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94下信號與系çµ&plusmn

94下信號與系統

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Signals and Systems, Midterm Exam Solutions (Draft) Spring 2006, Edited by bypeng 1. [10] A system with output signal ( ) y t given input signal ( ) x t as below: 2 ( ) ( ) t y t x d τ τ −∞ = Is this system memoryless [2] , time-invariant [2] , linear [2] , causal [2] , or stable [2] ? Justify your answers. Solution: (i) Memoryless property: No. Example : considering 1 ( ) ( ) x t u t = and 2 ( ) ( 1) x t u t = , which satisfies 1 2 ( ) ( ) x t x t = for any 1 t > . If this system is memoryless, we may find that 1 2 ( ) ( ) y t y t = for any 1 t > , but 1 ( ) 2 y t t = for 0 t > (and thus for 1) t > , and 2 ( ) 2 1 y t t = for 0.5 t > (and thus for 1) t > . [ Another example : see arguments for causality on the below.] (ii) Time-Invariance: No. Example : considering 1 ( ) ( ) x t u t = and 2 1 ( ) ( 1) ( 1) x t x t u t = = , and their outputs 1 ( ) 2 y t t = and 2 ( ) 2 1 y t t = . If the system is time-invariant, we may find that 2 1 ( ) ( 1) 2( 1) 2 2 y t y t t t = = = , which is a contradiction. (iii) Linearity: Yes. For any signals 1 ( ) x t and 2 ( ) x t , and their outputs 1 ( ) y t and 2 ( ) y t , we have that [ ] 2 2 2 2 2 1 2 1 2 1 2 ( ) ( ) ( ) ( ) ( ) ( ) t t t t t ax bx d ax d bx d a x d b x d τ τ τ τ τ τ τ τ τ τ τ −∞ −∞ −∞ −∞ −∞ + = + = + 1 2 ( ) ( ) ay t by t = + . (iv) Causality: No. Example : considering 1 ( ) ( ) x t u t = and 2 ( ) ( ) ( 1) x t u t u t = − + , which satisfies 1 2 ( ) ( ) x t x t = for any 1 t < . If this system is causal, we may find that 1 2 ( ) ( ) y t y t = for any 1 t < , but 1 ( ) 2 y t t = for 0.5 1 t < < , and 2 ( ) 1 y t = for 0.5 1 t < < . (v) Stability: No. Example : consider ( ) ( ) x t u t = , which is a bounded signal ( 1 B = ). The output is ( ) 2 ( ) y t tu t = , which is not a bounded signal. 2. [8] Use the following operational definition of ( ) t δ , ( )* ( ) ( ) x t t x t δ = , any ( ) x t , show the following properties of ( ) t δ : (i) [4] ( ) 1 t dt δ −∞ = (ii) [4] 0 0 0 ( ) ( ) ( ) ( ) x t t t x t t t δ δ = , any ( ) x t Solution: (i) Let ( ) 1, x t = then ( ) ( ) ( ) ( )* ( ) ( ) 1 d x t d x t t x t δ τ τ τ δ τ τ δ −∞ −∞ = = = = , or ( ) 1 t dt δ −∞ = . (ii) Since ( )* ( ) ( ) x t t x t δ = , then 0 0 0 ( ) ( )* ( ) ( ) ( ) x t t x t t t x t t d δ τ δ τ τ −∞ = = + . By letting 0 t = , we may find that P 0 0 0 0 0 0 0 0 ( ) ( ) ( ) ( ) ( ) t x t x t d x t d τ τ τ δ τ τ τ δ τ τ + −∞ −∞ = + = ± , or 0 0 ( ) ( ) ( ) x t x t t t dt δ −∞ = . Now 0 0 0 0 0 0 ( ) ( ) ( ) ( )* ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) x t t t x t t t t x t t d x t t d x t t t δ δ δ τ δ τ δ τ τ τ δ τ δ τ τ δ −∞ −∞ = = = =
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