2-11(c)求所示有源电路网络的传递函数 选取电容C上端电压为 uA , 则其拉式变换方程为: U A (s) U i ( s ) R R2 1 U A (s) U A ( s) U o ( s) U A ( s) 1 R R 2 4 Cs 消去 U A (s) ,得: 同理可推得: X o1 (s) G1G2G3 (1 G4 ) X i1 (s) 1 G1G2 G4 G1G4G5 H1H 2 G1G2G4 X o1 (s) G1G2G3G4G5 H1 X i 2 (s) 1 G1G2 G4 G1G4G5 H1H 2 G1G2G4 解:展开F(s)得: F ( s) 1 2 2 ( s 1) 2 s 1 s 2 f (t ) (t et 2et 2e2t ) 1(t ) 2-2(6) 解: F (s) 4 s2 s 4 15 2 15 2 ) 2 8 15 4 15 F (s) 1 2 15 1 (s ) (s ) 2 ( 2 4 2 2 对方程式进行拉氏变换得: U i ( s ) I ( s ) R1 U O ( s ) 1 U ( s ) I ( s ) I ( s ) R2 O Cs U o ( s) R2Cs 1 U i ( s) ( R1 R2 )Cs 1 消去I(s),得: 8 15 2 15 f (t ) e sin t 1(t ) 15 2 t 2-2(7) F (s) 1 3 解: s 3 F ( s) 2 s 32 s 2 32 f (t ) (cos 3t 1 sin 3t ) 1(t ) 3 s 1 s2 9 2-6(b)
4
) s s e 由延时定理可得:L[ f (t )] 6 2 s 9
4 4 6 se 2 s 9 s 2-1(7) 解 f (t ) e 6t (cos8t 0.25sin 8t ) 百度文库1(t ) f (t ) e6t cos8t 1(t ) 0.25e6t sin 8t 1(t ) X o ( s) G1 (G2G3 G4 ) X i ( s) 1 (G2G3 G4 )(G1 H 2 ) G1G2 H1 (C) X o ( s) G1G2G3 G4 X i ( s) 1 G2G3 H 2 G2 H1 (1 G1 ) X o1 ( s) G1G2G3 (1 G4 ) X i1 ( s) 1 G1G2 G4 G1G4G5 H1 H 2 G1G2G4 L[ f (t )] L[e6t cos8t 1(t )] L[0.25e6t sin 8t 1(t )] s6 1 8 s 8 2 2 2 2 2 ( s 6) 8 4 ( s 6) 8 s 12s 100 2-1(8) 解 f (t ) e X o ( s) k1 Ds X i ( s) (k1 k 2 ) Ds k1k 2 2-10(a) 试求题图2-10(a) 所示无源电路网络传递函数。 ui (t ) i(t ) R1 uo (t ) 解: 1 uo (t ) c i(t )dt i(t ) R s 2 1 3 6 L[ f (t )] 5 0 2 3 e s 20 ( s 20) 2 s2 9 2 2 5 9e s 20 ( s 20) 2 s2 9 6
s 2-2 试求下列函数的拉氏反变换 2-2(5) F (s) s ( s 2)(s 1) 2 k1 ( xi x2 ) Ds( x2 xo ) Ds( x2 xo ) k 2 xo Ds k 2 xo Ds k Ds k1k 2 k1 x1 1 xo k 2 xo Ds ( k1 Ds k 2 Ds k1k 2 ) xo k1 Dsxi x2 X o 2 ( s) G4G5G6 (1 G1G2 ) X i 2 (s) 1 G1G2 G4 G1G4G5 H1H 2 G1G2G4 2—9试求题图2~9所示机械系统的传递函数。 xo ) k1 ( xi x2 ) D( x2 xo ) k 2 xo D( x2 e 2 s 006 s 2-1(6) f (t ) 6 s in(3t ) 1 (t ) 4 4 解:由 L[af (t )] aF( s) 解:由 f (t ) 6 sin[3(t 6 cos[3(t