( nu

Gii: a. .).()]([ 1

=

==n

nznnZT

Chui hi t vi mi z, nn )]([ nZT xc nh vi mi z.

b. kn

n zznnZT kk

=

== ).()]([ .

Chui hi t vi mi z > 0, nn )]([ knZT xc nh vi mi z > 0.

c. kn

n zznnZT kk =+=+

=

).()]([ .

Chui hi t vi mi z < , nn )]([ knZT + xc nh vi mi z < .

d. 2112

1

..).().()( 523

=

=

++=== zzzznxznxzn

n

n

nX .

Hm X(z) xc nh trong min 0 < z < .

e. )()(

).()]([11

11

0 ====

=

=

z

z

zzznunuZT

n

n

n

n .

Dy nhn qu v hn )(nu c bin i Z bng ti z = 1.

f. )()(

).()]([1

1

133

23

0

)3(

3 =

====

=

+

=

=

zzz

zzzzznunuZT

m

m

n

n

n

n .

Ta i bin, t = mn )( 3 )( 3+= mn v khi 3=n th .0=m

Dy nhn qu v hn )( 3nu c bin i Z bng ti z = 1 v z = 0 .

g.)()(

).()]([11

334

3

0

)3(

3 =

===+=+

=

=

=

z

z

z

zzzzznunuZT

m

m

n

n

n

n .

Ta i bin, t =+ mn )( 3 )( 3= mn v khi 3=n th .0=m

Dy khng nhn qu )3( +nu c bin i Z bng ti z = 1 v z = .

h. )(

).().().()]([1

1

00

0

zzzmuznuznunuZT

m

m

m

m

n

n

n

n

=====

=

==

=

.

Ta i bin, t = mn khi =n th =m .

Dy phn nhn qu v hn )( nu c bin i Z bng ti z = 1.

2.1.1.2. Bin i Z mt pha nh ngha: Bin i Z mt pha ca dy x(n) l chui ly tha ca bin s

phc z:

=

=0

1 ).()(n

nznxzX . (2.2)

Min xc nh ca hm )(1 zX l cc gi tr ca z chui (2.2) hi t.

41

Bin i Z mt pha c ly theo tng vi n bin thin t 0 n . Chui (2.2) l biu thc ca bin i Z mt pha thun v c k hiu nh sau:

)()]([ 11 znxZT X= .

Hay: )()( 11

znx XZT

V d 2.2. Hy xc nh bin i Z mt pha ca cc dy v d 2.1 v so snh kt qu vi bin i Z hai pha tng ng.

a. )(n b. )( kn c. )( kn + d. }{)( 1,5,2,3

=nx

e. )(nu f. )( 3nu g. )( 3+nu h. )( nu .

Gii : a.

=

==0

1 1).()]([n

nznnZT .

Dy nhn qu )(n c bin i Z mt pha ging bin i Z hai pha.

b. .).()]([0

1 k

n

n zznnZT kk

=

==

Dy nhn qu )( kn c bin i Z mt pha ging bin i Z hai pha.

c. 0000

1 .).()]([ ==+=+

=

=

n

n

n

n zznnZT kk .

Dy phn nhn qu )( kn + c bin i Z mt pha lun bng 0.

d. 212

00

1 .).().()( 52

=

=

+=== zzznxznxzn

n

n

nX .

Dy khng nhn qu )(nx c bin i Z mt pha khc bin i Z hai pha.

e. 11

11

00

1 ).()]([

====

=

=

z

z

zzznunuZT

n

n

n

n .

Dy nhn qu )(nu c bin i Z mt pha ging bin i Z hai pha.

f. )1

1

133

().()]([

23

0

3

30

1

=

====

=

=

=

zzz

zzzzznunuZT

m

m

n

n

n

n .

Dy nhn qu )( 3nu c bin i Z mt pha ging bin i Z hai pha.

g. 1

3300

1 ).()]([

==+=+

=

=

z

zzznunuZT

n

n

n

n .

Dy khng nhn qu )( 3+nu c bin i Z mt pha khc hai pha.

h. 0000

1 .).()]([ ===

=

=

n

n

n

n zznunuZT .

Dy phn nhn qu v hn )( nu c bin i Z mt pha lun bng 0.

Nh vy, cc dy nhn qu c bin i Z mt pha v hai pha ging nhau, cc dy khng nhn qu c bin i Z mt pha v hai pha khc nhau, cc dy phn

nhn qu c bin i Z mt pha bng khng.

2.1.2. Min hi t ca bin i Z

42

nh ngha: Tp hp tt c cc gi tr ca bin s phc z m ti cc

chui (2.1) v (2.2) hi t c gi l min hi t ca bin i Z. Min hi t ca bin i Z c k hiu l: RC[X(z)] hoc RC.

