Tuto2_sol - 1 n n y n n x n x n y − = − − = − = ....

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EE3210 Tutorial 2 (Solution) 1. The output y [ n ] of the unit delay element is ] 1 [ ] 1 [ ] [ = n y n x n y . Rearranging, we obtain ] 1 [ ] 1 [ ] [ = + n x n y n y . 2. (a) No (b) 0 (c) No 3. Unit-delay system: a) Since the output value at n depends on the input values at n -1, the system is not memoryless. b) Since the output does not depend on the future input values, the system is causal. c) Let . Then ] [ ] [ ] [ 2 1 n bx n ax n x + = ] [ ] [ ] 1 [ ] 1 [ ] [ 2 1 2 1 n by n ay n bx n ax n y + = + = where y 1 [ n ] and y 2 [ n ] are the outputs corresponding to x 1 [ n ] and x 2 [ n ], respectively. The superposition property is satisfied and hence the system is linear. d) Let y 1 [ n ] be the response to x 1 [ n ] = x [ n - n 0 ]. Then ] [ ] 1 [ ] 1 [ ] [ 0 0 1
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Unformatted text preview: 1 n n y n n x n x n y − = − − = − = . Therefore, the system is time-invariant. e) If C n x ≤ ] [ for all n , then C n x n y ≤ − = ] 1 [ ] [ for all n . Hence the system is stable. 4. (a) No; e.g., ) ( ) ( x y = − π (b) Yes. The reason is shown below: Let and )) (sin( ) ( 1 1 t x t y = )) (sin( ) ( 2 2 t x t y = . Consider the following input signal: ) ( ) ( ) ( 2 1 t bx t ax t x + = . The corresponding output is: ) ( ) ( )) (sin( )) (sin( )) (sin( ) ( 2 1 2 1 t by t ay t bx t ax t x t y + = + = =...
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