連續(xù)時(shí)間信號(hào)與系統(tǒng)的S1 一、 變換 變x(t)=etu(t)>0

x(t乘以衰減因子e-F[x(t)et]

x(t)etejtdt

ete(j)t0 (s0

( -e令s若

1s

j 一、 變換 變F[x(t)et]令s=

x(t)etejtdtx(t)estdtX

x(t)e(j)t定義：X(s

x(t)estdt對(duì)x(t)e-t x(t)

12j

X(s)estds 一、 變換 變X(s)L[x(t) x(t) L1[ (s)x(t) X(s信信號(hào)x(t)可分解成復(fù)指數(shù)est的線性組4二、單 變單 變x(t)

u(t)

X(s)estdsXX(s)e和計(jì)算時(shí)可以直接利用起始給定的0狀態(tài)。5三 變換及其存在的條件變換存在的

|x(t)et|dt對(duì)任意信號(hào)x(t)x(t)limlimx(t)ettlimx(t)et6例計(jì)算下列信 變換的收斂。u(t)u(t。e3ttnu(t)etut(6)e2tu(t)e4tu(t)

|x(t)e

|dt或limx(t)et limx(t)et

的取值范圍7三 變換及其存在的條件變換存在的收 區(qū)

S平0稱收斂

0稱絕對(duì)收斂8三 變換及其存在的條件若x(t)是有限持續(xù)期的信號(hào)，其收斂域是整S如果x(t)是一個(gè)右邊信號(hào)，則X(s)Re{s如果x(t)是一個(gè)左邊信號(hào)，則X(s)的收斂域Re{s右當(dāng)x(t)是一個(gè)雙邊信號(hào)時(shí)左Re{s}9零極例:Xs

極點(diǎn)s1s2.s1s2左邊信號(hào)Res右邊信號(hào)Res(3)雙邊信號(hào),2 Res

四、常用信號(hào) 變指數(shù)型etutt

test

u(t)]

s

etu(t) s

e

u(t)

s

cos0t

0s20

四、常用信號(hào) 變階躍函數(shù) L[u(tlimL[etu(t 0Re(s) (n) L[(t)](t)est

Re(s)

'

st

(est

t L[(n)(t)]

stdt

d

sn

(

dsn

)t 四、常用信號(hào) 變tnu(t) n!,Re(s) (tn)estdtt

(est)

L[tu(t)]

ntn1estdts

nL[tsL[tnu(t)]nL[tn1u(t)]nn1L[tn2u(t)] n n L[t 五 變換 變換的關(guān)當(dāng)收斂域包含j軸時(shí) XjX當(dāng)收斂域不包含j軸時(shí)，

s當(dāng)收斂域的收斂邊界位于j軸時(shí)， X(j)X

s

πKn(nn例計(jì)算下列信號(hào) e3t e3t

變e3t

je3t

s s

π[(2)(2)]

(j)2 例X(sXjs(s

0(s29)s解：1)收斂域4包含jX(j)X s

(j收斂域的收斂邊界位于jX(j)X

s

πKn(nnX(s)1 1 1118s s 9X(j)

j(92

π[(3)(3)]π 線性 x(t)X(s)Re(s) x2(t)X(s)Re(s) 則ax(t)ax(t)X(s)aX(s)則 Re(s)max(1,2 時(shí)移特 X(t)(s) R0t)00t)0sRocX(s)ex(tRe(s)例試求如圖所示周期信號(hào)的單邊Laplacef2012345。f2012345tf(t)k

f1(tkT)u(tkTL[f(t)]

k

eskTF1(s)

1esT

Re(s)>。f2012345例試求如圖所示周期信號(hào)。f2012345t f1(t)2[u(t)u(tf(t)f1(t)f1(t2)f1(t4) F(s)L{f(t)}

2(1es

Re(s) F(s)F1(s)(1e2s

e4s

1e2s

Re(s)> 展縮特

x(t)

X(s Roaa X(s/ax(at) etu(t) s-e-tu(t) s

aR ss1tu(t)e 卷積x1(t) X1(s Rx2(t) (s x1(t)*x2(t)X(s) (s)