(RC l ch vit tt ca thut ng ting Anh: Region of Convergence). C th thy ngay rng cc dy x(n) hu hn c bin i Z l chui hu hn

nn s hi t trn ton b mt phng z, tr hai im |z|= v z = 0 l phi xt c th : ])([)( NnxZTzX = c 1.

s dng tiu chun hi t Cauchy xc nh min hi t ca chui (2.3),

phi tch )(zX thnh hai chui nh sau :

)()().().()( 210

1

zzznxznxz XXXn

n

n

n +=+=

=

=

.

Trong )().().()( 001

1 xznxznxzn

n

n

nX == =

=

(2.5)

v

=

=0

2 ).()(n

nznxzX .

Theo tiu chun Cauchy, chui )(2 zX s hi t nu tha mn iu kin:

1|)(|lim.||1.)(lim1

11

43

Nu x(0) hu hn th chui )(1 zX s hi t nu tha mn iu kin

1|)(|lim.||1.)(lim11

44

)(zX khng xc nh vi mi z, nn trong trng hp dy nhn qu x(n) khng

c bin i Z.

Khi x(n) l dy phn nhn qu th bin i Z ca n c 0)(2 =zX ,

nn )()( 1 zz XX = , do min hi t ca )(zX l min hi t ca )(1 zX theo (2.9),

nn +< xRXRC zz ||:)]([ , l min nm trong vng trn tm gc ta , ng

knh +xR nh Hnh 2.1c . Bn knh hi t +xR c xc nh theo (2.8). Nu 0=+xR

th )(zX khng xc nh vi mi z, nn trong trng hp dy phn nhn qu x(n)

khng c bin i Z.

Bin i Z mt pha c dng ging vi bin i Z hai pha ca cc dy nhn

qu, do min hi t ca bin i Z mt pha l:

> xRXRC zz ||:])(1[ .

l min nm ngoi vng trn tm l gc ta , ng knh xR . Bn

knh hi t xR c xc nh theo (2.6).

V d 2.3. Hy xc nh )(ziX v )]([ ziXRC ca cc dy )(nxi sau:

a. )()( 31 nrectnx = e. )()(5 nuanxn =

b. )()( 32 nrectnx = f. )()( 26 += nuanxn

c. )()( 133 += nrectnx g. )()( 27 += nuanxn

d. )()(4 nuanxn= h. nanx =)(8 .

Gii: a. 2

212

0

031

111).()(

zzzzzzznrectz

n

n

n

nX ++=++=== =

=

.

Dy nhn qu hu hn )(3 nrect c ZT vi 0[ ||:)](1 >zzXRC .

b. 12010

2

232 ).()( ++=++===

=

=

zzzzzzznrectzn

n

n

nX .

Dy phn nhn qu hu hn )(3 nrect c ZT vi

45

i bin, t n = - m - n = m v khi n = - th m = nhn c:

)()()()(

10

1

05

1

1

za

a

zazazaz

m

m

m

mmX

=

===

=

=

.

Theo (2.8) , bn knh hi t ||||lim1

aa nnn

xR == +, vy dy phn nhn qu v

hn )( nua n c ZT vi .||:)]( ||5[ az zXRC <

f.

=

=

=

++==+=0

122

26 ).()( 2

n

nn

n

nn

n

nn zazazazaznuazX

)()()()(

2

3

2

2

4122

6aza

z

az

z

a

z

a

zzzazaz XX

=

++=++= .

Theo (2.6) v (2.8), xc nh c dy khng nhn qu v hn )( 2+nua n vi

),[ 2 n c ZT vi .||:)]( ||6[

46

Nu

X1(z) = ZT[x1(n)], RC[X1(z)] X2(z) = ZT[x2(n)], RC[X2(z)]

x3(n) = ax1(n) + bx2(n) trong a, b l cc hng s th ZT[x3(n)] = X3(z) = a.X1(z) + b.X2(z),

RC[X3(z)] = RC[X1(z)] RC[X2(z)]. Chng hn, x1(n) = 2

nu(n), x2(n) = 3nu(n).

b. Tnh cht dch thi gian Nu

X(z) = ZT[x(n)], RC[X(z)] th ZT [x(n-k)] = z-kX(z)

Min hi t: + Nu k >0 th RC: l RC[X(z)]/0

+ Nu kz .

x(0) = 13

lim =

zz

z.

d. Tch chp trn min z.

Nu X1(z) = ZT[x1(n)], RC[X1(z)]

X2(z) = ZT[x2(n)], RC[X2(z)]

x3(n) = x1(n) * x2(n)

th ZT[x3(n)] = X3(z) = X1(z).X2(z), RC[X3(z)]=RC[X1(z)] RC[X2(z)]. Min hi t ca X3(z) c th rng hn min hi t ca X1(z) v X2(z).

