高二物理競(jìng)賽課件動(dòng)態(tài)網(wǎng)絡(luò)的復(fù)頻域分析法

動(dòng)態(tài)網(wǎng)絡(luò)的復(fù)頻域分析法y(0+)，y(1)(0+)，···，y(n–1)(0+)動(dòng)態(tài)網(wǎng)絡(luò)的復(fù)頻域分析1、動(dòng)態(tài)網(wǎng)絡(luò)的描述引言對(duì)正弦穩(wěn)態(tài),x(t),y(t),jddtX.Y.問(wèn)題：一般動(dòng)態(tài)網(wǎng)絡(luò)的分析(時(shí)域分析)[an(j

)n+an–1(j

)n–1++a1(j

)+a0]Y……=[bm(j

)m+bm–1(j

)m–1++b1(j

)+b0]X……??dnydtndn–1ydtn–1dn–2ydtn–2dydtyanan–1an–2a1a0+++???++dxdtdmxdtmdm–1xdtm–1dm–2xdtm–2bmbm–1bm–2b1b0x+++???++=*2、為什么要將拉普拉斯變換引入動(dòng)態(tài)網(wǎng)絡(luò)分析？

拉普拉斯變換拉普拉斯變換的定義0-

￡[f(t)]=

f(t)e–Stdt

=F(S)關(guān)于積分下限0–例0-

￡[K]=

Ke–Stdt=Ke–St–S10-

=KSS=+j￡[1(t)]=

1(t)e–Stdt0-

￡[

(t)]=

(t)e–Stdt0-

=

e–Stdt0+

=1S=

(t)dt0-0+=1￡[e–t]=

e–te–Stdt0-

e–(+S)tdt0-

=

e–(+S)t–(S+)1=0-

S+1=

￡[]=SF(S)–f(0-)

df(t)dt

￡[

1f1(t)+

2f2(t)]=

1F1(S)+

2F2(S)

拉普拉斯變換

拉普拉斯變換的基本性質(zhì)

設(shè)

￡[f1(t)]=F1(S)

￡[f2(t)]=F2(S)

1、線性性質(zhì)2、微分性質(zhì)￡[kcost]=￡[0.5k(ejt+e–jt)]=0.5k()S–j

S+j

11+=kS2+

2S

設(shè)

￡[f

(t)]=F

(S)

uCCR+-iLus(t)+-

￡[

f(t

)dt]=F(S)

0-t1S

拉普拉斯變換

拉普拉斯變換的基本性質(zhì)3、積分性質(zhì)

設(shè)

￡[f

(t)]=F

(S)

￡[i(t)]=I(S)￡[uS(t)]=US(S)Ri+L+uC(0–)+idtdidtC10–t=uS(t)

R￡[i

(t)]+L￡[]++￡[]=￡[uS(t)]didtC1

idt0–tuC(0–)S(R+SL+)I(S)–

Li(0–)+=US(S)SCuC(0–)S1I(S)=SCUS(S)+SLCi(0–)–CuC(0–)S2LC+SRC+1(R+SL+)I(S)–

Li(0–)+=US(S)SCuC(0–)S1

拉普拉斯變換

拉普拉斯變換的基本性質(zhì)

部分分式法求拉普拉斯反變換出發(fā)點(diǎn)￡[ke–t]S+k=￡–1[]=ke–tS+k集中參數(shù)電路中響應(yīng)變換式的特點(diǎn)F1(S)F2(S)F(S)=bmSm+bm–1Sm–1++b1S+b0???anSn+an–1Sn–1++a1S+a0???=變換式在一般情況下為S的實(shí)系數(shù)有理函數(shù)F1(S)F2(S)F(S)=bmSm+bm–1Sm–1++b1S+b0???anSn+an–1Sn–1++a1S+a0???=

拉普拉斯變換

部分分式法求拉普拉斯反變換F(S)=H0(S–zi)mi=1(S–pj)j=1nH0

實(shí)數(shù)常數(shù)ziF(S)的零點(diǎn)pjF(S)的極點(diǎn)(1)n>m(2)n

mF(S)=Q(S)+F2(S)R(S)F(S)可展開為部分分式之和例F(S)=S3+1S2+2S+2=S–2+S2+2S+22S+5其中，￡–1(S–2)=(t)2(t)F(S)的極點(diǎn)

單極點(diǎn)重極點(diǎn)實(shí)數(shù)復(fù)數(shù)復(fù)數(shù)實(shí)數(shù)1、F(S)只含實(shí)數(shù)單極點(diǎn)F(S)=S–p1A1S–p2A2S–pkAkS–pnAn+???+???+++f(t)=￡–1[F(S)]=Akepktk=1n問(wèn)題歸結(jié)為求F(S)的極點(diǎn)和確定相應(yīng)的常數(shù)Ak

拉普拉斯變換

部分分式法求拉普拉斯反變換Ak=(S–pk)F(S)S=pkF(S)=S–p1A1S–p2A2S–pkAkS–pnAn+???+???+++(S+1)(S+2)(S+3)S2+3S+5F(S)=例求的反變換S3+6S2+11S+6S2+3S+5F(S)=S+1S+2S+3A1A2A3++=A1=(S+1)F(S)=(S+2)(S+3)S2+3S+5S=–1A2=(S+2)F(S)=(S+1)(S+3)S2+3S+5S=–2=–3A3=(S+3)F(S)=(S+1)(S+2)S2+3S+5S=–3=(S+1)(S+2)(S+3)S2+3S+5F(S)=S+1S+2S+31.5–32.5++=

拉普拉斯變換

部分分式法求拉普拉斯反變換f(t)=￡–1–t–3e–2t–3tt02、F(S)除含實(shí)數(shù)單極點(diǎn)外，還含有復(fù)數(shù)單極點(diǎn)1、F(S)只含實(shí)數(shù)單極點(diǎn)(1)復(fù)數(shù)極點(diǎn)是共軛形式成對(duì)出現(xiàn)的F(S)=S–(+j)A1+S–(–j)+A2???(2)與復(fù)數(shù)極點(diǎn)對(duì)應(yīng)的兩個(gè)常數(shù)也互為共軛復(fù)數(shù)A2=A1~A1=A1ej

令A(yù)2=A1e–j

則

拉普拉斯變換

部分分式法求拉普拉斯反變換2、F(S)除含實(shí)數(shù)單極點(diǎn)外，還含有復(fù)數(shù)單極點(diǎn)F(S)=S–(+j)A1+S–(–j)+A2???A1=A1ej

令A(yù)2=A1e–j

則f(t)=A1ej

e(+j)t+A1e–j

e(–j)t+???=A1e

t[ej(

t+

)+e–j(

t+

)]+???=2A1e

tcos(t+

)+???注意A1是虛部為正的極點(diǎn)對(duì)應(yīng)的那個(gè)常數(shù)方程*￡S域代數(shù)方程（初始條件含在其中）（復(fù)頻域）Y(S)￡–1y(t)初始條件（時(shí)域）例求的反變換[(S+2)2+4](S+1)S2+3S+7F(