§2-1線性定常齊次狀態方程的解（自由解

若初始時刻t0x(t0)x0，則式（2-1） x(t)eA(tt0)x，t 若初始時刻從t0x(0)x0

0x(t)eAtx0

t

證：先假設式（2-1）的解x(t)為t的矢量冪級數形式，即x(t)bbtbt2btk

x(tb2bt3bt2kbtk1 代人式（2-1）xA(bbtbt2btk) 既然式（2-4）是（2-1）的解，則式（2-5）對任意時刻t都成立，故t的同次冪項的系數應bAb，b1

1A2b，b1Ab1A3b

b1 1Akb k 在式（2-4）中，令t0b0x(0)4，故得：x(t)(1At1A2t21Aktk

括號內的展開式是nn矩陣，它是一個矩陣指數函數，記為eeAt1At

A2t2

1AktkK01AktkK0 再用(tt代替(t0，即在代替tx(t)eA(tt0x 2-2矩陣指數函數——狀態轉移矩 X(teAtXX(teA(tt0 由上式可知，它反映了從初始時刻的狀態向量到任意t0或tt0

X(t0種向量變換關系。變換矩陣就是eAt，它不是常數矩陣，它的元素一般是時間t的函數，是一個nn所以eAt也稱為狀態轉移矩陣，記為(t，(t)eAtX(0X(t的轉移矩陣，而(tt)0X(t0X(tXAXX(t(tX(0X(t(tt0X(t0，在tX(0)X10以此為初始條件，且已知(t，那么在t

X X 20X

)X11(t)X

X X 21xt00若已知

，那么t

X

)X12

)X

X X

22X(0(t1或(t2X(t1X(t2若以tt1X(t1是初始狀態從t1轉移到t2的狀態將為：X(t2)＝(t2t1)X(t1將式（2-8）x(t2)(t2t1)(t1

x(0x(t1x(t2比較式（2-9）和（2-11）可知轉移矩陣（或矩陣指數）

)

或eA(t2t1e

e

x(t0，求得任意時刻t的狀)(t)()(t或eAte

eA(t

這就是組合性質，即從0，0t(t0)[0()][t()](t(tt)I或eA(tt)

它意味著狀態矢量從時刻t又轉移到時刻t，顯然狀態矢量是不變的[(t)]1(t或[eAt]1e

轉移矩陣的逆意味著時間的逆轉，利用這個性質，可以在已知x(t)的情況下求出小于時刻tx(t0（t0t(tA(t(tA

Ae

eAt

即(t)或eAt矩陣與A矩陣是可以交換的對于方陣AB（nn）ABBAeAte

eAB)tABBAeAte

eAB)t

0 AA＝∧＝

eAt(t)

n0 t enAT1ATeAt(t)T

0T

ent0A＝∧＝

0n0eAt1At1A2t21Antn X(t)(tt)X(t)xxtT1ITT1ATt1T1A2Tt2 1T1AnTtn 1211

12t

0 2! 1 1

2t2

2! I

t

n 0

12t21t12t2 1

2! 2!

n!

12t

12t2 22

21 2 1t12t2 1ntn

0

2!

n!

ent

entT[T1eAtT]T1

0

T T1T T ent

ent A 0 AJ

， 1

n1101 0101 00 00 eJt(t)et eAtIAt1At)21At)3

t

A

t

12!

1

(n1(n1t

(nt10t1

0 11

0 0

01 01 A

1

0

設t(n1)!t(n1)!2 (n1)!t1t

0

1t 1t

(n

1 1 t1(n1t1t1

(n2)! et

(n et 1ABBA的條件下eAB)teAteBt X 解：

0

0 00 0 000

ABBA 0

0又知，eAt ，注：A e

eBtIBt1B2t2

1 0

1 3

1 0 t

t2

t3

t4 0

2! 2

3! 0

4! 4112t214t4 t13t3

sin

sintcost t 3t3 2t2 4t4

sinxx 注：級數

cosx e

0

sint(t)eAteBt etsin costet etsint

t

sintetsin etcost sin cost 例2：已知X 解 sIAs

0 s

s s

s

s

(s)2

(s)22sI

(s)22 s s (s)22 (s)22 eatsin(sa)2k

s eatcos(sa)2k 1

1

et etsint

t

sint(t)

