Midterm_Slns_Fall_09

Midterm_Slns_Fall_09 - Midterm Solution ECE 161A Fall 2009...

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Midterm Solution ECE 161A, Fall 2009 1. (12 points) W [ k ] = X R [ k ] = 1 2 ( X [ k ] + X * [ k ]) . IDFT of X * [ k ] is 1 N N - 1 X k =0 X * [ k ] e j 2 π N nk = 1 N N - 1 X k =0 X [ k ] e - j 2 π N nk ! * = 1 N N - 1 X k =0 X [ k ] e j 2 π N ( - n ) k ! * = x * [( - n ) N ] Using this result w [ n ] = IDFT( 1 2 ( X [ k ] + X * [ k ])) = 1 2 (IDFT( X [ k ]) + IDFT( X * [ k ])) = 1 2 ( x [ n ] + x * [( - n ) N ]) 2. (a) (10 points) x 1 [ n ] : { 4 , 3 , 2 , 1 , 0 , 0 } x 2 [ n ] : { 2 , 1 , 0 , 1 , 0 , 0 } x 2 [( - n ) 6 ] : { 2 , 0 , 0 , 1 , 0 , 1 } y 6 [0] = 4 · 2 + 3 · 0 + 2 · 0 + 1 · 1 + 0 · 0 + 0 · 1 = 9 x 2 [(1 - n ) 6 ] : { 1 , 2 , 0 , 0 , 1 , 0 } y 6 [1] = 4 · 1 + 3 · 2 + 2 · 0 + 1 · 0 + 0 · 1 + 0 · 0 = 10 Similarly, y 6 [2] = 3 + 4 = 7 y 6 [3] = 4 + 2 + 2 = 8 y 6 [4] = 3 + 1 = 4 y 6 [5] = 2 y 6 [ n ] = { 9 , 10 , 7 , 8 , 4 , 2 } From class, the samples between ˜ N - N and N - 1 will not be aliased, where ˜ N is the length of the signal when performing linear convolution. Here, ˜ N = 4 + 4 - 1 = 7 and N = 6. Thus, the samples 1 n 5 will correspond to linear convolution. (b) (8 points) Here, since 10 7, all the points will correspond to linear convolution. The linear convolution x 1 [ n ] * x 2 [ n ] (extrapolated to 10 points) is x 1 [ n ] * x 2 [ n ] = { 8 , 10 , 7 , 8 , 4 , 2 , 1 , 0 , 0 , 0 } 1

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3. (a) (8 points) The simplest way to do this problem is to write cos( x ) and sin( x
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