Midterm Solution

x1 k 1 ku 3x2 k a 0 3 1 k 12 b k 1 c

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Unformatted text preview: u + ku y =E −2 E u − k 2 Ey + ku − 3y =E −1 u − E −1 (k 2 Ey ) + E −1 (ku − 3y ) =E −1 u − (k − 1)2 y + E −1 (ku − 3y ) x2 (k ) = y, x2 (k + 1) = u − (k − 1)2 x2 (k ) + x1 (k ) x1 (k + 1) = ku − 3x2 (k ), A= 0 −3 , 1 −(k − 1)2 B= k , 1 C= 0 1 , D= 0 , c) y + t2 y + 3y = u + tu = D2 y + t2 Dy + 3y = Du + tu ¨ ˙ ˙ y =D−2 Du − t2 Dy + tu − 3y =D−1 u − D−1 (t2 Dy ) + D−1 (tu − 3y ) D−1 (t2 Dy ) = t2 y − D−1 (2ty ) y = D−1 u − t2 y + D−1 (tu + 2ty − 3y ) x2 = y, A= x1 = tu + 2tx2 − 3x2 , ˙ 0 2t − 3 , 1 −t2 B= t , 1 x2 = u − t 2 x2 + x1 ˙ C= 0 1 , D= 0 , Problem 3 A fixed linear system has the zero state impulse response h(t) = t exp(−2t)1(t). Find the zero state response of the system to the input u(t) = 1(t) − 1(t − 2) Solution: The zero state step response is given by 2 0 0 −∞ 1 τ exp(−2τ )dτ = − exp(−2τ )τ 2 1 = − t exp(−2t) + 2 1 = − t exp(−2t) + 2 a( t ) = τ exp(−2τ )dτ τ exp(−2τ )1(τ )dτ = 1(t) h(τ )dτ = −∞ t t t t a( t ) = − t t 0 − 0 1 − exp(−2τ )dτ 2 1 − exp(−2τ )τ 4 t 0 1 (1 − exp(−2t)) 4 1 t exp(−2t) + (1 − exp(−2t)) 1(t) 2 4 The response to the input u(t) = 1(t) − 1(t − 2) is given by a(t) − a(t − 2), i.e., 1 (t − 2) 1 t exp(−2(t−2))+ 1−exp(−2(t−2)) − exp(−2t)+ (1−exp(−2t)) 1(t)− − 2 4 2 4 Problem 4 Consider the following system x= ˙ 1 02 u x+ 1 1 −1 y= 1 0 x a) Find the transition...
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This document was uploaded on 03/23/2014 for the course EL 6253 at NYU Poly.

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