(sI

etsin etcost sin cost 一.(t)或eAt的計算（狀態轉移矩陣的計算根據(t)或eAt的定義直接計eAt1At1A2t21Antn1：A

1解 eAt1At1A2t

Aktk k 1 12teAt t

13 3

3 1 1 1t 13 1 3t 3 32 33! 0 1 3t 3 1 3t 72 7 36 0 1

13t 1

76

3t 72

5t t3t27t3t3t27t37

22t3t2

t32612613t t2 t3 AT1ATTA變換為對角線矩陣的變換陣。eAtTetT2：AIA2

1

23212

1，

1TP P P12，APP01 P 1 22 1P11P11

P21，

P11

2，則P

3

P

P

1 1

21

21

1P122P12

P22，

取

1

3

2P

P

2 2

22

22

1

2 1TP P ，T1 2

1 1 1

1 1 0 2

2

3

20T1AT，1， 2

1

0

1 1

0 1eAtTetT1

2 2

2

e2t

2

e2t

1 2et

ete2t

2t

2

2t

P56JT1AT，根據式（2-18）eAtTeJtT3：A

1；求e 1 解：IA

1

1，

2，J

00

0p11

p11

1p21

＝

21

p31p21p11，p31p21，2p115p214p31

P1P

1P254PP1，所以P

21P21P2AP2p12

0p12

p11

p－

1

＝

＝22

22

21

p12p221，p22p321，p322p125p224p32取 0，

1，

2

32P23P3p13

0p132

＝

1p23 23

2p13p23，2p23p33，p332p135p234 1，

2，

4

T

2，det1

24＋0＋2－1－0－4＝1，adjT

2

1T1

1

1＝

1

1

1

1

0 JT1AT＝

11

1

2

＝

1

2＝ ∴eJt

et

000e2t

2

s 0eAt

0 0 s 0 s

ssIA1 s 0 s

(s (s (s1)(s

s

(s1)(s (s1)2(s2)2 (s (s1)2 (s2)1sIA1

0 0

e2tJ

e

還需求出變換矩陣T和T1eAtTeJtT

0p12 P

P，

1

1

22

22

22

p12p22 p2p5p4 當 1時，

0，

1

0

TT012，1＝531At121 t2(e2tet利用拉氏反變換法求eeAt(t)L1[(sIA)1

X(tAX(tX(0)X00sX(sX(0AX(s)(sIAX(s)X(0)X，左乘(sIA1000X(ssIA)1 X(tL1[(sIA)100

X(t)eAt

,t

eAtL1[(sIA102-4：A0

試用拉氏反變換法求e解 sIA

1s(s3)sssI(sIA)sI

s(s1)(s (s (s1)(s (s2e

e

e

e eAtL1[sIA]1

2e

e

2t應用-定理求e（1）f(A)An An1aAaI An An1 An2 aAaIAn1

AI 同 An1AAn An An1aA2a1100 1100

An1

An2aA

An1aA2a

)An1

)An2

)A

n1a0以此類推An1An2都可以

An1，

AI在定義式（2-7）中eAt1At

A2t2

Aktk，用（1）An次及neAt1At

A2t2

(n

An1tn1

Antn

(n

An1tn1

(t)An1

(t)An2a(t)A

1010已知狀態轉移矩陣(t)的情況下，如何確定系統矩陣A，方法有A證明：因為(t)A(t)A(t)(t)(t)1(t)，1(t)1(t)，故有A(t)(t)A證明：因為(t)eAt，當t0A(tt0由上式可以求出矩陣A例：已知系統的狀態轉移矩陣，試求系統矩陣A

(t) (1

(12t)e2t

A(t)(t) (4 (48t)e2t (1

(12t)e2t

A

(4 (48t)e2t t

(1

t

s

s

(sIA)1L(t)

s (s (s (s (s2)2 s4 (s s (s2)2 (s (s2)2 s s 0

(sIA)

s 4 0 (s

(s2)2

s

s

s4 (s (s2)2 A 12-5：A

求eAt表示式中的ai(tAIA2

3 定理有A23A2I A23AA3

A(3A2I)

2A3(3A2I)2A7AA4AA3A(7A6I)7A26A7(3A2I)6A15A…A22eAt1At1A2t21A3t31A4t4 1At1(3A2I)t21(7A6I)t31(15A14I)t4 (t3t27t315t4 )A(1t2t314t4 a1(t)Aa0∴a(t)＝t3t27t315t4，a(t)＝1t2t314t4 ai(t上例求ai(t只是對（2-22）加深理解，并說明ai是時間t的函數，實際上不宜計算ai(t，一則是得不到ai(t)的解析表達式，二則是當維數較高時，將造成計算上的繁瑣。下面給出計算ai(t)的一般公式。A0a(t)0

n11e1t1a1(t)1

n1

e2t

n n

n1

t

AA是可以互換的，因此也23，從而有：a(t)a

(t)n1

01a(t)a(t) (t)n101 a(t)a(t) (t)n1 上式對a0

a2

23A的特征值均相同，為1時，

t

1

e1

(t)

