Solution_System

# Solution_System - Problem(1 Solution 3(a The system...

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Problem (1) Solution: (a) The system described by 3 [ ] [ ] y n n x n = is linear, causal, unstable and time-variant. For the input 1 [ ] x n , the output is 3 1 1 [ ] [ ] y n n x n = ; For the input 2 [ ] x n , the output is 3 2 2 [ ] [ ] y n n x n = ; For the input 3 1 2 [ ] [ ] [ ] x n ax n bx n = + , the output is { } 3 3 3 3 3 3 1 2 1 2 1 2 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] y n n x n n ax n bx n n ax n n bx n ay n by n = = + = + = + ; So the system is linear. Since the current output does not depend on the future inputs, the system is causal. If [ ] x n is bounded, the [ ] y n is not bounded as n → ∞ . So the system is unstable. For the input 1 0 [ ] [ ] x n x n n = - , the output is 3 3 1 1 0 [ ] [ ] [ ] y n n x n n x n n = = - . While 3 0 0 0 1 [ ] ( ) [ ] [ ] y n n n n x n n y n - = - - , the system is time-variant. (b) The system described by 3 0 [ ] [ ] k y n b x n k = = + - is nonlinear, causal, stable and time-invariant. For the input 1 [ ] x n , the output is 3 1 0 [ ] [ ] k y n b x n k = = + - ; For the input 2 [ ] x n , the output is 3 2 0 [ ] [ ] k y n b x n k = = + - ; For the input 3 1 2 [ ] [ ] [ ] x n ax n cx n = + , the output is 3 3 3 3 3 1 2 1 2 0 0 0 [ ] [ ] [ ] [ ] [ ] [ ] k k k y n b x n k b ax n k cx n k ay n cy n = = = = + - = + - + - + ; So the system is nonlinear. (Note: the superposition holds only when a + c =1) Since the current output does not depend on the future inputs, the system is causal. If [ ] x n < ∞ , then the output [ ] y n < ∞ . So the system is stable. For the input 1 0 [ ] [ ] x n x n n = - , the output is 3 3 1 1 0 0 0 0 [ ] [ ] [ ] [ ] k k y n b x n k b x n k n y n n = = = + - = + - - = - . So the system is time-invariant. (c) The system described by [ ] [ 2] y n x n = - + is linear, noncausal, stable and time-invariant. For the input 1 [ ] x n , the output is 1 1 [ ] [ 2] y n x n = - + ; For the input 2 [ ] x n , the output is 2 2 [ ] [ 2] y n x n = - + ; For the input 3 1 2 [ ] [ ] [ ] x n ax n bx n = + , the output is 3 3 1 2 1 2 [ ] [ 2] [ 2] [ 2] [ ] [ ] y n x n ax n bx n ay n by n = - + = - + + - + = + ; So the system is linear. The outputs depend on the future value of input. For example,

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Solution_System - Problem(1 Solution 3(a The system...

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