Winter_2007_Midterm_Solution_1_8

# Winter_2007_Midterm_Solution_1_8 - EE 113 Midterm Solution...

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EE 113 Midterm Solution Winter 2007 Inst: Dr. C.W. Walker Problem Points Score 1 9 2 15 3 18 4 18 5 10 6 10 7 10 8 10 Total 100

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EE113 Winter 2007 Midterm Solution (2/14/2007) (a) Non-linear () () [] () ( ) () b a n bx n ax n x bT n x aT n bx n ax n bx n ax T + + + = + + + = + 2 1 1 1 2 1 2 1 1 Q Time-invariant ( ) k n y k n x k n x T = + = 1 Q BIBO stable ( ) bounded also is which N M n x n x n y M n x input bounded for < = + + + = < 1 1 ) ( 1 ) ( ) ( Q (b) Non-linear ( ) [ ] ( ) [ ] b a n x bT n x aT n bx n ax T + = + = + 1 1 2 1 1 Q Time-invariant [] () k n y k n x T = = 1 Q
BIBO stable ( ) bounded also is which n y M n x input bounded for < = < 1 ) ( Q (c) Linear () () [] ( ) ( ) () n x b n n x a n n x T b n x T a n x b n x a n n x b n x a T 2 1 1 1 2 1 2 1 ) ( + = + = + = + Q Time-variant [] () ( ) k n y k n x k n k n x n k n x T = = ) ( ) ( Q Not BIBO stable ( ) bounded not is which n when n x n n x n n y M n x input bounded for = = < ) ( ) ( ) ( Q (a) We can use unilateral Z-transform to solve this problem. Perform the unilateral Z-transform on the above LCCDE (linear constant coefficient difference equation), we got

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[ ] [ ] 4 1 4 3 3 1 9 26 2 1 4 15 6 11 4 1 4 3 3 1 9 26 2 1 4 15 6 11 4 1 4 3 3 1 9 26 2 1 4 15 6 11 6 1 6 5 1 4 6 1 6 5 ) ( , 4 6 1 6 5 ) ( ) 6 1 6 5 1 ( ) 1 ( , 4 ) ( 0 ) 2 ( , 1 ) 1 ( ) 1 ( ) ( ) 1 ( ) 1 ( ) ( 6 1 ) 1 ( ) ( 6 5 ) ( 1 1 1 2 1 1 1 2 1 2 2 1 + + = + + + = + + + = + + = + = + = = = = + + + + + + + + + + + Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Y therefore Z Z Z Z Y Z Z to rearranged be can Z Z Z X and y y Z X Z y Z y Z Y Z Z y Z Y Z Z Y Q Therefore, () 1 ] 4 1 4 3 3 1 9 26 2 1 4 15 [ ) ( 6 11 1 1 1
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Winter_2007_Midterm_Solution_1_8 - EE 113 Midterm Solution...

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