(n

(n1

n2

(t)

(n

e1 ＝

(n1)(n2)n3

aa

(n

1t2e 1 11 1

證明：同上（P60）a(t)a(t)a

(t)n1

上式對1求導數，有a(t)2a

(n

(t)n2 再對1求導數有

(t)

(n1)(n

(t)n3t211(n (t)t由上面的n個方程對ai(t)求解，即得式（2-24A 1

eAt2-

3， 解：已知1=-1，2=-2，為互異根a

1e1t 11et

1et 2ete2t101

2

e2t

e2t

1e2t ete2t (2ete2t) 2et 2et2et 2et

2e2t2et

2

A，(t) tA

2

et 12

2e2tet

t

2-7：A

t10

2 14 1,求eAt

1解 IA 12132重根部分按式（2－24）處理，非重根部分按式（2－23）a0 a 1

1 1

213213

2

1a

et t

2

adjA

2

2e

10 10

1323

1

12

e eAta(t)Ia(t)Aa(t

1 4

2tet 3tet2et tetete2t2(e2tettet 3tet5et tet2et2e2t 3tet8et tet3et4e2t

2tete2t，

3tet2et2e2t，

tetet1212XAXY(t)CX

d[eAtX(t)]eAtBu(t)d(eAtX

eAt

de

X

At

eAt

e

對上式從0到t積分dtd

teABu(0 t eAX()

eABu(0teAtX(t)X(0)eABu(0teAtX(t)X(0)eABu(0t兩邊同乘eAt，即eAte

1X(t)eAtX(0eA(tBu()d，而e0

(t)tX(t)(t)X(0)(t)Bu(0如果從t0tX(t)(tt0)X(t0)

(t)Bu(

x1

1x10u，Y 0x

3x

2 t

2

eax0X(t)eAt0

eA(tBu()d0

eaxdx C先求(t，已知(t)eAt

2et

et 2et

et2e2tu(t)1(tB0X(0)x1

x tX(t)(t)X(0)(t)Bu(02et

et

x(0)

(t)Bu(02et et2e2tx20 (2ete2t)x(0)(ete2t)x

t

2e(t)e2(t

e(t)e2(t

0(2et2e2t)x(0)(et2e2t)x 0

2e(t)2e2(t

e(t)2e2(t) (2ete2t)x(0)(ete2t)x

t

e(t)e2(t)

(2et2e2t)x(0)(et2e2 x(0)0

e(t)2e2(t)t e(t)de2(t)d

12t

x(t)

et

1 1

0 x2 t

t e(t)d2e2(t)d

0y(t)CX(t)

0x1(t)x(t)

1et1x x 脈沖響應時；當u(tK(tX(0X時X(teAtXeAt 階躍響應時；當u(tK1(tX(0X時X(teAtXA1(eAt 斜坡響應時；當u(tKt1(tX(0X時X(teAtX[A2eAt1 K——與u(tu(t)

tt

X(0) te(t)e2(t X(t)(t)B

e(t)2e2(t)0t＝

te(t)d0

0 e2(t

e(t)d

2e2(t)d et＝

ede2t0

e2d

e注 xeaxdxe

(ax1)

aet

ed

2t0

e2de te ＝

(1)t

2t4

(21) 00

te(t

t

00et00

(1)t

2t4

(21)t 00000e te

(t1)

[4

(2t1)1(01)]4＝(t1)et(1)2e2t

4

(2t1)

1(0 1t3

e2tet

X(t)＝

t x1(t)

31t

1xy(t)CX(t)x

0 作業：2.4（2，2.5(2)(4)，2.6附：伴隨方前面討論的是系統自由響應問題，假定初始條件已知X(0)(t0。現在是某一時刻tX(t)(t)X X(0)1(t)X設

T1

T(t)1(t)對t取導(t)

T(t)1(t)T(t)1(t)X(t)(t)X(t) [T(t)T(t)A](t)若(t)0T(t)T(t)(t)AT

乘法定理AT]T

(

BT

Z(tAZ(t,上式中的(t

Z(t)(t)Z

（

1(t)）33求得系統的轉移矩陣(t1tX(0Z(tATZ(tX33

例：

A 1

Z(t)

(t)L1{[sIA]1}

2et 2et2e2t＝et e2t

＝

et

et2e2t1(t)T

2et

et 2etX(0)1(t)X

et2e2ts

s 1

(s1)(s (s1)(s2)L1{

}L1 (s1)(s (s1)(s2)

