解:1=∑[(-3n)-(-3n-2) n=0 f2= sin tu(t) f()=∑[6(t-3n)-06(-3n-2) n=0 2( (1-cos rt)u(t) r(t)=f(t)*f2(t)dt 4
解: 0 1 [ ( 3 ) ( 3 2)] n f u t n u t n sin ( ) 2 f tu t 0 ' 1 ( ) [ ( 3 ) ( 3 2)] n f t t n t n (1 cos ) ( ) 1 ( ) 2 f d t u t t r t f t f d t ( ) ( ) ( ) 2 ' 1 2 3 5 ( ) ' 1 f t 1 2 4
r(t)=-(1-cosm)n()*>[6(t-3m)-(t-3n-2 n=0 =(1-cosm)()-(-2)*∑8(=3m) 2一 01234567t
0 0 (1 cos )[ ( ) ( 2)] ( 3 ) 1 (1 cos ) ( ) [ ( 3 ) ( 3 2)] 1 ( ) n n t u t u t t n r t t u t t n t n