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HW4 solution - Problem2 From figure P3.28 we can get(1 The...

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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 1 2 [ ]exp( ) 6 1 2 2 [1 2exp( 1) ( 1)exp( 2) 6 6 2 2 0 ( 1)exp( 4) 2exp( 5)] 6 6 1 2 2 {1 2[cos( ) sin( )] ( 1)[cos( ) sin( )] 3 3 3 3 4 4 5 ( 1)[cos( ) sin( )] 2[cos( ) sin 3 3 3 N k n a x n j kn N j k j k N j k j k k j k k j k N k j k k j       5 ( )]} (1) 3 4 2 5 cos( ) ( ),cos( ) ( ) (2) 3 3 3 3 4 2 5 sin( ) sin( ),sin( ) sin( ) (3) 3 3 3 3 (2) (3) (1) g 1 2 [1 4cos( ) 2cos( )] 3 3 k k Notice that k cos k k cos k k k k k take and into then you will et a k k N     and N = 6
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Problem3: From Figure P3.29(b): We can obtain 0 1 2 3 4 7 1 1 2, 1, , , 0, ... , 2 2 2 a a a a a a Thus, x[n] can be described as follows. 7 0 2 [ ] exp( ) 8 2 2 1 2 1 2 2 exp( 0 ) 1 exp( 1 ) exp( 2 ) exp( 3 ) 8 8 2 8 4 8 1 2 1 2 2 0 exp( 5 ) exp( 6 ) 1 exp( 7 ) 4 8 2 8 8 1 1 1 2 1 exp( 1 ) exp( 2 ) exp( 3 ) 4 2 4 4 4 1 0 exp 4 k n x n a j kn j n j n j n j n j n j n j n j n j n j n           1 ( 5 ) exp( 6 ) 1 exp( 7 ) 4 2 4 4 1 2
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