Chng hn, x1(n) = 2nu(n), x2(n) = 3

nu(n). e. Nhn vi hm m

Gi s c dy x(n) c ZT[x(n)] =X(z), RC : 21 RzR

47

1

0

1

1

1)()()(

=

=

===

zzznuzU

n

n

n

n vi RC: 11 Z

Vy 11 21

1

21

1

2)(

=

=

=

zz

zUzX vi RC: 22.1 =>Z .

2.1.4. Bin i z hu t

Gi s X(z) l hm hu t

=

=

==N

k

kk

M

or

rr

za

zb

zD

zNzX

0

)(

)()( .

a. Cc khi nim cc v khng.

+ im cc ca X(z) l cc gi tr z ti X(z) = , k hiu l zpk Khi D(zpk) = 0.

+ im khng ca X(z) l cc im ti X(z) = 0, k hiu l zor Khi N(zor) = 0.

b. Biu din X(z) di dng cc v khng

Gi s N(z) l a thc bc M ca z khi :

N(z) = bM(z- zo1) (z- zo2) (z- zo3).... (z- zoM)= =

M

rorM zzb

1

)( .

Gi s D(z) l a thc bc N ca z khi :

D(z) = aN(z- zp1) (z- zp2) (z- zp3).... (z- zpN)= =

N

kpkN zza

1

)( .

Khi X(z) c vit li nh sau:

=

=

=

N

kpk

M

ror

N

M

zz

zz

a

bzX

0

1

)(

)()( hay ta c th vit di dng hm ca z-1 nh sau:

=

=

=

=

=

=

N

kpk

M

ror

NM

N

kpk

M

ror

N

M

zz

zzcz

zz

zz

z

zczX

1

1

1

1

)(

0

1

1

1

)1(

)1(

)1(

)1()( .

Vi c = bM/aN X(z) c M im khng v N im cc. biu din trn th

cc im cc c nh du bng (x) v cc im khng c nh du bng (o).

48

Hnh 2.2: Khng v cc ca X(z).

49

2.2. Bin i z ngc

2.2.1. nh l Cauchy

nh l Cauchy l mt nh l quan trng trong l thuyt bin s phc, n l c

s chng ta xy dng cng thc ca bin i z ngc.

nh l Cauchy c pht biu nh sau:

=

=

C

kn

nk

nkdzz

j .0

1

2

1 1

Trong C c ly theo chiu dng trn ng cong khp kn bao quanh gc

ta v nm trong min hi t ca bin i Z.

2.2.2. Bin i z ngc

T biu thc

=

=n

nznxzX ).()( ta c

=

=n

nkk znxzzX 11 ).()( ly tch phn trn min hi t ROC ca n ta c:

=

=

==n n RC

knnk

RC

k

RC

dzznxdzznxdzzzX 111 )().()( .

p dng nh l Cauchy ta c:

=

=n

k

RC

nxdzzzXj

)()(2

1 1

vi k = n.

Hay )()()(2

1 1 kxkxdzzzXj n

k

RC

==

=

.

vy: dzzzXj

kx k

RC

1)(2

1)( = hoc ta c th vit:

dzzzXj

kx k

RC

1)(2

1)( =

Hay =C

n dzzzj

nx X )1()(1

)(2

. (2.11)

Biu thc (2.11) c gi l biu thc ca bin i Z ngc ( IZT - Invert Z

Transform ). Tnh trc tip tch phn (2.11) l kh phc tp, v th thng s dng cc

phng php gin tip tm bin i Z ngc. Khi ng dng bin i Z gii cc bi ton phn tch v tng hp h x l s,

cn s dng bin i Z thun chuyn dy x(n) sang min bin s Z. Sau khi thc hin nhng bin i cn thit trong min Z, cn s dng bin i Z ngc nhn

c kt qu trong min thi gian.

50

2.2.3. Cc phng php tm bin i z ngc Tm bin i Z ngc xc nh dy x(n) bng cch tnh trc tip tch phn

(2.11) thng rt phc tp, v th ngi ta xy dng cc phng php gin tip sau

tm bin i Z ngc: - Phng php thng d.

- Phng php khai trin X(z) thnh chui ly tha. - Phng php phn tch X(z) thnh tng cc phn thc n gin.

2.2.3.1. Phng php thng d Trong l thuyt hm bin s phc, phng php thng d dng tnh tch

phn: C

dzzj

Q )(1

2. (2.12)

Tch phn (2.12) c ly theo chiu dng trn ng cong khp kn C bao

quanh gc ta v nm trong min hi t ca hm Q(z).

Nu Q(z) c mt cc bi bc q ti pzz = th c th phn tch Q(z) thnh:

q

pzz

zz

NQ

)(

)()(

= .

Trong , cc nghim ca phng trnh 0)( =zN phi khc cc bi pz .