（利用部分分式解2

22

t t

2tL1s

s

s

s2

1s1

s

1s1

2s2

et

et2e2t＝ s ＝1(s1)(s

s(s(s3)(s(s1)(s

s

s

2a2

(s(s(s

11s∴(s1)(s2)

s1

s2 ＝

(s1)(s s

s2(s2(s(s1)(s

s1

b2

2(s2(s(s1)(s

21 (s1)(s

2s1

s3 (s1)(s

s

sc1c2

s1(s1(s(s1)(s1(s2)

111∴(s1)(s2)

s1

1s (s1)(sd

ss(ss(s(s1)(s

ds

s1

1d2

s(s2)s(s2)(s1)(s2)

21 (s1)(s

1s1

s§2－4狀態轉移矩陣的性一.討論狀態轉移矩陣的性質就是討論矩陣指數eAt的性 (t2t1)(t1t0)(t2t0 (tt)(tt)eA(t2t1eA(t1 1(t)

A(t2t1t1t0)＝eA(t2t0)＝(tt(t)eAt，(t)e (t,t)1(t,

eAt(eAt)1

1(0)

t,te0(t,t0)A(t)(t,t0

XAX設t00X(0)

yX(s)(sIA)1X(0)(sI-A)1Bu(s)(sIA)1XsIA)1But注：x(teA(tt0)eA(tBu(

x(t)ty(t)CeA(tBud（拉氏反變換得出）輸入為(ty的輸出為系統的脈沖響應h(t0

h(t)(t)＝

h(t)＝0

(t)h(t)d＝(t)h(t)n1h(iT)T(tiT n當n0 n

i1,2,3,n1∴h(t)CeAt h(t)L1[C(sIA)1f(t在處連續，如果有另一個函數(t(t)(t)＝lim0

tt

(t)df(t)(t)d

t[t1,t2則稱(t為Dalta1，故又稱為單位系統在t0時刻的r個輸入均為單位脈沖函數(tt0，即Ui(t)ei(tt0

ei1第ii＝1，2…，X(t00時，在tt0H(tX(t)(tt0)X(t0)

t(t)Bu(X(t00D0

X(t)(t)Bu(0 Y(t)C(t)Bu()d＝C(t)Bei(tt0)d＝C(tt0

hi(tt0令t

h(t)C(t

t

thi(t當t

hi(t)系統t時刻的輸出僅與t以前的輸入有關，而不取決于t以后的輸入H(t)＝h1(t

h2(t

hr(t

C(t

C(t)Ber＝C(tH(t)

er]

C(t

tt設0CeAt

tH(t)

t令t0

Y(t)CeAtX(t0)H(t)u((t0)(t)etY(t)H(t)u(H(t)ceAtk1u(t)

2

K是與u(tk kX(t)X(t)eAtXeAt0Y(t)CeAttX(t)eAt Y(t)CeAtX0CH(t)()KdCH(tX(t)eAt0tX(t)eAtX0eA(t)Bu(tX00X(t)eA(tBu()deAtB(t)KeAt0rr

u(t)K(t)

tt

K

kT與u Y(t)CeAtX0H(t)u()dCeAtX0[CeA(t)BK tCeAtX0CeAteA0te0t

AdIt

1At2

1A2t3 AA1

eAdA1[At0

A2t2

A3t3A1[(IAt1A2t21A3t3)I A1eAt0A1(eAtIX00

Y(t)CeAteA0

CeAt[A1(eAtI[CeAtA1eAtCeAtCA1[IeAt]BKCA1[eAtI0或 X(t)eAtXA1(eAtI0Y

u(t)

ttK

k

X(t)XrtrY(t)C(tt0)X(t0)C(t)Bu(tCeA(tt0)X(t0)H(t)u(tCeAtX0H(t)u(0tCeAtX0CeA(t)Bu(0tCeAtX0CeAteA0

eAteABKdeAt(IA A22

eAt t

1 1 33244A3t5 1A2t2)A21A2t2A3t33A4t44

At若A2存在，將上式改寫teAteABKd(IAt0

A5t5A2(1A2t21A3t31A4t4)BKA2(eAtI 0Y(t)CeAtXCA2(eAtI000X(t)eAtXA2(eAtIAt)BKeAtX[A2(eAt1)00

u(t)2

t

tK

k

X(0)r0Y(t)CeAtr0

CA3(eAtIAt1A2t22假 H(t)CeAtH(t000H(s)LH(t)H(t)estdt[CeAtBestdt]C[eAtBestdt]C(sIA)1B000L(eAt)eAtestdt(sI0

