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 tx d 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 tu t = and 2 ( 1 ) xt u t =− , which satisfies 12 x t = for any 1 t > . If this system is memoryless, we may find that y ty t = for any 1 t > , but 1 () 2 y tt = for 0 t > (and thus for 1) t > , and 2 () 2 1 y for 0.5 t > (and thus for t > . [ Another example : see arguments for causality on the below.] (ii) Time-Invariance: No. Example : considering 1 x t = and 21 ) ) xt xt u t = −= − , and their outputs 1 y = and 2 y = . If the system is time-invariant, we may find that ) 2 ) 2 2 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 [] 22 2 2 2 1 2 1 2 t t t ax bx d ax d bx d a x d b x d τ τ ττ −∞ −∞ −∞ −∞ −∞ + =+=+ ∫∫ ay t by t =+ . (iv) Causality: No. Example : considering 1 x t = and 2 () ( 1 ) tu t = −+ , which satisfies x t = for any 1 t < . If this system is causal, we may find that y t = for any 1 t < , but 1 y = for 0.5 1 t << , and 2 () 1 yt = for 0.5 1 t < < . (v) Stability: No. Example : consider ( ) ( ) x t = , which is a bounded signal ( 1 B = ). The output is () 2 () tut = , which is not a bounded signal. 2. [8] Use the following operational definition of ( ) t δ , ()* () x x t = , any x t , show the following properties of ( ) t : (i) [4] 1 td t −∞ = (ii) [4] 00 0 ( ) ( ) x t x t t t −= , any x t Solution: (i) Let , xt = then ( )() dx t d x t t x t δτ τ τδτ τ ∞∞ −∞ −∞ = = = , or 1 t −∞ = . (ii) Since ( )* ( ) ( ) x x t = , then 0 * ( ) ( ) ( ) x x t x t t d τδ −∞ = +− . By letting 0 t = , we may find that P 0 0 0 0 ( ( )( ) t x t d x t d τ δτ + −∞ −∞ = ± , or ( )( ) x t t t d t −∞ . Now 0 0 ( ) ( ) * ( ) ()( ) ) ) t t t x ttd x d δδ τδτ δτ −∞ −∞
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3. [12] An echo generating environment can be modeled by the following system: or expressed as: () ( ) y tx ta y t T = +− . Assuming initially at rest, i.e., ( ) 0, 0 y tt = < if ( ) 0, 0 xt t = < . (i) [4] Find the impulse response ( ) ht of this system. (ii) [8] Does there exist an inverse system which can cancel the echoes? If yes, write down its impulse response gt and a block diagram as in the above. If no, explain why. Solution: (i) We have that ( ) ( ) ( ) t aht T δ =+− . Since the input is an impulse and the output delays the impulse again and again for time multiple of T , the impulse response needs to be 0 ( ) k k h t kT = =− . Now we have 00 ( ( 1 ) ) ( ) kk htk Tahtk T t δδ ∞∞ == −− + = ∑∑ , so k k ha = , and then 0 ( ) k k a t kT = .
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This note was uploaded on 02/21/2012 for the course EE 101 taught by Professor 張捷力 during the Spring '07 term at National Taiwan University.

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94下信號與系統

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