Roc

乘積特x(t)(s) x(t)(s) 2121Rx(t)x(t)1[X(s)*2121R1

乘積特 指 性質(zhì)(S域平移若x(t)X(s 1則etx(t)(s RocR1Re(s0 性質(zhì)(S域微分tx(t) R 指 性質(zhì)(S域平移u(t) Re{s}sse2tu(t)e2tu(t)

sss

Re{s}2Re{s}2 線 性質(zhì)(S域微分ssRe[ s]1etu(t)11tetu(t)11tetu(t)d

(s)2t2etu(t) t2

Re[s] (s)t(n1

etu(t) Re[s](n

(s 微分

x(t)X ROCdx(t)sX(s) 包括Rdnx(t)X(s)dx(t)(s) Re(s) 6.微分

dx(t) st

x(t)e

x(t)(sestx(0)sx(t)estdtsX(s)x(0 微分特 d2x(t)s2X(s)sx(0)x'dnx(t)snX(s)sn1x(0)sn2x'(0)...xn1(0)snX(s) snr1xr(0x(t0t<0,則有xr(0dnx(t)

X 積分

x(t)X ROCtt

x()d1Xs

R{Re{s}x(t)(s) Re(s)0

(0X(s) (0X(s)tLssx()d Lsstf()dF 例5試求如圖所示信號(hào)的Laplace例5試求如圖所示信號(hào)的Laplacef(t)f1(t)f1F(s)F1(s)F1(s) )s

Re(s) f(t)r(t)2r(t1)r(tr(t) s

F(s)1

se2ss

Re(s)例試求如圖所示信號(hào)的Laplace解：d2f(t)s2F(s)sf(0)f'(0s2F12e

e2s12e

e2sF(s) Re(s)s8、初值定當(dāng)t<0x(t)=0，而且在t=0時(shí)x(t)不包含任何沖激。則x(0limx(tlimsX(s)9、終值定

若t<0x(t)=0t=0x(t)不包含任何則limx(t)x() sX

s§

etu(t)etu(t)

1s1s

Re(s)Re(s)-etu(t -etu(-t

s1s

Re(s)Re(s)§ee

t

1s1ss

Re(s)Re(s)Re(s)0sint

s200

Re(s) s

00§(t)(n)

Re(s)-Re(s)-u(t s

Re(s)tu(t

Re(s)stnu(t

Re(s)§tetu(t)

L

1(s)s

Re(s)tcos0

Re(s)0(s22)0tsin0tu(t

200(s22)0

Re(s)e0t

cos0

tu(t)

s(s0)20

e0t

tu(t

(s0)20

§e0t

cos0

tu(t)

s(s0)20

e0t

tu(t

(s0)20

§7.5 ——部分分式展開(kāi)2πj2πjf(t) jF(s)est計(jì) 反變換方法部分分式N bsm

sm1 bsX(s) sn sn1 a X(s)為有理真分式m<n)，極點(diǎn)為X(s)N(s)D(s

N(s(sp1)(sp2 (spnssssX(s)kki(spi)X(s)sii1,,x(t)(kep1tkep2t kepnt)u(t) 部分分式N bsm

sm1 bsX(s) sn sn1 a X(s)為有理真分式mn)，極點(diǎn)為rX(s)N(s)

N(s)D(s) (sp)r(s (sp r

kr

s

(sp1)

(sp1)

spr s11drr,X(s j1,2[(sp1j)!dsr(rkki(spi)X

s

ir1,r ,部分分式N bsm

sm1 bsX(s) sn sn1 a 、當(dāng)X(s)的分母多項(xiàng)式D(sX(s)N(s)D(s

N(sD1(s)(s2asbs2as (sa)2配方 的形式然后再用比較系數(shù)法求出它的系數(shù)部分分式N

bsm

sm1 bsX(s) sn sn1 as XX(s)N(s)BBs01smnN1

為真分式，根據(jù)極點(diǎn)情況按1或2 B Bs B

m sm

m

(mn)(t)例 X(s) ss34s2

X(s)

s (sX(s)

s32ss24s解：(1)X(s) s s34s2X(s)

s

s

k1

2s34s2 s(s1)(s s s2k1(s)X

s(s1)(s

s0k2(s1)X

s1

ss(s

2k3(s3)X

s3

ss(s

6x(t)2u(t)1etu(t)1 例 解(2)X(s) s (s

2X(s)k1

(s (s (sk1sX

s0

s(s

k4(s1)3X

ss

s1k3

d(s1)3X(s)

(s2s

s1 d2(s1)3X(s) s2

2

s1例 (3)X(s)

s32ss24sX(s)s4

20ss24sx(t)'(t)4(t)L1[20s12s24sx(t)'(t)4(t)20.6e4.45tu(t)例 求下列F(s)的反變換(1)F(s

s F(s)

13s2(s2

F(s)

(s4)1e2s(s2(1)F(s)

