第五章|连续时间信号的数字处理 例 xa (t)=cos Q2t x(j2) x,()=∑cos(9m7)6(t-m7) X(2)=X(92)*S(92) 2丌 2丌 r[6(g-g2)+(2+_20 2丌 ∑6 k=-0 ∑[(2-90-kg2)+δ(9+s x(2) 丌 (23-g2)-92-90g29293-92939 92-2+994-92092920 Q Q 数字信号处理精品课程2 2 0 0 s s T T   > =   < =   [ ( ) ( )] ( ) 2 [ ( ) ( )] 2 1 ( ) ( ) 2 1 ( ) ( ) cos( ) ( ) ( ) cos . 0 0 0 0 0 0 s k s k s p c k p a k k T k T X j X j S j x t nT t nT x t t =  − −  +  +  −  = •  − +  +    −   =    =  − =      =−  =−  =−            − 0 0  X ( j) a −  X ( j) p -(s − 0 )- s /2 − 0 0 s /2 s - 0 s  T  X ( j) p − 0 - s /2 - s + 0 s - 0 s /2 0 s  T  例：