Page 5 - PowerPoint Presentation

P 1 (  2X  1 ) 4  P 0 (  2X  ) 3

3
 ( P 0  X  ) K1

2

2
 F ( )

3

  3  2 

 2   

2  2   3   2 
  4       K1

3   3   2  3 
 

 

 . 0 73125 J1