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Midterm Solution

# Midterm Solution - EL6253 Linear Systems Midterm Solutions...

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EL6253 — Linear Systems — Midterm (03/23/2012) — Solutions Problem 1 Consider the system described by the following zero-state response y ( t ) = S { u ( t ) } = exp( t ) i t +2 t - 2 u ( τ 1) exp( τ 1) Determine whether the system is or is not (justify your answer): a) zero-state linear b) causal c) Fxed (time-invariant) Solution: y ( t ) = exp( t ) i t +2 t - 2 u ( τ 1) exp( τ 1) = exp( t ) i t +1 t - 3 u ( β ) exp( β ) a) It is zero-state linear since for u ( t ) = α 1 u 1 ( t ) + α 2 u 2 ( t ) y ( t ) = α 1 p exp( t ) i t +1 t - 3 u 1 ( β ) exp( β ) P b ² y 1 ( t ) + α 2 p exp( t ) i t +1 t - 3 u 2 ( β ) exp( β ) P b ² y 2 ( t ) b) It is non-causal, since y ( t ) depends on the evaluation of u ( β ) for β [ t 3 ,t + 1] c) It is time-varying, because for u ( t T ), y ( t ) = exp( t ) i t +1 t - 3 u ( β T ) exp( β ) = exp( t + T ) i t - T +1 t - T - 3 u ( λ ) exp( λ ) n = y ( t T ) Problem 2 ±ind the A , B , C , and D matrices for the following linear systems a) ¨ y + 3 ˙ y + y = 5¨ u + 7 ˙ u + 9 u b) y ( k + 2) + k 2 y ( k + 1) + 3 y ( k ) = u ( k + 1) + ku ( k ) c) ¨ y + t 2 ˙ y + 3 y = ˙ u + tu Solution: a) ¨ y + 3 ˙ y + y = 5¨ u + 7 ˙ u + 9 u = D 2 y + 3 Dy + y = 5 D 2 u + 7 Du + 9 u 1

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y = D - 2 ( 5 D 2 u + 7 Du 3 Dy + 9 u y ) = D - 1 ( 7 u 3 y + D - 1 (9 u y ) ) + 5 u ˙ x 1 = 9 u y, ˙ x 2 = 7 u 3 y + x 1 , y = x 2 + 5 u ˙ x 1 = x 2 + 4 u, ˙ x 2 = x 1 3 x 2 8 u A = b 0 1 1 3 B , B = b 4 8 B , C = ± 0 1 ² , D = ± 5 ² , b) y ( k + 2) + k 2 y ( k + 1) + 3 y ( k ) = u ( k + 1) + ku ( k ) = E 2 y + k 2 Ey + 3 y = Eu + ku y = E - 2 ( k 2 + ku 3 y ) = E - 1 ( u E - 1 ( k 2 ) + E - 1 ( ku 3 y ) ) = E - 1 ( u ( k 1) 2 y + E - 1 ( ku 3 y ) ) x 2 ( k ) = y, x 1 ( k + 1) = ku 3 x 2 ( k ) , x 2 ( k + 1) = u ( k 1) 2 x 2 ( k ) + x 1 ( k ) A = b 0 3 1 ( k 1) 2 B , B = b k 1 B , C = ± 0 1 ² , D = ± 0 ² , c) ¨ y + t 2 ˙ y + 3 y = ˙ u + tu = D 2 y + t 2 Dy + 3 y = Du + tu y = D - 2 ( Du t 2 Dy + tu 3 y ) = D - 1 ( u D - 1 ( t 2 Dy ) + D - 1 ( tu 3 y ) ) D - 1 ( t 2 Dy ) = t 2 y D - 1 (2 ty ) y = D - 1 ( u t 2 y + D - 1 ( tu + 2 ty 3 y ) ) x 2 = y, ˙ x 1 = tu + 2 tx 2 3 x 2 , ˙ x 2 = u t 2 x 2 + x
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Midterm Solution - EL6253 Linear Systems Midterm Solutions...

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