Khi tch phn (2.12) s c dng

pC

qpC

sdzzz

z

jdzz

j

NQ Re

)(

)(1)(

1

22=

= .

Vi psRe c gi l thng d ca hm Q(z) v c tnh theo biu thc :

.)()()(

))(22 1

1

12

1

12

3(

=

=

zzzz

z zH (2.13)

Trong trng hp ring, nu pz l nghim n th 1=k nn:

)()(Re ppp zzzzs NN === (2.14)

tm bin i Z ngc theo tch phn (2.11), p dng phng php thng

d cho hm )1().()( = nzzz XQ . Gi s Q(z) c m cc bi bc iq th c th phn tch

Q(z) thnh tng:

=

==

m

iq

pi

in

izz

zzzz

NXQ

1

)1(

)(

)().()( .

Khi , biu thc bin i Z ngc (2.11) c a v dng:

=

==

C

m

iq

pi

i

C

n dzzz

z

jdzzz

jnx

i

NX

1

)1(

)(

)(1)(

1)(

22 . (2.15)

V ng cong khp kn C nm trong min hi t ca hm )1().( nzzX nn tch

phn v phi ca (2.15) c th ly trn tng s hng ca chui, v th c th i v tr ca du tng v du tch phn:

51

==

=

==m

ipi

m

i Cq

pi

i

C

n sdzzz

z

jdzzz

jnx

i

NX

11

)1( Re)(

)()()(

2

1

2

1

Vy: =

==m

ipi

C

n sdzzzj

nx X1

)1( Re)(1

)(2

(2.16)

Cc thng d pisRe ng vi cc cc piz ca)1().( nzzX . pisRe ca cc n tnh

theo (2.14), pisRe ca cc bi bc q tnh theo (2.13).

V d 2.5. Hy tm

==)(

)]([)(az

zzIZTnx X vi ||||:)]([ azzXRC > .

Gii: C )()(

.).(

)1()1(

az

z

az

zzzz

nnnX

=

=

.

Vi 0n , hm )(

).( )1(az

zzz

nnX

= c mt cc n az p = v

nzzN =)( . T

biu thc thng d (2.14), vi 0n tm c: np aas N == )(Re .

Theo (2.16) th: 0Re)( == nkhiasnx np .

V ||||:)]([ azzXRC > nn )(nx l dy nhn qu, do kt qu l

)()(

)( nuaaz

zIZTnx n=

= .

2.2.3.2. Phng php khai trin X(z) thnh chui ly tha

V X(z) l hm gii tch ca z, nn trong min hi t ca n, c th khai trin

X(z) thnh chui ly tha ca nz theo dng:

=

=n

nn zazX .)( . (2.17)

Mt khc, theo nh ngha ca bin i Z c:

=

=n

nznxzX ).()( . (2.18)

Trong min hi t ca X(z), c hai chui trn u hi t nn khi ng nht

cc h s ca hai chui (2.17) v (2.18), tm c dy:

nanx =)( . (2.19)

Vy khi khai trin X(z) thnh chui lu tha (2.17), s tm c dy x(n)

theo cc h s ca chui.

V d 2.6. Hy tm dy x(n) ca hm nh )(

)(az

zzX

+= .

a. Vi ||||:)]([ azzXRC > b. Vi ||||:)]([ azzXRC < .

52

Gii: a. Chia c t s v mu s cho z nhn c:

).()()(

11

1+

=+

=zaaz

zzX .

V ||||:)]([ azzXRC > nn )(nx l dy nhn qu, do hm nh phi l chui ly

tha ca nz . khai trin X(z) thnh chui ly tha ca nz , chia t s cho a

thc mu s (1 + az-1) :

1 | 1 + az-1 _

1 + az-1 1 - az-1 + a2z-2 - a3z-3 + a4z-4 - ......

- az-1

- az-1 - a2z-2

+ a2z-2

+ a2z-2 + a3z-3

- a3z-3

- a3z-3 - a4z-4

+ a4z-4

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

Mt cch tng qut nhn c nn

n zazX

= =

0

)()( .

Theo (2.19) nhn c : )()()( nuanx n= vi ||||:)]([ azzXRC > .

b. Vi ||||:)]([ azzXRC < th x(n) l dy phn nhn qu, nn hm nh phi l

chui lu tha ca nz . khai trin X(z) thnh chui lu tha ca nz , chia t s

cho a thc mu s (az-1 + 1) :

1 | az-1 + 1 _

1 + a-1z a-1z - a-2z2 + a-3z3 - a-4z4 + ......

- a-1z

- a-1z - a-2z2

+ a-2z2

+ a-2z2 + a-3z3

- a-3z3

- a-3z3 - a-4z4

+ a-4z4

53

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

Mt cch tng qut nhn c : mm

m zazX

=

=1

)()( .

a chui v dng (2.17), i bin t n = (- m + 1) m = (- n + 1),

khi m = 1 th n = 0 v khi m = th n = - :

nn

nn

n

n zazzazX

=

+

=

==0

)1()1(

0

)1( )(.)()( .

