文档介绍:第一章作业参考答案: 求下列积分值: ( a )解: 26 24 2 ) 2 ( ) 2 ( 2 ) ( ) 0 ( ) 2 ( ) ( 2 ) ( ) ( )] 2 ( 2 ) ( [ ) ( )] 2 ( 2 ) ( )[ 2 3 ( 4 4 4 4 4 4 4 4 4 4 2 = + = ?+ = ?+ = ?+ = ?+ + + ∫∫∫∫∫????? dt t x dt t x dt t t x dt t t x dt t t t x dt t t t t δδδδδδδδ(b) 解: 6 5 1 0 ) 2 ( ) 2 ( ) ( ) 0 ( ) 5 ( ) 5 ( ) 2 ( ) ( ) ( ) ( ) 5 ( ) ( )] 2 ( ) ( ) 5 ( [ ) ( )] 2 ( ) ( ) 5 ( )[ 1 ( 4 4 4 4 4 4 4 4 4 4 4 4 2 = + + = ?+ + + ?= ?+ + + = ?+ + + = ?+ + + + ∫∫∫∫∫∫?????? dt t x dt t x dt t x dt t t x dt t t x dt t t x dt t t t t x dt t t t t δδδδδδδδδδδδ( C )解: 1 ) 2 ( ) cos 1 ( ) 2 ( ) cos 1 ( 2 = ??= ??∫∫??ππππππδπδ dt t dt t t ( d )解: 4 2 3 1 2 1 2 1 2 3 1 ) (cos ) 2 3 ( ) (cos ) 2 ( ) (cos ) 2 ( ) (cos ) 2 3 ( ) (cos ) 1 ( 2 0 0 2 2 2 = + + + + ?+ ?= + + ?+ ?= + ∫∫∫∫∫?????ππππδπδπδπδπδππππππππ dt t x dt t x dt t x dt t x dt t t 解: 0 1 2 4 t x (3-2t) -1 0 1 2 4 t x (-2t) ← x (-2(t-3/2)) -3-2 -1 0 1 2 4 t x (2t) ← x (-2t) 4 x (t) ← x (2t) -5 -3 -1 0 1 2 t 课后答案网 .21 判断下列每个信号是否周期的?如果是周期的,是求它的基波周期。( a )解: 3 2 , / 2 3 ) cos( 2 ) 4 3 cos( 2 0 0 ππω?ωπ= = = + = + T T t t 基波周期为: 是周期信号(b) 解: e e e T e e e t j T t j T j T j t j T t j ) 1 ( ) 1 ) ( ( ) 1 ( ) 1 ) ( ( 1 2 ??± ± ± ??± = = = = ππππππ, 时, 当是周期信号,基波周期是 T 0 =2 (c) 解: 互质与是有理数,且 7 4 , 7 4 2 7 8 2 ) 2 cos( ) 2 7 8 cos( = = ?+ ?= + ππππ n n 所以原式是周期信号,基波周期 N 0 =7. (d) 解: 不是有理数, , 8 1 2 4 1 2 cos 4 cos πππ= = ??= n n 所以原式不是周期信号( e )解: 。有为整数, 其中则令