信号与系统奥本海默答案

Chapter 1 Answers 1.1 Converting from polar to Cartesian coord      2 cos( ) sin( ) 2 2 j e j j       2 c o s ( ) s i n ( ) 2 2 j e j j         5 2 2 j j e e j     4 2 ( c o s ( ) s i n ( ) ) 1 2 4 4 j e j j        9 2 2 4 4 1 j j e e j      9 2 2 4 4 1 j j e e j        2 4 1 j e j    1.2 converting from Cartesian to polar coordinates: 0 5 5  e j ,   2 2e j ,   3 3 j e j2 2 1 3 2 2 j j e     , 1 2   j e j4 , 1 j2  2e j2 j j (1 ) e 4    , 1 4 1 j j e     12 2 2 1 3 j j e      1.3. (a) E  = 4 0 1 4 t e dt     , P  =0, because E    (b) (2 ) 4 x e 2( ) t  j t , x2( ) 1 t  .Therefore, E  =  x2( ) t 2dt =  dt = , P  = lim lim 1 1 2 2 T T   2 2 T T     T T T T x ( ) t dt dt   Tlim1 1   (c) x 2( ) t =cos(t). Theref 2 2 1 3 j j e      1.3. (a) E  = 4 0 1 4 t e dt     , P  =0, because E    (b) (2 ) 4 x e 2( ) t  j t , x2( ) 1 t  .Therefore, E  =  x2( ) t 2dt =  dt = , P  = lim lim 1 1 2 2 T T   2 2 T T     T T T T x ( ) t dt dt   Tlim1 1   (c) x 2( ) t =cos(t). Theref