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20071ee102_1_HW-2.SOLS

# 20071ee102_1_HW-2.SOLS - W 2007 EE102 SYSTEMS SIGNALS HW#2...

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W 2007 EE102: SYSTEMS & SIGNALS HW#2: SOLUTIONS 0. Let x ( · ) and y ( · ) denote input and corresponding output of a system S , respectively. Given the following IPOP relation: y ( t )= Z −∞ e t ( t τ ) U ( τ t ) x ( τ ) dτ, t ( −∞ , ) Verify whether S is: L, TI, Causal . Solutions a) S is clearly L. b) S is TV since y ( t Z −∞ e t ( t τ ) U ( τ t ) x ( τ ) = T [ x ( t )] ,t ( −∞ , ) From which it follows that y ( t A Z −∞ e ( t A ) ( t A τ ) U ( τ ( t A )) x ( τ ) dτ, t ( −∞ , ) Next z ( t T [ x ( t A )] = Z −∞ e t ( t τ ) U ( τ t ) x ( τ A ) dτ, t ( −∞ , ) Therefore it is easy to check that z ( t ) 6 = y ( t A ) S is TV. c) We have U ( t τ )=0 , for t τ< 0 t<τ. Therefore y ( t ) can be rewritten as y ( t Z t e t ( t τ ) x ( τ ) dτ, t ( −∞ , ) This shows that y ( t ) depends on all (future) values of input x ( τ ) — for τ in [ t, ). Hence S is NC. 1

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1. (i) Use BT to write down the IRF h ( t, τ ) of the system of Problem 0. (ii) Given the IRF h ( t, τ ) of a L,TI,C system: h ( t, τ )= e ( t τ ) ( t τ ) U ( t τ ) , for t, τ ( −∞ , ) , and the input x ( t δ ( t 2) e t U ( t ) ,t ( −∞ , ) .
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20071ee102_1_HW-2.SOLS - W 2007 EE102 SYSTEMS SIGNALS HW#2...

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