Theo (2.19) v tnh cht tr ca bin i Z nhn c:

)()()( 1)1( += nuanx n vi ||||:)]([ azzXRC < .

2.2.3.3. Phng php phn tch X(z) thnh tng cc phn thc ti gin

y l phng php s dng bng bin i Z c bn (Bng 2.3). tm dy x(n) ca cc hm X(z) phc tp, ch cn phn tch X(z) thnh tng ca cc hm nh

c trong bng bin i Z, v p dng tnh cht tuyn tnh tm c hm gc bng tng ca cc hm gc thnh phn.

Trong a s trng hp, c th a hm X(z) v dng phn thc hu t:

)...(

)...(

)(

)()(

11

1

11

10)(.

.NN

NN

MN

MM

MM

azazaz

zbzbzbzb

z

zz

A

D

BAX

++++++++

==

. (2.20)

Trong A l hng s v a thc mu s D(z) c a0 = 1 c gi l a

thc c trng ca hm X(z). Phng trnh c trng D(z) = 0 c N nghim zpk, chng l cc cc im ca hm X(z).

Nu hm X(z) (2.20) c bc ca a thc mu D(z) ln hn bc ca a thc t B(z), tc l N > M th n c gi l hm X(z) dng chnh tc. Trong trng

hp hm X(z) (2.20) c N M th n l hm dng khng chnh tc. Khi , bng cch chia a thc t cho a thc mu hoc bng bin i ton hc, s nhn c hm X(z) dng:

)()(0

.)(

)()(. zzcz XA

zD

zNzCAX

NM

r

rr + +=

=

=

.

Trong X(z) l hm dng chnh tc. V C(z) l a thc ly tha ca z, nn

c th d dng tm c bin i Z ngc ca n:

=

==NM

rr rnczIZTnc ACA

0

)()]([)( .. .

V vy, trong mi trng hp ch cn nghin cu phng php tm bin i

Z ngc ca hm X(z) (2.20) dng chnh tc. C th biu din hm X(z) chnh tc (2.20) qua cc cc im zpk :

))....()((

)....(

)(

)()(

21

22

110

)(..

N

MNM

M

ppp zzzzzz

zbzbzbzb

z

zz

A

D

BAX

++++

==

. (2.21)

54

Cc cc im zpk ca hm X(z) (2.20) v (2.21) c th l cc cc n (cc c

gi tr khc nhau), hoc cc cc bi bc q (q cc c gi tr ging nhau), hn na zpk c th l cc s thc hoc s phc. Trc ht chng ta nghin cu trng hp X(z)

c nghim n gin. 2.2.3.3a Trng hp hm X(z) ch c cc cc n l s thc

Khi X(z) l hm (2.20) hoc (2.21) dng chnh tc v c N cc n zpk l s thc (N cc thc n), th c th phn tch X(z) thnh tng ca cc phn thc n

gin dng:

)(

...)()()()(

)()(

2

2

1

1

1

.N

NN

pppk pk

k

zzzzzzzzz

zz

BBBB

D

BAX

++

+

=

==

=

(2.22)

xc nh h s kB , nhn c hai v ca (2.3-12) vi (z - zpk ):

)(

)(......

)(

)(

)(

)())((

2

2

1

1

N

N

p

pkk

p

pk

p

pkpk zz

zz

zz

zz

zz

zzzzz

BB

BBX

++++

+

= .

Ti z = zpk th tr kB , cn tt c cc s hng khc v phi ca biu thc

trn u bng khng, do c:

[ ]pkpkk zzzzzXB == ))(( . (2.23)

Ly bin i Z ngc hm X(z) (2.3-13) , tm c dy x(n) :

=

==

== )(.)()]([)(

1

11 pkkk

k pk

k

zz

zzIZT

zzIZTzIZTnx

NN

BB

X .

Theo tnh cht tr v, vi ]max[||:)]([ pkzzzXRC > , nhn c

.)( 01 , nhn c

55

=

==N

k

npkk nuzzIZTnx BX

0

)(..)]([)( . (2.28)

V d 2.8. Hy tm hm gc nhn qu ca )(

)()(

682

52 +

+=

zz

zzX .

Gii: Hm X(z) l phn thc dng chnh tc. V a thc c trng c 120 =a nn

phi nhm tha s 2 ra ngoi. nhn c hm gc x(n) dng khng tr, phn tch hm:

)()())((

)(

)(

)()(

31312

5

342

5 2102

+

+=

+=

++

=zzzzzz

z

zzz

z

z

z BBBX .

Theo (2.27) xc nh c cc h s 0B , 1B , v 2B :

6

5

312

50312

5

))(())((

)(00 =

==

+= BB zzzz

zz .

