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