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