2

3

4

6

311.2

511312

15

).(

)(

))((

))((11 =

=

+==

+

= BB zzzzzz .

3

2

12

8

133.2

533312

35

).(

)(

))((

))((22 ==

+==

+= BB zzzz

zz .

Vy: )()(

)(

3

1

3

2

1

1

2

31

6

5

+

=

zzzz

zX .

Suy ra )()(

)(33

2

12

3

6

5

+

=

z

z

z

zzX .

V dy x(n) l nhn qu nn 3[ ||:)]( >zzXRC , theo (2.28) nhn c:

)()()()]([)( 33

2

2

3

6

5nununzIZTnx nX +== .

2.2.3.3b Trng hp hm X(z) c nhiu cc dng phc tp

n gin v d hiu m khng lm mt i tnh tng qut, gi s X(z) l

hm (2.20) hoc (2.21) dng chnh tc v c r cc thc n zpk , mt cc thc bi

zpq bc q, mt cp cc phc lin hp pez v *pez , khi c th phn tch X(z) thnh

tng ca cc phn thc dng:

==

+

+

+

=q

ii

pq

ir

k pk

k

pepe zzzzzzzzz

z CBEEX

10* )()()(

*

)(

)(oo

. (2.29)

Trong thnh phn ng vi r cc thc n zpk l:

=

=r

k pk

kb

zzz

z BX

0 )(

)(. (2.30)

Thnh phn ng vi cc thc bi zpq bc q l

56

=

=q

ii

pq

ic

zzz

z CX

1 )(

)( . (2.31)

Thnh phn ng vi cp cc phc lin hp pez v *pez l

)(

*

)(

)(*pepe

e

zzzzz

z EEX

+

=

oo

. (2.32)

Tng t trng hp hm X(z) ch c cc nghim thc n, cc h s kB ca

(2.30) c xc nh theo (2.27). Vi pkb zzzXRC max||:)]([ > , t hm Xb(z), theo

(2.28) nhn c thnh phn xb(n)

=

==N

k

npkkbb nuzzIZTnx BX

0

)(..)]([)( . (2.33)

Cc h s iC ca (2.31) ng vi cc thc bi zpq , c xc nh nh sau:

.)()(

)!( )(

)(1pq

qpqiq

iq

i zzzzz

z

dz

d

iq

XC =

=

(2.34)

Vi ||||:)]([ pqc zzzXRC > , t hm Xc(z) nhn c thnh phn xc(n)

( )=

==q

i

inpqicc nuzi

innnz CXIZTnx

1

)( )(..)!(

))....((.)](

1

11[)( . (2.35)

Cc h s phc o

E v *o

E ng vi cp cc phc lin hp pez v *pez . Ta ch cn

xc nh o

E theo biu thc:

ejpepe

ezzz

zE

XE zz

.)(

)(=

= =

o

. (2.36)

v theo l thuyt hm bin s phc th )(*)( * zz ff = , nn c:

ejpe

pe ezzz

zEE

Xzz

==

= .*)(

)(*

*o

.

Do c: )()(

)(*

..

pe

j

pe

je

zz

e

zz

e

z

z ee EEX

+

=

.

Vy: )()(

)(*

..pe

j

pe

je

zz

ze

zz

zez ee EEX

+

= .

vi ||||:)]([ pee zzzXRC > , t hm Xe(z) nhn c hm gc xe(n) :

)()()()()]([)( *.. nuzenuzezIZTnx npejn

pej

eeee EEX

+==

)().|(|)().|(|)( .. nuezenuezenx njpejnj

pej

epepe EE

+=

)..()(||)( .. zeze jnjjnjnpee eeeenuznx E += .

57

+=

++

2.2

)()(

)(||)(epep njnj

npee

eenuznx E

.

Vy: )cos().(||)]([)( .2 epn

peee nnuzzIZTnx EX +== . (2.37)

Trong h s phc ejeEE .=o

c xc nh theo biu thc (2.35).

T , theo tnh cht tuyn tnh ca bin i Z nhn c:

)()()()]([)( nxnxnxzIZTnx cbeX ++== . (2.38)

Trong , xb(n) c xc nh theo (2.33), xc(n) c xc nh theo (2.35), v xe(n) c xc nh theo (2.37).

V d 2.9. Cho 22 ba < , hy tm hm gc x(n) ca hm nh:

).()(

22 2 bzaz

zzX

++= vi ||||:)]([ bzzXRC > .

Gii: y s phn tch X(z) thnh tng ca cc a thc n gin. nhn c

dy x(n) dng khng tr, phn tch hm:

).(

)(22 2

1

bzazz

zX

++= .

V 22 ba < , nn phng trnh c trng 02 22 . =++ bzaz c hai nghim l cp

s phc lin hp: pjpp ezabjaz

.||)( 22 =+= .

v pjpp ezabjaz== .||)( 22* .

