已知:R=3,L=0.5HC=0.25F输入为冲激画飘,求h(t)=v(t) vn(t)+v/()+V2(t)=6(1) t VI R ∥)d8+()=0(0)① 2+rc dt 因t=0电路处于零状态:故 (0)=00)=0 8t>0时,()=0 dv (t di C +rC-y+v()=0 (0+) 8 d t C (O+)=0 ±/() 3±1 v(t)=ke 2t 4t +ke v2(0+)=0=k1+k2 h()=(4e2-4e4)u()(0)=8 2k1-4k
L 已知:R=3,L=0.5H,C=0.25F输入为冲激函数,求h(t)=vc(t)? vc (t) R + vl - c + - v (t) v (t) v (t) (t) r l c + + = ( ) ( ) ( ) ( ) 2 2 v t t dt dv t rc dt d v t lc c c c + + = 因t=0-电路处于零状态：故 (0 ) = 0 − vc 0 (0 ) = − dt dvc 因t>0时, (t) = 0 ( ) 0 ( ) ( ) 2 + + v t = dt dv t rc dt d v t lc c c c 3 1 1 ) 2 ( 2 2 1,2 = − − = − l lc r l r p t t c v t k e k e 4 2 2 1 ( ) − − = + 8 1 (0 ) ' = = + lc vc (0 ) = 0 + vc ( ) (4 4 ) ( ) 2 4 h t e e u t − t − t = − 1 2 ' 1 2 (0 ) 8 2 4 (0 ) 0 v k k v k k c c = = − − = = + + +
结语:(对于右边只有δ(1)的情况) 1.写出激励和响粒系的微分方程 2t<0,h(t)=0;t>0,h(t)是一个特淼的zi 3.t=0时的初始条件是由子七0时冲激信号作用的结 h")(O+) h10(0)=…=h(O)=h(0+)=0 P69;(§28用算子符号表示微分方程)p83:2-9 a,hon(t)+am-hn-(t)++a, h(t)+ao=bm 8(t)+ .8(t)+b8(t) dt 用算子符号表示上式 (anp"+an1p+…+a1p+a0)h(t)=(bnp"+…+bp+b)6() 设 np +41p+ P+…+b1p+
*.结语:(对于右边只有δ(t)的情况) 1.写出激励和响应关系的微分方程. 2.t<0,h(t)=0;t>0,h(t)是一个特殊的z.i.r. 3. 时的初始条件是由于t=0时冲激信号作用的结果. n n a h 1 (0 ) ( ) = + (0 ) ..... (0 ) (0 ) 0 ( 1) ' = = = = − + + + h h h n P69;(§2.8用算子符号表示微分方程）p83;2-9 ( ) ( ) ... ( ) ( ) ... ( ) ( ) 0 ' 0 1 ' 1 ( 1) 1 ( ) a h t a h t a h t a b t b t b t m m n n n n + + + + = + + − − dt d p = 用算子符号表示上式 ( ... ) ( ) ( ... ) ( ) 1 0 1 0 1 1 a p a p a p a h t b p b p b t m m n n n n + + + + = + + + − − 设: 1 0 l a p ... a p a n = n + + + 1 0 l b p ... b p b m d = m + + + = 0+ t