HW4 solution - Problem2: From figure P3.28, we can get (1)...

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Problem2: From figure P3.28, we can get (1) The period N = 6. (2) x[0]=1, x[1]=2, x[2]= 1, …., x[6]=1 Therefore, we can obtain a k as follows: 1 0 12 [ ]exp( ) 6 2 [1 2exp( 1) ( 1)exp( 2) 66 22 0 ( 1)exp( 4) 2exp( 5)] 2 {1 2[cos( ) sin( )] ( 1)[cos( ) sin( )] 33 3 3 44 5 ( 1)[cos( ) sin( )] 2[cos( ) sin 3 N k n ax n j k n N jk N kj k k N k        5 ( )]} (1) 3 42 5 cos( ) ( ),cos( ) ( ) (2) 3 3 5 sin( ) sin( ),sin( ) sin( ) (3) 3 3 (2) (3) g [1 4 cos( ) 2 cos( )] k k Notice that kc o s k o sk kk k k take and into then you will et ak k N   and N = 6
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Problem3: From Figure P3.29(b): We can obtain 012 34 7 11 2, 1, , , 0, . .. , 2 22 aa a a   Thus, x[n] can be described as follows. 7 0 2 [] e x p ( ) 8 1 2 1 2 2 exp( 0 ) 1 exp( 1 ) exp( 2 ) exp( 3 ) 8 82848 12 2 0 exp( 5 ) exp( 6 ) 1 exp( 7 ) 4828 8 1 2 1 exp( 1 ) exp( 2 ) exp( 3 ) 42 4 4 4 1 0e x p 4 k n xn a j kn jn j n j n j n           1 ( 5 ) exp( 6 ) 1 exp( 7 ) 424 4 213 3 2 1 [cos( ) sin( )] [cos( ) sin( )] [cos( ) sin( )] 44 2 4 4 4 4 4 15 516 6 0 [cos( ) sin( )] [cos( ) sin( )] 4 24 4 77 1[ c o s ( ) s i n ( ) ] ( 1 ) nj n n n n n n N    
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HW4 solution - Problem2: From figure P3.28, we can get (1)...

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