Vi babaz p =+= )(||222 .

v

=

a

abarctgp

)( 22 . (2.39)

s dng cng thc (2.37), theo biu thc (2.36) tm c

)())((

)(**

1

ppp

pp

p

zzzz

zzzz

zzEE

==

=

oo

)()()( 222222 2

1 2

ab

e

abjaabja

j

E

=

+

=

o

.

Vy: )(

||222

1

abEE

==

o

v 2

][ == eEArg

o

.

Theo (2.37) nhn c dy x(n)

=

2.

2

12 .cos).(

)()(

22

pn nnu

abnx b .

58

Bin i lng gic v xc nh p theo (2.39), nhn c kt qu

Vi 22 ba < v ||||[ bzRC > th

=

++ aab

arctgnab

nub

bzaz

zIZT

n )(.sin.

)(

)(

).(

22

2222 2. (2.40)

Cc cng thc (2.37) v (2.40) thng c s dng nh mt cp bin i Z thng dng tm bin i Z ngc ca cc hm X(z) c hai nghim n l cp s

phc lin hp.

V d 2.10. Hy tm dy x(n) ca hm nh:

))((

)()(

1441

3222

2

+++

=zzz

zzzX vi 1[ ||:)]( >zzXRC .

Gii : V a thc mu c 140 =a nn phi nhm tha s 4 ra ngoi. nhn

c hm gc x(n) dng khng tr, phn tch hm :

2

2

22

2

)5,04

32

25,014

32

)()((

)(

))((

)()(

++

=++

+=

zjzjzz

zz

zzzz

zz

z

zX . (2.41)

Phng trnh c trng 0)5,0 2)()(( =+ zjzjzz c:

- Mt nghim n ti 00 =pz ,

- Mt nghim bi bc 2 ti 5,01 =pz ,

- Hai nghim phc lin hp ti jz p =2 v jz p =*

2

1|| 2 = pz v 2

2 =p .

Theo cc cc im trn, c th phn tch hm (2.41) thnh dng:

)(

*

)()()(

)(2

21

5,05,0 jzjzzzzz

z EECCBX

++

+

+

+=

oo

. (2.42)

Trong cc h s c xc nh nh sau:

35,014

305,014

320 222

2

).(.))((

)()(=

==

++

==

= zzz

zzzzz

zXB

5,014

325,0

5,0)(

)()(

)(2

22

1 =

++

==

= zzz

zz

dz

dzzz

z

dz

d XC

44,35,014

3213141222

222

1)(

))(())((==+

++++= zzz

zzzzzzC

8,05,014

325,0

5,0)(

)()(

)(2

22

2 ==

++

==

= zzz

zzzzz

zXC

59

1,12

2

.5,05,04

32

))((

)()(

)( jejzzjzz

zzjzjzz

zXE =

++

==

=

o

.

Thay gi tr cc h s vo (2.3-33), nhn c:

)()()()(

)( 1,11,1

2

.5,0.5,0

5,0

8,0

5,0

44,33

jz

e

jz

e

zzzz

z jjX

++

+

+=

,

1,11,12

5,05,05,0

5,0

5,0

8,0

5,044,33

)()()()()( jj e

jz

ze

jzz

z

z

zz

zX

++

+

+= .

Theo Bng 3.2 v cng thc (2.37), vi 1[ ||:)]( >zzXRC , nhn c

++= 1,1

21.5,0.25,0.6,15,0.44,3)(3 .cos)(||)().()()()( nnununnunx nnnn

.

Hay )(31,12

.6,144,32 .cos)()).(()( nnnunnunx n

+= .

2.3. Phn tch h thng ri rc trn min z Chng ta bit trn min n mt HT-TT-BB c c trng bi p ng

xung hoc phng trnh sai phn tuyn tnh h s hng. Nhng vic phn tch h

thng nhiu khi gp phi s kh khn ca vic tnh tch chp, gii PT-SP.... Trong phn trc chng ta biu din tn hiu sang min bin s z, by gi ta s phn

tch h TT-BB trn min z, trc tin ta tm hiu khi nim hm truyn t ca h thng.

2.3.1. Hm truyn t ca h thng TT-BB

Min n Min z y(n) = x(n)*h(n)

=

=

k

knhnx )().(

h(n) = IZT[H(z)]

X(z) = ZT[x(n)], Y(z) = ZT[y(n)]

H(z) = ZT[h(n)]

)(

)()(

zX

zYzH =

Nh vy hm truyn t ca h thng TT-BB chnh l bin i z cu p ng xung ca n. Hm truyn t c k hiu l H(z) v n cng c trng hon ton

cho h thng trn min z.