即eAtL1[(sI1：

X

1X

y

eAt

2et

1et12et

et2e2t

At1

2et

et

11 11

112et

et2e2t＝2ete2t2et

ete2tet2e2t＝e2t

e2t2e2t＝e2t2e2te2t

＝T

1C~CTC

1

1

3B B

11AT1AT

0~e

0

作業 e2t ~

~

e

03 3

H CeAt 1

e2t H(t)h(t)

e2t

2x10(t

y

0x1

H(t

3x

2x2

2

2

2

2(1e3teAt(t)L1[(sIA)1]L1

L1

0s(s3) 0 s

s3

0

3tH(t)CeAtB

3

030

3(1 2

§2－5離散時間系統狀態方程的先介紹用遞推法（迭代法X(k1)GX(k)

X(k Xk這是一階矩陣差分方程的解， 一差分方X(k1)0.2X(k) k＝0，1，2，…，X(0)0，u(k)k

X(1)0.2X(0)2u(0)21k

X(2)0.2X(1)2u(1)0.222k

X(3)0.2X(2)2u(2)0.21.6210.322k

X(4)0.2X(3)2u(3)0.21.68210.3362這種算法適用于計算機計算，應該X(k)已知u(k)已知

X(k1)GX(k)y(k1)CX 當k0當k X(1)GX(0)當k X(2)GX(1)Hu(1)G2X(0)GHu(0)當k X(3)GX(2)Hu(2)G3X(0)G2Hu(0)GHu(1)從k1

X(k)GX(k1)Hu(k1)GkX(0)Gk1Hu(0)GHu(k2)Hu(kkX(k)GkX(0) GkJ1Hu(j

（k1,2,3,）（j1,2,3,,k1）很明顯，X(kX(0ujj1,2,3,k1k個時刻的狀態，只與此采樣時刻以前的輸入采樣X(k1)GXF(k)y(k)CTX(k)X

GG2

0 0

u(0)u(1)注：用矢量矩陣形式表示

X(0)

X k

k

k

G

Hu(kX1(k1)X2(k)X2(k1)X3(k)…Xn1(k1)Xn(k)Xn(k1)a0X1(k)a1X2(k)an1Xn(k)y(k)X1(k)

hn1

n2X(k1) X(k)u(k0001 0001 h1

an1

y(k) hn

0X(k)hn1bn1

h0b0a0bna1bn1an1b1b00式中Gk（或Gkh）相當于連續系統中的(t)eAt（或(tt)eA(tt0)0陣(t)ek，它是滿足(k1)G(k)，(0)I的唯一的矩陣。利用狀態轉移矩陣(k將前式可寫成kX(k)(k)X(0) (kj1)Hu(j

性質有1(k)(k)k X(k)(k)X(0) (j)Hu(kjj

ky(k)C(k)X(0) (kj1)Hu(jk y(k)C(k)X(0) (j)Hu(kjjkh

(h)[(GkhkX(k)GkhX(h) Gkj1Hu(j或z

kX(k)(kh)X(h) (kj1)Hu(jX(k1)GX(k)

zzX(z)zX(0)GX(z)(zIG)X(z)zX(0)X(z)(zIG)1zX(0)(zIG)1

zXX(k)L1[(zIG)1zX(0)]L1[(zIG)1(k)X(0)GkX(0)L1[(zIG)1zXGk(k)L1[(zIG)1k(kj1)Hu(j)L1[(zIG)1jk X(k)GkX(0) Gkj1Hu(j要獲得采樣瞬時之間的狀態和輸出，只需在此采樣周期內，即kTtk

內，令tkTT(k)T,(0X(k)T(T)X(kT)

(T)dHu(kTk首先， Gkj1Hu(j)的z變換，kjk

k

k[Gkj1Hu(j)]

Gkj1Hu(j)zk

zkjGkj1Hu(j)zj

k

j

k

j[Gk1Hu(0)zkGk2Hu(1)zkGk3Hu(2)zkk(Hz1GHz2G2Hz3)[u(0)u(1)z1u(2)z2(IGz1G2z2)Hz1[u(k)zkk(IGz1)1Hz1u(k)zk

(IGz1)1z1[zI[zIG]1k∴L1[zIG]1Hu(z)] Gkj1Hu(j

k0GkX(k1)GX(k

G 1

H

1當k1

u(k)1X(k

X(k)(k)X(0)L1[zIG]1u(z)

z

X(0)1

u(k)

（k0

11

z 先求：(k)L1[(zIG)1z]L1

zL1

z 1

(z0.2)(z0.8)

z(k)L1zz

z z

z3

z

z

z0.2

z 1 4(0.2)k

5(0.2)k5(0.8)k(k)30.8(0.2)k

(0