(sF(s)18s81 (s (s sk1(s4)2F

s4 (8s

s

d(s4)2F

(8s8)'

f(t)(t)8te4tu(t)24e4t F(s)解：令

3s2(s2則F(s)

1(

1 (q1 q

q k2 (q4)

q

4于 F(s)1(1 4(s2f(t) 1(t1sin 例 1e2F(s)

s(s2 先用部分分式求F(s)

的反 s(s2e2再利用時(shí)移特性求F2 s(s2 4)的反變F(s)

k1k2s

k2,k3 s(s2 f(t)1(1cos

f2(t)1[1cos2(t2)]u(t 例3、X(s)10(s2)(s5)c1 s(s1)(s s sc1X(s)

10(s2)(s(s1)(s

10251 cX(s)(s1)10(s2)(s

1014 s(s

c3X(s)(s

10(s2)(ss(s

310 1 (s

2

3 0(1Re{s}10100

0

20s

s

x(t)x(t)10020et10e3tu(t)-

(s0

se-atu(t

1

Re{s}s(2)Re{s}10100

0

20

,

s s則x(t)為負(fù)時(shí)域信x(t)10020et10e3t

Re{s X(s-3- 5.-e-atu(t

s

Re{s}3 310010, s s

x(t)10e3tut10020etu(t)0 0 1Re{s}

1001

0,20 s

103sx(t)

u(t)

復(fù)頻域分析主要用于線性系統(tǒng)的分析§ 微分方程描述系統(tǒng)的S域析時(shí)域差分

時(shí)域響應(yīng)S域代數(shù)方 S域響應(yīng)解代數(shù)方 二階系統(tǒng)響應(yīng)的1dtd2y(t)1dt

dy(t)

y(t)

d2x(t)

dx(t)

x(t)21dt02x(t)，y(0)，y'(0，求y21dt02求解s域代數(shù)方程，求出Y0(s二階系統(tǒng)響應(yīng)的域求 a2y[s2Y(s)sy(0)y'(0)]a1[sY(s)y(0)]a2Ybs2X(s)bsX(s)bX(s)

b0x"(t)b1x'(t)Y(s) sy(0)y'(0)a1y(0)b0s2b1s

F as as y(t)

(t)

L1{Y(s)Y 例1y''(t)+5y'(t)+6y(t)=2f'(t)+8f(t)f(tetu(t)，初始狀態(tài)y(0)=3，解：對(duì)微分方程取拉氏變換可s2Y(s)sy(0)5[sY(s)y(0)]6Y(s)2sF(s)Y(s)

2s 5s

F(s)

(s5)y(0)y'(0)(s25s6)Y(s)

3s

11 s25s s s 例1y''(t)+5y'(t)+6y(t)=2f'(t)+8f(t)f(tetu(t)，初始狀態(tài)y(0)=3，解Y(s)

2s

2s

5s6s (s2)(s3)s s s (sy(t)L1{Y(s)}(3et4e2te3t) y(t)y(t)y(t)3et7e2t7e3t,t 例2

d2y(t)3dy(t)dt2 dt

2y(t)etu(t)y(0) y(0)方法一、對(duì)方程兩邊進(jìn)行單邊拉氏[s2Y(s)sy(0)y(0)]3[sY(s)y(0)]2Y(s) s

Re{s}(s23s2)Y(s)

s

)

)3y(0Y(s)

3s(s1)2(s (s1)(s

1＋1＋

(s s s s s

(s syx(t)

e2

u(t)y0(t)

2e2t

u(t)

u(t)解：方法二、零狀232 0零輸0

y(t)aet

t y(0) y(0) y0(t)

2e2t

u(t)零狀 s2Y(s)3sY(s)2Y(s)X(s)H(s)Ys X s23s

s

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

(s1)2(s (s s syx(t)

e2tet

u(t)XX(s)L[x(t)]H(s)L[yx(t)]Yx(s)H(s)與h(t) yf(t)=(t)*h(t)HH(s)h(t)L1[H求零狀態(tài)

求H(s)

H(s)L[yxL[x(t)]③由系統(tǒng)的微分方程寫(xiě)出零極點(diǎn)分H(s)

bm

b1ss an1s

a1s

(sr1)(s (srm(ss1)(ss2 (ssn 1111Hs 2s 2e2t

Hs2A系統(tǒng)穩(wěn)定時(shí)，令H(s)s=j，則得系統(tǒng)H(j)H(s)sH(j)H(j)幅頻響 相頻響 系統(tǒng)頻響當(dāng)系統(tǒng)穩(wěn)定時(shí)，令s=j，則

H(s)

m(