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EE 20N - Fall 2002 - Lee - Final Solutions

# EE 20N - Fall 2002 - Lee - Final Solutions - EECS 20 Final...

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EECS 20. Final Exam Solutions December 19, 2002. 1. (a) d (b) a (c) none (d) c (e) e (f) b 2. (a) The solution is: ( t , f ) ( f , f ) ( t , t ) ( f , t ) { f }/ t { f }/ t { t }/ f { t }/ t { t }/ t { t }/ f { t }/ f { t }/ t { f }/ f { f }/ f { f }/ f { t }/ t { t }/ f { f }/ f { f }/ t { f }/ t (b) ( f,f,f,t,f,f, ··· ) . (c) The solution is: 1

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t f { f }/ f { t }/ t { t }/ f { f }/ t The bisimilarity relation is { (( t, f ) ,t ) , (( t, t ) ) , (( f,t ) ,f ) , (( f,f ) ) } . 3. (a) false (b) true (c) true (d) false (e) true (f) true (g) true 4. (a) p =8 and ω 0 = π/ 4 radians/sample. (b) K =4 , A 0 = A 1 = A 2 =1 , A 3 = A 4 =0 , φ 1 , φ 2 = 2 , and the values of φ 3 and φ 4 do not matter. (c) X 0 , X 1 / 2 , X 2 / 2 i = i/ 2 , X 3 = X 4 = X 5 , X 6 = 1 / 2 i = i/ 2 , and X 7 / 2 . (d) The output is y ( n )=1+sin( πn/ 4) cos( 2) . (e) The frequency response of the composite is ω Reals ,H 0 ( ω )=1+ iH ( ω ) , 2
by linearity, or H 0 ( ω )= 2 if 0 <ω<π 1 if ω =0 or ω = π 0 if π<ω< 0 For other values of ω , the value of H 0 ( ω ) is determined by the fact that it is periodic with period 2 π . (f) The output is y ( n )=1+ e iπn/ 4 ie iπn/
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EE 20N - Fall 2002 - Lee - Final Solutions - EECS 20 Final...

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