W08-102-HW-1 - y t as y t = Z 1 −1 dσ where is TBD(To Be...

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WINTER 2008: Put First Letter * of LAST Name in the corner →→%% (* Otherwise Your HW will be LOST) PRINT: (LAST, Middle, First):——————————————————– EE102: SYSTEMS & SIGNALS HW: # 1 A LATE HW IS NEVER A HW! Posted: January 9 Hand In To NL: January 16 Attach This Sheet To Your HW 1. Consider the di ff erential equation: dy ( t ) dt + y ( t ) = dx ( t ) dt x ( t ) , t 0 , y (0) = 0 , x (0) = 0 . (i) Solve for y ( · ) in terms of x ( · ). (ii) Calculate y ( · ) given that x ( t ) = (1 e 2 t ) U ( t ) . 2. A system S is described by the following IPOP description: y ( · ) = T [ x ( · )] , y ( t ) = Z 1 −1 ( t τ ) U ( τ t ) x ( τ ) d τ , t > −1 , where x ( · ) is input and y ( · ) is output. Is S : TI or TV? C or NC? 3. Compute: I 1 ( t ) = Z 1 −1 ( t τ ) e ( t τ ) U ( t τ ) U ( τ 1) d τ , I 2 = Z 5 1 e 4 t U ( t ) δ ( t 2) dt. 1
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4. A system has the IPOP description: y ( · ) = T [ x ( · )] , y ( t ) = x ( t ) 2 Z 1 t e t e σ x ( σ ) d σ , t > −1 . (i) Rewrite y ( t ) as
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Unformatted text preview: y ( t ) as y ( t ) = Z 1 −1 [ ? ] dσ where [ ? ] is TBD (To Be Determined). (ii) State all properties of the system (L, NL, TI, TV, C, NC). (ii) Find y ( · ) given that x ( t ) = t U ( t ) . 5. Given dy ( t ) dt + 1 t + 1 y ( t ) = x ( t ) , t > , y (0) = . (i) Solve for y ( · ) in terms of x ( · ) . (ii) If the solution you found in (i) is the IPOP description of a system S . Verify that S is TV. 6. A system admits the IPOP description: y ( · ) = T [ x ( · )] , y ( t ) = Z t −1 ( t − σ ) e − t e σ x ( σ ) dσ, t ≥ −1 . (i) Is this system TV or TI? (ii) Find y ( · ) given that x ( t ) = e − 2 t U ( t + 1) 2...
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