S2+3S+5 例：求F(S)-S3+6S2+11S+6 的反变换 S2+3S+5 K1,K2,K3 FS)S+1S+2S+3)s++s+2+s+3 K=(S+1)FS)= S2+3S+5 =1.5 (S+2)S+3)s--1 K2-(S+2)FS)=S2+3S+5 =-3 (S+1)S+3)川s=-2 K,=(S+3)FS=S2+3S+5 =2.5 (S+1)S+2)S=-3 F(S) 8篇+品+篇 S2+3S+5 ft)=f-1[F(S)]=1.5e-t-3e-2t+2.5e-3t (t≥0)
(S+1)(S+2)(S+3) S 2+3S+5 F(S)= S+1 S+2 S+3 1.5 –3 2.5 = + + f(t)=£–1[F(S)]=1.5e–t–3e–2t+2.5e–3t (t t 0) (S+1)(S+2)(S+3) S 2+3S+5 F(S)= ֻ˖≲ Ⲵ৽ਈᦒ S 3+6S2+11S+6 S 2+3S+5 F(S)= S+1 S+2 S+3 K1 K2 K3 = + + K1=(S+1)F(S)=(S+2)(S+3) S 2+3S+5 S= –1 =1.5 K2=(S+2)F(S)=(S+1)(S+3) S 2+3S+5 S= –2 = –3 K3=(S+3)F(S)=(S+1)(S+2) S 2+3S+5 S= –3 = 2.5
(2)D(S)除含实数单根外，还含有复数根 (a)复数根是共轭形式成对出现的 K K F(S)-S(a+jB)+S-(a-jB) (b)与复数根对应的两个常数也互为共轭复数 K2=K1 令Kj=Kleie 则K2=Ke- ft)=|Klej0e(a+-it+lKle-jea-jBt+··· =Klet[ejBt+)+e-jBt+)]+··· =2 Keat cos(βt+O)+··· 注意K是虚部为正的极点对应的那个常数
f(t)= K1 e jTe (D+jE)t + K1 e –jTe (D–jE)t +••• = K1 e Dt [ej(Et + T) + e–j(Et + T) ] + ••• =2 K1 e Dt cos(Et+ T) + ••• ⌘K1ᱟ㲊䜘Ѫ↓Ⲵᶱ⛩ሩᓄⲴ䛓њᑨᮠ (a) ༽ᮠṩᱟޡ䖝ᖒᔿᡀሩࠪ⧠Ⲵ F(S)=S–(D+jE) K1 + S–(D–jE) + K2 ••• (b) о༽ᮠṩሩᓄⲴєњᑨᮠҏӂѪޡ䖝༽ᮠ K2=K1 * K1= K1 e Ԕ jT K2= K1 e ࡉ– jT ˄2˅D(S)䲔ਜ਼ᇎᮠঅṩཆˈ䘈ਜ਼ᴹ༽ᮠṩ
例：求FS)F S2+3S+7 [S+2)2+4S+1) 的反变换 KI K2 K3 FSs+22+s+22)+s1 X S2+3S+7 KFS+(2+j21S+1)S-2+2 0.25ej90° S2+3S+7 K,1s+221S+10）s-22-0.25e0 K、 f0)=0.25ei0°e2+i2f+0.25e-90°e2-i2f+et(t≥0) =0.5e-2tc0s(2t+90°)+e-tt≥0
ֻ˖≲ Ⲵ৽ਈᦒ [(S+2)2+4](S+1) S 2+3S+7 F(S)= F(S)= S + (2-j2) S + (2+j2) S +1 K1 K2 K3 + + K1 = [S + (2+j2)](S+1) S= –2+j2 S 2+3S+7 =0.25ej90° (S+2)2+4 S 2+3S+7 K3 = S= –1=1 =0.5e–2tcos(2t+90°) + e–t t t 0 K2 = [S + (2-j2)](S+1) S= –2-j2 S 2+3S+7 =0.25ej-90° (t 0) t o o j90 (-2+j2)t j-90 (-2-j2)t -t f(t) = 0.25e e + 0.25e e + e
(3)D(S)有重根p:,含有(s-p)的因式 设含有(s-p1)3的因式 s+s品+ K 求K、K12和K13 K13=(S-Pi)3F(S)s=pI 例： F(s)=_ +2 s+1)3 (s+1)2F(S)= S+2 =KL+K2-+K3 (s+1) (s+1)(s+1)2”(s+1)3 6+I)FS)= S+2 (s+1)2 K13=(s+1)3F(s)s=1=1 K12=?K1=?
䇮ਜ਼ᴹ(s-p1 ) 3Ⲵഐᔿ ≲K11ǃK12઼K13 K13 = (S –p1 ) 3 F(S) s=p1 F(S)= S –p1 K11 (S –p1 ) 2 K12 S –pn Kn + + + • • • + (S –p1 ) 3 K13 + S –p2 K2 11 12 13 2 3 K K K =+ + (s +1) (s +1) (s +1) 1 3 K = (s +1) F(s) 13 s=-1 3 s+2 F(s) = (s +1) ֻ˖ 2 2 12 11 s+2 (s +1) F(s) = (s +1) s+2 (s +1)F(s) = (s +1) K = ?K = ? ˄3˅D(S)ᴹ䟽ṩpi ˈਜ਼ᴹ(s-pi ) nⲴഐᔿ
(s-p1)F(s) -Kn(6-P.)+Kr(-P)+Kp+K:(-p)+..K.-(s-p) (s-P2) S-Pn d(s-Pi)F(s) ds s=PL -kek s-P2 s-P2)2 =K2 d2(s-p1)F(s ds2 s 弧，+66卫，6水，p+2 S-P2 (s-P2)2 (s-P2)2 =2K1=(3-1)K1
13 2 33 2 n 11 1 12 1 1 1 2 1 n 3 (s K K = K (s - p ) + K (s - p ) + + (s - p ) + ... (s - p ) (s - p ) s K - - F( p p ) s) 2 2 ) ) 1 1 23 23 21 21 2 12 3 1 s=p 1 21 11 1 s=p 22 nn 12 3K (s -p ) K (s -p ) 3K (s -p ) K (s -p ) = 2K (s -p ) + + d(s -p ) F(s) ds - ... + - s - p (s - p s - p (s - p = K K 22 3 2 )) ) 1 1 22 3 21 21 21 21 s=p 2 2 22 2 11 11 11 3 1 2 s=p 6K (s - p ) 3K (s -p ) 3K (s -p ) K (s -p ) = + - ... s -p (s -p (s -p (s -p = 2K d (s -p ) F(s) (3-1 ds )!K 2K