2.3.2. Hm truyn t ca h c m t bi PT - SP - TT - HSH Quan h gia u vo v u ra ca mt HT - TT - BB c m t bi PT

sau:

= =

=N

k

M

rrk rnxbknya

0 0

)()( ; ly bin i Z 2 v ta c:

[ ] [ ] nN

k

M

rr

n

nk

n

zrnxbzknya

= =

=

= =

0 0

)()( .

60

p dng tnh cht tr v tuyn tnh ta c

rrn

n n

M

rr

kknN

kk zzrnxbzzknya

=

==

= = )(

0

)(

0

)()( .

Suy ra )(0

)(

0

)()( rn

n n

M

r

rr

knN

k

kk zrnxzbzknyza

=

==

=

=

.)(

)()(

)(.)(.

0

0

00

=

=

=

=

==

=

N

k

kk

M

r

rr

M

r

rr

N

k

kk

za

zb

zX

zYzH

zXzbzYza

Nu a0 = 1 th ta c:

.1

)(

1

0

=

=

+=

N

k

kk

M

r

rr

za

zbzH

Ch : Ta cng c th biu din H(z) di dng hm ca z-1, hoc cc cc v khng ca n.

2.3.3. Gii phng trnh sai phn TT HSH s dng bin i z 2.3.3.1. Tm hm truyn t h thng H(z) t phng trnh sai phn V d 2.11. Cho h x l s TTBBNQ c phng trnh sai phn

)()()()()( 1322142 =+ nxnxnynyny .

Hy xc nh hm truyn t h thng H(z) v c tnh xung h(n) ca h.

Gii: Ly bin i Z mt pha c hai v ca phng trnh sai phn trn, ta c:

)()()()()( 121 3242 zzzzzzzz XXYYY =+ .

Hay: ))(())(( 121 31242 =+ zzzzz XY

)(

)(

)(

)(

)(

)(

)(

)()(

122

3

122

3

242

31222

1

21

1

+

=+

=

+==

zz

zz

zzz

zz

zz

z

z

zz

X

YH .

Vy hm h thng l: 22 )(

)(

)(

)(

)(

)()(

12

3

122

3

=+

==

z

zz

zz

zz

z

zz

X

YH .

Ly bin i Z ngc hm h thng H(z), tm c c tnh xung h(n):

==2)(

)()]([)(

12

3

z

zzIZTzIZTnh H vi 1[ ||:)]( >zzHRC .

tm h(n), phn tch hm: 2

212 )()()(

)()(

1112

3

+

=

=

zzz

z

z

z CCH .

Trong 12

31112

1322

2

2)(

))((=

==

= CC zz

zz

61

2

1112

1312

2

1)(

))((==

= CC zzzz

dz

d .

Vy 22 )()()(

)()(

1

1

12

1

12

3

=

=

zzz

z

z

zH

2)()(

)(112

1

=

z

z

z

zzH .

Vi 1[ ||:)]( >zzHRC , theo Bng 2.3 tm c c tnh xung h(n):

)(.)()( 5,0 nunnunh = ,

hay: )().()( 5,0 nunnh = .

2.3.3.2. Tm phn ng y(n) ca h x l s qua bin i Z

Khi bit c tnh xung h(n) ca h x l s TTBBNQ v tc ng x(n), c

th tm c phn ng y(n) ca h x l s theo tch chp:

h(n)*x(n)ny =)( .

Cc phng php tnh trc tip tch chp c trnh by chng mt u

kh phc tp, v trong nhiu trng hp khng th tm c biu thc ca phn ng y(n). C th tm phn ng y(n) ca h x l s TTBBNQ d dng hn bng

cch tnh tch chp qua bin i Z, cc bc thc hin nh sau:

1- Tm cc bin i Z thun: )]([)( nxZTzX = v )]([)( nhZTzH =

2- T xc nh c: )().()( zzz HXY =

3- Tm bin i Z ngc: )]().([)]([)( zzIZTzIZTny HXY == .

Trong a s cc trng hp, hm h thng H(z) v tc ng X(z) c dng phn

thc hu t:

)(

)()(

z

zz

A

BH = v

)(

)()(

z

zz

Q

PX = .

Do phn ng Y(z) ca h x l s TTBBNQ l

)().(

)().()().()(

zz

zzzzz

QA

PBHXY == .

Trc ht xt trng hp hm h thng H(z) c N cc im n zpk l

nghim ca phng trnh c trng 0)( =zA , cn tc ng X(z) c m cc im n

zpi l nghim ca phng trnh c trng 0)( =zQ , trong cc cc zpk zpi vi mi

k v i. ng thi, cc cc im ca hm h thng H(z) khng b loi tr bi cc

khng im ca tc ng X(z) v ngc li. tm phn ng y(n) ca h x l s TTBBNQ, phn tch hm:

==

+

=m

i pi

kn

k pk

k

zzzzz

z QAY

1